Mathematics Contentfor Elementary Teachers

This page intentionally left blank

Mathematics Contentfor Elementary Teachers

Douglas K. BrumbaughPeggy L. Moch

MaryE Wilkinson

LAWRENCE ERLBAUM ASSOCIATES, PUBLISHERS2005 Mahwah, New Jersey London

Copyright © 2005 by Lawrence Erlbaum Associates, Inc.All rights reserved. No part of this book may be reproduced inany form, by photostat, microform, retrieval system, or any othermeans, without the prior written permission of the publisher.

Lawrence Erlbaum Associates, Inc., Publishers10 Industrial AvenueMahwah, New Jersey 07430

Cover design by Kathryn Houghtaling Lacey

Library of Congress Cataloging-in-Publication Data

Brumbaugh, Douglas K., 1939-Mathematics content for elementary teachers / Douglas K. Brumbaugh, Peggy L. Moch,

MaryE Wilkinson.p. cm.

Includes bibliographical references and index.ISBN 0-8058-4247-0 (acid-free paper)

1. Mathematics—Study and teaching (Elementary). I. Moch, Peggy L. II. Wilkinson,MaryE . III. Title.

QA135.6.B768 2004327.7-dc22 2004050674

CIP

Books published by Lawrence Erlbaum Associates are printed on acid-free paper,and their bindings are chosen for strength and durability.

Printed in the United States of America1 0 9 8 7 6 5 4 3 2 1

In loving memory of Pat and Web BrumbaughTo Shawn, Mike, Jennifer, and Laura, Doug's kids

To Linda Brumbaugh, simply the best

In loving memory of Elma McGrew Moch,Flossie Jenkins McGrew, and Mary Wheeler Jenkins,

who all cherished and encouraged their children to learndespite being denied the opportunity for much

of a formal education of their own.To Julie, Peggy's favorite and only child

To Grady

To all our teachersTo students: past, present, and future

This page intentionally left blank

Contents in Brief

Preface xv

Personal Acknowledgments xxi

About the Authors xxiii

1 Guiding Principles 1

2 Number and Operations 9

3 Algebra 117

4 Geometry 133

5 Measurement 151

6 Data Analysis and Probability 173

7 Problem Solving 203

8 Reasoning and Proof 209

9 Communication 219

10 Connections 223

11 Representation 229

Index 231

Solutions Manual SM-1

VII

This page intentionally left blank

Contents

Preface xv

To the Student xvFocal Points xvi

To the Instructor xviiOverview of Content xviiiPedagogical Features xixReferences xix

Personal Acknowledgments xxi

About the Authors xxiii

1 Guiding Principles 1

The Equity Principle 2The Curriculum Principle 2The Teaching Principle 3The Learning Principle 4The Assessment Principle 5The Technology Principle 5The Challenge 6References for Guiding Principles 8

2 Number and Operations 9

Sets 9Set Definitions 9Set Operations 75Special Sets 16Venn Diagrams 16Properties 18Factors and Multiples 22Prime and Composite Numbers 22Sieve of Eratosthenes 24Divisibility Rules 25Greatest Common Factor and Least Common Multiple 28Conclusions 32

Whole Number Addition 32Terminology 34Standard Algorithm 34Partial Sum 35Denominate Numbers 36Horizontal and Vertical Writing 36Expanded Notation 37

ix

CONTENTS

Left to Right Addition 37Scratch Method 38Any Column First 38Low Stress Addition 39Conclusions 40Bibliography 40

Whole Number Subtraction 40Terminology 41Concrete Subtraction 42Denominate Numbers 44Expanded Notation 44Standard Algorithm 44Borrow Pay Back Method 45Left to Right Subtraction 46Scratch Method 46Any Column First 47Integer Subtraction 47Conclusions 48

Whole Number Multiplication 48Terminology 48Beginnings 49Standard Algorithm 49Partial Product Method 50FOIL 51Lattice Multiplication 51Left to Right Multiplication 53Horizontal and Vertical Writing 53Russian Peasant (Simple Halving/Doubling Method) 54Try This 55Conclusions 55Bibliography 55

Whole Number Division 55Terminology 57Division as a Rectangle 58Repeated Subtraction Division 58Standard Division Algorithm 60Technology 61Remainders 61Say What? 62Conclusions 63

Equivalent Fractions and Multiplication of Fractions 63Concrete Beginnings 63Equivalent Fractions 64Converting a Mixed Number to an Improper Fraction 66Converting an Improper Fraction to a Mixed Number 67Product of a Whole Number and a Fraction 68Product of Whole Number and Mixed Number 68Product of Two Fractions 69Product of a Fraction and a Mixed Number 70Product of Two Mixed Numbers 71Conclusions 71

Addition of Fractions 72Concrete Beginnings 72Adding Fractions When Denominators Are the Same 72Adding Fractions When the Denominators Are Related 74Adding Fractions When the Denominators Are Relatively Prime 75Adding Fractions When the Denominators Are Not Relatively Prime

and One Is Not a Multiple of the Other 76Conclusions 77

x

CONTENTS xi

Subtraction of Fractions 77Same Denominators 77Related Denominators 78Relatively Prime Denominators 79Denominators That Are Not Relatively Prime or Related 80Mixed Numbers 80Conclusions 81

Division of Fractions 81Concrete Beginnings 82Whole Number Divided by a Fraction 82Fraction Divided by a Fraction 83Mixed Number Divided by Mixed Number 84Common Denominator Division 84Conclusions 85

Addition of Decimals 86Concrete Beginnings 86Denominate Numbers 87Adding With the Same Number of Places 87Zeros at the End 88Lining up the Ones 88Expanded Notation 89Conclusions 89

Subtraction of Decimals 89Concrete Beginnings 90Subtracting With the Same Number of Places 90Zeros at the End 91

Multiplication of Decimals 91Concrete Beginnings 91Moving Beyond the Concrete 92Relating to Old Ways 94Conclusions 94

Division of Decimals 95Concrete Beginnings 95Whole Number Divided by a Whole Number 96Decimal Divided by a Whole Number 97Whole Number Divided by a Decimal 98Decimal Divided by a Decimal 98Conclusions 99

Addition of Integers 99Models 99Rules for Adding Integers 707Conclusions 702

Subtraction of Integers 703Models 703Conclusions 705

Multiplication of Integers 705Multiplying Integers 705Conclusions 707

Division of Integers 707Division Using Signed Numbers 708Signed Numbers in Inverse Operations 708Conclusions 709

Ratios and Proportions 709Rational Numbers as Ratios 709Percents 777Equivalents 772Proportions 772

xii CONTENTS

Cross Products 774Solving Problems Using Proportions 114Conclusions 776

3 Algebra 117

Historic Underpinnings 118Laying the Foundation for Algebra 119Having Fun With Algebra 119Integrating Algebra 121Patterning 722Representing Situations With Algebra 723Using Models 124Rate of Change 126Sequences 728Formulas 730Conclusions 737References 732

4 Geometry 133

Undefined Terms 733Angles 736Simple Closed Curves, Regions, and Polygons 738Polygons 739Circles 742Constructions 743Third Dimension 744Coordinate Geometry 746Transformations and Symmetry 747Conclusions 749

5 Measurement 151

Terminology 752Attributes 752Systems 753Units, Tools (Instruments), and Precision 755Linear Measure 755Area 756Volume 767Capacity 764Weight 765Time 765Money 766Temperature 767Angle Measure 767Dimensional Analysis 777Conclusions 772

6 Data Analysis and Probability 173

Data Collection and Representations 173Data and Where to Get It 7 73Representations of Data 174

CONTENTS xiii

Venn Diagrams 775Percentages 175Circle Graph 7 75Line Plot 776Bar Graph 176Line Graph 177Histogram 778Frequency Polygon 178Box and Whisker Plot 779Scatter Plot 780Stem and Leaf Plot 787Conclusions 782

Data Analysis and Statistics 782Measures of Central Tendency and Scatter 783

Mean 783Median 784Mode 785Range 786Variance 786Standard Deviation 787Quartiles 188

Statistics 189Assumptions 190Generalizations 190

Conclusions 197Counting and Probability 197

Factorial 197Permutations 793Combinations 794Independent 794Dependent 797Conditional Probability 799Lottery 799Odds 200Conclusions 207

7 Problem Solving 203

What Makes a Problem a Problem? 203Polya's Steps 204Strategies 205Collection of Problems 207Conclusions 208Bibliography 208

8 Reasoning and Proof 209

And 209Or 270If, Then 277Negations 272Tautologies 272Logical Equivalence 272Informal Proofs 273Beyond the Informal Proofs 274

Two-Column Proof 275

xiv CONTENTS

Paragraph Proof 217Indirect Proof 217

Conclusions 277

9 Communication 219

10 Connections 223Independent or Interconnected Topics 223How Are Things Connected? 224Conclusions 227

11 Representation 229

Different Ways of Saying the Same Thing 229Conclusions 230Bibliography 230

Index 231

Solutions Manual SM-1

Preface

TO THE STUDENT

What is mathematics? You are probablywondering just what sort of question thatis for someone who has completed thegeneral education mathematics prerequi-sites. It is an important question becauseyour own opinions and attitudes will influ-ence your teaching of mathematics foryour entire career. Whether or not you areconscious of the process, you are alreadydefining and redefining your personalpractical theories about teaching. Thesetheories are deep-seated beliefs that mayresult from your nonteaching experiencesas a student or parent and be influencedby your practical experiences in design-ing and implementing curriculum as youteach (Cornett, 1990, p. 188). You mayfind it difficult to articulate your personalpractical theories. However, they will sig-nificantly influence your students' atti-tudes about the mathematics that theywill learn.

The dictionary defines mathematics as"the study of quantity, form, arrangement,and magnitude; especially the methodsand processes for disclosing, by rigorousconcepts and self-consistent symbols,the properties and relations of quantitiesand magnitudes, whether in the abstract,pure mathematics, or in their practicalconnections, applied mathematics" (Funk& Wagnalls, 1968, p. 835). How does thisformal, dictionary definition of mathemat-ics compare with your own opinion? Howwill you define mathematics to others, es-pecially children?

As an elementary educator, you will beexpected to teach children in a variety ofsubject areas. In this text, the focus is onone content area—mathematics. To teachmathematics, you will need a significantcommand of the foundations of elemen-tary school mathematics content. If ateacher is not competent and confidentwith the subject matter, then there may bebarriers to creating a positive experiencebest suited for each student. We want youto understand what some of these barriersare and how you can begin to prepareyourself to avoid them by building a strongfoundation in mathematics content.

As you go through this text, you will seethat we present you with multiple ways ofdoing arithmetic operations. We do that tohelp you help your future students. Youwill find that not all students think thesame way. Furthermore, you will learn thatnot all students will understand how youdo some arithmetic. Having command ofseveral ways to do a given arithmetic op-eration, combined with your knowledge ofyour students, enables you to select amethod of doing the work that has a betterchance of connecting with students.

Our main purpose is to introduce the"what" and the "why" of teaching mathe-matics; in other words, elementary schoolmathematics content. In many of thechapters, you will find hints about the"how" of teaching mathematics to chil-dren, but we do not develop instructionalconcepts in depth. It is expected that,later in your educational program, you willenroll in a mathematics methods course.In such a course, the groundwork we es-

xv

XVI PREFACE

tablish will be expanded and you will learnto use effective methods to teach childrenmathematics. The hints we provide are in-tended to help you with your quest to gainthe content proficiency that will equip youto make the most of your mathematicsmethods course.

Focal Points

Teaching Mathematics to ALL Students.The National Council of Teachers of Math-ematics (NCTM) has developed a docu-ment called Principles and Standards forSchool Mathematics (NCTM, 2000), whichpresents the position that an effectiveteacher of mathematics will be able to mo-tivate all students to learn. We agree withthis position. It is important for all studentsto develop a strong basis in elementaryschool mathematics because it is the keyto many opportunities. Mathematicsopens doors to educational opportunitiesand careers, enables informed decisions,and helps us compete as a nation. Wewant to help you develop the ability to ab-sorb new ideas, adapt to change, copewith ambiguity, perceive patterns, andsolve unconventional problems (Mathe-matical Sciences Education Board[MSEB], 1989). This is an important lessonbecause we know you will need to passthese skills along to your students in therapidly changing world of the 21 st century.

Understanding How Children Learn. Ourown opinions about child developmenthave been influenced greatly by the semi-nal works of visionaries such as Dewey,Piaget, Bruner, Vygotsky, and Montessori.We believe that children of all ages benefitfrom learning experiences at each of fourdevelopmental levels. Children at the"concrete" level need a physical object tomanipulate. At the "semi-concrete" level,

they are able to understand a connectionbetween the concrete object and a picto-rial representation of that object. At the"semi-abstract" level, they might move toa visual aid, such as a tally, which is not anactual picture of a real object. At the "ab-stract" level, they will understand mathe-matics using only numbers and symbols.We will not establish ages at which youshould expect children to move from onelevel to the next. You will find that childrenneed a spiraling strategy, with the initial in-troduction of each concept at the "con-crete" level. Children learn by experienc-ing mathematics and there are windows ofteaching opportunities that must be rec-ognized and used effectively. Finally, webelieve that children must take centerstage in the classroom. Teachers shouldgive up the role of sage on the stage andconcentrate on what children need to be-come active participants in the learningenvironment. Frequently, teacher telling,which has children being passive learn-ers—like they are sponges who should besoaking up every word the teacher ut-ters—is the model used.

Learning by Doing. In this book you willfind that we introduce mathematics con-tent concepts in ways that you may nothave seen before. An effective teacher ofmathematics continues to investigate newmathematical concepts and teaching strat-egies. Such a teacher is a lifelong learner,always ready to investigate new ideas or analternate way of doing something familiar.Children think in different ways as they de-velop, so we will be presenting concepts atthe four distinct developmental levels: con-crete, semi-concrete, semi-abstract, andabstract. Your own elementary schoolteachers may have skipped one or more ofthese levels, teaching most often at thesemi-abstract or even abstract level. If yousometimes feel as if we are treating you like

PREFACE XVII

the children you will someday teach, thenyou are right! It is not that we want to treatyou like elementary schoolchildren, but webelieve you must experience mathematicsat each level if you are to understand howchildren learn and then help them masterthe tasks placed before them. At the sametime, you will find relatively few algorithmsin this text as we focus on the "what" and"why." We don't really think that you needhelp with algorithms. This text will mentionmanipulatives as learning tools. Anythingthat helps a person learn mathematicsmight be called a manipulative. We believethat you will be aided in your use of manip-ulatives in your classroom if we introducesome of them and ask you to solve prob-lems using them. If you were not taught us-ing manipulatives, then you may find theiruse not easy or natural for you—but keep inmind that children do not learn as you weretaught. Use your experiences as you learnto exercise your "concrete" or "semi-con-crete" brain to understand how difficult it isfor children to learn an abstract concept.

We Believe That Every Child Can Learn.Children will not learn at the same rate or inthe same way. We hope that you will setthe following goal for yourself: I will do all Ican, including learning all the mathematicscontent I will need, to help each child learnto value mathematics as well as how to domathematics. This text contains many fea-tures to help you accomplish this goal. Wehave tried to keep the reading at an infor-mal level and have provided a glossary forthe terms with which you may not be famil-iar. Rather than write about general schoolmathematics, we have concentratedstrictly on the mathematics that is taught inelementary schools. We have spaced theactivities and exercises throughout eachsection; rather than assignments at theends of chapters, you will find brief setsthat immediately follow discussions.

TO THE INSTRUCTOR

The major objective of this text is to helpthe prospective teacher of elementaryschool mathematics learn content beyondthe rote level; to stimulate your students tothink beyond "getting the problem right."Our experiences have shown that if theideas expressed in this text are used withprospective teachers of elementary schoolmathematics, then the likelihood of theirdevelopment into thoughtful, reflective,self-motivated, lifelong learners is muchgreater. We encourage you to challengeyour students to achieve that status.

Although there are many excellent textsavailable that combine mathematics con-tent and methods in creative ways, we be-lieve there is a need for a text that stressesthe "what" and "why" of elementaryschool mathematics content. We providehints about "how" elementary schoolmathematics should be taught, but weleave most of the methods part for latercourse work and a text that is dedicated tothat purpose. It is important to reverse thecurrent trend of combining elementary andmiddle school mathematics content. Theaudience for this text is preprofessionalteachers in elementary education pro-grams. We focus all of our attention on themathematical needs of prospective teach-ers at this level, giving us the luxury of suf-ficient space to carefully develop conceptsstarting at the concrete level and buildingto the abstract level.

To gain and hold both the attention andthe trust of preprofessional elementaryschool teachers, we write informally. Oftenthe needs of prospective teachers at thislevel are better addressed when they areallowed to feel as comfortable as possiblewith the discussions. We provide brief ac-tivities and exercises immediately follow-ing short discussions and easy access tocomplete solutions. We have spaced them

XVIII PREFACE

more frequently, so that your students willpractice for shorter times but completemany activities.

The text is based on several fundamen-tal premises. First, we, as teachers ofmathematics, do not have the right tomark the work of any student "wrong," ifthat student does not use exactly the pro-cedure we recommend. We believe thatthe focus of mathematics needs to be onthe process, not the answer. Thus, thestudents using this text need to becomeproficient in understanding the founda-tions behind doing a problem so they canadapt their thinking to interpret the workthey will see from their students. To thatend, they must move beyond thinking thatthe way they do a problem is the only wayit can be approached.

Second, we believe that teachers ofmathematics should consider three axi-oms (Dr. Joby M. Anthony of the Univer-sity of Central Florida mathematics de-partment, personal communication, May26, 2003):

1. Know the content being presented.2. Know more than the content being

presented.3. Teach from the overflow of knowl-

edge.

Third, being a teacher who is flexibleenough to allow a student to use differ-ent procedures, who knows the content,and who teaches from the overflow ofknowledge implies knowing how to do agiven operation more than one way andbeing willing to examine many differentways. Thus, it is imperative for prospec-tive teachers of elementary school math-ematics to learn how to carefully coverthe topics to be taught, to reflect onthem, and to be able to organize them.Further, this thinking and organizingneeds to be done well in advance, to al -

low time for the ideas to germinate andblend in the prospective teacher's sub-conscious mind so that they can connecttopics from different lessons throughoutthe course. To help your students con-centrate on the mathematics content,they will be expected to teach and beginto build the groundwork for the methodsthey will use, we have not included mid-dle school concepts.

Based on these premises, in this textwe do not focus solely on "pure" mathe-matics content, but also include hintsabout topics that are typically found inmathematics methods texts. We do thisbecause we know that, in reality, mathe-matics content and methods of teachingare closely connected. Thus, in additionto mathematics content, you will find:

• Different approaches to present tostudents for doing operations andproblems

. Methodology• Suggested problems and classroom

activities• Use of technology. References to current school theory

and practices

OVERVIEW OF CONTENT

The text is built around the standards ex-pressed by NCTM in "Principles andStandards for School Mathematics"(2000). Standards 2000 will dictate thebasic sections for the text. For example,the first major section is headed "Num-ber and Operations" (Standard 1 in Stan-dards, 2000). Within each section, appro-priate specific topics will be developed,intertwined with technology, problemsolving, assessment, equity issues, plan-ning, teaching skills, use of manipulatives,sequencing, and much more. The text isorganized into 11 chapters:

PREFACE XIX

1. Guiding Principles2. Number and Operations3. Algebra4. Geometry5. Measurement6. Data Analysis and Probability7. Problem Solving8. Reasoning and Proof9. Communication

10. Connections11. Representations

PEDAGOGICAL FEATURES

Questions, exercises, and activities areinterspersed within the sections. We be-lieve students will progress more effec-tively if they are given several small andvaried opportunities to practice as op-posed to larger ones like those typicallyfound at the end of a chapter.

Pedagogical features of the text:

. Informal reading with a focus on thepreprofessional teacher

• Purpose: The "what" and "why" of el-ementary school mathematics, withbrief hints about "how" to teach

• Concentration on the mathematicstaught in elementary school grades

• Brief sets of activities or exercises im-mediately following discussions

• Complete solutions for exercises• Discussions organized according to

NCTM standards. A focus on multiple methods of problem

solving at four developmental levels

REFERENCES

Bruner, J. S. (1986). Actual minds, possible worlds.Cambridge, MA: Harvard University Press.

Bruner, J. S. (1996). The culture of education.Cambridge, MA: Harvard University Press.

Cornett, J. (1990). Utilizing action research ingraduate curriculum courses. Theory into Prac-tice, 29(3), 187-195.

Dewey, J. (1997). How we think. Mineola, NY: Do-ver. (Original work published 1910)

Dewey, J. (1948). Experience and education. NewYork: Macmillan. (Original work published 1938)

Funk & Wagnalls. (1968). Standard college diction-ary (text ed.). New York: Harcourt, Brace, &World.

Mathematical Sciences Education Board. (1989).Everybody counts. Washington, DC: NationalAcademy Press.

Montessori, M. (1965). The Montessori method;scientific pedagogy as applied to child educa-tion in "the Children's Houses" with additionsand revisions by the author (A. E. George,Trans.). Cambridge, MA: R. Bentley. (Originalwork published 1912)

Montessori, M. (1967). The discovery of the child(M. J. Costelloe, Trans.). Notre Dame, IN: FidesPublishers.

National Council of Teachers of Mathematics.(2000). Principles and Standards for SchoolMathematics. Reston, VA: Author. [http://www.nctm.org/tcm/tcm.htm]

Piaget, J. (1952). The child's conception of num-ber (C. Gattegno & F. M. Hodgson, Trans.).London: Routledge & Kegan Paul.

Piaget, J. (1999). The child's conception of geom-etry (E. A. Lunzer, Trans.). London : Routledge.(Original work published 1960)

Vygotsky, L. S. (1978). Mind in society: The devel-opment of higher psychological processes.Cambridge, MA: Harvard University Press.(Original work published 1934)

Vygotsky, L. S. (1987). The development of scien-tific concepts in childhood. In R. W. Rieber &A. S. Carton (Eds.), The collected works of L. S.Vygotsky (N. Minich, Trans., Vol. 1, pp. 167-241). New York: Plenum.

This page intentionally left blank

Personal Acknowledgments

Dr. Gina Gresham came into our lives nearthe end of this writing. Dr. Gresham left theState University of West Georgia andjoined the mathematics education facultyof the University of Central Florida in Au-gust 2003. Dr. Gresham read the entiremanuscript in its final stages and offereda plethora of useful suggestions. We aregrateful to her for her contribution andlook forward to working with her.

Dr. Michael Reynolds has provided abrief description of a very real conditioncalled math anxiety, which hinders math-ematics achievement in people of allages. We invite you to take a careful lookat this summary (see p. 7) and to reviewhis ongoing work, as well as the work ofother researchers, in this important area.We thank Dr. Reynolds for his valuablecontribution to our effort.

We offer a special thanks to the follow-ing undergraduate students who read themanuscript in its formative stages: CrystalJackowski (University of Central Florida),Wendy Lane (Valdosta State University),

Michelle Backes (Valdosta State Univer-sity), and Emily M. Hunter (Valdosta StateUniversity).

Naomi Silverman has again served asthe supportive, cooperative, encouraging,helpful, guiding editor. Such a joy to workwith! Thanx again Naomi.

We once again thank Dr. Gina Gresh-am. She prepared the Index section ofthis book. Thanks again Gina! Well done!

We are grateful to the wonderful staff atLawrence Erlbaum Associates, particular-ly Lori Hawver. Lori was always availableto answer our many production ques-tions. Class act! Thanx again Lori. EileenMeehan has provided invaluable guid-ance and assistance as our productionsupervisor. Thanx Eileen.

Last, but certainly not least, are our re-spective families. They sacrificed a lot sowe could create this text, and it is appre-ciated. To some extent we have becomeone big blended family, which is a won-derful experience. Thanks to each of youfor your support in this adventure.

XXI

This page intentionally left blank

About the Authors

DOUGLAS K. BRUMBAUGH

I am a teacher. I teach college, in-service,or K-12 almost daily. I received my BSfrom Adrian College, and went on to theUniversity of Georgia for my master's anddoctorate degrees in mathematics edu-cation. As I talk with others about teach-ing and learning in the K-20 environment,my immersion in teaching is beneficial.Students change, classroom environ-ments change, the curriculum changes, Ichange. The thoughts and examples inthis text are based on my experiences asa teacher. Classroom-tested successstories are the ideas, materials, and situa-tions you will read about and do. Thistext's exercises and activities will stretchyou while providing a beginning collectionof classroom ideas. Learn, expand yourhorizons, and teach.

PEGGY L. MOCH

I had the good fortune to have wonderfulteachers as I was growing up that encour-aged me to be the best at whatever I wasinterested in at the time. I became a medi-cal laboratory technologist in the late1970s, but in the early 1990s I decided togo back to school to become a teacher ofmathematics. I studied medical technol-ogy in Aurora, Colorado, obtained a BS

and MEd in mathematics education and aPhD in curriculum and instruction with anemphasis in mathematics education atthe University of Central Florida in Or-lando, Florida. I have always loved mathe-matics and took the opportunity affordedme by this book to share the passion myteachers ignited in me with all you in thehope that zeal will continue to spread.Mathematics is in everything, it is un-avoidable. Embrace the challenge, be en-thusiastic, and watch your own love formathematics grow!

MARYE WILKINSON

I am a teacher because of the wonderfulteachers I had, including Dr. Brumbaugh.My degrees were earned at the Universityof Central Florida. I taught mathematicsto high school students for 9 years, thenbegan work on my doctorate and my newfound love for teaching mathematics andmathematics methods to preprofessionalelementary school teachers. I am hum-bled when I realize how rich and deep theconcepts of mathematics for children are!I am always delighted when a young per-son teaches me something new. Myteaching days are spent with college stu-dents; my field days are spent with manyof those same college students and thechildren who will be their students. Whocould ask for a better job?

XXIII

This page intentionally left blank

1Guiding Principles

FOCAL POINTS

• The Equity Principle• The Curriculum Principle. The Teaching Principle. The Learning Principle• The Assessment Principle• The Technology Principle• The Challenge• Math Anxiety• Where to From Here?

Although we hope you will be a lifelonglearner, we know that your focus will soonturn from learning mathematics to the artof teaching mathematics to children. It ispossible that this will be the last formalcourse you will take in mathematics con-tent. Because you are enrolled in thiscourse, we know that you have success-fully learned a significant amount of math-ematics content, either from your own ele-mentary, middle school, high school, andcollege teachers or on your own.

Much of your study during this term willreview material that is familiar to you. Wewant you to look at concepts from differ-ent points of view. In order to successfullyhelp children learn mathematics, you willneed more than the ability to do mathe-matics. You will need a profound under-standing of the mathematics you will beexpected to teach. You will need to knowwhich concepts are easy to learn, whichare difficult to learn, and just what makeseach concept easy or hard. This means

that you must accept the challenge toprobe deeply into elementary schoolmathematics content, expanding and so-lidifying your knowledge into a powerfulbase of content knowledge from whichyou may draw as you help children learnand understand mathematics. We call thispedagogical content knowledge (Shul-man, 1986) because it is a class of knowl-edge that is held almost exclusively byexpert teachers.

In Principles and Standards for SchoolMathematics, the National Council ofTeachers of Mathematics invites you to"imagine a classroom, a school, or aschool district where all students haveaccess to high-quality, engaging mathe-matics instruction" (NCTM, 2000, p. 3).The environment in which such a dreamcan come true will exist only if you, as theclassroom teacher, have a deeply reflec-tive pedagogical content knowledge soyou can provide rich curricula and helpyour students learn and understand theimportant concepts of elementary schoolmathematics. It is on these central con-cepts that your future students will rely asthey find their ways through their highereducations. You must prepare for a futurethat will require an ever greater under-standing of mathematical concepts inyour personal life, workplace, and scien-tific and technical communities. In thePrinciples and Standards for SchoolMathematics, NCTM identified six princi-ples for school mathematics (pp. 10-27).

1

CHAPTER 1

THE EQUITY PRINCIPLE

Excellence in mathematics education re-quires equity—high expectations and strongsupport for all students. (NCTM, 2000, p. 12)

We can no longer consider mathematicsto be an enrichment subject for the mostable students. We cannot allow mathe-matics to be the separator between thehaves and the have-nots. Individuals whohold a limited mathematical backgroundand understanding are severely handi-capped as they select a career. Further-more, studies now indicate that individu-als entering the 21st-century workforcewill change careers up to five times overtheir working years. Changing careers im-plies a revamped education with eachnew selection. Persons who do not pos-sess fundamental mathematical skills andunderstandings will have difficulty findingoptions that allow for a limited mathemat-ical background. Every elementary schoolteacher must be ready to help all studentslearn and understand mathematics. In ev-ery classroom, you will find studentsranging from those with deep and abidinginterests and talents in all academic sub-jects to those with special educationalneeds in some or all subjects.

Traditionally, certain groups of studentshave been viewed with reduced expecta-tions. Some of these children live in pov-erty, aren't fluent in English, or have learn-ing disabilities. Additionally, female andnon-White students have been victims oflowered teacher expectations. Your ped-agogical content knowledge must be suf-ficient to facilitate learning for all.

Simply being able to do arithmetic willbe insufficient to the task you have set foryourself. Your goal must be to have theability to teach mathematics as compo-nent parts and as an integrated whole, toidentify each student's academic need,

and to provide the exact amount of sup-port needed by that student. You will beresponsible for helping each student ob-tain a common elementary foundation ofmathematics. To do this, you must solid-ify your own pedagogical content knowl-edge so you can maintain high expecta-tions and provide strong support for eachand every one of your future students. Inyour mathematics methods course, youwill learn a variety of techniques for put-ting your pedagogical content knowledgeto use in lesson planning and teaching.

THE CURRICULUM PRINCIPLE

A curriculum is more than a collection of ac-tivities; it must be coherent, focused on im-portant mathematics, and well articulatedacross the grades. (NCTM, 2000, p. 14)

To say that the mathematics curriculum inelementary schools is arithmetic is insuffi-cient because it gives the impression thatyou only need to be proficient in arithme-tic to teach it to children. The foundationsof mathematics must be formed in ele-mentary schools, which means that chil-dren will need the underpinnings of manyinterwoven strands of mathematics. Arith-metic is certainly one of those strands,but it is not the only one.

Some preprofessional elementaryschool teachers have told us that theywant to teach kindergarten, first, or sec-ond grade because the early grades re-quire less arduous mathematics. This issimply not true. The foundations of impor-tant mathematics strands, such as alge-bra and geometry, are formed very earlyin the lessons of the elementary schoolmathematics curriculum. Mathematicsmust not be broken into discrete, discon-nected bits. You may not be involved indesigning the curriculum for your schoolor grade level, but you will most certain-

2

GUIDING PRINCIPLES

ly be responsible for organizing lessonsthat teach the power of interconnectedstrands of mathematics. You will find thatyou must understand and be proficient inthe elementary school mathematics of allgrade levels in order to be an effectiveteacher. A reflective knowledge of theconcepts taught in the classes before andafter the grade level to which you are as-signed will help you focus on the impor-tant mathematics your students alreadyknow and will need to learn before theymove on. You will be one rung of their lad-der of learning and you must help themdeepen and extend their understanding ofmathematics.

Whether you are assigned to kindergar-ten or sixth grade, your curriculum will in-clude foundational concepts such as theBase 10 numbering system, place value,sets of numbers, proportionality, and func-tions. These foundational concepts willsupport your students as they work toconnect and extend their ideas throughmathematical reasoning, conjecturing, andproblem solving. Additionally, conceptssuch as symmetry will help your studentssee life's beauty. Modeling problems onreal-world phenomena will help them rec-ognize the importance of quantitative liter-acy. You can see that the daily mathemat-ics curriculum of your classroom willsignificantly depend on the extent of yourpedagogical content knowledge.

THE TEACHING PRINCIPLE

Effective mathematics teaching requires un-derstanding what students know and needto learn and then challenging and supportingthem to learn it well. (NCTM, 2000, p. 16)

Your decisions and actions in your class-room will determine what your childrenlearn. Consider your own elementaryschool experiences: Which memories are

rich and satisfying and which memoriesare disappointing and lacking? It is verylikely that each memory is linked to ateacher and that teacher's ability to chal-lenge and support you as you learned.You may not remember whether yourteachers possessed profound knowledgeof the subjects they were teaching, butyou learned the most from those teacherswho knew the most. Your ability, confi-dence in that ability, and your attitudestoward mathematics were shaped byyour elementary school teachers. Take afew moments to reflect on how you willshape the abilities, confidence, and atti-tudes of your students.

Many elementary school teachersspend a great deal of time trying to under-stand the mathematics they are teaching.With college courses and certification ex-aminations completed, they find that theysimply don't have knowledge that is deepenough or flexible enough to teach math-ematics to children. Some take years tobecome confident in their understandingof and ability to teach mathematics.These teachers reach high and go far bycontinuing to learn and grow in theirknowledge of mathematics and mathe-matical pedagogy. Perhaps the best giftyou can give your future students is tostart today to be a self-reflective lifelonglearner—a person who loves to learn andloves to help others learn.

What of the children who sit in a class-room while their teacher is striving to in-crease personal knowledge of the subjectto become a more effective teacher? Wecannot afford to ignore the needs of chil-dren while their teachers study to becomeproficient. We invite you to work duringthis term to construct and solidify a baseof pedagogical content knowledge onwhich you can build during your mathe-matics methods course. That way, youwill become a teacher who is capable of

3

CHAPTER 1

reaching the mathematical curricularneeds of each student—and a lifelonglearner who will continue to grow into aneven stronger teacher.

THE LEARNING PRINCIPLE

Students must learn mathematics with un-derstanding, actively building new knowl-edge from experience and prior knowledge.(NCTM, 2000, p. 20)

For many decades, there have been dis-agreements about what concepts in math-ematics are important and how theyshould be taught. Traditionally, elementaryschool teachers have guided students inhow to perform operations, isolating basicskills into discrete segments, and buildingincrementally toward higher order skills.This reflects a behavioral or information-processing point of view in which theteacher conveys the knowledge and thestudents record (supposedly memorize—without understanding what is going on)the steps involved. Whereas this may leadto proficiency in each skill, it does not nec-essarily lead to understanding or flexibilityof knowledge. Students, who can followmemorized steps with precision and accu-racy, often lack the ability to connect basicskills to problem situations; they simplydon't understand when or why to applythese basic skills. That is why many stu-dents find it so difficult to solve word prob-lems in which the required operations arenot explicitly defined.

The daily experiences provided byteachers influence what and how stu-dents learn. The mathematics contentprovided in a classroom depends on thecontent knowledge of the teacher. If theteacher doesn't know the concepts re-quired by the curriculum, those conceptsmay be poorly taught or even ignored.Additionally, teachers who are insecure

about their own knowledge of mathemat-ics often provide activities that areteacher centered—focusing only on fol-lowing a given rule to get a single right an-swer. Students in such a classroom maylearn algorithms without applications andwill certainly learn teacher-dependent be-haviors. Students adopt the teacher'smathematical insecurities. In contrast, ateacher who is secure and confident ofthe content —not only what is beingtaught, but also that which precedes andfollows the particular concept being ad-dressed—encourages student independ-ence and delivers less rule-based, moreheuristically based lessons, enabling stu-dents to recreate their thinking and findconnections to applications of mathemat-ics. To create autonomous learners, ateacher must have the content knowledgeand confidence to select tasks that areappropriate for encouraging students totackle hard material, explore alternatepaths, and become tenacious, independ-ent, problem solvers.

Foundational concepts are best taughtat the concrete, or hands on, level. Manip-ulatives are important tools for buildingunderstanding of foundational concepts.When you think about how you learn, youmay realize that manipulatives are neededwhen learning something new, even withinyour university courses. In this text, ma-nipulatives such as Base 10 blocks, alge-bra tiles, Cuisenaire rods, number lines,colored chips, buttons, and coins are dis-cussed and used. Some of your studytime will be spent learning to use thesetools.

Anything, at any age, that helps you un-derstand can be considered a manipula-tive. Concrete learning experiences arecertainly not limited to elementary agechildren. Think about your experiences asyou learned to drive a car. Of course, youlearned the rules of the road first so that

4

GUIDING PRINCIPLES

you would not be a danger to yourself orothers. But to learn to drive you actuallygot into a car or a simulator, grabbed thesteering wheel, put your foot on the ped-als, and drove. A car as a learning tool?Sure! We hope that you will make it apoint to look for learning tools wheneverand wherever they might be useful.

THE ASSESSMENT PRINCIPLE

Assessment should support the learning ofimportant mathematics and furnish useful in-formation to both teachers and students.(NCTM, 2000, p. 22)

One might think that a discussion of as-sessment has no place in a contentcourse. Indeed, the discussion of howand when you will assess the mathemat-ics learning of your students will be left foryour methods course and classroom ex-periences. However, an early understand-ing of the purposes of assessment mayhelp as you develop a focus on this im-portant part of a teacher's responsibility.

The assessments to which you are sub-jected as a college student are summa-tive. They are designed to determine whatyou have attained with respect to the in-structional goals of the course and maybe used to establish your grade for thecourse. However, you can use all of theactivities of the course as personally for-mative assessments. As you examineyour progress, you can make decisionsabout which concepts need to becomethe basis of further investigations andwhich only need polishing efforts.

Feedback from formal and informal as-sessment tasks provided by your instruc-tor is intended to assist you in setting yourlearning goals and becoming a more in-dependent learner. Even if feedback fromthe instructor is not provided for a task,you can examine your own completed

work to determine if it represents yourbest effort, if it completely satisfies thetask requirements, and if improvementsare desirable. A good strategy is to enlistthe support of a critical friend who is in-volved in the course with you. The folkswho are taking this course with you mayvery well become your colleagues and es-tablishing a rapport with them now mayprovide a support system for you later.Critical friends encourage and challengeeach other based on examinations of oneanother's work and behaviors. In thisway, you can gain formative assessmentsfrom several points of view.

Making personal assessment a vital,productive part of your education now willprovide an understanding of and practicewith summative and formative assess-ment that will aid you as you begin to de-termine assessment tasks for children.

THE TECHNOLOGY PRINCIPLE

Technology is essential in teaching andlearning mathematics; it influences themathematics that is taught and enhancesstudents' learning. (NCTM, 2000, p. 24)

While you were in elementary school, didyou have daily access to calculators andcomputers? Even if you were luckyenough to have significant exposure toclassroom technology during elementaryschool, it is likely that your students willhave access to technology that will bemuch more sophisticated than the de-vices you used. Now is the time for you toexperiment and learn about some of thetools that will become commonplace asthe 21st century continues.

Calculators and computers will not elim-inate the need for you to be proficient inthe basic skills of elementary school arith-metic. But the use of technology can savetime and effort that can then be focused

5

CHAPTER 1

on decision making, reflection, reasoning,and problem solving. Using the tools oftechnology alongside your review of basicconcepts will help you deepen and extendyour understanding of when and why touse those concepts to make importantconnections within elementary schoolmathematics. It is important for studentsto memorize number facts and the func-tions involved in operational algorithms.Those certainly are a part of the curricu-lum. However, they are not the only part.Technology can provide opportunities toopen a variety of windows for your stu-dents. Technology should not be used justto get answers. Skillfully used, technologycan stimulate student learning and inquisi-tiveness into the world of mathematics.

An important aspect of learning mathe-matics is conjecture posing, an activitythat is supported when more examples,representative forms, and possibilitiescan be investigated quickly and efficient-ly. You and your students will no longerbe limited to easy problems because cal-culators and computers can providecomputational and graphic power to sup-port your inquiries into number sense,measurement, geometry, algebraic think-ing, and statistics. With this support, youcan examine real-life problems and con-nections much more efficiently and effec-tively than using paper and pencil.

We hope that you will take the time toexamine as many different calculators aspossible. You should begin your investi-gation with the four-function calculatorsthat are generally available in the earlygrades, followed by scientific calculators,and, finally, graphing calculators. If possi-ble, you should not confine your investi-gation to the machines commonly used inelementary schools, but rather experi-ment with different models. As the level ofsophistication continues to increase, themore complex and powerful models will

appear in ever-earlier grades. You maybegin your inquiry by visiting various Websites provided by the manufacturers ofdifferent calculators:

Casio: http://www.casio.com/

Hewlett Packard: http://www.hewlettpackard.com/ (handheld/cal-culators)

Sharp: http://www.hewlettpackard.com/ (business/calculators)

Texas Instruments: http://education.ti.com

These Web sites are commercial, but pro-vide a great deal of information about fea-tures and activities. Many of the manufac-turers are willing to loan new models forreview and classroom sets for examina-tion and experimentation.

Visiting various Web sites will also pro-vide a great deal of information about thesoftware that is available for elementaryschool mathematics. Many programs canbe described as plug-and-chug, simplytransferring worksheets to the computerscreen. Such programs are useful only aspractice or review and we question thevalue of using a valuable tool like a com-puter as a drill machine. However, thereare powerful programs available that al-low and, indeed, encourage experimenta-tion and learning of important concepts inall areas of mathematics. These are thecomputer programs on which we hopeyou will focus your investigation.

THE CHALLENGE

In this chapter, we have provided briefoverviews of the six principles delineatedin NCTM's Principles and Standards forSchool Mathematics (2000) to provide youwith important information about the fea-tures of high-quality mathematics instruc-

6

GUIDING PRINCIPLES

tion. We have focused this discussion ofequity, curriculum, teaching, learning, as-sessment, and technology on your role asa learner of the elementary school mathe-matics that you will be expected to teach.You should read and study this importantdocument as you continue your education,extending your pedagogical contentknowledge of mathematics with knowl-edge of teaching methods.

An earlier NCTM document, Curriculumand Evaluation Standards for SchoolMathematics, published in 1989, was thebasis for the standards established inmany states. This document, along withtwo other NCTM publications, Profes-sional Standards for Teaching Mathemat-ics (1991), and Assessments Standardsfor School Mathematics (1995), repre-sented an important and historical at-tempt to explicitly articulate the goals ofmathematics education in America. Prin-ciples and Standards for School Mathe-matics (2000) was written to revise andupdate the earlier documents.

Math Anxiety

Teachers who lack adequate mathemati-cal preparation can foster math anxiety intheir students. Math anxiety is a term thatis often used to describe feelings of help-lessness and nervousness that many indi-viduals experience when they are expect-ed to complete a mathematical task orassignment. Math anxiety is not limited toelementary students—it can become a life-long condition for many people. Peopleare not born with math anxiety; indeed,many educational specialists believe thatmost people develop math anxiety duringelementary school. It is important, there-fore, that you be aware of the threat ofmath anxiety and take active steps toavoid its inception in your future class-room.

The consequences of math anxiety arefrequently detrimental. Those with mathanxiety tend to perform at lower levels inmathematics classes and many adopt thebehavior of math avoidance—choosing toenroll in the fewest (and easiest) mathe-matics courses. A single statistics courserequirement or business mathematicscourse can be the difference between astudent selecting a major they feel drawnto and choosing a major simply becauseit does not require an extra mathematicscourse. For these and other reasons,mathematics has been labeled a criticalfilter into the job market (Sells, 1978).

Cognitive and external factors such asintelligence, aptitude, and the quality ofone's mathematical instruction may havea greater impact on achievement thanmath anxiety. However, recent findingsindicate that math anxiety is more detri-mental to achievement than any otherpsychological factor, including self-con-cept, self-efficacy, or the value that oneplaces on mathematics (Reynolds, 2003).Motivation may be an extremely powerfulfactor in mathematics learning and teach-ing. Given the overpowering impact thatmath anxiety can have on mathematicalmotivation, the impact of math anxietyshould not be underestimated.

Some common elementary classroompractices have been shown to causemath anxiety, including excessive drilland practice, a focus on answers ratherthan processes, excessive criticism of mi-nor errors, embarrassment as a means ofdiscouraging questions or mistakes, andinadequate class time spent helping stu-dents with mathematics. Many sufferersof math anxiety are able to recall the hu-miliation of a single negative episode asthe reason for their math anxiety.

Although math anxiety is not a learningdisability, it can disable those who sufferfrom it. As educators, we have a respon-

7

CHAPTER 1

sibility to be aware of the condition, toshow respect and consideration, and torealize that math anxiety is not terminal —with patience and care, it can be re-versed. According to Stuart (2000), "Peo-ple like to do something if they think theyare good at it, and to feel good aboutmathematics, you have to believe thatyou are good at it. Therefore, as teachers,we must be the mathematics coaches—the ones to build that self-confidencewhile refining the skills needed to be suc-cessful" (p. 334).

Where to From Here?

The success of this continuing effort toimprove mathematics education in theUnited States will depend on you, the pre-professional classroom teacher, who willteach mathematics to children in the fu-ture. Your ability to fulfill this dream willdepend to a great extent on your peda-gogical content knowledge of elementaryschool mathematics. Increasingly, ac-countability is discussed in educationalcircles. Many states administer highstakes tests at several levels in publicschools. Many teachers across theUnited States are providing sound educa-tional experiences for their students—andmany of those students are doing just fineon high stakes tests. Widespread suc-

cess on these tests seems illusive. Thereis no definitive information on how to pre-pare students for mandated evaluations,even though the results have serious im-plications for both students and teachers.Your challenge is to take the opportunityprovided by this course to deepen andextend your knowledge of this vital andwide-ranging subject.

REFERENCES

National Council of Teachers of Mathematics(NCTM). (1989). Curriculum and evaluationstandards for school mathematics. Reston, VA:Author.

NCTM. (1991). Professional standards for teachingmathematics. Reston, VA: Author.

NCTM. (1995). Assessment standards for schoolmathematics. Reston, VA: Author.

NCTM. (2000). Principles and standards for schoolmathematics. Reston, VA: Author.

Reynolds, J. M. (2003). The role of mathematicsanxiety in mathematical motivation: A pathanalysis of the CANE model (Doctoral disserta-tion, University of Central Florida, 2003). Dis-sertation Abstracts International, 64, 435.

Sells, L. (1978). Mathematics a crucial filter. Sci-ence Teacher, 45(2), 28-29.

Shulman, L. (1986). Those who understand:Knowledge growth in teaching. Educational Re-searcher, 15(2), 4-14.

Stuart, V. B. (2000). Math curse or math anxiety?Teaching Children Mathematics, 6, 330-335.

8

2Number and Operations

SETS

FOCAL POINTS

• Set Definitions• Set Operations• Special Sets• Venn Diagrams• Properties• Factors and Multiples• Prime and Composite Numbers• Sieve of Eratosthenes• Divisibility Rules. Greatest Common Factor and Least

Common Multiple

A firm understanding of the concept ofsets establishes a critical background forthe study of mathematics. We start bylooking at collections of things. Fromthere, we focus on how many things arecontained in the set. Once we have thatcardinal number (total number of elementsin the set), the emphasis may shift to per-forming different operations with the num-bers associated with sets. The basics ofarithmetic are established through ourwork with sets. Once one has arithmeticunder control, the sky is the limit.

Set Definitions

You have probably already encounteredthe concept of sets, so you undoubtedlyhave some idea of what the word meansin a mathematical context. We could be

talking about a set in a tennis match, a setfor a play, the set for a lesson you are go-ing to teach, or a set of dishes. Actually,any set of dishes begins very naturally toestablish mathematical leanings becauseyou might ask if the "service" is for 4, 6, 8,10, or 12. Useful considerations are givento the numbers of different types ofpieces in the set, including or excludingthe service pieces. This is only one exam-ple of how we use the ideas of sets in ourdaily lives.

Mathematically, we may speak of a setin terms of a collection, a group, a flock, agaggle, a herd, a bunch, a pile, somethings, and so on. We can use a variety ofways of saying the same thing and, aslong as the people in the conversation un-derstand the desired meaning, the dis-cussion continues without confusion.

Mathematicians are concerned that ev-ery set be well defined. If a set is well de-fined, then it is easy to tell whether or notsomething belongs to it. If we talk aboutthe names of the people in your family,you might wonder if you should includegrandparents, first cousins, second cous-ins, and so on. If one person in the con-versation is not sure whether secondcousins should be included or excluded,then the set is not well defined. At thesame time, if you were asked to list thenames of the people in your immediatefamily, typically you would interpret theset to include only your parents and sib-lings. That is fairly well defined, but whereshould a married person with children

9

10 CHAPTER 2

draw the line? Perhaps the definition"your immediate family" is not completelydescriptive. The set of all the stars in oursolar system is a clearly defined set, isn'tit? How many items are contained in thatset? Is the Earth in that set? Think of theset of clothes in your closet or drawers. Isit clear which items fit and which do not?

It is not reasonable to try to define a setwithout discussing the items contained inthe set. Whether the items in a set areconcrete (e.g., plates or bowls) or ab-stract (e.g., numbers), they are called ele-ments. An element of a set is a member ofthe set, something that is in the set. Mostof the time it is easy to tell what is andwhat is not a member of a set. More thanlikely, the sets you've seen have beenmade up of things with common themes:a set of dishes, a set of silverware, a herdof sheep, a flock of birds, a gaggle ofgeese, and so on. Those sets lead to theimpression that the elements of a setmust have something in common. Thisconclusion may seem reasonable, but itsimply is not required.

Think of a classroom. If you were askedto list everything in it, what would you in-clude? Desks (or tables and chairs), over-head projector, books, pads, pens andpencils, chalk (or markers), boards, eras-ers, carpet, and who knows what else.Would you list the people too? Supposeyou do include people. If that whole col-lection of elements makes up the set weare discussing, then what is the common-ality? One answer might be, "All of thosethings are in this room." Although thatwould be a shared characteristic, it is nota typical use. You could give a unifieddefinition for the set of desks in yourroom, even if several different sizes ortypes of desks are present. You mightalso talk about the group of people andthe connection is made. But, putting all ofthe things in one big set seems strange

because of the lack of any shared charac-teristic (except being in the same room).Still, this is technically a well-defined set.There is no need for the elements of a setto have common attributes.

In the discussion about the members ofyour family and immediate family, the am-biguity of wording was easily corrected sothe words gave an adequate description ofthe set. The set was described verbally,which is an important way of communicat-ing the required image. Effective discus-sions may focus on verbal descriptions ofall sorts of things that are common in peo-ple's lives: pets, things in a toy box, differ-ent kinds of shoes, or a myriad of other im-portant real-life objects. Most of the time,these discussions are clear and the de-sired images are projected.

Another common way of discussing theelements of a set is to list all of the ele-ments by name. Every teacher has a list ofstudents in a grade book. Look at that listand you quickly know who is and is notsupposed to be in the class. Santa Clausfrequently calls his reindeer by name.Rudolph was added later in the story, so,for a long time, he was not listed as amember of the set of Santa's reindeer.

Sometimes sets are discussed using ti-tles or names. An example of a set de-fined by a title is "Students." This is use-ful, but sometimes the issue of whetherthe set is well defined creeps back in.When you think of students, do you thinkof your current class and the people in it,all the students who matriculate at yourcollege, children in elementary school, orsome adult who might be a student oflife? Even a name as simple as "Students"might not deliver a clearly defined mes-sage. The set A = {1, 2, 3, 57} tells youwhat is being described. Set A has fourelements: 1,2,3, and 57. You know whatdoes and does not belong to this well-defined set and you have a quick, easy

NUMBER AND OPERATIONS 11

way of referring to the set—just call it SetA. Braces are commonly used when nam-ing sets this way. Customarily, the begin-ning capital letters of our alphabet areused for naming sets in mathematicsclasses, but this is not mandatory. Later,you will encounter R, C, Z, and W as thenames of sets of numbers.

Sometimes we can list all the elementsof a set, as with A = {1, 2, 3, 57}, but thatis not always possible or even desirable.When discussing the counting numbers,we generally call the set, N. However, wecannot possibly list all of them as we didwith Set A. Please do not try this at home;even professionals on closed tracks don'tattempt this chore. Why not? A solutionfor this dilemma is to use an ellipsis andwrite, N = {1, 2, 3, 4, . . .}. A couple ofthings happen right there. First, enough ofa pattern is established to clarify what setof numbers is being discussed. Further-more, the impression that things keep onfollowing that pattern is established bythose three dots, or ellipsis. When yousee those, it means that the pattern con-tinues. The continuation could be infinite,but sometimes it terminates.

In a large, but finite, set where the pat-tern stops, the ellipsis is still used, asshown by: B = {2, 4, 6, 8, . . . , 50}. Whatdo you suppose that means? Do you seethe even counting numbers, starting thepattern with 2, 4, 6 and continuing until itstops at 50? Did you notice that in the setnotation you see 2, 4, 6, 8 and yet in thediscussion we gave only 2, 4, and 6? Thatis not a big deal, as long as the pattern isclear. If in doubt, then you may insert ad-ditional elements of the pattern to avoidany possible confusion.

Sets can be written in a variety of for-mats. Often the environment and back-grounds of the participants determinehow the set is expressed. The exampleswe discussed have included formal set

FIG. 2.1.

notation, using mathematical sentencesand braces. Figure 2.1 gives ways ofshowing sets of things using looping. Fig-ure 2.2 depicts things that are not exam-ples of sets of things being looped.

FIG. 2.2.

Your Turn

1. Based on the examples in Figs. 2.1and 2.2, write a definition of what loopingthe elements of a set means.

Notice that we switched tactics on you.Initially we told you that a set was a col-lection, a group, a flock, a gaggle, a herd,a bunch, a pile, some things, and so on.

12 CHAPTER 2

We told you about elements of a set andhow a set would be well defined. That is acommon way of doing business in theworld of teaching mathematics, but it isnot always the best way for you to learnthings. We know that having you be anactive participant in your learning is muchbetter than if you are constantly beingtold. Research shows that you will retain50% of the material if you participatewithin group discussions, 75% if you ac-tually practice the skill, and 90% if you aregiven the opportunity to teach the mate-rial to others (National Training Labora-tories, Bethel, Maine; http://www.gareal.org/learningpyramid.htm). We have pro-vided an answer section and you couldcertainly look at it, which is equivalent tobeing told the answers. We hope you donot do that. The more you think aboutwhat is being discussed, the easier it willbecome, the better it will be for you, andultimately, your future students will benefit.

Now that you know what sets and ele-ments are and how to show things in dif-ferent ways, we can deal with some of theother common terminology that accom-panies sets. The following sets are finite:

A = {1,2, 3, 57}.B = {2, 4, 6, 8 50}The hairs on your head

These sets are not finite:

N = {1,2, 3, 4, . . . }The set of multiples of 3The set of integers

Your Turn

2. Define the term finite set.3. Define the term infinite set.

These last two exercises can be used toraise some interesting questions. Is the to-tal population of the earth finite or infinite?

Most people would say finite, even thoughit is a very big number that changes be-cause of births and deaths. Is the numberof blades of grass on your campus finite orinfinite? This gets a little more obscure, butmost people would say it is finite. It mighttake a while to determine how manyblades there are, but it is still finite. Extendthat idea to all the campuses in the world,and you still have a finite number. Throw inall the yards, fields, pastures, and what-ever other name you can come up with fora grassed area and the number will still befinite—huge, but finite. Well, how aboutthe grains of sand at Daytona Beach? Fi-nite. Put in all of the beaches in the worldand the number of grains of sand is still fi-nite—again, huge, but finite. Almost any-thing you name will be finite whencounted. It is only when you shift to ab-stract ideas, like sets of numbers, that youget into the realm of the infinite.

Now that set is defined, we need to ex-amine some other aspects of the topic. Ineach situation, we will give you examplesand nonexamples of situations. Your taskwill be to extrapolate a definition from theinformation. Each of the following exam-ples shows a set as a part of another set,or a subset:

{2, 7} is a part of {1, 2, 5, 7, 19}.{Jo, Chris} is a part of {Jo, Chris, Pat}.Dogs and cats are a part of the set ofmammals.

{2, 4, 7} is a part of {7, 2, 4}.

The following are situations where the firstset is not a subset of the second:

{Honda, Chevrolet} is not a part of{Historically American-made cars}.

{H, O, W} is not all part of {V, O, W, E, L}.{2, 4, 6, 8, 11, 12} is not part of the set

of even numbers.{2, 4, 7, 8} is not part of {7, 2, 4}

NUMBER AND OPERATIONS 13

Your Turn

4. Define the term subset.

Think about a set that has no elementsin it, in other words, an empty set. Al-though that might sound strange, theempty set is an important part of the toolsnecessary to build our number system.Historically, the formalization of the ideaof the empty set took a long time to de-velop. When a cave man looked at thecave walls of the next cave man over, itwas easy to tell what the neighbor had.The pictures of three horses, four bronto-sauruses, and two chickens told thestory. No pictures of cows on the wallsmade it clear that the neighbor had nocows. Before the idea of the empty set de-veloped, nobody gave a second thoughtto that which was not listed. Thus, at leastinitially in our mathematical developmen-tal history, the concept of an empty setwas not even considered!

As times changed, there was a need tobe able to say, "I do not have any cows."This was the beginning of the develop-ment of the empty set. Although youknow the cave man discussion is a fig-ment of our imaginations, you should seethe major idea—the empty set depicts aset with no elements in it.

The empty set plays an important partin our lives. Have you ever passed a con-struction site and noticed a 55-gallondrum with "MT" written on the side inlarge letters. Say those letters fast, oneafter the other, and you get empty. Theletters mean the drum is empty. Thatnaming method is not normally seen inthe school curriculum, but it accuratelydescribes the situation for the membersof the construction crew. In mathematicsclasses, we use either {} or fy to depict theempty set. A very common error is to use{<))} to represent the empty set. Use either

{ } or <j), but NEVER use the two symbolstogether. As soon as something is placedbetween the "{" and the "}", the set is notempty, even if the only element of the setis the symbol §.

Another definition needed in the frame-work for using sets is demonstrated in thefollowing examples of proper subsets:

{2, 7} is a part of {1, 2, 5, 7, 19}.

{Jo, Chris} is a part of {Jo, Chris, Pat}.Dogs and cats are part of the set of

mammals.The empty set is a part of every set.

These are nonexamples of proper sub-sets:

{2, 4, 7} is a part of {7, 2, 4}.{1, 2,3,4} is not a part of {2,4,6, 7}.

Your Turn

5. Define the term proper subset.

Improper subsets are generally dis-cussed along with proper subsets. Thefollowing example of an improper subsetcoupled with your definition of propersubsets should help you create a defini-tion for improper subsets:

{2, 4, 7} is an improper subset of{7, 2, 4}.

Your Turn

6. Define the term improper subset. Howmany improper sets may any set have?

We are now at a point in our discussionwhere we need to take the first steps to-ward developing the number system youuse every day. One initial interest in a setoften revolves about what is called the

14 CHAPTER 2

cardinality, or cardinal number, of a set.We would say:

The cardinality of {2, 4, *, A} is 4.The cardinal number of (3, v, V, ®, #}

is 5.The cardinality of the letters of our

first names {M, a, r, y, E, P, e, g, g, y,D, o, u, g} is 11 because letters are notrepeated when elements of a set arelisted.

Your Turn

7. Define cardinality.8. What is the cardinality of the empty

set?9. What is the cardinality of ft}?

10. What is the cardinality of {BMW}?

A lowercase n is placed in front of theletter name of a set to express the cardi-nality of that set. If D ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, then the cardi-nality of D is 10 and would be writtenn(D) = 10.

With the basic definitions established,we can investigate some interestingmathematical ideas. The set {3} has twosubsets: { } and {3}. Of those two subsets,one is improper and one is proper. Theset {7, $} has four subsets: { }, {7}, {$},and {$, 7}. Of these four subsets, one isimproper and three are proper. The set{13, #>, &} has 8 subsets: { }, {13}, {&},{&}, {13, p}, {13, &}, {$>, &}, and the set it-self or {13, £>, &}. Of these, one is im-proper and seven are proper subsets. Aset with four elements has 16 subsets.You should list them and observe the pat-terns that are forming:

Your Turn

11. Generalize the pattern establishedby the total number of subsets as thenumber of elements in the set is in-creased one element at a time.

12. A set with 11 elements will havesubsets, of which are

of which are

# elementsin set

1234

# impropersubsets

1111

# propersubsets

137

15

# subsets248

16

improper, andproper.

13. Is there a case where the numberof proper and improper subsets will beequal? Is there more than one case? Whyor why not?

Now that cardinality of sets has beenintroduced, the definitions of finite andinfinite sets should be revised. Finitesets have cardinality, whereas infinite setsdo not have a defined cardinal number.Sometimes it is said that the cardinality ofany infinite set is infinite. That may seemlike saying a rose is a rose and yet, in ourlanguage, infinite is how we name never-ending situations if elements are beingcounted.

Two more ideas grow directly from theconcepts of sets and their cardinality:equal and equivalent.

{$, 5, *} is equal to {$, 5, + }{3, &, Z, } is equal to {3, &, Z, }

{3, BMW} is equal to {BMW, 3}{$, 5, *} is equivalent to {B, M, W}{3, &, Z, } is equivalent to {5, *, Z, }

{3, BMW} is equivalent to {BMW, 8}

Your Turn

14. Define equal sets and equivalentsets.

15. Which tells you more about twosets: equal or equivalent? Why?

NUMBER AND OPERATIONS 15

For the next few exercises, we will pro-vide statements and ask you to formulatemore definitions:

{3, &, Z, } and {5, + , Z, } are overlap-ping sets.

{3, &, Z, } and {5, *, 83, 409, Z, } areoverlapping sets.

{3, BMW} and {3, 4, 5, 6} are overlap-ping sets.

{3, BMW} and {3, &, Z, } are disjointsets.

{A, B, C, D, E, F} and {312, 409, 513}are disjoint sets.

{1, 2}, {Buckle my shoe}, {5, 6}, and{Pick up sticks} are disjoint sets.

{1, 2, 3, 4, 5,. . .} can be partitioned into{1, 3, 5, . . . }and {2,4, 6, . . .}.

The shoes I wear can be partitionedinto right or left.

Triangles can be partitioned into equi-lateral, isosceles, and scalene.

Your Turn

16. Define overlapping sets.17. Define disjoint sets.18. Define a partitioned set.

Another subset idea that needs to becarefully defined is when we want totalk about everything BUT the subset. LetA = {s, t, a, r}, where two subsets could beformed like B = {a} and C = {s, t, r}. Thesubset B is all the vowels in A and C is allof the consonants in A. Together, B andC, would form A. So far, that isn't any-thing new. Mathematicians look at the re-lation between B and C and say C is ev-erything in A BUT the elements in B. Theycall a set with such a set with such a cor-relation the complement of B. This can bewritten symbolically in a variety of ways:

We will use the notation B' to

represent the complement of a subset. IfC is the complement of B, or C = B', thenwhat is implied about B to C? The com-plement of C is B, or B = C'. This associa-tion will exist anytime a set is split into twodisjoint parts such that all of the elementsof the original set are accounted for bythe subsets.

Your Turn

19. Define complements of sets.

Set Operations

We operate (add, subtract, multiply, di-vide, raise to a power, .. .) on numbers.We can also operate on sets, but the oper-ations are different from those applied tonumbers. We talk about 3 + 4, but it wouldbe inappropriate to talk about A + B,where A and B are names of sets. Withthat in mind, consider the following exam-ples of set operations where u, n, and Xrepresent union, intersection, and set mul-tiplication, respectively:

The set operation of subtraction is re-lated to the complement of a set idea andis called the relative complement of a set.

16 CHAPTER 2

The relative complement of A to B is writ-ten B - A and can be read, "The comple-ment of A relative to B" or "The set differ-ence between B and A." This operation isused between any two sets that intersect.Revisit the two previous statementsabout intersection:

Observe the following examples of rela-tive complements:

Your Turn

20. Define u (the union of sets).21. Define n (the intersection of sets).22. Define X (set multiplication or set

product).23. Define relative complement (set

subtraction).24. State a generalization about the

cardinalities of sets and their union.25. State a generalization about the

cardinalities of sets and their set product.

Special Sets

Some sets will occur often throughoutthis text. They are commonly used in theworld of mathematics and, because thedefinitions are generally agreed on forthese, we choose to list them for you.Please take the time to refresh your mem-

ory about the following sets because theneed for lengthy reflections about whatthey mean as you move on will slow youdown and may limit your understanding:

Digits, or D = {0,1, 2, 3, 4, 5, 6, 7, 8, 9}

Counting numbers, orN = {1,2, 3, 4, . . .}

Whole numbers, or W = {0,1,2, 3,. . .}

Integers, orZ = {. . .-3,-2, -1,0, +1,+2, +3, . . .}

Rational numbers, orr ~i

Note that some people use Q to repre-sent the rational numbers. We will useRa in this text.

Irrational numbers, or IRa = {numbers

that cannot be written as such as n

or }

Real numbers, or

Complex numbers, or C = {a + b/ wherea and b are real numbers, [

Universal set, or U = contains all the el-ements needed to describe a situation

Venn Diagrams

You have probably heard the expression,"A picture is worth a thousand words."Venn diagrams are pictures that clarifysome of the mathematical situations youmight encounter. Figure 2.3 shows exam-ples of disjoint sets in a Venn diagram. Dothe pictures show what you believe aredisjoint sets? Figure 2.4 depicts intersect-ing or overlapping sets. The darkenedsection in each example is common to allthe sets involved. Both Fig. 2.3 and Fig.2.4 are generic and although they depictthe situation, they are not very helpful indetermining exactly what is meant by the

NUMBER AND OPERATIONS 17

FIG. 2.4.

two situations. Suppose you had the setof odd counting numbers, O, and the setof even counting numbers, E. Both ofthose sets are infinite so all the elementscannot be shown. Some elements of thedisjoint sets are shown in Fig. 2.5. An el-lipsis could perhaps be inserted in eachset to indicate their infinite natures butthat could be confusing. Some peoplemight consider using "etc." That toocould be confusing. One method that issometimes used to indicate infinite sets inVenn diagrams is to make the loopdashed, which we have done.

Similar diagrams could be generatedusing the rational and irrational numbers;

positive and negative numbers; countingnumbers less than 100, the counting num-ber 100, and counting numbers greaterthan 100; as well as many others. Noticethat one of the examples (counting num-bers less than 100,100 itself, and countingnumbers greater than 100) involved threedisjoint sets. The only stipulation for dis-joint sets is that you need at least two sets,with no common elements. Later, in statis-tics, you will see disjoint sets referred to asmutually exclusive sets.

Suppose you had setA = {1, 2,3,4,25,35, 57} and the digits (D).Figure 2.6 shows how these overlappingsets might appear in a Venn diagram.

FIG. 2.5.

FIG. 2.6.

Venn diagrams may have more thantwo sets; the possibility for examples islimited only by your imagination. For ex-ample, you could use counting numbersin one set and whole numbers in anotherset; even counting numbers in one setand digits in another set; or countingnumbers in one set, digits in a second set,and integers in a third set. The possibili-ties are limitless. A fine detail you shouldnotice is that the overlapping, or commonelements, appear only once in the Venndiagram. In Fig. 2.6, the numbers in theoverlapping part, 1,2,3, and 4, are shownonly in the overlapping part because they"belong" to each set.

18 CHAPTER 2

Your Turn

26. Describe what the universal setmight be for each of these examples ofunions and then construct a Venn dia-gram that accurately depicts each state-ment:

27. Describe what the universal setmight be for each of these examples of in-tersections and then construct a Venn di-agram that accurately depicts each state-ment:

28. Describe what the universal setmight be for each of these examples ofset multiplication and then construct aVenn diagram that accurately depictseach statement:

29. Let the universal set for each ofthese examples be the set of digits and

then construct a Venn diagram that accu-rately depicts each statement:

Properties

There are properties, rules, axioms, andideas that are consistently true as westudy numbers. There is a basic collectionof 11 properties, sometimes called fieldaxioms, which are developed, used, andreferred to throughout the mathematicalcurriculum. These properties are funda-mental to the development of mathemati-cal knowledge and can even save time asnumber facts are learned. As one pro-gresses to the study of more advancedmathematical topics, some of the proper-ties are removed to create groups, loops,rings, and other topics in the study of ab-stract algebra. So, what are these proper-ties?

Commutative property of addition on aset

Commutative property of multiplicationon a set

Associative property of addition on aset

Associative property of multiplicationon a set

Identity element for addition on a set

Identity element for multiplication on aset

Inverse element for addition on a set

NUMBER AND OPERATIONS 19

Inverse element for multiplication on aset

Closure property of addition on a set

Closure property of multiplication on aset

Distributive property of multiplicationover addition on a set

As we discuss these important propertiesin detail, we will use proper terminologyas opposed to slang (yet commonly ac-cepted) terms. Any discussion of a prop-erty must begin with a set of elementsand an operation.

Undoubtedly, you are aware that theorder in which you add two numbers is oflittle consequence: 3 + 5 = 5 + 3. This isan example of the commutative propertyof addition on the set of counting num-bers; the property is true for all the setswe will discuss (digits, wholes, integers,rationals, reals, or complex). The signifi-cant thing is that an operation (addition)and a set of elements have been identi-fied. In our discussion of 3 + 5 = 5 + 3, wedid not merely mention the commutativeproperty. It was called the commutativeproperty of addition on the set of countingnumbers. Although you might know whatis meant when someone mentions the com-mutative property, or commutativity, there isa need for precision of language in theworld of mathematics. We must be carefulto differentiate between the commutativeproperty of addition on the set of countingnumbers and the commutative property ofmultiplication on the set of integers.

In our example, we could have usedthe commutative property of multiplica-tion on the set of whole numbers, sayingthat 3x5 = 5 x 3 . You know that multiply-ing these two factors will always give youthe same product, regardless of the orderin which you choose to use the factors.The gist of the commutative property of

either addition or multiplication for the setyou use is that the order of the elementscan be switched and the result will be thesame. You have probably seen these twoproperties defined together as:

a + b = b + a and a x b = b x a, where a and bare elements from the chosen set and theoperation is either addition or multiplication.

Stating the properties with letters ratherthan specific numbers generalizes theidea. This is a convenient way of summa-rizing the statements for all numbers be-cause any number could be substitutedfor either a or b.

There is a difference between the com-mutative property of addition on the set ofwhole numbers and the commutative prop-erty of multiplication on the set of wholenumbers. This is generally not difficult tosee with 2 + 3 = 3 + 2 and 2x3 = 3 x 2 .The addition gives a sum of 5 and themultiplication gives a product of 6, sosaying whether or not you are commut-ing 2 and 3 over addition or multiplicationis important. However, you could have2 + 2 = 2 + 2 and 2x2 = 2 x 2 , each ofwhich gives an answer of 4 and impliesthat saying commutative alone is suffi-cient. Generally, it does make a differ-ence, however, as you deal with additionor multiplication.

Is there a commutative property forsubtraction on the set of integers? If yousaid yes, then you need to revisit thethinking process that led you to that con-clusion. For that statement to be true, itmust be the case that 6-2 = 2-6. Noticethat, with subtraction over the set of inte-gers, 6 - 2 = 2 - 6 is not a true statementbecause 6-2 = 4 whereas 2 - 6 = ~4.There is a BIG difference between 4 and~4. In general then, there is not a commu-tative property for subtraction on the setof integers. We could provide an example

20 CHAPTER 2

to show that there is not a commutativeproperty for division on the set of inte-gers. However, we ask you to consider it,using our subtraction example as amodel, to give you a chance to exerciseyour reasoning skills. Can you think of anyexceptions to these general rules?

The associative properties for additionand multiplication on a given set are simi-lar to those for the commutative propertyof an operation with some set. The orderof the elements does not change, only theplacement of the parentheses, which indi-cates the operation to be performed first.Some examples of the associative prop-erty of addition and multiplication on theset of counting numbers are(2 + 4) + 7 = 2 + (4 + 7) and(2 x 4) x 7 = 2 x (4 x 7). The more generalstatement is typically given as:

(a + b) + c = a + (b + c) and (a x b) x c = a x(b x c), where a, b, and c are elements fromthe chosen set and the operation is eitheraddition or multiplication.

This last statement generalizes what isgoing on. You should be able to describeit in words as well as give specific exam-ples or a generalized version involving let-ters. Why do you think our discussion re-volves about addition and multiplicationonly? Investigating this property for sub-traction and division will enhance yourunderstanding of the associative propertyof addition or multiplication on a givenset.

The concept of closure within a set forany operation is difficult for some tograsp. Fraternities and sororities are ex-amples of closed groups. Althoughguests are welcome to attend many func-tions, only members are permitted tovote. In such an exclusive situation, whois and who is not a member is clearlyknown. The closure property is quite simi-

lar, actually. Think of any two countingnumbers. Add them. Is the sum a count-ing number? Of course, you say. Is thereany example of adding two counting num-bers where the sum is not a countingnumber? No? It is said that the countingnumbers are closed under addition (no-tice the set and operation are both namedthere). True or false—the digits are closedfor multiplication? False. At first glance,you might ask, "What about 2 x 3 = 6?"This statement is true because the prod-uct is a digit. However, 5 x 7 = 35 gives aproduct that is not an element of the set(digits). The conclusion must be that thedigits are not closed for multiplication.

The identity elements for addition andmultiplication are special numbers. Zerois the identity element for addition andone is the identity element for multiplica-tion within any set that contains these re-spective elements. Consider 3 + 0 = 3 =0 + 3. The commutative property for ad-dition on the set of wholes is implicit inthe example. We cannot generalize froma single example, but we assume thatyou are already familiar with the idea thatzero is the additive identity (identity ele-ment for addition). You may have seena + 0 = a, where a is an element of any setthat contains 0. Similarly, one is the iden-tity element for multiplication starting withthe counting numbers, typically shown asa x 1 = a, where a is an element of any setthat contains 1.

Whereas the inverse element for an op-eration on some set is easy to conceptu-alize, it is often difficult to express. First,look at some examples where elementsare the operative (either additive or multi-plicative) inverse for a given element fromsome appropriate set:

NUMBER AND OPERATIONS 21

might express specific preferences.When you are in Rome

Your Turn

Given these examples, you should havethe idea that the inverse element needs tocommute for the given operation in theselected set, and when the operation iscompleted, the result is the identity ele-ment for the respective operation.

The remaining property is unique in thatit is the only one that combines two oper-ations. You have seen it many times inyour prior work and yet might not haverecognized it. Perhaps you recall seeing3 x (4 + 5) = (3 x 4) + (3 x 5). Notice thatthe order of the elements is maintained,which is significant. You probably call thisthe distributive property even though it isproperly called the left distributive propertyof multiplication over addition because the3 is being distributed via multiplicationfrom the left over the sum of 4 + 5. Didyou remember seeing this concept whilelearning to factor in your algebra class?Perhaps you saw it with the setup re-versed. Also, because the times signlooks so much like a variable, you mighthave seen an expression or equation in-volving juxtaposition, or implicit multipli-cation, such as 4yz +12z = 4z(2y + 3)where multiplication is indicated, but themultiplication symbol is absent, but im-plied or understood to be there. Still, youcan recognize this as the left distributiveproperty of multiplication over addition onthe set of numbers.

We will use all of these propertiesthroughout the text. We encourage you tobecome comfortable with concepts aswell as the different ways of referring tothem. You are invited to use the phraseol-ogy that is most comfortable for you, butbe advised that other teachers or writers

30. We know 2 + 2 = 2 x 2, as dis-cussed with the commutative property ofaddition on the set of counting numbers.Are there any other pairs of numbers forwhich the sum and product are equal?

31. Is there a situation where commu-tativity of subtraction on some set wouldexist?

32. Is there a situation where commu-tativity of division on some set would ex-ist?

33. Generalize the idea of commuta-tivity of operations for some set in yourown words.

34. We know 2 + 2 = 2x2 as dis-cussed with the commutative property ofaddition on the set of counting numbers.Are there any examples that will work likethis for associativity?

35. Is there a situation where associa-tivity of subtraction on some set wouldexist?

36. Is there a situation where associa-tivity of division on some set would exist?

For each of 38 through 41, select True orFalse and explain the reason for yourchoice.

37. Generalize the idea of associativityof operations for some set in words.

38. The even counting numbers areclosed for addition.

39. The odd counting numbers areclosed for addition.

40. The even counting numbers areclosed for multiplication.

41. The odd counting numbers areclosed for multiplication.

22 CHAPTER 2

42. Give an example of a set that isclosed for addition and describe why it isso.

43. Give an example of a set that isclosed for multiplication and describe whyit is so.

44. Would the commutative property foraddition on a given set have a negativeimpact on the situation if we insisted thatit hold true while we discussed closure?

45. Generalize the idea of closure ofoperations for some set in words. Don'tforget about division and subtraction.

46. Generalize the idea of the identityelement for an operation for some set inwords.

47. Generalize the idea of the inverseelement for an operation for some set inwords.

48. In general, is there a left distributiveproperty of multiplication over subtractionin the real numbers?

49. In general, is there a right distribu-tive property of multiplication over addi-tion or subtraction on the set of reals?

50. In general, is there a left distributiveproperty of division over addition in thereals?

51. In general, is there a right distribu-tive property of division over addition inthe set of real numbers?

52. Generalize the idea of the distribu-tive property for multiplication over addi-tion for some set in words.

Factors and Multiples

Concepts involving the properties ofwhole numbers, or number theory, pro-vide the foundations for much of thestudy of higher mathematics. Factors andmultiples of numbers appear early in thestudy of numbers as aspects of divisionand multiplication. The emphasis of multi-ples and factors is a central part of work

in beginning algebra, but often receivestoo little attention in our elementaryschools. Yet, two important conceptssurrounding factors and multiples, oftenreferred to as LCM (least common multi-ple) and GCF (greatest common factor),are confusing for many students. Usually,the definitions are known, but peoplestruggle to know which definition goeswith which concept.

Your Turn

53. Using the following examples, writea definition of multiple and a definition offactor:

Multiples of 12 are 12, 24, 36, 48, ...Multiples of 5 are 5,10,15, 20, 25, . . .Multiples of 7 are 7,14,21, 28, 35,...Multiples of 2 are 2, 4, 8, 16, 32, 64,

128, . . .Factors of 12 are 1, 2, 3, 4, 6, 12Factors of 7 are 1, 7Factors of 100 are 1, 2, 4, 5,10, 20, 25,50, 100

Factors of 36 are 1,2,3,4,6,9,12,18,36

Prime and Composite Numbers

Can you state definitions for the set ofprime numbers and the set of compositenumbers? Traditionally, these definitions,with examples, are provided by teachersand memorized by students. An approachusing patterning allows you to discover thesets of prime and composite numbers.This approach requires a basic under-standing of finding areas of rectangles, butit is well worth the effort to develop.

Using only counting number dimen-sions, how many different rectangles arethere with area 2 square units? You might

NUMBER AND OPERATIONS 23

wonder if the position of the rectanglematters in this discussion. Most peoplewould say there are two rectangles that fitthis requirement, as shown in Fig. 2.7.

FIG. 2.7.

It should be noted that both the 1 unitby 2 unit and the 2 unit by 1 unit rectan-gles in Fig. 2.7 are in standard positionand that there are infinitely many rectan-gles that would have dimensions of 1 unitby 2 units and areas of 2 square units asindicated in Fig. 2.8.

FIG. 2.8.

Although there is an infinite number ofrectangles with areas of 2 square units,given counting number dimensions, therectangles in Fig. 2.8 are all rotations ofthe same 1 unit by 2 unit (1 x 2) rectangle.A convenient way to organize this discus-sion is to give only one of the infinite set ofrectangles with counting number dimen-sions as representative of the entire set,and list the smaller dimension first.

We talked about figures with a givenarea, 2 square units. Some mathemati-cians say that the word area implies thepresence of square units whereas othersfeel the idea of square units should alwaysbe included in the discussion. The easyanswer to the situation is to always listsquare units with a discussion about area.

Develop a three-column table in whichyou list the areas given in the following

questions in the left column, the countingnumber dimensions of all rectangles thatgive each area in the middle column, andthe total number of different dimensionpairs in the right column. For example, forarea two, you will have 2 in the left col-umn, 1 x 2 in the middle column, and 1 inthe right column. It would be helpful if yousketch your rectangles on square grid ordot paper.

Using only counting number dimen-sions, how many different rectangles arethere with area 3 square units?

Using only counting number dimen-sions, how many different rectangles arethere with area 4 square units? For thisone, you must remember that all squaresare rectangles, so you count the 2x2square, in addition to the 1x4 rectangle,giving you two rectangles.

Using only counting number dimen-sions, how many different rectangles arethere with area 5 square units? With area6 square units? With area 7 square units?With area 8 square units? With area 9square units? With area 10 square units?With area 11 square units?

Your Turn

54. What conclusions can you drawfrom your table?

An example like Fig. 2.9 may provehelpful in leading to the desired conclu-sions about your table in Exercise 51.Your table should indicate two distinctsets of numbers. Some rectangles haveareas that are found only one way(2, 3, 5, 7, 11), whereas others can befound two ways (4, 6, 8, 9, 10). Do yourecognize these sets? They are the primeand composite numbers. Given the infor-mation you have so far, you might definethese sets as, "Prime numbers give onlyone rectangle" and "Composite numbersgive two rectangles."

24 CHAPTER 2

FIG. 2.9.

Now think about a rectangle with area12. You can get 1 x 12, 2 x 6, and 3 x 4 asdimensions—or three different rectangles.The definition of a composite number hasto be refined to say two or more rectan-gles. This discussion is valuable as ameans of emphasizing that functional defi-nitions sometimes need to be expanded oraltered as horizons are expanded.

What about 1 ? You can sketch a rectan-gle that is 1 x 1, so is it prime? In fact, 1 isneither prime nor composite. The Funda-mental Theorem of Arithmetic assures thatany number may be expressed as aunique product of prime numbers, exclud-ing order. Thus, 6 = 2x3 or 3x2 and weare assured that no other product ofprimes will give an answer of 6. If 1 wereprime, then we would have 1 x 2 x 3 = 6,1 x 1 x 2 x 3 = 6, and so on. Although it isstill the case that the product is 6, theuniqueness part of the definition, which iscritical to our number system, would belost. We need the assurance that a spe-

cific set of prime factors will give only oneproduct, and conversely, that any com-posite number is generated by a uniqueset of primes. Without that uniqueness,we do not have the assurance that 12 = 2 x2x3 because there might be some otherprime lurking out there somewhere that wedo not see, or know about. To avoid allthose complications, 1 is left in a class byitself and is neither prime nor composite.Some texts call 1 the generator.

Sieve of Eratosthenes

The Sieve of Eratosthenes is used to de-velop a list of primes by applying the pat-terns associated with multiples of count-ing numbers. A multiple of a number isdefined as the number times a countingnumber. For example, the multiples of 7are 7, 14, 21, 28, 35, 42, . . . , which isalso 1 x 7, 2 x 7, 3 x 7, 4 x 7, 5 x 7, 6 x 7,. . . . In this example, 7 is the number un-der consideration and the counting num-bers are 1, 2, 3, 4, 5, 6 , . . . . Some peopledefine multiple to be a whole numbertimes the value in question, making 0 avalid multiple of 7 as well. For this exer-cise, we will use the counting number def-inition. Use the following chart to com-plete the instructions given below it.Although 1 is neither prime nor compos-ite, and therefore will not be part of theexercise, it appears in the chart.

NUMBER AND OPERATIONS 25

Loop 2. Cross off all multiples of 2 inthe chart.

Loop 3. Cross off all multiples of 3 inthe chart.

Loop 5. Cross off all multiples of 5 inthe chart.

Loop 7. Cross off all multiples of 7 inthe chart.

Continue looping each prime andcrossing off all multiples in the chartuntil all primes on the chart are shown.

Your Turn

55. Complete the Sieve of Eratos-thenes following the directions given inthe text immediately before this exercise.Although this chart stops at 100, it couldbe continued to any desired value.

56. What is the greatest number ofprimes in any given row?

57. Would the answer in Exercise 53change if the chart were extended indefi-nitely to include more rows or more col-umns?

58. Next consider the least number ofprimes in any given row?

59. Would the answer in Exercise 55change if the chart were extended indefi-nitely by adding more rows or columns?

60. Create a sieve on a 6-columnchart, then answer the questions that fol-low (the same questions asked for the 10-column sieve). Even though the samequestions are asked, the answers willchange, thus enhancing understanding.

61. What is the greatest number ofprimes in any given row of the 6-columnsieve?

62. Would the answer in Exercise 58change if the 6-column sieve were ex-tended indefinitely?

63. Now, what is the least number ofprimes in any given row of the 6-columnsieve?

64. Would the answer in Exercise 60change if the 6-column chart were ex-tended indefinitely?

The sieves open the door to an interest-ing question. Assume you have the 10-column sieve. Is it possible to have 10consecutive composite numbers? Theanswer is yes, but the table would need tobe extended vertically a lot farther. Sup-pose you consider11 x 1 0 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1(39,916,800). The next counting numberafter that is 39,916,801 and the countingnumber after that is 39,916,802. Focusingon 39,916,802, it can be written as39,916,800 + 2. We know that 2 divides 2and we know that 2 is one of the factorsof 39,916,800. Using the distributiveproperty of multiplication over addition onthe set of counting numbers, 39,916,802= 2 x 19,958,400 + 2 x 1 or 2(19,958,400+ 1), showing that 2 is a factor of39,916,802. A similar approach can beused for 39,916,803, 39,916,804,39,916,811, showing 10 consecutivecomposite numbers. When dealing withsuch large numbers, there is a more con-cise way to write them;11 x 1 0 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ,or 39,916,800, is also written 11!. You areseeing correctly. The 11 is followed by anexclamation point—in mathematics, 11!means11 x 1 0 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ,and is read 11 factorial.

Divisibility Rules

The discussion about factors and multi-ples provides the background for a veryimportant concept that is built on an-other definition. When a mathematiciansays a number is divisible by another, theassumption is that, when the division iscompleted, the remainder will be zero.

26 CHAPTER 2

For example, 36 divided by 9 is 4, andthe remainder is zero. This fact might bestated as 36 is divisible by 9 (or 1, 2, 3, 4,6, 9, 12, 18, and 36, depending on thesituation) or 9 divides 36. Notice that di-vides means there will be no remainder inthe answer. As we consider divisibilityrules and extensions of the concepts ofmultiples and factors, this definition isimportant.

Think about 15. It is divisible by 1, 3, 5,and 15. For now we will focus on the ideathat 5 divides 15. Because 5 divides 15,we can be assured that 5 will also divide30. There are two ways to certify that—one way would be to divide 30 by 5 anddetermine there is no remainder; the otherway would be to express 30 as 2 x 15and, because we know 5 divides 15, wecan see that 5 must also divide 30. Thissecond approach will prove more usefuland efficient than the first as we proceedwith our discussion. In general, we saythat, if a is divisible by b, then we are as-sured that any multiple of a is also divisi-ble by b. This is an essential concept forthe following discussion.

The number 27 can be rewritten as3 + 9 + 15. The common factor, 3, can beused to rewrite 3 + 9 + 15 as the productof 3 and the sum of the residue of each ofthe terms, 3(1 + 3 + 5). This shows thatthe original number is divisible by 3. Ex-amples such as this will aid in under-standing and verifying for divisibility rules.

What is the rule for showing a numberto be divisible by 2? "The last digit iseven." or "The last digit is a 2, 4, 6, 8, or0." The real question is, "How do youknow that to be true?" or "Can you showit?" One of the goals of the Standards iscommunication, which involves explain-ing why things work, and producing con-vincing arguments.

Consider any number xy, where x is anyinteger and y is any digit. In 7,354, x is 735and y is 4. The number xy can be written in

a quasi-expanded notation as (x)(10) + y.No matter what integer is used for x, (x)(10)must be divisible by 2, because 10 is divis-ible by 2. Any multiple of 10 will also be di-visible by 2. One term of the expandedform is guaranteed divisible by 2. If theother term is also divisible by 2, then a 2can be factored out of the expanded formand the original number written as 2 timessomething. Because xy can be written as 2times something, this verifies that the orig-inal number is divisible by 2. But, whencan 2 be factored out of both terms? Onlywhen y is even, or 0, 2, 4, 6, or 8, whichgives the rule statement.

The divisibility rules for 5 and 10 aresimilar to that for 2. The digit in the unitsplace must be either 5 or 0 for divisibilityby 5. For divisibility by 10, there must be azero in the units column. You should de-velop a convincing argument to supportthese statements.

Understanding divisibility rules for 2, 5,and 10 facilitates the development of thedivisibility rule for 4. Because 10 is not di-visible by 4, the expanded expression,(x)(10) + y, cannot be used. The numberbeing considered needs to be written asxyz, where x is any integer and y and z areany digits. The number is then written as(x)(100) + yz (note y and z are not multi-plied here—they represent the tens andones places of the number). Because 100is divisible by 4, attention is turned to thelast two places, yz. If this number is divisi-ble by 4, then so is the original number.For example,

5,732 = 57(100) + 32

= 57 (4)(25) + 4 (8)

= 4[57(25) + 8].

If yz is not divisible by 4, neither is theoriginal number, because 4 cannot befactored out of both terms. For example,

NUMBER AND OPERATIONS 27

5,731 is not a multiple of 4, because 4cannot be factored out of 31.

For divisibility by 8, the last 3 digits ofthe number would be set off and added tosome multiple of 1000. A rule for 16 canbe established, but there is some ques-tion about the value of it. In the divisibilityrules for 2, 4, 8, and 16, there is a parallelbetween the exponents of 2 and 10. Fordivisibility by 2, use 21 and 101; for divisi-bility by 4, use 22 and 102; for divisibilityby 8, use 23 and 103; and for divisibility by16 use 24 and 104. Why do you think thereis a question about the value of a divisi-bility rule for 16?

The divisibility rule for 3 uses expandednotation, along with the idea of having acommon factor in each term, so the origi-nal number can be expressed as a multi-ple of 3. Suppose you want to know if the3-digit number xyz, where x, y, and z aredigits, is divisible by 3. The rule says to findthe sum of the digits. If that sum is divisibleby 3, then the original number is also.

Using expanded notation,xyz = (x)(100) + (y)(10) + z. For divisibilityby 3, we want to rewrite (x)(100) + (y)(10) + zso that x + y + z is a part of the expres-sion. The challenge is to find a way to re-write (x)(100) so it is a sum of x and some-thing. If (x)(100) is written as (x)(99 + 1),distributing the x yields (x)(99) + (x)(1) or99x + x. The x is now an addend. The 99xis a multiple of 3. Using the same tech-nique for (y)(10),

It is known that 99x and 9y are divisi-ble by 3. If x + y + z is divisible by 3, then99x + 9y + x + y + z can be written as

3(33x + 3y + w), where 3w = x + y + z, andthe original number is a multiple of 3. Ifthe sum of the digits is not a multiple of 3,then the original number cannot be ex-pressed as a multiple of 3.

Divisibility by 9 works like that of 3 ex-cept that the sum of the digits must be di-visible by 9. Divisibility by 6 uses a combi-nation of the 2 and 3 divisibility rules. Theeasiest way to use the 6 rule is to see ifthe number in question is even. If it is not,then the number cannot possibly be divis-ible by 6. If the number in question iseven, then apply the 3 divisibility rule.Consider the prime factorization of 6 for aclue about how this rule works.

Divisibility by 7 provides explorationopportunities for more inquisitive individ-uals. The rule states, "Double the last digitin the number and subtract that productfrom the original number after the lastoriginal digit has been deleted. If the newnumber is divisible by 7, the original num-ber is. If the new number is not divisibleby 7, the original number is not. The proc-ess may be repeated." For example,

Your Turn

65. Develop an argument that showshow the divisibility rule for 9 would workwith a 4-digit number wxyz.

66. Why does the 6 rule break into aneven 3 rule? Explain why a similar rulecould or could not be devised for divisi-bility by 15.

67. Describe a divisibility rule for somenumber other than those discussed andshow why it works.

68. Are divisibility rules limited to inte-gers?

28 CHAPTER 2

Divisibility by 11 revolves around renam-ing 10 as 11 - 1 and using the concept ofexpanded notation. Consider (11 - 1)2. Ex-panding yields 112 - 2(11)(1) + 12. Be-cause 11 is a factor of two of the terms inthe expansion, that part of the sum mustbe divisible by 11. Expanding (11 - 1)3

gives an expression in which all terms ex-cept 13 are multiples of 11. The situationis similar for all cases of (11 - 1)n, where nis any counting number—all terms exceptfor 1n will be multiples of 11. The sign of1n will be negative for odd values of n, andpositive for even ones.

Using this information to check for di-visibility by 11 on the number uvwxyz,where u, v, w, x, y, and z are any digits,uvwxyz would be rewritten asu(10)5 + v(10)4 + w(10)3 + x(10)2 + y(10)1 + z(10)°or as u(11 - 1)5 + v(11 - 1)4 + w(11 - 1)3 +x(11 - 1)2 + y(11 - 1)1 + z(11 - 1)°. Fromthe earlier discussion, expansion of thispolynomial will yield a set of terms thatare multiples of 11 plus some residue.The only terms in question are - u, + v,-w, + x, - y, and + z. Inspection showsthe rule to be the rightmost digit minus itsleft neighbor, plus the next left neighbor,and so on, until all digits are considered.In other words, the sum of every otherdigit is subtracted from the sum of therest of the digits (for divisibility, the sign isnot important). If that missing addend is amultiple of 11, then the original number isdivisible by 11.

Greatest Common Factorand Least Common Multiple

As we embark on this discussion, vocab-ulary is critical. Looking first at greatestcommon factor (GCF), examine each ofthe words:

greatest means biggest, largestcommon implies something shared be-tween at least two numbers

factor divides the number being con-sidered

GCF can lead to confusion because weare looking for the greatest common fac-tor of at least two numbers, which endsup being less than at least one of thenumbers. Similarly, when consideringleast common multiple (LCM) we mustexamine each of the words:

least means littlest, smallest

common implies something shared be-tween at least two numbers

multiple is a counting number times thenumber being considered

With LCM, we are looking for the smallestvalue that is a multiple of each numberunder consideration. It will be bigger thanall but perhaps one of them. Intertwined inall of this is the potential confusion aboutwhich numbers are factors and which aremultiples. Before proceeding with thisdiscussion, we advise you to be certain ofyour definitions.

The mathematics you learned mayhave been pretty much self-contained.Each topic was treated at the necessarylevel and then the curriculum moved onto something else. At times topic wereconnected, for instance, prime and com-posite numbers being followed by GCFand LCM. These connections help builda more coherent picture of the world ofmathematics.

There is a variety of ways to find LCMand GCF. We start with one you may nothave seen. Suppose the exercise is tofind the LCM and GCF of 16 and 24. Whatis a factor of both 16 and 24? You couldsay 8 or 4, but 2 will do for now. You areleft with 8 and 12 from the original num-bers and the process is repeated. 2|16 24

NUMBER AND OPERATIONS 29

What is a factor of both 8 and 12?

Two could have been selected again, in-stead of 4. Eventually, the final pair ofnumbers will be relatively prime.

Two and 4 are factors of the GCF (8),and the factors of the LCM (48) are 2, 4, 2,and 3. This is much easier to see if thewhole presentation is consolidated:

Notice how the 2, 4, 2, and 3 form an "L,"indicating the factors of the least com-mon multiple. Notice also that the factorsin the vertical part of the "L" are the fac-tors of the GCF of 16 and 24. Notice howthis process connects GCF and LCM.

We now turn to methods of finding GCFand LCM that are perhaps familiar to you.As these are discussed, think in terms ofhow they connect to other mathematicaltopics. Suppose the task is to find theGCF of 12 and 18. One way to attack theproblem would be to list all of the factorsof each number:

the factors of 12 are 1, 2, 3, 4, 6, 12

the factors of 18 are 1, 2, 3, 6, 9, 18

Looking for common factors, you have 1,2, 3, and 6, but the task is to determinethe greatest common factor, so the an-swer is 6.

Listing, although it works, can be cum-bersome if the numbers are large. Thereare other approaches that will shorten

and simplify the task. First, we need to in-troduce a few more terms. Consider thefollowing:

the prime factors of 12 are 2 and 3

the prime factor of 7 is 7

the prime factors of 72 are 2 and 3

the prime factors of 105 are 3, 5, and 7

the prime factor of 16 is 2

the prime factors of 35 are 5 and 7

the prime factorization of 12 is 2 x 2 x 3or 22 x 3

the prime factorization of 7 is 7

the prime factorization of 72 is2 x 2 x 2 x 3 x 3 o r 2 3 x 3 2

the prime factorization of 105 is3 x 5 x 7

the prime factorization of 16 is2 x 2 x 2 x 2 or 2 4

the prime factorization of 35 is 5 x 7

You should be able to build definitions offactorization and prime factorization fromthese examples and use exponents inthem. When considering the prime fac-torization of a number, all of the factorsare listed, and sometimes exponents canbe used to shorten the presentation.When thinking of the prime factors of anumber, all prime factors of the numberare listed, but each is listed only once.

Although there are different ways of de-termining the prime factorization of anynumber, the factor tree is probably themost common. We show some of theways of determining the prime factoriza-tion of 36:

30 CHAPTER 2

There are other ways this could be done(36 = 6 x 6), but these are enough to meetthe needs for this discussion. In eachcase, we want to list all of the prime fac-tors of 36. How we proceed from the be-ginning is not important. It is significant,however, that no matter how we initiallyfactor 36, we ultimately end up with thesame set of prime factors. Notice that 12could have been factored as 4 x 3, butbecause we know 4 can be factored into2 x 2, we expressed the 12 as a productof three primes. You should also noticethat although the prime factors can belisted in any order, it is most common andconvenient to list them in increasing or-der. Arranging the prime factors makes iteasier to list the prime factorization usingexponents (22 x 32).

The GCF can be found via primefactorization where spacing of the primefactors makes it easier to determine theanswer. The prime factors of 18, 24, and39 are:

Although the spacing may seem strange,it helps us see how to determine the GCF.If a factor is going to be common, it mustappear in each of the numbers being con-sidered. Looking at the example, 18 and24 share 2 as a factor, but 39 does not.Thus, 2 cannot be a common factor of 18,24, and 39. Only 39 has a factor of 13,which cannot be a common factor of 18,24, and 39. However, 3 is a factor of allthree numbers, 18, 24, and 39, hence, 3 isa common factor for them. This examplealso can be used to show that once all theprime factors of each of the numbers arearranged and listed with appropriatespacing, one need only look for any col-

umns that have an entry for each number.That column will be a common factor. Inthis case, the only candidate is 3, which isthe GCF of 18, 24, and 39. Consider only18 and 24:

Using the column idea, there is a commonfactor of 2 and another common factor of3. Because 2 and 3 are common factorsfor 18 and 24, 6 is the GCF. Another wayof explaining this is to say that, if 6 divides18, its factors divide 18. This could beconfirmed by using the listing technique:

the factors of 18 are 1, 2, 3, 6, 9, 18the factors of 24 are 1, 2, 3, 4, 6, 8, 12,

and 24

Checking the list, the common factors are1,2,3, and 6, and the greatest is 6. Noticethat 1 is a factor, but not a prime factor.

The exponential expression of theprime factorization provides another wayof determining the GCF. Using 18 and 24:

The GCF is 6. How can we come up withan explanation from this to match theknown answer? Notice that each primefactor is in one of the columns and thatthe smallest power is 1 in each case. Oneshould conclude that 21 x 31 is yet anotherway to express the GCF of 12 and 18.Using the other example of 18, 24, and39:

NUMBER AND OPERATIONS 31

The column idea works again and theGCF of 18, 24, and 39 is 3, which agreeswith the previous result. Look at one moreexample:

Here again, the column idea is present,but which exponent is selected? Becausewe are looking for the GCF, what is thelargest multiple of 2 found in each num-ber? The answer to that question is 23,which indicates that the smallest expo-nent in each column is the one to be se-lected when looking for the GCF.

We use a mentality similar to that ofGCF to find the LCM of numbers. We canlist the multiples:

the multiples of 18 are: 18, 36, 54, 72,90, 108, 126, 144, . . .

the multiples of 24 are: 24, 48, 72, 96,120, 144, . . .

Common multiples include 72 and 144,and there is an infinite number of them.The quest is for the LCM, so the choice is72. The idea of prime factorization couldbe used as well:

In this case the LCM is 72, the primefactorization of which is 2 x 2 x 2 x 3 x 3.The question becomes one of match-ing the prime factorization with the col-umn idea. Notice there are 5 columns,three for 2s and two for 3s. Each columnvalue is listed. The LCM of 18 and 24 is2 x 2 x 2 x 3 x 3, or 72. The exponentialapproach could be used, too:

Knowing the answer is 72 comes inhandy: 72 = 2 x 2 x 2 x 3 x 3, or 23 x 32.Whereas the smallest exponent is used infinding the greatest common factor, thelargest is used for the least common mul-tiple. If the task is to find the LCM of 18,24, and 39:

The LCM is 23 x 32 x 131, or 936. Pick 3numbers and find the GCF and LCM ofthem using each method described here.

Showing how things are intertwined,consider finding all the factors of 72. Thedivisibility rules help complete the task.

Notice the factor list for 72 can be brokeninto a top half and bottom half, where eachpair of factors is reversed in the other half.Produce a similar chart for 96 and you willsee the same pattern emerge. Focus onwhere the reversals start and you shouldnotice that the square root of the numberwhose factors are being found is between

32 CHAPTER 2

the two central reversed factors (8x9 and9 x 8 in the case of 72). The square root of72 is approximately 8.49, which is be-tween 8 and 9. When searching for all fac-tors of a number, list the factor pairs, start-ing with 1 times the number, and proceedto the counting number before the squareroot of that number. At this point, all thefactor pairs are listed.

The counting numbers are traditionallypartitioned into primes, composites, andone. Other partitions of the counting num-bers provide interesting practice with fac-tors and multiples. Three potentially newterms are perfect numbers, deficientnumbers, and abundant numbers. In allthree cases, all factors except the numberitself, are added. The following are defi-cient numbers:

Studying these examples should lead tothe conclusion that the sum of the factorsof a deficient number (excluding itself) isless than the number.

The definition of a deficient numbershould provide a clue to the definition ofan abundant number. Some examples ofabundant numbers are:

These examples, plus knowledge of defi-cient numbers, should lead to the conclu-sion that abundant numbers have a factor

sum that is greater than the number itself,even when the number is omitted fromconsideration.

There is one additional possibility forthe sum of the factors of a number (ex-cluding the number itself). A number issaid to be perfect when the sum of thefactors equals the number. The first per-fect number is 6 (1 +2 + 3 = 6) and thesecond one is28(1 +2 + 4 + 7 + 14 = 28).The next perfect number is 496. The nextfew after that are 812, 33550336,8589869056, 137438691328, and2305843008139952128. Perfect numbersget large in a hurry, yet they add an inter-esting twist to how mathematics can belearned and connected to prior work.

Conclusions

There is more to sets and properties thanwe have given here. This is a beginningfor you. Concepts involving the propertiesof whole numbers, or number theory, pro-vide the foundation for much of the studyof higher mathematics. You have seen afew of those ideas discussed and devel-oped. These topics are a part of the basicbuilding blocks for a more extensivestudy of number theory. As your mathe-matical growth continues, we encourageyou to keep looking for extensions andapplications of these ideas while addingto the collection of building blocks. Restassured, you will see them used through-out this text. Enjoy.

WHOLE NUMBER ADDITION

FOCAL POINTS

. Terminology• Standard Algorithm. Partial Sum. Denominate Numbers

NUMBER AND OPERATIONS 33

• Horizontal and Vertical Writing• Expanded Notation. Left to Right Addition• Scratch Method. Any Column First• Low Stress Addition

Most people believe the standard algo-rithm they learned in elementary school isthe most efficient way to perform additionby hand. Many people think the algorithmthey learned is the only way to do addition.The typical addition algorithm, althoughefficient, is not the most convenient way toapproach addition for beginners. The ad-dition algorithm you probably learned formultidigit addition dictates working fromright to left, which is not natural for chil-dren who are learning to read from left toright. In this section, we will discuss sev-eral methods for adding and show youthere is no single method that is betterthan any other.

Some of the methods we will discussmight seem cumbersome to you, but youwill find that learning them will increaseyour understanding of the inner workingsof addition. We know you want to be-come an effective teacher of mathemat-ics. We hope that we can help you dealwith the inner workings of arithmetic op-erations by reminding you of conceptsthat have become automatic for you overthe years since elementary school. Ourdiscussion includes a variety of ways toadd whole numbers and is intended toprovide insight into the complexities be-hind the simple idea of "how many alto-gether."

Whereas full discussions of concreteand semi-concrete approaches will be thefocus of your mathematics teachingmethods course, they will not be empha-sized in this book. You might remembersolving simple story problems, perhapseven before you started kindergarten. "If I

have four buttons and you have three but-tons, how many buttons would we have ifwe put them all together?" As a child, youmight have counted out four buttons forthe first group and three buttons for thesecond group. After sweeping all of thebuttons together as shown in Fig. 2.10,perhaps you counted all the buttons tofind that there were seven altogether.

FIG. 2.10.

Number lines are very useful in under-standing the inner workings of addition.Because the unit markings on a numberline are evenly spaced, the idea of adding"like" terms is introduced before the word"algebra" enters our discussion. Thewords "start," "first addend," "second ad-dend," and "sum" are helpful, as shown inFig. 2.11. "Start" will always be at zero.The first vector represents the first ad-dend and placing a second vector in "tipto tail" fashion with the first shows the

FIG. 2.11.

34 CHAPTER 2

second addend. "Sum" depicts a point onthe number line that represents the an-swer for the two addends. In summary, toadd on a number line, use a vector to rep-resent each addend, always start at theorigin, place the tail of the second vector(addend) exactly on the tip of the first vec-tor, and mark the sum clearly.

Your Turn

1. Use the grouping method describedin the aforementioned button example tofind the total of two sets of elements as alittle child might. Use buttons, pennies,chips, or any other manipulative.

a) A group of three and a group of fiveb) A set of four and a set of eightc) An array of two and an array of six

2. Using only addition, write out a sim-ple word problem that would be appropri-ate for an early childhood addition prob-lem. Use your manipulatives to solve theproblem.

3. Use the number line method de-scribed previously to show:

Terminology

The language of mathematics is very richand can seem complicated. You shouldalways use the correct terminology whenyou discuss mathematical concepts. Forthis section, two major terms you usewithout thought are addend and sum.Numbers that await addition are calledaddends and the answer for an additionproblem is called the sum.

An addition fact involves three num-bers, two addends and a sum. Another

way of thinking of an addition fact is thatthree numbers are involved and at leasttwo of them are members of the set ofdigits, {0, 1,2,3, 4, 5, 6, 7, 8, 9}. What isthe largest possible addend in an additionfact? What is the largest possible sum foran addition fact? Is it possible to have anondigit addend in an addition fact? Theanswer to the first question is 9. The an-swer to the second question is 18. It is notpossible to have a nondigit addend in anaddition fact because the sum would alsobe nondigit, which would violate the defi-nition of having at least two digits in theaddition fact.

Please notice that a more in-depth un-derstanding of this definition is possible ifwe carefully consider the basic conceptsthat are at the heart of the definition. Theability to sort through definitions in thisway is an essential ingredient in the learn-ing of mathematics. As a matter of fact,this addition fact definition can be gener-alized to cover any operation (+, -, x, -=-)by saying that a fact involves three num-bers, at least two of which must be digits.

The Standard Algorithm

The following discussion of the standardalgorithm and the alternate ways of add-ing that follow are not presented in anyparticular order. They simply representdifferent ideas about adding. We encour-age you to become adept at using differ-ent ways of doing mathematics to in-crease your mental flexibility.

You may have learned the Standard Al-gorithm with very little emphasis on un-derstanding—in other words, you weretaught only "how to do it." Referring to thefollowing example, you were probablytold to first add the numbers in the unitscolumn and get 26; write the 2 above thetens column and the 6 beneath the units

NUMBER AND OPERATIONS 35

column. Did you understand that the 2represents two groups of ten, which iswhy it is placed at the top of the tens col-umn? Many students are taught to addthe digits in the tens, hundreds, thou-sands, and ten thousands columns as ifthey were simply adding digits with no ex-planation or expectation of understandingof place value. As long as they can do theproblem and get the right answer, there islittle emphasis on knowing why thingswork as they do, for example:

Take a closer look at the problem anddetermine what happens as you stepthrough the standard algorithm. First, un-less the addition facts have been memo-rized, completion of the problem is diffi-cult. One could count fingers, use tallymarks, or some other method, but, if effi-cient performance of the addition algo-rithm is expected, the addition facts mustbe second nature.

Consider the values in the ones column.As you add them, from the top down, thebottom up, or in any other order, you se-lect two addends and get a sum. Once youhave that sum, you use it with another ad-dend, creating a new sum. At this point inthe process, you may have moved beyondthe facts and into the realm of adding atwo-digit addend and a one-digit addend.You continue this binary process until alladdends have been used. At that point,you have a sum for the units column anduse your place value knowledge to writethe digits of the sum as you were taught.This process is repeated for the tens col-umn, and then for the other columns, inturn, until the sum is completed.

The columns themselves are quite sig-nificant because we often ignore placevalue as we add. Looking at the units orones column, you find the sum of 6, 4, 7,and 9 (or 9, 7, 4, and 6 if you add from thebottom up) to be 26. The 6 is written in theones place of the sum and the 2 is writtenat the top of the tens column. The 9, 9, 0,and 7 are digits found in the tens column.The task in the problem is to add thosedigits representing groups of tens, with-out forgetting the 2 tens generated as anexcess of ones. Thus, you add 2, 9, 9, 0,and 7, which yields a sum of 27, repre-senting 27 tens or 270.

Whoa! That looks strange, yet, what wejust described is what is going on behindthe scenes as you add whole numbers.We will leave this discussion of how youdo the problem via the standard algorithmfor you to think through on your own. Aswe look at some other methods of findingthe sum, you will gain more insight intowhat goes on in the background as youadd.

Partial Sum

Consider the example we used earlier:

The objective is still to find the sum of fourmultidigit addends, but we are going tochange emphasis. Focus on the placevalue associated with each column andthen treat each column as an individualproblem. This approach is dependent onideas generated in the sequence for add-ing whole numbers, in particular, the con-cept of adding multiples of tens to multi-ples of tens, multiples of hundreds to

36 CHAPTER 2

multiples of hundreds, and so on. Eventhough the format is going to appearstrange to you, try to focus on the stepsinvolved and how using partial sums ac-tually simplifies multidigit addition:

The total of the partial sums(26 + 250 + 1400 + 17000 + 160000) cannow be found rather easily. The differencebetween the partial sum method and thestandard algorithm is that, because of theemphasis on place value, regroupings arenot listed above the respective columns.Aside from rearranging the order of stepssomewhat, the problem is essentiallydone the same as you would do it usingthe standard algorithm. Although it mightlook strange in the partial sum format, theproblem is easier to do because the re-groupings are more logically arranged.Yes, all those zeros might look strangetoo, but they do clarify what is going on.

Denominate Numbers

Denominate numbers use a numeral anda word that describes the place value ofthe number. An example of this notationis to write 247 as 2 hundreds 4 tens 7ones. We can make a quick associationfrom 3 oranges plus 4 oranges to 30 + 40by saying 3 tens plus 4 tens. You may findthis trivially simple, never realizing what agiant leap is involved. You even make thetransition from 7 tens back to 70 intu-itively. Like partial sums, denominate

numbers are vital to understanding theconcepts behind whole number addition.

Using denominate numbers, the addi-tion problem 4567 + 319 + 208 = ? be-comes:

Adding in a column and remembering the3 oranges plus 4 oranges example, wehave 4 thousands 10 hundreds 7 tens 24ones. You should immediately call onprior knowledge about place value to re-group the 10 hundreds to 1 thousand 0hundreds and the 24 ones to 2 tens 4ones. Finally, combine the 1 thousandwith the 4 thousands and the 2 tens withthe 7 tens. The sum is 5 thousands 0 hun-dreds 9 tens 4 ones, or 5094. This mayseem no less cumbersome than the par-tial sums method, but it provides a re-minder of how important place value is inour arithmetic.

Your Turn

Complete each exercise twice, first usingpartial sums and then using denominatenumbers. Explain how the two algorithmsare essentially the same:

Horizontal and Vertical Writing

You should be comfortable with both verti-cal and horizontal writing while solving ex-ercises. We used a vertical format forshowing how to work with partial sumsand denominate numbers, and then wrotethe exercises in a horizontal format. Whichformat did you use as you solved the exer-

NUMBER AND OPERATIONS 37

cises? We could have presented the de-nominate number example as "4 thou-sands + 5 hundreds + 6 tens + 7 ones + 3hundreds + 1 ten + 9 ones + 2 hundreds +0 tens + 8 ones." This can be convenientlyrewritten to group place value names to-gether. Writing "4 thousands + 5 hundreds+ 3 hundreds + 2 hundreds + 6 tens + 1ten + 0 tens + 7 ones + 9 ones + 8 ones"makes the problem very similar to the ex-ample in vertical format and touches, onceagain, on the algebra concept of combin-ing like terms. A person who understandsthe place value names can add the ones orhundreds first and still correctly completethe exercise.

Your Turn

Use the "other" format for denominatenumbers to complete the addition prob-lems. If you used the vertical format be-fore, then try the horizontal format now, orif you used the horizontal format before,then try the vertical format now:

Expanded Notation

Expanded notation is a fast way to em-ploy the concept of denominate numberswhile using the power of our place valuesystem. Instead of writing a word for theplace value, zeros are used to hold theplace value. A vertical or horizontal formatmay be used, and you decide which col-umn to add first. The exercise2981 + 306 + 247 = ? would be expandedas:

and addition within columns, gives re-spective sums of 2000 + 1400 + 120 + 14.The respective column sums could thenbe expanded giving 2000 + (1000 + 400) +(100 + 20) + (10 + 4). Next, grouping likeplace values together yields (2000 + 1000)+ (400 + 100) + (20 + 10) + 4 or 3000 + 500+ 30 + 4, which is 3534 in standard nota-tion.

Your Turn

Complete the exercises using expandednotation. Do not skip steps. Writing out allthe steps will help to solidify the con-cepts:

Left to Right Addition

Adding from left to right makes a lot ofsense to some people because they readfrom left to right. Consider the problemwe used earlier to introduce partial sums,but this time work it from left to right.

The total of the partial sums(160000 + 17000 + 1400 + 250 + 26) cannow be found, only this time the exercisewas worked from the larger numbers tothe smaller numbers, which is often theway real-life problems are approached(depending on the problem situation, we

38 CHAPTER 2

may ignore the smaller partial sums, in ef-fect rounding to a significant value).

You might experience a little difficultywith alignment in this format. Typically,we do not write all those zeros but, if therehas been any emphasis on adding multi-ples of tens, hundreds, and so on, thenthe idea is fairly simple. One quick solu-tion is to use standard notebook paperrotated 90° from the normal position as ameans of keeping values aligned withinrespective columns.

Your Turn

Complete the exercises using left to rightaddition. Be sure to use partial sums, de-nominate numbers, and expanded nota-tion at least once each in this set. Can youdetermine which exercise is easiest inwhich method?

Scratch Method

The scratch method is very similar to the"left to right" method. Rather than listingthe partial sums as shown in the "left toright" method, digits are scratched outand replaced by new values as needed.Start working from the left and proceed tothe right, one column at a time. In the fol-lowing example, the sum of the ten thou-sands column is 160000. The sum of thethousands column is 25000, which is 2ten thousands plus 5 thousands. The 2ten thousands are grouped with the 6 tenthousands already present, so scratchout the 6 and replace it with an 8 in the tenthousands column. Continue to the right,scratching out the old value and insertingthe new one as needed. The final sum is

read starting from the left and using thenew values along with any that are notscratched out.

Your Turn

Complete the exercises using the scratchmethod.

Any Column First

Anyone with a good understanding ofplace value might start by adding any col-umn first. The next example is completedin a seemingly random order, but the or-der could be driven by familiarity with cer-tain facts, groups of digits to be added, orrelevance based on a real-life problem.You will notice similarities between thismethod and partial sums or left to rightaddition. Without a clear understanding ofplace value, this method would be veryconfusing:

NUMBER AND OPERATIONS 39

Your Turn

Examine the exercises and determine ifthere is a column addition order that isoptimal. State the reason for each deci-sion. Then complete the exercise usingyour column order.

Low Stress Addition

Suppose the task is to find the sum of 9,8, and 9, in column addition. Considerwhat happens in your head as you findthis sum. The first part, 9 + 8 is easy be-cause it is a fact, yielding a sum of 17. Thenext step is to find the sum of 17 and 9,which is not a fact. You have worked thisproblem, but it probably was formattedas:

As 17 and 9 are added, what really hap-pens in your head is the 17 is expressedas 10 + 7. The "10" is remembered and thesum of 7 and 9, which is a fact, is deter-mined to be 16. That 16 is actually 10 + 6,so now the aggregate is 10 + (10 + 6) =(10 + 10) + 6 = 20 + 6 = 26. Granted, youdo the problem much more reflexively,and much more quickly, than is shown inthis example. But, if you think about it,what we described is exactly what youmust ask your brain to do for you.

The low stress algorithm eliminates theneed to remember all those multiples often and keeps the problem as a collectionof addition facts. The demonstration be-low shows the addition of 9 + 8 + 9 + 7 + 9:

Does that look strange to you? As withmany procedures that are quite simple inuse, it takes a bit of explaining. Even ifyou have to read through the explanationtwice, it will be worth your time to learnthis simple method. The initial partial sumof 9 + 8 is written below the 8, but spacedout as shown in the example. Actually,you are just writing your scratch work in adifferent format, as you go. Once the 17 iswritten, a new problem is done but theemphasis is only on the ones digit fromthe 17 and the next addend, 9. This main-tains the idea of dealing only with additionfacts and the sum 16 is listed below thesecond 9, using the same format as wasused for the 17 initially. Once again, theones digit (6 in this case) from the partialsum, and the next addend (7) are consid-ered, giving a sum of 13, which is writtenlike the 17 and 16 before it were. For thefinal addition fact, there is a slight differ-ence in the presentation. The ones digit(3) from the last partial sum is added tothe next addend (the final 9). The onesdigit from the new partial sum is written inthe ones column of the answer, where itwould normally be found. The tens digitfrom this new partial sum is written belowand to the left of the last addend. Now thetens from each of the partial sums arecompiled and, in this example, there arefour of them. That total of 4 tens is listedin the sum in its normal location. Typically,

40 CHAPTER 2

the 40 is not shown as a part of the partialsum as we have done, for clarity, in theexample. Rather, the problem would beshown as:

Although this method might look strangeto you, it is extremely appealing to individ-uals who struggle with column addition.What could appear to be a formidable taskis reduced to dealing with a collection ofaddition facts. The low stress method canbe used with more than single digit ad-dends. You only need a little more spacebetween columns and any regroupingfrom a value is placed at the top of the re-spective next column. After that, the proc-ess is exactly as discussed here.

We are not going to formally assignproblems to be done using the low stressmethod for addition. Rather, we want youto take the responsibility for doing someproblems on your own to practice usingthe method. Without that practice, lowstress addition will not become a part ofyour repertoire.

Conclusions

We hope we have tweaked your interestin the different ways to do addition. Wehave shown you some of them and en-courage you to learn to add using a vari-ety of methods. Of course, there is tech-nology available that provides a fast wayof doing all arithmetic. We believe in usingtechnology when appropriate, but we

also believe you must have an under-standing of how the operations are done.Your responsibilities may include deci-sions about when technology is appropri-ate to use and when it is not; these deci-sions must not be made lightly. As youreflect on this issue, remember that thereis more to the world of mathematics thanbeing good at doing arithmetic, but beinggood at doing arithmetic will help smoothyour way into the world of mathematics.

Bibliography

Hutchings, B. (1976). Low-stress algorithms. Meas-urement in school mathematics, 1976 yearbook(D. Nelson, Ed., & R. E. Reys, General Ed.).Reston, VA: NCTM.

WHOLE NUMBER SUBTRACTION

FOCAL POINTS

• Terminology« Concrete Subtraction• Denominate Numbers• Expanded Notation• Standard Algorithm. Left to Right Subtraction. Scratch Method. Any Column First. Borrow-Pay Back Method. Integer Subtraction

Subtraction follows addition in most con-texts. Most of the methods used for addi-tion can be altered and used in subtrac-tion. The sequence starts with learningthe subtraction facts, which is really just adifferent way of thinking about the addi-tion facts. Remember that when we say"facts," we imply an operation involvingthree numbers, two of which are digits.For example, in addition, the two digits

NUMBER AND OPERATIONS 41

are addends and the third number (whichmay or may not be a digit) is the sum. Thesubtraction procedures we will discusshere require subtracting a whole numberfrom a whole number, so that a wholenumber is the result.

manipulative with the take away model asshown in Fig. 2.12. In this model, you startout with 8 buttons and literally take threeaway. Counting how many buttons areleft after the removal of the three givesyou how many buttons remain.

Terminology

Subtraction and addition are inverse op-erations, meaning that subtraction un-does addition and addition undoes sub-traction. Connections such as this inverserelation can help you become more com-fortable with subtraction. Consider thefollowing arrangement where a subtrac-tion exercise is written vertically andnames for the parts are written beside therespective numbers:

Increasing your mathematics vocabularyis one of the goals of this text, but we feelthat vocabulary words should be mean-ingful. Whereas the words minuend, sub-trahend, and difference are descriptive,and historical, they do not strengthen theconnection between subtraction and ad-dition. These three words continue to beused. We believe it makes more sense tohave a verbal connection between sub-traction and addition problems so we willuse sum, addend, and missing addend.We will discuss three commonly acceptedversions of subtraction: take away, com-parison, and add-up.

Focus on the subtraction problem 8-3= *. What is the *? If you know your sub-traction facts, then you can do the prob-lem. If you do not know your subtractionfacts, then you could figure it out using a

FIG. 2.12.

Subtraction is often used when com-paring the cardinality sets. Whereas noth-ing is literally taken away in such situa-tions, the process involves subtracting thecardinal number of the smaller set fromthe cardinal number of the larger set. Thisis often seen in simple word problemslike, "Pat has a collection of 8 buttons.Chris has 3 buttons. How many more but-tons than Chris does Pat have?" Theproblem would be done as 8 - 3 = 5, eventhough nothing has been taken away.This is a different description of the prob-lem in Fig. 2.12.

The missing addend, or add up, ap-proach can be used with buttons too, butit is dependent on knowledge of additionfacts, helping to perhaps open your mindto a slightly different way of thinkingabout subtraction of whole numbers.Suppose the problem is 11 - 6 = <8>. Youknow the sum is 11 and that one addendis 6. The task is to determine the missing

42 CHAPTER 2

addend. Asking, "What number, whenadded to 6 gives 11?" implies that weneed to recall an addition fact. You arenot starting at 11 and taking away in thisscenario. Rather, you are searching yourmemory banks to determine the specificaddition fact that will meet all the require-ments of the problem. By doing that, youare identifying the missing addend. Thinkcarefully about how you do some sub-traction problems and you might find thatthis is a technique you subconsciouslyuse. We hope it is one you will now con-sciously endeavor to employ as you ap-proach subtraction problems in whichneither take away nor comparison is in-volved. A simple word problem of thistype might be, "Jesse has saved $14.00and wants to buy a $21.00 shirt. Howmuch more must Jesse save to be able tobuy the shirt?" In this example, the moneyneeded does not exist, so we can't usetake away or comparison. We must useour knowledge of addition and determinehow much must be added on, we needthe missing addend.

Figure 2.13 shows what happens as8 - 3 = * is done on the number line. Youhave the starting point of zero and thelength of the vector that represents the ad-

1. Sketch a vector from zero to the sum and label the sum.2. Sketch a vector from zero to the addend and label it.3. Sketch a dashed vector from the addend to the sumto show the missing addend.1. The direction of the motion from the addend to the sumndicates the way the arrowhead should point.

FIG. 2.13.

dend three. The vector from the origin orzero to eight shows the sum. The dashedvector shows the distance between theaddend and the sum, or the missing ad-dend. If you think in terms of the missingaddend terminology, then you have aknown sum of eight and a known addendof three. The thing you do not know is themissing addend. Using the number linemodel, you can determine the missing ad-dend by counting from the arrowhead ofthe first vector to the arrowhead of thesum. Right now the arrowheads on thesegments might not seem necessary, butthis model is very handy when looking atinteger operations and we are laying thegroundwork for using arrowheads later toindicate positive and negative values.

Your Turn

Do the following problems using the ap-propriate model:

1. Jo has 14 marbles and loses 8. Howmany are left?

2. Shawn owns 4 high value stamps,and Sean owns 11 high value stamps.How many more high value stamps doesSean own?

3. Chris has 5 cards left but startedwith 17. How many are missing?

Concrete Subtraction

Whereas addition is typically presentedusing a single model, a couple subtrac-tion models are frequently used. We showwhat is probably the most common onenow, but remember, not everyone sub-tracts this way. This developmental dis-cussion will help you gain a better under-standing of what is going on in your headas you subtract. Remember, there areseveral valid ways of subtracting. Con-sider the exercise 312 - 147 = *. Rarely

NUMBER AND OPERATIONS 43

would it be solved in the horizontal for-mat. Rather you would see it vertically asshown here:

Base 10 blocks have three fundamentalpieces, as shown in Fig. 2.14: a squarewith 100 units, usually called a Hundred,or Flat; a bar with 10 units, usually calleda Ten, or Long; and a single unit, usuallycalled a One, or Unit.

FIG. 2.14.

We will show several stages to com-plete 312 - 147 = ?. The first step isshown in Fig. 2.15. The one Long in thesum (312) needs to be traded for ten Unitsso there will be enough Units in the sumto allow subtracting 7. Notice how theLong is traded for 10 Units and then aone-to-one correspondence is estab-lished between the 7 Units to be takenaway and any 7 of the 12 Units in the sumafter the trade. Now 5 Units are left. Ob-serve how the regrouping is quite similarto what you would do using a pencil and

paper subtraction procedure where youwould cross out the one Ten, write a zeroabove it, and then write a 12 above the 2.

Next, consider the Tens column. Afterthe regrouping, the sum has no Longsand the addend has a 4 in the Tens col-umn. Figure 2.16 shows how one of the

FIG. 2.16.

Flats in the sum is regrouped to give 10Longs so the procedure can be com-pleted. Once the regrouping is finished, aone-to-one correspondence is estab-lished between any 4 of the Longs in thesum and the 4 Tens in the addend, leav-ing 6 Longs.

Finally, we will deal with the hundreds.Figure 2.17 shows how that would be ac-complished with the Base 10 blocks byestablishing a one-to-one correspon-dence between any one of the Flats in thesum and the Hundred in the addend.Finally, Fig. 2.18 models the missing ad-dend. Granted, you could quickly do thisproblem abstractly or on a calculator, but

FIG. 2.15. FIG. 2.17.

44 CHAPTER 2

FIG. 2.18.

it is important that you investigate thisprocedure using the Base 10 blocks be-cause it shows clearly what is going on inyour head as you regroup to subtract. Re-membering our discussion about addingany column first, you may wonder why wedid not begin by taking away one Flat. Byrearranging the discussion and accompa-nying figures, you'll find that the model isvalid whether you begin the exercise withthe Longs, Flats, or Units.

Denominate Numbers

With denominate numbers, the problem312-147 would be expressed as (3 Hun-dreds 1 Ten 2 Units) - (1 Hundred 4 Tens7 Units). With paper and pencil, the prob-lem could be done with denominate num-bers in a manner very similar to that of theBase 10 blocks. Its appearance will alsoremind you of the standard algorithm forsubtraction.

A little investigation should show this pro-cedure to be quite familiar. The one Ten inthe sum is traded to make 12 Units, leav-ing no Tens in the sum. Once that trade ismade, the 7 Units are subtracted from the12 Units. Because we need 4 Tens andthere are now no Tens in the sum, one ofthe Hundreds in the sum is exchanged for

10 Tens, leaving 2 Hundreds in the sum.The 4 Tens in the addend are now sub-tracted from the 10 Tens in the sum.Finally, the 1 Hundred in the addend issubtracted from the 2 Hundreds in thesum, giving a missing addend of 1 Hun-dred 6 Tens 5 Units.

Expanded Notation

Expanded notation can also be used toexpress the sum and addend as(300 + 10 + 2) - (100 + 40 + 7). Thescratch notes are similar to the denomi-nate number work:

Notice that the regroupings are shownabove the crossed out values, all in onerow, getting ever closer to the standardalgorithm used by most people.

Standard Algorithm

Finally, we arrive at the standard algo-rithm, which is probably the most com-mon way of subtracting:

You should be able to now connect thestandard algorithm back through the se-quence of expanded notation, denomi-nate numbers, and the Base 10 blocksthemselves.

We spent a lot of time developing thatsequence with you. We did it to help youunderstand what your brain is doing as

NUMBER AND OPERATIONS 45

you subtract. We have assumed that theproblem we used for our example is rep-resentative of several problem types andthat you will transfer the development toall other aspects of whole number sub-traction.

Your Turn

4. Shawn scored 512 on a video gameand Reggie scored 178 on the samegame. How much higher was Shawn'sscore than Reggie's? Do this problem us-ing each of the stages: concrete, denomi-nate numbers, expanded notation, andstandard algorithm. Write a concludingparagraph explaining how the differentsteps in the various stages are connectedacross the stages.

Borrow Pay Back Method

This method is another candidate for astandard subtraction algorithm, but it willseem strange to you if you have not doneit. You will find it rather easy to do, andquite natural, if you just practice it a little.Consider again 312 - 147 and examinehow the borrow pay back method works.Although we will write replacement valuesin the example, people who use thismethod rarely write any scratch work:

Here, the 2 Units are "made to be" 12Units. Essentially, 10 Units have beenadded to the sum. However, that wouldchange the problem so, to compensate,an additional Ten is added to the 4 Tensthat are to be subtracted, implying that 5Tens must be subtracted. The problem

has not been changed because both thesum and the addend have been increasedby 10 giving a net change of zero. Theonly thing is, we have opted to write thatzero a little differently, as shown in the ex-ample. Similarly, the 1 Ten in the sum ismade to be 11 Tens and the 1 Hundred tobe subtracted is changed to 2 Hundredsto be subtracted. Again, we increasedboth the sum and the addend by thesame quantity, so the net resultantchange is zero. So, in the Tens columnyou have 11-5, which is 6 Tens—asshown in the sum. Finally, 2 Hundreds aresubtracted from 3 Hundreds and themissing addend is 165:

As we said, that might look like astrange way to subtract, but it is com-monly taught in some areas of the UnitedStates and it is the dominant proceduretaught in Australia. Check out another ex-ample:

The 1 in the Units place of the sum ismade to be an 11 to accommodate sub-tracting 8. However, that essentially adds10 Units to the sum and to compensate,another Ten must be subtracted. That isaccomplished by changing the 7 Tens tobe subtracted to 8 Tens. Once that changeis made, the Tens column subtraction is2 - 8 . Again, there is a need to changethose 2 Tens to 12 Tens so the subtractioncan be completed. That change to 12 Tens

46 CHAPTER 2

mandates that the 9 Hundreds to be sub-tracted be changed to 10 Hundreds so theproblem is not changed. Next, the Hun-dreds column now asks for 10 to be sub-tracted from 3, something that is not feasi-ble within the parameters of the take awaymodel of subtraction. So, the 3 Hundredsis made to become 13 Hundreds and thesubtraction in the Hundreds column canbe completed. Those extra 10 Hundreds inthe sum must be compensated for bychanging the 3 Thousands that are to besubtracted to 4 Thousands. Now, in theThousands column, the problem asks for 4to be subtracted from 5, and because thiscan be done within the realm of subtrac-tion facts, no compensation or changesare necessary. Similarly, the Ten-thou-sands column subtraction can be donewith no further complications and themissing addend is 31343. As we said ear-lier, once you try a few, you will get thehang of it; the limitations of a written expla-nation imply greater complexity than is ac-tually involved in practice.

Left to Right Subtraction

You do not have to start in the right-handcolumn when subtracting. Starting fromthe left may seem more natural becausealmost everything students do excepttheir arithmetic is done working from theleft to the right. Doing the problem312 - 147 in expanded notation (or withthe blocks, denominate numbers, etc.)would be started in this manner if you wereusing the method of starting from the left:

Now, dealing with the Tens column, thereis a need for additional Tens in the sum soone of the 2 Hundreds in the missing ad-

dend must be regrouped into 10 Tens toenable the subtraction. It would look like

and now you would focus on the Units.Again, there are not enough Units in thesum, so one of the 7 Tens in the missingaddend has to be regrouped to makeenough Units so the subtraction may becompleted. The completed exercisewould look like:

Scratch Method

The left to right procedure is quite similarto the scratch method for subtraction andcould be considered a developmentalstep to doing scratch subtraction. Theproblem would be done as shown, with165 as the missing addend:

Notice how the 2 from the Hundreds inthe missing addend is scratched out. It isreplaced with the 1 Hundred below it.Similarly, the 7 from the Tens is scratchedout and replaced with 6 Tens below it.This looks very much like left to right sub-traction.

NUMBER AND OPERATIONS 47

Any Column First

This subtraction is quite similar to left toright and yet different because it permitsbeginning in any column. Regroupingsare done as needed. This procedure em-phasizes, once again, how important theunderstandings of place value, regroup-ing, and subtraction facts are to any sub-traction exercise. Starting with the Tenscolumn first would look like:

If we move to the Hundreds next, then thesituation is simple because no regroupingis necessary. Then, if we do the Units last(although it could have been done secondbecause we are not concerned with thesequence of columns being used), thenregrouping is necessary and would looklike this:

Integer Subtraction

Folklore has it that an elementary-age stu-dent developed this method of subtract-ing. Card games were common in thehome and the child often experienced nothaving enough points on the books to"cover" when someone else "went out."For example, the child might have 3 pointson the books when someone else wasable to "go out," "catching" the child with7 points. So, the child was "down" 4, 4 "inthe hole," or had a "negative" 4. Similar re-

suits would be generated if the child had40 points on the books and got caughtwith 60 when another player "went out."This child was doing subtraction problemsin school and the teacher noticed the un-usual manner in which the problems weredone. The child's work looked like this:

At first glance, it seems as if the little num-ber is subtracted from the big number,but, somehow, the right answer evolves.The child consistently did problems thisway, getting them correct each time. In-vestigation revealed that the child was infact "subtracting the little number fromthe big number" in a column, but was "re-cording" the values as negative if the ad-dend in a column was greater than thesum value in the column. In this example,the child compiled a positive 200 with anegative 30 and a negative 5, giving a netresult of 165 for the missing addend. Thename for this method is integer subtrac-tion, because the child was using bothpositive and negative numbers.

Your Turn

Do the following subtraction problems us-ing each of the following methods: bor-row pay back, left to right, scratch, anycolumn first, and integer. Show yourregroupings in each style in a manner thatwould justify a complete explanation:

5. 8314 - 27566. 703 - 1647. How would a problem like

8152 - 1936 impact each of borrow pay

48 CHAPTER 2

back, left to right, scratch, any columnfirst and integer subtraction?

Conclusions

You have seen a variety of ways to sub-tract. We encourage you to practice usingthe different methods. Each one will en-hance your understanding of subtractionand build a stronger mathematical back-ground for you.

WHOLE NUMBERMULTIPLICATION

FOCAL POINTS

• Terminology. Beginnings• Standard Algorithm• Partial Product Method. FOIL. Lattice Multiplication• Left to Right Multiplication• Horizontal and Vertical Writing• Russian Peasant

Few adults think beyond the standard al-gorithm and multiplication facts memo-rized in elementary school. It is not un-common for people to think that thestandard algorithm is the only correct wayto multiply, even though they may usetheir own invented shortcuts in their dailylives. We will discuss several methods formultiplying whole numbers and show youthat no one method is better than anyother, although one may be more practi-cal than others in any given situation.

Terminology

The language of mathematics uses manyeveryday terms. Times is no longer linkedsolely to the idea of events, but is associ-ated with the operation called multiplica-

tion. Product is now more than somethingto be associated with a brand name orstore item. Factor is something beyond anidea or point to be considered. In multipli-cation, a factor times a factor yields aproduct. For example, in 3 x 4 = 12, boththe 3 and 4 are factors and 12 is the prod-uct. To express the idea of multiplying the3 and 4, we say 3 times 4. The product of 3and 4 could also be written as (3)(4), 3*4, or3»4. In algebra, the product of 7 and thevariable v would normally be written as 7v,but it could also be written as (7)(v), 7(v), oreven v7 (v7 is not commonly seen and isusually considered poor notation). In gen-eral, numerals are placed before variables,that is, 7v. Rarely will you see 7v written as7 x v because of the potential confusionbetween whether the x is a variable or thetimes symbol. Once you grasp the con-cepts behind the language and writingstyles, you will better understand the won-derful world of mathematics.

Multiplication depends on an under-standing of place value and assumes acommand of multiplication facts. Theproducts of all the combinations of digitsmake up the 100 multiplication facts. Tenof the multiplication facts are doubles (adigit times itself). Remove the doublesfrom the hundred multiplication facts and90 are left. However, if the commutativeproperty of multiplication on the set ofwhole numbers is known, the need tomemorize 90 facts is reduced to 45, giv-ing a total of 55 facts that need to bememorized (the 45 plus the 10 doubles).

If you know and understand the role ofzero as a factor (zero times any number iszero), then you can eliminate 19 facts.Starting with a new fact table, knowing therole of the multiplicative identity (one timesany number is the number itself) decreasesthe number of multiplication facts to bememorized by another 19. Thus, althoughthere are a total of 100 multiplication factsto be memorized, some basic knowledge

NUMBER AND OPERATIONS 49

can trim that number significantly. Figure2.19 shows a completed multiplication facttable with a dashed line segment passingalong the doubles or major diagonal.

FIG. 2.19.

Beginnings

The basic concept of multiplication can bemodeled in straightforward, everyday con-texts. We start with sets of shirts andpants. If you have three shirts and twopairs of pants, how many different outfitscan you make (we are not going to be styleconscious here, so we do not care aboutmixing things that might not go well to-gether)? Figure 2.20 shows that you can

create six different combinations or outfits.This set representation can be extended tolarger numbers and different situations,but the mentality is still the same.

The number line is another way thatmultiplication can be modeled. If we thinkin terms of repeated addition, we can de-scribe multiplication as a shortcut way ofadding. Perhaps you have heard some-one say four threes, when referring to fourtimes three. That verbiage is a clue thatfour threes are being added together. Fig-ure 2.21 shows how the situation wouldappear on a number line. We have shownthe addends as stops along the way asthe product is approached.

FIG. 2.21.

Standard Algorithm

The standard multiplication algorithm isanother process you may have learnedwithout understanding the concept. Con-sider the following example and thencarefully examine our explanation of theprocess involved in finding the product.You might determine that many of thesteps we list are things you do reflexively.Yet, as you investigate what it means tomultiply, you need to be aware of each ofthese steps:

FIG. 2.20.

What happened as 58 is multiplied by 79?First you multiply the number in the Units

50 CHAPTER 2

column of the second factor times thenumber in the Units column of the firstfactor to get a product of 72. That 72 isreally 7 Tens and 2 Units. The 7 Tens arewritten above the Tens digit of the topfactor of the problem and the 2 is writtenbeneath the Units digit of the lower factor.Granted, you know all of that, but did youknow or had you ever thought about theterminology and rationalizations? Nextyou multiply the number in the Units col-umn of the lower factor times the numberin the Tens column of the upper factor.This is really 9 x 50 (although you proba-bly say 9 x 5 ) and the product is 450. Weknow that zero is the additive identity andso, adding it to the 2 you already placedbelow the Units digit of the lower factorwould not change anything. You may nothave thought about it in this way before.The product of 450 could be consideredas 45 Tens (because 450 = 45 x 10), butthere is an additional 7 Tens from the ear-lier multiplication. Add the 7 Tens thatwere listed above the Tens digit of the topfactor to the 45 Tens you just generatedand you get 52 Tens. Because that is re-ally 520, write the 52 next to the 2 in theUnits column. Then multiply the digit inthe Tens place of the lower factor timesthe digit in the Units column of the topfactor and get 560 (it was really 70 x 8).Perhaps you were taught to write a zerounder the 2, being told it was a place-holder. It is a placeholder, but there is areason for the existence of the zero andyou now either know or recall the expla-nation. Write the 5 above and to the left ofthe regrouped 7 from before and placethe 6 beneath the Tens column of the 522.This happens because the product of 560is really 500 + 60 and although you havethe ability to write the 60, the 500 willneed to be added to more Hundredsabout to be generated as the multiplica-tion is continued. Multiply the digit in the

Tens column of the bottom factor timesthe digit in the Tens column of the topfactor and get 3500 (70 x 50 even thoughyou probably only say 7 x 5 to yourself),which can be expressed as 35 Hundreds.Add the 5 Hundreds that were regroupedto the 35 Hundreds, getting a total of 40Hundreds. We know that 40 Hundredswould also be 4000 and adding the 60 toit will not change things, which allows usto write the 40 to the left of the 6. The mul-tiplication problem has been converted toa problem that involves adding 522 and4060, giving an answer of 4582, which isthe product of 58 and 79.

Even today, many students are taughtto multiply the digits in the Tens columnsas if they were simply multiplying digitswithout any explanation or expectation ofunderstanding. As long as you can do theproblem and get the right answer, littleemphasis is placed on knowing why orhow things work. How wrong is that? Wetook the time to carefully explain thesteps in a standard multiplication algo-rithm to help you understand the impor-tance of each of them. It is imperative thatyou master the behind the scenes work-ings of arithmetic operations.

Partial Product Method

Partial products are user friendly and pro-vide an intuitive background of what hap-pens during the operation of multiplication.We delayed introducing the partial productmethod to help you understand the com-plexities of what you do as you use thestandard algorithm for multiplication. Thepartial product approach to multiplicationrelies on the two basic concepts: multipli-cation facts and place value. Although thepartial product method of multiplication re-quires a little more writing, it eliminates re-grouping, and could ease confusion. Con-sider the following example:

NUMBER AND OPERATIONS 51

Each of the four products needed to com-plete this exercise are present and thereis an emphasis on place value; the ideaof multiplying a digit by a multiple of ten(9 x 50 and 70 x 8), and the need to findthe product of a multiple of ten times an-other multiple of ten (70 x 50). This is an-other instance where you could rotateyour notebook paper 90° to keep track ofyour columns.

These partial products could be listedin any order as long as they are all pres-ent, but we opted to list them in the sameorder they would typically appear if thestandard multiplication algorithm hadbeen used with the problem. Take a mo-ment to compare this procedure to thestandard algorithm, step by step.

Perhaps you are thinking that we shouldhave introduced the idea of partial prod-ucts before the standard algorithm. Wehope you are, because that would indicateyou are beginning to pay attention to whatgoes on behind the scenes. It is importantthat you understand how multiplicationworks. If the goal is only to get the prod-uct, then we could simply use a calculator,because they do not give wrong productsif the correct buttons are pressed. How-ever, calculators cannot think—that con-tinues to be your responsibility.

FOIL

An important bonus of partial products isthat they demonstrate the distributiveproperty of multiplication over addition on

the set of whole numbers, an importantand direct link to algebraic thinking. Em-phasize place value by rewriting the exer-cise we have been doing and apply thedistributive property on these factors:

You have probably used the FOIL shortcut(Firsts, Outers, Inners, Lasts) in algebra.Partial products require the same proce-dure as FOIL! If you had learned to multiplywhole numbers using partial products, alot of valuable groundwork for algebrawould have been established early.

Your Turn

Use the partial product method of multi-plication (in either the vertical or horizon-tal format) to find the products in the fol-lowing exercises:

Lattice Multiplication

So far, we have presented multiplicationmethods that could be sequenced to as-sist in learning the standard multiplicationalgorithm. We think a case can be built forapproaching multiplication developmen-tally by starting with the facts followed bya careful staging of steps, followed bydigits times multiples of ten, multiples often times multiples of ten, digits timesmultiples of hundreds, multiples of tenstimes multiples of hundreds, and so onuntil the role of place value in products is

52 CHAPTER 2

understood. That staging is necessary ifone is to understand why each subse-quent row is indented one space to theleft as multiplication is performed usingthe standard algorithm.

Whereas lattice multiplication may ormay not be a part of the developmentalsequence, you will see that it is closelyaligned with partial products and, assuch, could be included in the growthprogression. Lattice multiplication is aninteresting process if you know your mul-tiplication facts and can add. The exer-cise 436 x 29 is set up for lattice multipli-cation in Fig. 2.22. The digits of one factorare written above the array of squares,

FIG. 2.22.

starting at the left and going right until yourun out of digits. The digits of the otherfactor are placed to the right of the rowsof squares, starting at the top and goingdown until you run out of digits. Each col-umn and row intersects to form a square,called a cell. The diagonals divide eachcell into two triangles. Each cell contains aproduct generated by the digit from its re-spective row and column. The Tens digitof each product is placed above and to theleft of the diagonal and the Units digit ofeach product is placed below and to theright of the diagonal. Notice that the prod-uct 4 x 2 has no Tens. To avoid confusion,a zero is written in the Tens position andthe 8 is written in the Units position.

One beauty of lattice multiplication isthat only multiplication facts are neededand they can be taken in any order. Onceall the cells are filled, ignore the digits out-

side the array (the factors 426 and 39) andadd the elements of each diagonal. In theexample, starting from the right, the firstsum is 4. The sum of the second diagonalis 14 (7 + 5 + 2 or 2 + 5 + 7). Careful exam-ination will show that the actual value is140, not 14. The 4 from the 14 is written atthe base of the diagonal and the 1 is re-grouped to the next diagonal. The thirddiagonal sum is 6 + 2 + 6 + 1, or 15, butthe regrouped 1 must be added as well.Write the 6 and regroup the 1. The nextsum is 3 + 8 + 0, or 11, plus the regrouped1, giving 12. The 2 is written at the base ofthe diagonal and the 1 regrouped to thenext diagonal. Finally, the last diagonal is0 plus the regrouped 1 and the sum iswritten at the base of the last diagonal asshown in Fig. 2.23.

FIG. 2.23.

Look at that same problem done in partialproduct mode:

Examine the columns in the partial sumsection and you will see the diagonalsfrom Fig. 2.23 except that the zeros arenot written. Thus, a connection is made

NUMBER AND OPERATIONS 53

between lattice multiplication and partialproducts. Because partial products con-nect directly to the standard algorithm,lattice multiplication is closely related tothe standard algorithm. Are you wonder-ing if there is a connection with FOIL, too?We hope you are. If you were wonderingabout that, then you are beginning to in-vestigate the behind the scenes workingsof the arithmetic procedures you perform.

Yes, there is a connection between lat-tice multiplication and FOIL and it can bedemonstrated with this same problem bywriting it as (400 + 30 + 6)(20 + 9). Re-member that FOIL is really a shortcut forthe distributive property of multiplica-tion over addition on the set of wholenumbers. Each term from one factor mustbe multiplied by each term of the otherfactor. You have (400)(20) + (400)(9) +(30)(20) + (30)(9) + (6)(20) + (6)(9), which is8000 + 3600 + 600 + 270 + 120 + 54, andthere are all the values in the partial prod-uct and also in the diagonals of the latticepresentation of the same problem. Howabout that?!

You should notice that the only differencebetween this left to right method and thepartial product style is the order in whichthe partial products are written.

Your Turn

Use lattice, left to right, and the distribu-tive property of multiplication over addi-tion on the set of whole numbers methodsto find the products in:

Left to Right Multiplication

Because of concerns about place value,we have been taught to perform arithme-tic operations starting at the right side ofthe exercise. Partial products make thisunnecessary. Think about the exampleusing expanded notation and the distribu-tive property of multiplication over addi-tion on the set of whole numbers andyou'll realize that the partial products canbe done in any order. Because we readfrom left to right, working from the left ofan exercise has a natural feel or logic to it.The vital roles of place value and multipli-cation facts continue to be present in thismethod:

Horizontal and Vertical Writing

You should become proficient with bothhorizontal and vertical formats whileworking arithmetic exercises. We usedthe vertical format for demonstrating thepartial product and left to right methods,then wrote the exercises in a horizontalformat. As in addition, the argument con-cerning multiplying from left to right ver-sus multiplying from right to left is one ofconvention rather than of correctness. Wecould have presented the last example as58 x 79 = (50 x 70) + (50 x 9) + (8 x 70) +(8x9 ) = 3500 + 450 + 560 + 72 = 4582.You can multiply partial products exer-cises in any order, sum them, and still getthe correct answer.

54 CHAPTER 2

Russian Peasant (Simple Halving/Doubling Method)

Consider someone who is struggling tolearn how to multiply. Offer the statement,"If you can multiply by 2, divide by 2, andadd, you can do any whole number multi-plication problem in the world." That chal-lenge seems almost too good to be trueand can be used as a diversionary ap-proach for someone looking for an alter-nate way to multiply. Russian peasantmultiplication requires the two factors tobe written side by side. One factor ishalved (any remainder is dropped), andthe other factor is doubled. Each of theseresults is listed vertically under the re-spective factors. The halving and dou-bling continues until the bottom numberof the halving column is one. To completethe exercise, even values in the halvingcolumn and the corresponding doubledvalues are crossed out. The numbers thathave not been crossed out in the doublingcolumn, including the original factor if ap-propriate, are added to get the product.Finding the product of 78 and 51 usingRussian peasant multiplication wouldtempt one to start by halving 78. How-ever, the goal is to get to 1 in the halvingcolumn, and starting at 51 will requirefewer steps:

The 312 and 624 are not added into thesum because they are associated witheven values (12 and 6, respectively) in thehalving column. The initial factor of 78 is

included in the sum because its associ-ated value (51) is odd.

How in the world does that work, youmight ask (we hope this is becoming apart of your routine thought processesnow). The answer is rooted in Base 2 no-tation, which is used to express the halv-ing factor and the problem is representedas 78 x 1100112. That is, 78 x (32 + 16 + 2+ 1). Eight and four are not consideredbecause there is a zero in the 23 and 22

places, respectively, in the Base 2 nota-tion for 51. Therefore, the rows containing312 and 624 are eliminated because theirrespective numbers in the halving col-umn, 12 and 6, are even. The product be-comes the sum of 78, 156, 1248, and2496, which is 3978. The following is a fullexplanation of the process:

Somewhere in your education you haveworked with bases other than 10 and in-vestigated multiplication, although proba-bly not in the Russian peasant format.You are aware of the concepts involved inmultiplication and you have seen proofs.Yet, with all that background, each of theideas about multiplication seems to re-main isolated from the others. If you de-compartmentalize the things you havelearned and think about the processes,then the proof of Russian peasant multi-plication is accessible. Much as we wouldlike to, we cannot show you all of theseintricacies. You must capitalize on yourbackground and develop the connectionsbetween the various mathematical expo-sures you have had. Only then will your

NUMBER AND OPERATIONS 55

knowledge and understanding of mathe-matics begin to grow.

Your Turn

Use the Russian peasant method of multi-plication to find the products in:

Try This

Here is an interesting mental multiplica-tion trick you can try. Tell me any 2-digitnumber; I will pick a second number andget the product mentally before you can,using a calculator. Suppose you choose27. I will choose 23. The product is 621.Try again with a larger number? Youchoose 74, I choose 76, and the productis 5624.

Magic? No, here's the trick. Whatevernumber you choose, I choose my numbersuch that the digits in the Tens placesmatch, and the Units digits sum to 10.Multiply the digit in the Tens place timesitself + 1. Multiply the Units digits. Nowplace your answers side by side (Unitsproduct on the right) and you are done.

You choose 27, so I choose 23, right?The Tens digits match, and the Units dig-its sum to ten. The Hundreds digits is theproduct of (2 + 1) and 2 or 3 x 2 yielding aproduct of 6. The product of the Unitsdigits is 7 x 3, or 21. Put them together toget 621. The same thing happened forthe second example 74 x 76, first think7 + 1=8 and 7 x 8 = 56 hundreds. Then5600 + 4x6 = 5624. Practice and you,too, can Beat the Calculator (a free sub-scription to Beat the Calculator can beobtained by writing to beatcalc@aol.comand requesting a subscription)!

Conclusions

Combining the multiplication proceduresdiscussed here with the addition proce-dures discussed earlier provides a robustset of choices for reviewing multiplication.Because people who lack technical skillswill be at a disadvantage in our data richsociety, you should review the calculatorsthat are available. We believe that calcu-lators should be used judiciously. Peoplewho do not understand arithmetic opera-tions are at a disadvantage in our ever-growing technological society, even ifthey own the best computing equipmentavailable. You will need to decide when touse and when to withhold technology.

Bibliography

Beat the calculator. (2000). Available online atbeatcalc@aol.com

WHOLE NUMBER DIVISION

FOCAL POINTS

• Terminology. Division as a Rectangle• Repeated Subtraction Division. Division Algorithm• Remainders. Say What?. Conclusions

You know how to do longhand division byhand, right? Given that, you might wonderwhy we want to talk with you about it. Aswith the other whole number operationswe have discussed with you, we knowyou know how to use the algorithms butwe want to help you look behind thescenes as you divide and think aboutwhat is really going on there. When doingan exercise such as 72)25704, the basic

56 CHAPTER 2

question is, "How many sets of 72 thingsare contained in 25704 things?"

We want to ask you to do somethingright now that might help you gain someinsight into the difficulties associated withdivision. Please do the exercise 72)25704by hand and focus on the various placeswhere errors can be made. We will tell youthe answer is 357 because the answer isnot the focus. We would like you to do theexercise while concentrating on lookingfor places where something could goawry within the various steps needed toachieve the answer.

There should be no question that thedivision facts must be mastered beforeattempting an exercise such as72)25704.

Otherwise, how could we justify askingyou to do it? Skillful use of and under-standing of whole number multiplicationand subtraction are crucial elements ofthe standard long division algorithm,along with place value, estimation skills,the ability to round numbers, as well asthe use of unusual new formats for theoperations being performed.

The first potential error occurs in theability to estimate the initial partial answerto the problem. You may have beentaught to round the 72 to 70 for ease ofestimation. If so, rounding is another po-tential area of difficulty. Most of us wouldhave little problem rounding 72 to thenearest ten and calling it 70, but thatprocess could lead to trouble if we end upestimating too low. Whether or not weround the 72 to 70, our next task is to de-termine what to take the 70 (or 72) into.We know we cannot divide 2 by 70 in thisformat. Next consider 25 as a candidate,but that will not work because 25 is lessthan 70, also. That last statement identi-fies another potential misunderstandingin the area of place value, a skill that mustbe present in order to successfully divide.One must know when a number is greater

than or less than another. Moving on, weknow that 257 (which actually represents25700) is greater than 70, so division cannow take place. However, we now face anew difficulty. How many 70s are in25700? Here comes the need for estima-tion skills. There are approximately threehundred 70s in 25700.

For the rest of the discussion, we willabandon the place value discussion in-volving how many 70s are in 25700, anduse the terminology that is much morecommon. It should be noted here that,whereas dealing with 70 x 3 to generatean estimate is much easier than using72 x 3, you must realize that there arecases where the rounded 70 would leadto difficulties. For example, if we were try-ing to determine how many 72s (with 72rounded to 70) are contained in 213, wewould get a false reading. That is, 3 x 70is 210, but 3 x 72 is greater than 213. Ifyou have used the rounding method of di-vision, then you are undoubtedly wellaware of this potential difficulty.

Moving along, there are three 70s (or72s) contained in 257. That implies a newdilemma. Where should the 3 be placedon the vinculum? Of course, it goes abovethe 7 in the product, but it is rather com-mon to have it incorrectly placed above

3the 4, as in 72)25704. Thus, another po-

tential source of error has been located —where to place the first digit of the esti-mated answer.

Correctly placing the 3 above the 7 in25704, a new dilemma arises and it in-volves multiplication. As stated earlier, weassume the multiplication skills are pres-ent, but look at the configuration of the

3multiplication exercise: 72)25704 which is

3really 72) , which is unlike any "regular"

NUMBER AND OPERATIONS 57

multiplication exercise—3 x 72 may beold material, but certainly not in thissetup. Such an exercise almost always

would have been written as almost

never as and this format of is

really unusual. On top of all that, the prod-

uct is to be placed, thus which

is unusual to say the least. Look at howthe multiplication problem, 3 x 72 now

appears: If an individual does

not have command of multiplication skills,this exercise would be next to impossible.

For the sake of discussion, let us sup-pose that the multiplication can be doneand correctly placed. Now a new problemarises. All of the difficulties associatedwith subtraction requiring regroupingcome charging to the forefront. The set-

ting is now and we are lucky

because there is no regrouping requiredperforming this subtraction. But the nextdifficulty appears to be, "How many digitsdo we bring down after the subtraction iscompleted?" Do we bring down just thezero or do we bring down 04? There aretwo schools of thought on that question.Our purpose is not to debate this ques-tion, so we will bring down only the zero,

making the problem become

Now a new monster raises its head.The task is to determine how many 70s(or 72s if we do not round) are containedin 410. No big deal? It is just another divi-

now, but you should have the picture. Di-vision is difficult because it involves somany skills from earlier work and a brandnew format making previously familiar op-erations take on a very different look.

Terminology

Earlier we connected addition and sub-traction by abandoning traditional termi-nology and using addend, addend, andsum for addition and sum, addend, andmissing addend for subtraction. We willuse that same approach here. Many ofyou have probably learned the division

terms to be . One of the

difficulties you might have faced is decid-ing which number is the divisor and whichis the dividend. Connecting the terminol-ogy with multiplication makes decidingwhich number is which much simpler—

. In multiplication, we

use factor times factor to get the product.In division, we divide the product by afactor to get the missing factor. The dis-cussion could be summarized by:

sion exercise? But it is a big deal. Look at

the format without all of the

other writing in there. What if we were justlearning to master division and we get thisstrange setup. If the multiplication ex-

pressed as caused any trouble,

then how much worse will the division

be? We will end this discussion

58 CHAPTER 2

Increasing your mathematics vocabularyis one of the goals of this text, but we feelthat vocabulary words should be mean-ingful. Whereas the words divisor, divi-dend, and quotient are descriptive, theydo not strengthen the connection be-tween division and multiplication. Paral-leling the discussion about minuend, sub-trahend, and difference, we believe itmakes more sense to have a verbal con-nection between division and multiplica-tion problems so we will use product, fac-tor, and missing factor.

Division as a Rectangle

What do you do when you find the area ofa rectangle? You multiply the length timesthe width. Can't all positive number multi-plication problems be thought of as find-ing the areas of rectangles? In a divisionproblem, you would know the area of therectangle and one dimension; divisionhelps you determine the other dimensionof the rectangle. That might seem trivial toyou, but you now have the ability to gen-erate a mental image of what divisionproblems look like. Sometimes that helpswith understanding what is going on be-hind the scenes in division. Figure 2.24shows how knowing the area of the rec-tangle and one dimension could be usedto gain understanding of division. The leftrectangle has squares inserted to showhow they can help determine the missingfactor. The right rectangle, while involving

FIG. 2.24.

larger numbers and without the squares,still presents the same idea.

Repeated Subtraction Division

With addition, subtraction, and multiplica-tion, sequences of steps are followed tobuild what many call standard algorithms.Division is no exception, but often the in-termediate step or steps remain hidden.Certainly here the sequence starts withthe learning of division facts, which is re-ally just a different way of thinking aboutthe multiplication facts. The investigationof a problem like would prove helpful

here. This is not a division fact and yet itreveals a lot of insight about division. Asimpler problem, also not a division fact,should be considered first Our

earlier discussion about the role of zero inmultiplication now comes into play, alongwith the idea of the area of a rectangle.Here, the area is 20 square units and onedimension is 2 units. The missing dimen-sion (missing factor) is determined to be10 units. That looks a lot like and the

missing link revolves around the role ofzero, much like what was done in multipli-cation. Skill can be developed with prob-lems like using some help

from expanded notation and area. Wecan deal with the area as 24 square unitsand one dimension of the rectangle as 2units, and look for the missing dimension.However, we could also consider theproblem as which we could

solve as . Base 10 blocks help

create a mental image of what is happen-ing. Two longs and four units representthe total area. The configuration in Fig.2.25 also shows that half of the area (we

NUMBER AND OPERATIONS 59

are dividing by 2, remember) would beone Long and 2 Units. Shift the emphasisto place value as shown by the Base 10blocks and you have 10 + 2 as a dimen-sion of the rectangle. Voila! This is the an-swer to the division problem, too.

FIG. 2.25.

You might be thinking we went over-board with that last explanation of

Could this process be repeated over andover until the initial stack is depleted?Certainly. However, it would take a lot oftime and be quite cumbersome. Theprocess can be streamlined somewhat byasking, anywhere along the line, if 10 bun-dles of 72 sticks could be removed fromthe stack, or 100 bundles of 72, or 30bundles of 72, continuing the process byselecting convenient ways of groupinguntil the stack is gone. This is the idea be-hind repeated subtraction division—takeaway conveniently sized groups. As longas you estimate fewer than the maximumnumber of groups, or exactly right, thingsare fine. Notice this is significantly differ-ent from the typical standard division al-gorithm, which mandates that all esti-mates be exactly correct, no more and nofewer. Here are two different looks at

done via repeated subtractiondivision:

You should realize that in each of the twoexamples, the estimation for the numberof bundles of 72 is below the actualamount, until the last step in each prob-lem. Add the values listed to the right ofeach problem and you get 357, which isthe number of bundles of 72 contained ina stack of 25704 sticks. Looking at thetwo examples should convince you that

but we felt compelled to draw your atten-tion to the division sequence and its asso-ciated missing links. Although might

appear similar to it is quite different.

Solving the problem as might

make sense to you, but it is not sequen-tially sound to go that route. A more sen-

sible configuration would be

because of the association with the earlieridea o and 16 divided by 2, which is

a division fact. We hope that we havedrawn your attention to some steps in-volved in developing the concept of divi-sion, some of which you may not havethought about before.

One very significant step in the sequen-tial development of division is the idea ofrepeated subtraction division. Visualize apile of 25704 sticks. Are there enoughsticks to permit the removal of 72 sticks?You probably laughed at that one, but it isan essential question and the fundamen-tal idea behind the notion of repeatedsubtraction division. Of course, 72 stickscould be removed from 25704, leaving25632. Could another bundle of 72 sticksbe removed from the stack? Of course!

60 CHAPTER 2

the convenient and easily spotted multi-ples could be used any time during theprocess. The value of being comfortablewith multiplying digits by multiples of tensor hundreds should be apparent as well.The "taken out" estimations could bewritten on top rather than on the side.This is called the pyramid, or stacked,style, but the process is the same. Theadvantage to stacking the numbers ontop is that as estimation skills get better,the collapsed values look very similar tothe standard algorithm:

permit moving on to the standard divisionalgorithm.

Your Turn

Do each of the following problems usingrepeated subtraction division. The groundrule is that you are not permitted to esti-mate exactly right each time as you dothe problems. Write the answers to atleast one of the problems in the pyramidformat and at least one down the side:

Whereas the whole process involved inrepeated subtraction division might seemcumbersome, one huge advantage shouldbe jumping out at you. As long as you es-timate low or exactly right, the problemcan be completed. Contrast that with thestandard division algorithm and youshould see the advantage of using re-peated subtraction division until the exactestimation skills are refined enough to

Standard Division Algorithm

Did you notice that we wrote the prob-lems in different formats in the exercises?That has not been a discussion point, butwe assume you are aware of differentways to write division problems. We men-tioned the standard division algorithmearlier. We know that you know how to dolong division by hand and are confident inyour use of the standard algorithm. Ourdiscussions should have given you someinsight into the strengths and weaknessesof the standard algorithm and, more im-portantly, what is going on behind thescenes as a problem is done. We will dothe problem we used as the example forrepeated subtraction division via the stan-dard algorithm so you have it as a refer-

ence point That should look

NUMBER AND OPERATIONS 61

familiar with the possible exception ofbringing down only the zero rather than04 after the first subtraction. Things arefine as long as you estimate exactly cor-rect. Did you check your work on the exer-cises by working the exercises using thestandard algorithm? We know that yourtime is limited, but suggest that you prac-tice by reworking a couple of them usingthat tried and true tool.

Technology

Did you do the exercises simply to prac-tice your long division skills? We are bet-ting you did not. Furthermore, we are will-ing to think that you would have grabbeda calculator to do the exercises if you hadbeen asked simply for the answers. Whywould you reach for the calculator? Mightyou have rationalized that long division istoo painful to do, the calculator is faster,and the calculator is a lot more accurate?All of that makes sense to us. Still, a verycritical issue has been raised.

When do you use a calculator? Wehave repeatedly said you need to knowthe number facts and the basics of theoperations. OK, when do you say that asufficient level of competency has beenreached and it is time to use technology?We cannot answer that for you, but weare willing to bet it is somewhere betweenthe division facts and s far

as you are concerned. In fact, the shiftpoint is probably long before here and wesympathize with your decision. Our pointhere is that technology has a place in allof this. We cannot tell you when to use acalculator. That is your decision. How-ever, when you decide to use or not touse technology, you must be able to de-fend your position. Saying technology isinappropriate because you had to do a

certain kind of problem by hand is unsuit-able. At the same time, deciding to usetechnology for generating a number factbecause it has not been memorized isequally inappropriate. Somewhere be-tween those two is the answer and we areattempting to help you make that into aninformed decision, rather than a blanketstatement made without forethought.

Remainders

Many division problems do not result inwhole numbers. Sometimes a problemwith a remainder makes sense and some-times it does not. For example, if you aredividing a dozen dolls between eight chil-dren, a doll and a half does not makemuch sense; you have four dolls remain-ing. On the other hand, if you are dividinga dozen cookies between eight children,then each child can get a cookie and ahalf (if you can get a half a cookie ©). Wewill assume that problems posed thathave remainders as we discuss decimalswould have an application somewhere inthe real world.

As you know, the missing factor in divi-sion is not always an integer. You have

seen the answer for xpressed as

1 1.67, and perhaps a host of otherways, including 1 r 2, meaning a missingfactor of 1 with a remainder of 2. The ideais that after dividing out as many knownfactors as possible from the product, thereare still some parts left over. Initially, thisleftover collection is expressed as a re-mainder. Later they are expressed as frac-tions, decimals, or repeating decimals.

Remainders can occur naturally. Sup-pose 21 people want to take a trip as agroup. Assume that all individuals are li-censed drivers and any car they elect touse will carry the driver plus three others.Do the division 21 -f 4 and you get five,

62 CHAPTER 2

but there is one person left. If you are thatremaining person and you want to go onthe trip, then you do not want to be omit-ted. Thus, in this case, the remainder issignificant and a sixth car is needed.Hopefully, although not germane here,that 21st person will also have some oth-ers ride along so there could be three carscarrying three people and three cars car-rying four people.

Suppose you have 21 cookies andwant to share them with 3 friends soeach of you has an equal amount. Doingthe problem arithmetically would yield21 ÷ 4 = 5 r 1. As things stand, there is anextra cookie. Certainly it could be dividedinto four equal parts and shared, whichwill happen when we get to fractions. Theeasy solution would be to set that extracookie aside, or better yet, do the arith-metic ahead of time and simply start with20 cookies, so that the dilemma isavoided. How you solve the quandary ofthe extra cookie is not the issue here. Theidea of having an extra cookie is our con-sideration. The solution here is much dif-ferent than the previous one where a tripwas taken. Would you want to be the per-son who was divided into parts to makethe trip? That does not make sense, andyet, dividing the cookie might makesense. Nonetheless, both solutions cameout of the same arithmetic exercise.

Certainly remainders could be avoidedwith careful planning, much like was donein the cookie example. All of the discus-sion about division of whole numbersprior to this part about remainders actu-ally did that because each missing factorwas an integer. Still, the possibility of a re-mainder does exist and you need to thinkabout how to deal with it. Later you willexpress them as fractions or decimals.

Some people do not think remaindersshould be considered. However, as youhave seen with the trip and the cookies,

Say What?

A final concern about division involves theway problems are worded. There are actu-ally two very different questions we mightask: 1) "How many sets of items?" or 2)"How many items per set?" We mightthink about this while packaging cookiesfor a bake sale and the stem of the prob-lem could be, "We have 48 cookies for theclass sale." If we have a good supply ofplastic bags, then we might arbitrarily de-cide to put three cookies in each bag. Thismethod would be a good example of re-peated subtraction division, as we meas-ure out three cookies for each bag. Thequestion would be, "How many plasticbags do we need if we want to have threecookies per bag?" However, if we haveonly 12 plastic bags, this would not work.In this situation, we know how many setswe need to make, but would need to figureout how to share the cookies among thepredetermined number of bags. The ques-tion would be, "How many cookies shouldwe put in each of the 12 plastic bags wehave?" Although there is no difference inthe operation, there is an important differ-ence in the problem solving.

there are situations when they are signifi-

cant. Consider and

They imply that because we

get the exact same answer for both exer-

cises. But, and

you know that; are not equal.

So what should we do with remainders?You know what they are and how theycan show up in real-world situations. Younow also see how remainders can con-nect to nonintegral answers in division.

NUMBER AND OPERATIONS 63

Conclusions

You should have a better understandingof what is going on behind the scenes indivision. Of our basic operations, it is eas-ily the most complex because it relies onprevious knowledge of both multiplicationand subtraction for mastery and requireslearning a new format. Still, with carefulpresentation and thoughtful reflection, di-vision can be understood and managed.Because of the complexities related to di-vision, the question of when to use tech-nology seems worth serious reflection onyour part.

EQUIVALENT FRACTIONSAND MULTIPLICATION

OF FRACTIONS

FOCAL POINTS

• Concrete Beginnings• Equivalent Fractions• Converting an Improper Fraction to a

Mixed Number• Product of a Whole Number and a

Fraction• Product of Whole Number and Mixed

Number• Product of Two Fractions. Product of a Fraction and a Mixed

Number• Product of Two Mixed Numbers• Conclusions

Our approach to working with fractions isprobably going to be different from whatyou have seen in the past. We are going todiscuss equivalent fractions and then mul-tiplication. Typically, addition of fractionsis done after dealing with equivalent frac-tions. Multiplication of fractions is mucheasier than addition of fractions. That,coupled with the idea that multiplication is

visited as equivalent fraction concepts aredeveloped, prompted us to discuss multi-plication of fractions before addition offractions. Stick with us on this. We areconfident you will like what you see.

Concrete Beginnings

People often struggle with the concept offractions because of fundamental miscon-ceptions. We all have been taught theterms numerator and denominator, alongwith their respective definitions. Still, ifthose definitions are presented only in theabstract, false impressions can be createdand the development of mal rules can oc-cur. These are rules people make fromtheir observations that seem to makesense. Because they are generally basedon misunderstandings of the ideas pre-sented, they may work only in one situa-tion or perhaps never. Once established,mal rules are often difficult to purge. Forexample, Fig. 2.26 shows a situation that

FIG. 2.26.

many people accept as a representation ofone half. We know that is not a half and yetwe participate in situations that often stim-ulate such impressions. Have you everbeen asked if you want the big half or thelittle half of something? That is a silly ques-tion if you think about it. Aren't two halvesof something supposed to be equal? Folda sheet of paper in half. Once you havedone that, describe how you know youhave created halves.

More than likely, you folded your paperin either hamburger (top) or hot dog (bot-tom) format as shown in Fig. 2.27. Fur-

64 CHAPTER 2

FIG. 2.27.

thermore, you probably rationalized thatyou knew you had folded the paper in halfbecause the two parts were identical. Ifwe pressed you to discuss the situationmore fully, then you would probably saythat you knew you had folded the paper inhalf because the edges of the piecesmatched. By this process, you refinedyour definition to make it clearer. Then, ifwe showed you Fig. 2.28 you might pause

FIG. 2.28.

before agreeing that the paper had beenfolded in half. A discussion about cuttingalong the diagonal and rotating one of thetriangles would lead to the conclusionthat the edges would then match, againconfirming that the paper had, in fact,been folded in half. Perhaps reflectionwould lead to a further revision of the defi-nition, but eventually we get to a workingdefinition of what a half is—a unit is di-vided into two equal parts and one ofthose two equal parts is under consider-ation. The discussion could also includeideas about matching edges yielding ar-eas that are the same.

The representation of a half is where

the numerator is 1 and the denominator is2. The definition clarifies that the denomi-nator represents the number of equal-sized pieces something is divided intoand the numerator indicates the numberof those equal pieces that is being con-sidered. Notice the operative word is"equal" here. Although we have only de-

veloped the idea of concretely, a similar

approach could be used to consider addi-tional fractions if you are uncertain of thegeneralized definition for a fraction.

Equivalent Fractions

The idea of concrete beginnings does notstop with the definition of fractions and ageneralized idea of fractions. Exactlywhat are equivalent fractions? We couldtell you, but we are opting to have you seefor yourself. Fold a sheet of paper in halfeither hamburger or hot dog style (but nota diagonal fold for this activity). Open thesheet once you have folded it and shadein one of the two equal-sized pieces.What part of the sheet is shaded? You

should say because you shaded one of

two equal pieces.Now close the paper along your fold so

the shaded half is inside. Fold the paper inhalf again, either hamburger or hot dogstyle. Your model could look like either ofthe two presentations in Fig. 2.29, al-though other shading or folding optionscould occur. There is a significant pointbrought out by Fig. 2.29. The unit (sheetof paper) has been divided into four equalparts and two of those four are shaded. In

other words, of the paper is shaded.

But, no additional paper was shaded, so it

NUMBER AND OPERATIONS 65

FIG. 2.29.

must be that are different names

for the same quantity. Similarly, if an addi-tional fold is made after refolding the pa-per twice, as shown in Fig. 2.30, you end

FIG. 2.30.

up with a representation that should con-

vince you that are different

names for the same quantity or, saying it

another way,

With that background, we can look at

something like and recog-

nize the rule you learned some time ago;multiply both the numerator and denomi-nator by the same value to get an equiva-lent fraction.Take a huge leap and look at

. Multiply both the nu-

merator and denominator by the samevalue to get an equivalent fraction is therule we used. Granted this is unusual, butit is mathematically correct. In the previ-

ous example- we found an equivalent

fraction to be The result we generated

involved larger values for both the numer-ator and denominator and yet the result,

, is equivalent to In this example, we

have found to be an equivalent ex-

pression for but we have expressed

the equivalent fraction using smaller val-ues in the numerator and denominatorthan were used in the original fraction.

Whereas is certainly equivalent to

it is often preferred that an equivalent

fraction to be expressed so that the

numerator and denominator use thesmallest possible values (also referred toas being relatively prime, which meansthe largest common factor shared by thenumerator and denominator is one). Forexample, you may have noticed that moststandardized tests mandate that all frac-tions be expressed in simplest terms,even though this might insert the oppor-tunity for additional errors.

For this case, the mandated equivalentfraction would be even though it is only

one of an infinite number of fractions that

are equivalent to In this example, is

the fraction obtained when is simpli-

fied to its lowest terms. Sometimes sim-plified to lowest terms is called the re-

duced version of but this leads to the

notion tha is a smaller fraction thar

which is false. We recommend that youuse the phrase simplified to lowest termsrather than reduced to simplest terms orreduced version.

66 CHAPTER 2

Your Turn

1. Find an equivalent fraction to eachof the following; being sure to show thesteps that assure your result is correct:

Converting a Mixed Numberto an Improper Fraction

If we gave you nine quarters and askedyou to count them, you could come upwith several different responses, the firstof which might be, "One quarter, twoquarters, three quarters, four quarters (orone dollar), five quarters, six quarters,seven quarters, eight quarters (or two dol-lars), nine quarters." Perhaps you wouldsummarize the situation by saying thatyou counted two dollars and a quarter.Figure 2.31 models this on a number line.You could count on the number line justas you did with the quarters. When you

get to another name for that point is 1.

Similarly, at an equivalent name is 2.

FIG. 2.31.

We believe that the phrase "divide outall common factors," is better than either"reduce" or "simplify." Divide out all com-mon factors is more descriptive of what is

happening when a fraction such as is

changed to an equivalent value in whichthere are no common factors between thenumerator and denominator. An under-standing of the process of dividing outcommon factors will be extremely usefulin algebra. In this case, you would have

a fraction in which the

numerator and denominator have nocommon factors.

To divide out all common factors of

we could have multiplied both the numer-ator and denominator by the same con-venient value and ended up at the same

place: Although

we have not talked with you about howto multiply whole numbers and frac-tions, your prior experience should con-vince you that this is an acceptable, al-though complex, way to simplify thefraction.

The result of dividing out all common

factors of which represents the

smallest possible values for the numera-tor and denominator. Another way of say-ing this is that the numerator and denomi-nator are relatively prime. This statementdoes not necessarily mean the both val-ues are prime numbers. It simply meansthat their greatest common factor is one.

In the numerator and denominator are

relatively prime even though 4 is not aprime number.

NUMBER AND OPERATIONS 67

9 9The point - could be called just that, -, or4 4

it could be called 2 and —, interpreted as

two whole things and — more. This could4

be written 2 + —, but it is generally written

as 2-, with the plus sign understood,4

much like when we discussed implicitmultiplication (4y rather than 4 x y). Weread the understood plus sign as and.This mixed number is two and one fourth.The assumed plus sign is a key elementto understanding the process of convert-ing from mixed numbers to improper frac-tions (a fraction in which the numerator isbigger than or equal to the denominator)or from improper fractions to mixed num-bers.

Consider the following conversion:

tor and denominator of a fraction by thesame value to get an equivalent fraction.The rest of the process should seem fa-miliar to you as you recall the basics offraction addition. The result of this proc-

21ess is —, an improper fraction.

Converting an Improper Fractionto a Mixed Number

The procedure for converting an improperfraction to a mixed number follows di-rectly from our example of converting amixed number to an improper fraction,because you work the process in reverseorder. Consider the following:

38 = 35 + 35 ~ 5

Notice how the first step in the process5

of converting 2- to an improper fraction8

involves inserting that understood plussign. Next, we give an equivalent fractionfor 2, using the appropriate denominator.

pWe remind you that 2 is equivalent to -,

which lets us apply the generalization wedeveloped about multiplying the numera-

The trick is to know how to split — ini-5

tially. Mastery of basic multiplication anddivision facts is essential here. We needto determine the largest multiple of 5 con-

38 3tained in 38, and we find that —=7-.

5 5Mixed numbers are often useful in helpingus understand the quantities involved inproblems, whereas improper fractions aregenerally used when multiplying mixednumbers. Improper fractions are espe-cially useful in algebra.

Your Turn

2. Convert the following mixed num-bers to fractions. Please note we have not

68 CHAPTER 2

discussed any shortcuts you might havelearned in the past, so the expectation isthat you will not apply them at this point.You need to practice the skills of convert-ing at a basic level so you will gain a fullunderstanding of the shortcuts you mighthave learned previously.

3. Convert the following improper frac-tions to mixed numbers. Once again, weask that you practice the skills of convert-ing at a basic level; leave the shortcuts forlater.

Product of a Whole Numberand a Fraction

Recall from multiplying two whole num-bers that you might have said that 3x4was three fours. Similarly, we could say

that is nine quarters. We write nine

quarters as and connect

Remember that 9 can be written so

could be written , Connect

the ideas here and conclude that, to get

t h e product o f w e could insert a n

intermediate step to show exactly whatis going on. With that step inserted,

which is . To get the prob-

lem done, we place the product of the nu-merators over the product of the denomi-nators. Another way of generalizing aboutthe product of a whole number times afraction would be to put the product ofthe whole number and the fraction's nu-merator over the denominator. Althoughthis second generalization is an efficientshortcut, the first generalization will workwith all fraction products.

Your Turn

4. Find the following products:

Product of Whole Numberand Mixed Number

You have all the tools for extending theprocess of multiplying a whole numbertimes a fraction to finding products thatinvolved mixed numbers. Suppose you

need to find the product of 4 and You

already know how to convert o an im-

NUMBER AND OPERATIONS 69

proper fraction, getting and the prob-

lem becomes Multiplying the

whole number by the fraction's numerator

gives , If you need to convert the prod-

uct to a mixed number, then you must de-cide the largest multiole of 7 that is con-

tained in 88, and

There is another way to find the product

of 4 and that introduces an important

concept. Remember tha

So. the problem can be rewritten as

. Using the distributive property

of multiplication over addition on the set

of rational numbers, we get

or which is The distributive

property of multiplication over additionon the set of rational numbers might seemcumbersome to you in this problem,but this is great background work for al-gebra.

Your Turn

5. Find the following products first byconverting the mixed number to an im-proper fraction and then by using the dis-tributive property of multiplication overaddition on the set of rational numbers:

Product of Two Fractions

We have already developed the idea ofputting the product of the numeratorsover the product of denominators whenmultiplying fractions. That same rulecan be used to deal with a variety ofproblems. The simplest involves two

factors, something like Multi-

plying the numerators and denomina-

tors would give Fraction

multiplication exercises are not alwaysthat easy to do, but the process is thesame each time.

C o n s i d e i T h e process could be

the same as before, giving

At this point, all of the prior discussionabout equivalent fractions comes intoplay. Whereas getting the product of thenumerators and denominators and thendividing out common factors is a validway to complete fraction multiplication,there is another way. Dividing commonfactors out first makes the arithmetic sim-pler, but yields the same final answer.

That is, could be done

which becomes . This could

have been done as » which

would become , To decrease

the chances of arithmetic errors, we en-courage you to adopt the practice of di-viding out common factors at the begin-ning of the problem.

70 CHAPTER 2

Multiplication of fractions could involveone or more factors that are improperfractions, but the process is the same. For

example, becomes

Depending on the situation, you mayneed to convert the resultant improperfraction to a mixed number.

A remaining issue with fractions is thesituation where more than two factors are

involved. Consider , There is

a hard way and an easy way to do thisone. You might show the product as

and then divide out

common factors until the answer is Al-

though that method works, it is more effi-cient to divide out the common factors

or . In the example, we identified the

pairs of common factors that were dividedout by using several types of slashes. Thereare other pairings that could have beenused to divide out common factors, but

they all would lead to a final answer of -. If

the factors were improper fractions, theprocess would be the same. The problem

Your Turn

c) Make up an interesting exercise andcomplete it.

Product of a Fractionand a Mixed Number

The process of multiplying a fraction anda mixed number is not significantly differ-ent from finding the product of a wholenumber times a mixed number. In fact,the problem becomes the product of twofractions if the mixed number is convertedto an improper fraction. For example,showing all steps, we might see

though you might want to skip some ofthese steps, this process will alwayswork. Let's examine the distributive prop-erty of multiplication over addition on theset of rational numbers once again. Ifwe introduce the understood plus sign

and use juxtaposition, becomes

Using equivalent

fractions, becomes

o r , Using your skills t o a d d frac-

tions with the same denominators.

giving a sum of . The choice of what to

do next often depends on the statementof the problem. As we have said before,there are advantages and disadvantages6. Find the following products:

NUMBER AND OPERATIONS 71

Why, you say, should you have to gothrough ail those steps when we end upwith the same partial products? Exam-ining this process will help you under-stand what you have been doing all theseyears. An equally significant consider-ation is that you are establishing a power-ful algebraic background.

Your Turn

7. Use both the partial product anddistributive methods to complete the fol-lowing exercises:

Conclusions

So, there you have it, equivalent fractionsand multiplication involving fractions. Wehope this experience with fractions hasbeen insightful and helpful in refining yourunderstanding. As you studied the topicsin this chapter, fundamental patterns andideas should have been forming that willprovide you a powerful background foryour future work.

We suggest that you investigate a vari-ety of calculators, not just brand names,but all levels offered by a manufacturer.You will find that some of them operatewith fractions, giving fractions as an-swers. Some will convert between frac-tions, decimals, and mixed numbers.

involved in simplifying this improper frac-

tion, or changing it to

the mixed number

Product of Two Mixed Numbers

Our final discussion will involve findingthe product of two mixed numbers. In

converting to improper fractions

makes this a problem similar to those we

have considered before. That is,

b e c o m e s w h i c h i s

The following model shows this problemdone in much the same manner as youdid whole number multiplication, usingthe distributive property of multiplicationover addition on the set of real numbers:

To complete the exercise, we need onlyadd the partial products,

The problem could also be shown an-other way, if we use the distributive prop-erty of multiplication over addition on theset of rational numbers.

72 CHAPTER 2

Some use pretty print (vinculum is pres-ent), whereas others have special codingto indicate fractions. When and how to in-troduce these calculators is somethingyou will need to decide.

ADDITION OF FRACTIONS

FOCAL POINTS

• Concrete Beginnings• Fraction Sums. Adding Fractions When Denomina-

tors Are the Same. Adding Fractions When the Denomi-

nators Are Related• Adding Fractions When the Denomi-

nators Are Relatively Prime• Adding Fractions When the Denomi-

nators Are Not Relatively Prime andOne Is Not a Multiple of the Other

We reviewed equivalent fractions andfraction multiplication first because only afew rules are required. Addition of frac-tions has more rules, but is not difficultto master if you approach the operationcarefully and in a logical manner.

Concrete Beginnings

If we gave you three U.S. quarters andasked you what you had, you would prob-ably say, "Three quarters" or "Seventy-five cents." Because a quarter is onefourth of a dollar, the problem could be

written The answer is known

to be so it must be the case that

Here we used money toAadd fractions. Consider the number line inFig. 2.32, which shows this sum. Each of

FIG. 2.32.

the three vectors represents Starting at

zero, you move of a unit, then you add

another followed by a final , so that

you end up a1 on the number line. Simi-

lar results could be generated with Cuise-naire rods, fraction bars or circles, eggcartons, or other manipulatives that areused with teaching fractions. For a coupleof examples, check out:

http://pegasus.cc.ucf.edu/~mathed/crods.html

http://pegasus.cc.ucf.edu/~mathed/egg.html

http://pegasus.cc.ucf.edu/~mathed/fk.html

We encourage you to investigate usingmanipulatives to add fractions. They willenhance your understanding and makeoperating with fractions easier for you.

Adding Fractions WhenDenominators Are the Same

When adding the three quarters, we know

the sum of the addends is , however the

intermediate step of

was omitted earlier and it holds an essen-tial clue to what is happening in the prob-lem. As you look at the example, you

NUMBER AND OPERATIONS 73

should conclude that, when the denomi-nators of the addends are the same, thenumerators of the addends are addedand their sum is placed over the commondenominator.

This same statement is true if the ad-dends were not unit fractions (fractionswith numerators of one). For example, if

you had you would just add the nu-

merators, (note again that the

denominator remains constant). On thenumber line in Fig. 2.33, you see the two

FIG. 2.33.

addends shown with vectors and that the

second vector ends at . This supports

the idea of adding the numerators andputting the sum over the common de-nominator as a means of expressing the

sum. Figure 2.34 shows addends ol and

FIG. 2.34.

, giving a sum of , Notice that

equivalent values for are shown

below each.We suggest that you limit your initial

practice of skills to addends that givesums less than one until you are comfort-able with this concept, then deal with

sums greater than or equal to one. Whena sum is greater than one, it is often nec-essary to convert an improper fraction toa mixed number. Although that is a skilllearned at this stage, it is another step inthe problem, and thus, another chance foran error. This step is generally not neces-sary in algebra classes, but often helps usget a feel for the value of the sum. For ex-ample, you may be quite comfortable with

a value such as but, depending on the

problem situation, might be easier to

understand.There is another group of addition exer-

cises to be considered when like denomi-nators are involved, and it deals withmixed numbers. If you add a whole num-ber and a mixed number, the task is

simple. Consider Because can

be written the problem becomes

Thanks to the associative

property of addition on the set of ration-

al numbers, the problem is or

We write and the exer-

cise is finished.Suppose the problem involves the sum

of a mixed number and a fraction when

the denominators are the same:

The problem can be expressed as

which is

If the problem is

there is an improper fraction in

the sum and the procedure will includethe additional step of converting the im-

74 CHAPTER 2

proper fraction to a mixed number. The

problem is written as and,

with applications of the commutative andassociated properties of addition on theset of rational numbers, it is changed to

The equivalent value is

but we would not write that as

rather we would convert the improper

fraction changing the exercise to

and the sum is

Adding several mixed numbers with likedenominators in the fractional parts, like

is an extension of the

process. /-\|j|jiying the definition of mixednumbers and the commutative and asso-ciative properties of addition on the setof rational numbers, the problem be-

comes but this is

Granted, we

skipped some intermediate steps.You might be wondering why we are

taking such small increments as we lookat different problem types for adding frac-tions. We are trying to build a slow, care-ful sequence that will help you under-stand and master the whole process ofadding fractions.

Your Turn

1. Find each of the following sums,showing the intermediate steps:

Adding Fractions Whenthe Denominators Are Related

Related denominators occur when onedenominator is a multiple of the other. Ad-dition with related denominators involvesexpressing one fraction in equivalentterms so that both fractions have the samedenominator. Because one denominator isa multiple of the other, the common de-nominator is easy to determine. For ex-

ample, if you wanted to add the

first step would be to determine that 10 isa multiple of 5 and that the missing factoris 2. At that point, the rule you establishedearlier for generating equivalent fractionsis applied and the equivalent expression

of - replaces changing the problem to

The idea of equivalent fractions is anessential basic theme in proportional rea-soning. Part of what we are doing here,and in other related sections of this book,will help you gain a better understandingof proportional reasoning.

Each of the following examples requiresdetermining which denominator to use asthe common denominator and setting upthe equivalent fraction, then completingthe exercise using the procedure for add-ing fractions with the same denominator.The first example is short, but the secondis quite involved, with many steps. We willnot provide a detailed discussion. If you donot find the examples easy to follow, thenrefer back to the section on how to get anexpression equivalent to a given fraction,then revisit the discussion about addingfractions or mixed numbers when the de-nominators are the same:

NUMBER AND OPERATIONS 75

We skipped steps in these examples. Ifthis causes you to struggle with either ofthe examples, then you should write outeach exercise and insert the missing stepsneeded to enhance your understanding.

Your Turn

2. Do the following problems, showingthe intermediate steps you made to getthe sum:

Adding Fractions When theDenominators Are Relatively Prime

Next, look at fractions with denominatorsthat are relatively prime. An example of anexercise with denominators that are rela-

tively prime is The greatest com-

mon factor of 3 and 4 is 1, so these num-

bers are relatively prime. To complete theexercise, we need a common denomina-tor that is a multiple of 3 and a multiple of4. Figure 2.35 shows one way that the

FIG. 2.35.

multiple can be determined using thenumber line. The vectors above the num-ber line show multiples of 3, whereas thevectors below the number line show mul-tiples of 4. Notice that the left ends ofboth start at zero, but the tips at 3 and 4do not match. Because the first 3 vectoris short of the end point of the first 4 vec-tor, the second 3 vector is added, giving 6as the multiple. The second 4 vector isadded, giving 8. The process is repeateduntil the arrowheads of the vectors matchexactly. The 3 vectors and 4 vectorsmatch at 12 (and at 24, 36, 48, . . .). Weuse 12 because it is the least commonmultiple of 3 and 4. By using 12, we keepthe arithmetic simpler, but any multiple of12 could be used.

Examine Fig. 2.35 and notice that ittook four 3 vectors and three 4 vectors toget the arrowheads to match. When wediscussed whole number multiplication,we talked about three 4s as another wayof expressing 3x4 . Aha! We have just de-veloped a rule for finding the least com-mon denominator when the two denomi-nators of the fractions being added arerelatively prime. Putting that to work,

is expressed as or

This is now an exercise with like

denominators, resulting in a sum of

Another example is

76 CHAPTER 2

Your Turn

3. Do the following problems, showingthe intermediate steps you made to getthe sum:

Adding Fractions When theDenominators Are Not RelativelyPrime and One Is Not a Multipleof the Other

Suppose the denominators are 4 and 6,which are not relatively prime becausethey have a common factor of 2. Certainlythe rule about multiplying denominatorscould be used here, but the productwould not be the least common denomi-nator. Depending on the problem, wemay elect to go that route, using 24 as thecommon denominator. The thing is, whenthe problem is finished, the numeratorand denominator of the sum will have acommon factor. As we have said before, itis generally accepted that the common

factors are factored out of a fraction.Thus, an additional step would be neededto simplify the denominator.

Most people prefer to work with theleast common denominator initially, andwe will give you a user-friendly method tofind it. In the case of 4 and 6, we list theprime factorization of each, getting 2x2for 4 and 2 x 3 for 6. All of the factors areused, but any common factor is used onlyonce. In this case, you have 2 x 2 x 3 = 12,the least common denominator for frac-tions having 4 and 6 as the denominators.Once equivalent fractions are determined,the exercise is completed exactly as be-fore:

Your Turn

4. Do the following problems, showingthe intermediate steps you made to getthe sum:

NUMBER AND OPERATIONS 77

Conclusions

Now you have the whole story on addingfractions and mixed numbers. We pre-sented the sequence the way we did tohelp you build the rules you have probablyused for some time. Our goal is to help yourealize why you make some of the movesyou do as you add fractions. We encour-age you to create your own additionalproblems or seek an alternate source withsolutions if you are still a little shaky aboutadding fractions and mixed numbers inany of the situations presented.

As usual, we advocate the appropriateuse of technology. When adding frac-tions, we ask you to resist the temptationto use technology prior to developing thenecessary skills and understandings, be-cause there is so much more than justgetting the answer. You need to be moreinterested in the process at this stage ofyour development.

SUBTRACTION OF FRACTIONS

FOCAL POINTS

• Same Denominators• Related Denominators• Relatively Prime Denominators• Denominators That Are Not Relatively

Prime or Related• Mixed Numbers

Addition and subtraction are inverse op-erations of one another, so all the denom-inator rules for dealing with addition offractions apply to subtraction involvingfractions. You may want to review theserules and the procedures for dealing withdifferent types of denominator combina-tions. We will restate the appropriate rulesand provide some helpful examples withlittle discussion. We will provide a more

elaborate discussion when dealing withmixed number situations.

Subtraction can be modeled on thenumber line. Figure 2.36 shows how an

FIG. 2.36.

exercise involving like denominators,

such as would be done using the

missing addend approach. The sum is

and the known addend is . The task is to

determine the missing addend, which is

as indicated by the dashed vector.

Similar results could be generated withCuisenaire rods, egg cartons, or othermanipulatives. We encourage you to usemanipulatives to investigate the conceptsof fraction subtraction. You will find thatthe experiences will enhance your under-standing and make operating with frac-tions easier for you.

Same Denominators

The rule for adding fractions when the de-nominators are the same allows you toadd the numerators and put the sum overthe common denominator. The rule forsubtracting fractions with like denomina-tors is similar, except that you must dealwith the numerators in the order in whichthey are given. Because the commutativeproperty does not hold for subtraction inany set, you must remember that the firstnumerator is identified with the sum and

78 CHAPTER 2

the second numerator with the given ad-dend. Return to Fig. 2.36 and notice that,

in the case of , 6 is the numerator of

the sum and 2 is the numerator of the

given addend, so you write

Subtraction problems with mixed num-bers can involve fractions with denomina-

tors that are the same. In no

regrouping is required. Expanding themixed numbers is one step in completing

the exercise. Because is being sub-

tracted, Fig. 2.37 shows that the distribu-

We realize that this looks like an algebraproblem, but feel that showing all thesteps this way should clarify the process.

Your Turn

1 . Do each of these problems, showingbasic intermediate steps:

FIG. 2.37.

tive property of multiplication over addi-tion in the set of rational numbers causesthe given addend to be rewritten. Fo-cusing on each looped section in Fig.2.38, you see two separate problems. The

FIG. 2.38.

first is whole number subtraction and thesecond is fraction subtraction where thedenominators are the same. In a final

step, the missing addend of would

be expressed as . This problem could

also be done in a horizontal format:

2. Discuss what happens if a fractionsubtraction problem involves two unitfractions with the same denominator.

Related Denominators

As in addition, subtraction with relateddenominators involves using equivalentfractions to provide denominators thatare the same. Because one denominatoris a multiple of the other, determine whatfactor is needed to provide a multiple ofthe smaller denominator. For example,

NUMBER AND OPERATIONS 79

7 by 2 to get a denominator of 14. Be-cause the denominator is multiplied bytwo, the numerator is also multiplied by 2

to get . Take a moment to read through

the following subtraction examples:

Relatively Prime Denominators

As in addition, subtraction with relativelyprime denominators involves findingequivalent fractions for both the sum andgiven addend so they have the same de-nominator. Because the denominatorshave no common factors, you can makequick work of this by multiplying the nu-merator and denominator of the first frac-tion by the denominator of the secondfraction, and multiplying the numeratorand denominator of the second fractionbv the denominator of the first fraction.

Your Turn

3. Do each of these exercises, showingbasic intermediate steps:

4. Discuss what happens if a fractionsubtraction problem involves two unitfractions with related denominators.

Once the common denominator is deter-mined, you complete the problem as inprevious exercises.

Your Turn

5. The examples we have shown in thetext involved two fractions with relativelyprime denominators. If the problem in-volved three (or more) fractions and theonly common factor shared by the de-nominators is one, how would you workthe problem?

6. Do each of these problems, showingbasic intermediate steps:

80 CHAPTER 2

Denominators That Are NotRelatively Prime or Related

Certainly the rule about multiplying de-nominators could be used when unlike de-nominators are not relatively prime, but theproduct would not be the least commondenominator. The solution would be a frac-tion with a numerator and denominator thathave a common factor. If you insist that allcommon factors be divided out, a simplifi-cation step would have to be performed.The following examples should help:

Your Turn

7. Do each of these problems, showingbasic intermediate steps:

Mixed Numbers

One type of subtraction exercise involvingfractions deserves special considerationbecause of the regrouping that must bedone. Consider . Writing this exer-

icise vertically introduces an alignment

that may appear strange You need

to express 4 in a different manner so thatthe problem will look like one you knowhow to handle. Although 4 can be writtenin a multitude of ways, we are going towrite it as 3 + 1 and express the 1 as ,

We want to use as an equivalent fraction

for 1 because it ensures that the fractionshave the same denominator. Thischanges the problem to a familiar form,

The problem could be solved in a hori-zontal manner, as follows. How to com-plete an exercise is often a matter of per-sonal preference. We hope that you willpractice several different methods, ratherthan rely on the method you find mostcomfortable:

NUMBER AND OPERATIONS 81

There is another way to look at thisproblem. Your experience with missing

addends should help you reason that is

less than one , or that must be

added to to generate . This line of

reasoning is extended when the sum (4

in this case) is made smaller, meaning

that the missing addend would include

the missing and the remaining 3 from

the sum, or Although it is unlikely

that someone would use this processwhen working with pencil and paper, itis a common technique for mental arith-metic.

A final situation involves an exercise inwhich both the sum and the given addendare mixed numbers with like denomina-tors. In this special case, the fraction partof the sum is less than the fraction part ofthe given addend. You will need to dosome regrouping before the subtractioncan be completed. If the exercise is

the first thing to do is rename the

sum so that you can regroup,

which is . Now the exer-

cise can be expressed as . Where-

as this might look strange, that improperfraction makes the subtraction of the frac-tional parts possible.

Your Turn

8. Do each of these problems, showingbasic intermediate steps:

Conclusions

Subtraction of fractions and mixed num-bers is not difficult once you have mas-tered the concepts involved with equiva-lent fractions, addition of fractions, andmixed numbers. We encourage you tothink about when and how technologyshould be used in subtraction of fractions.

DIVISION OF FRACTIONS

FOCAL POINTS

• Concrete Beginnings• Whole Number Divided by a Fraction. Fraction Divided by a Fraction. Mixed number divided by Mixed

Number• Common Denominator Division

Understanding the question is the key tofraction division. If you have 8 divided by4, the question is, "How many 4s arethere in 8?" In mathematics, we often dis-cuss the value of real-world applications.Division involving fractions has real-worldapplications, but sometimes they are diffi-cult to recognize. Halving a recipe is oftenmentioned as an example of dividing by afraction. In reality, when dividing a recipein half, you are dividing by 2, which al-

82 CHAPTER 2

though technically is the fraction , is not

dividing by a fraction as people typicallythink of a fraction.

Concrete Beginnings

If we have a problem like a real-life

interpretation could be to have a 3-footlong piece of ribbon that is to be dividedinto 6-inch, or half-foot sections. Figure2.39 shows the cutting. You see that you

FIG. 2.39.

get 6 pieces of ribbon, each of which is

foot long. Translating the situation, we

want to know how many there are in 3.

The question now becomes, "How canwe get the numbers to match the pictureof the cutting?"

Whole Number Dividedby a Fraction

Figure 2.39 shows the ribbon has been

cut into 6 pieces or . Asking how

to get the numbers to match the pictureinvolves another question: "How do weget an answer of 6 when we start with 3?"One answer could be that we find 3 + 3,which would certainly work. However, weknow that multiplication is a short formof addition so we have two threes or2 times 3. We could also express 2 times

3 as 3 times 2 because of the commuta-tive property of multiplication on the set ofrational numbers. Now we have to relate

to 3 x 2. What has happened?

Actually two things have happened. Weknow that 2 is the multiplicative inverse or

reciprocal of and we see that as one

thing that is different. The other thing isthat the operation has changed from divi-sion to multiplication. Because we know

that and 3 x 2 = 6, the transitive

property of equality tells us that

3 x 2 . So, as we analyze what has hap-pened in the problem, the conclusion isthat if you have a division problem in-volving a fraction doing the dividing, in-vert the second number in the problemand then follow the rules for multiplica-tion. Often this rule gets shortened to in-vert and multiply. We caution you to saythe rule properly, invert the second frac-tion and multiply, which could help avoidmisinterpretations. It is important to notehere that mathematicians normally do notarrive at a conclusion like invert the sec-ond fraction and multiply from so few ex-amples. We assume you are familiar withthe idea though, but we did not want tobelabor the issue.

Your Turn

1. Do each of these problems, showingbasic intermediate steps:

NUMBER AND OPERATIONS 83

Fraction Divided by a Fraction

If you have divided by , the question

really is, "How many fourths are in ahalf?" With the Cuisenaire rods, the proc-ess becomes one of determining the unit.Figure 2.40 shows that the first rod that is

FIG. 2.40.

"halfable" and "fourthable" at the sametime is Purple. Red is a half and White isa fourth. The question was, "How manyfourths are in a half?" or, "How manyWhites are in a Red?" Two! That is it!There is no discussion of inverting thesecond fraction. Doing more problemslike this shows a pattern, which aid in un-derstanding the rule developed earlier.That is,

In each of the previous examples, theidea of inverting the second fraction andmultiplying gives the correct answer.There is another way to generate that rule.

Multiplying or dividing the numeratorand denominator of a fraction by the

same value gives an equivalent expres-

sion. Consider written in the format

of a fraction over a fraction, Multiply

both the numerator and denominator of

this fraction by 6 or . Although that might

seem like a strange choice, look at what

happens: . Focus on the denomina-

tor and see that you have the product of acounting number and its multiplicative in-

verse, which is always 1. So, be-

comes or more simply, , This

is a problem we have done before and theanswer is 3. Thus, we have another wayof showing that the idea of inverting thesecond fraction and multiplying for a divi-sion problem involving two fractions gen-erates the correct answer. Given the

problem it is expressed as

which is and becomes

or 5 x 28, which is 140.So far, the answers to the problems

have always been whole numbers. Theanswer to a fraction divided by a fractionwill not always be a whole number and yetthis process will work as a means of solv-ing the problem. For example,

84 CHAPTER 2

Your Turn

2. We have shown the reasoning be-hind inverting the second fraction and fol-lowing the rules for multiplication in divi-sion problems involving fractions. Dothese exercises using the fraction over afraction routine.

becomes , which could be ex-

pressed as You will not al-

ways have a common factor like the 23 inthis problem, but that should not bother

you. Consider which could be

expressed as That would be

. It so happens that the

numerator and denominator of are

relatively prime, so there are no commonfactors to be divided out.

Your Turn

3. Do each of these problems, showingbasic intermediate steps:

Mixed Number Dividedby Mixed Number

It is important to see patterns in the math-ematics you do. Although a problem like

might look intimidating, you need

only apply one procedure you alreadyknow. After converting the mixed num-bers to improper fractions, you have a

fraction divided by a fraction. So,

Common Denominator Division

Before leaving division of fractions, wewant to discuss one other way of dealingwith the problems. You learned to do divi-sion using a format that looks like

Fractions can also be placed in that for-mat for division. Use old skills—finding aleast common multiple (LCM), equivalentfractions, and using the familiar divisionformat to look at the situation. Suopose

the problem is , We know . The

NUMBER AND OPERATIONS 85

exercise , which is , could be

written as The question is still, "How

many are in " or, because the de-

nominators are the same, "How manyones are in four?" The completed fraction

division would be

The equivalent fraction process can beused for any fraction problem. An exercise

like becomes: . Do

a few like this and you realize you couldjust do Think about finding the least

common multiple first and then dividingthe numerators as we just described. Forsome of you, this method seems easier;for others, it may be more complicated.Our point is that there are different waysto view a problem.

An exercise like becomes

but at this point, the development could

vary. Rather than expressing it as

you could express it as

which becomes Notice that this is

similar to multiplication of fractions. Ex-pressing the two fractions with the samedenominator allows you to divide the nu-merator of the product by the numeratorof the factor. The division in the denomi-nator will always yield one. That mightsound strange, but you can see this pro-

cedure at work in

Your Turn

4. Use the least common multiplemethod to do the following problems:

Conclusions

Here are examples of the types of exer-cises we have covered in this section:

Whole divided by fraction

Fraction divided bv fraction

Mixed number divided by mixed number

Least common multiple

•

Dividing the numerators and denomina-tors when using a common denomi-nator

86 CHAPTER 2

When all is said and done, these are allvery similar. We caution you to do exer-cises by hand to strengthen your skillsand understanding before reaching foryour fraction calculator.

ADDITION OF DECIMALS

FOCAL POINTS

. Concrete Beginnings• Denominate Numbers• Adding With the Same Number of

Places• Zeros at the End• Lining Up the Ones• Expanded Notation

Addition is a familiar topic by the time youget to decimals. The addition facts shouldbe mastered at this juncture and youshould have had many experiences add-ing. You should also understand the basicworkings behind addition. If you do notunderstand what is going on behind thescenes when adding whole numbers,then you may be able to survive addingdecimals by sheer force of memory. How-ever, we believe that an understanding ofthe foundations of addition will serve youbetter.

Any decimal addition problem could beconverted to a fraction addition problemand then the answer converted back to adecimal. However, it is simpler to learn todeal effectively with decimals in their ownright.

Concrete Beginnings

A single manipulative can be used formany purposes. You have seen the Base10 blocks used with whole numbers. Herewe will suggest another, equally powerful,application of this manipulative. As you

FIG. 2.41.

refer to Fig. 2.41, notice that the Flat nowrepresents 1 whole, the Long represents

of a whole, and the tiny Cube repre-

sents of a whole. This transition re-

quires that you accept the Flat as the Unitand the Long and tiny Cube as fractionalparts of that Unit.

Now consider Fig. 2.42. The Flat is

FIG. 2.42.

used as 1 whole, whereas the Long is la-beled 0.1 and 0.01 is the label for the tinyCube. These two figures should establish

that 0.1 is the same as and 0.01 is the

same as and the foundation for add-

ing decimal numbers through fractions isestablished. Although the model we haveused here can only deal with units, tenths,and hundredths, it provides a good toolfor understanding decimal numbers. Theset of Base 10 blocks could be extended

NUMBER AND OPERATIONS 87

FIG. 2.43.

whole number addition exercises such as4567 + 319 + 208, becomes:

If that problem is changed to 45.67 + 3.19+ 2.08, it is still done the same way, ex-cept the place values for the respectivedigits have changed and become:

to make a Big Block as shown in Fig.2.43, which would be equivalent to 10 ofthe Flats glued together. If the Big Block

is used as a unit, the Flat represents

the Long represents and the tiny

Cube represents

As adults, we often read decimal num-bers as if they have no connection to frac-tions. If we see 3.14, we say, three pointone four, when the number is actuallythree and fourteen hundredths. The firstreading is quick and easy, but it obscuresthe fact that 3.14 is another name for amixed number. Try randomly grouping ahandful of Base 10 blocks and readingthe result as a mixed number. For exam-ple, if you had a group of 3 Flats, 4 Longs,and 7 tiny Cubes when the Flat is the Unit,then you could say, three and four tenthswith seven hundredths. You could alsosay, three and forty-seven hundredths, ifyou recognize it.

Denominate Numbers

Denominate numbers are handy with dec-imals, but there is a need for extendingthe concept of place value. The first placeto the right of the decimal point is calledtenths. Using denominate numbers to do

Once these accommodations for placevalue are completed, all addition applica-tions from earlier apply. When we use de-nominate numbers, we don't need decimalpoints, the numbers are fully described.

Adding With the Same Numberof Places

Many decimal addition exercises will lookjust like whole number addition. Look at thefollowing examples and decide how theyare the same as and how they differ fromexercises involving only whole numbers:

What did you conclude? You probablysaid something like, "The decimals arelined up in the problem and the decimalgoes in that same column in the answer."That is a reasonable and expected con-clusion for the examples given. Anotherway of saying, line up the decimals is lineup the ones or units. The advantage isthat you learned this while adding whole

88 CHAPTER 2

numbers. Use the same rule here and youeliminate the need to learn a new one.

Zeros at the End

Consider 6.4 + 8.95. This does not looklike the examples in the last paragraphbecause the numbers of digits followingthe decimal points are not the same. Re-fer to the Base 10 blocks when the Flat isthe unit and consider 6.4. This could beexpressed by using 6 Flats and 4 Longs,as shown in Fig. 2.44. But, we know thateach Long is equivalent to, or made up of10 little Cubes. Thus, whereas the 4Longs represent 0.4, we could trade themfor 40 tiny Cubes. This means we couldalso express 6 Flats and 4 Longs as 6.40,using the tiny Cubes rather than theLongs. We could extend this model byusing a Big Block like the one in Fig. 2.43

FIG. 2.44.

as the Unit; then a Flat would represent0.1, a Long 0.01, and the tiny Cube 0.001.This would allow us to use a similar ap-proach to show that 6.4 is equivalent to6.400. It would then be a logical extensionto say that 6.4 is equivalent to 6.4000,6.40000, and so on. This should lead youto a generalization about zeros after the

last nonzero digit to the right of the deci-mal point. What would you say? If yousaid it is acceptable to add as many zerosas you want after the last digit to the rightof a decimal point, then you are correct.

The advantage of that last conclusion isthat, if an exercise such as 6.4 + 8.95seems strange or troublesome because thenumbers of digits to the right of the deci-mal points are not the same, a zero can beinserted so that 6.4 + 8.95 is expressed as6.40 + 8.95. Similarly, the problem couldbe 6.400 + 8.950, 6.4000 + 8.9500, andso on. Whereas, in general, there is no ad-vantage to putting more zeros to the right,it certainly could happen and the impor-tant thing is that the value of the problemwould not be changed. Sometimes thezeros after a number indicate precision.For example, in science, there is a reasonfor putting more zeros to the right. Thevalue 0.010 mm indicates a measurementwas accurate to the nearest 0.001 mm in-stead of 0.01 mm. Additionally, zeros canplay a role for significant figures in calcu-lations. For our purposes in this section,zeros to the right of the last nonzero deci-mal digit only serve as placeholders.

Lining Up the Ones

Perhaps you are more comfortable seeingthe problems listed vertically:

. It is com-

mon to leave out the extra zeros and just

do the exercise as , Either format is

acceptable. The important thing is to re-member that ones are added to ones,tenths are added to tenths, hundredthsare added to hundredths, and so on. Inthis particular exercise, there are 5 hun-dredths in one addend, and no hun-dredths in the other addend. We could

NUMBER AND OPERATIONS 89

elect to either represent the no hun-dredths in that one addend with a zero, oromit the zero and let it be understood thatthe absence of a digit in the hundredthscolumn means that there are zero hun-dredths. This same idea is then extend-able to any situation, as long as the zerosare placed to the right of the last digit af-ter the decimal point. The rule of lining upthe ones is just as applicable as, andmore explanatory than, the one you mayhave memorized about lining up the deci-mal points. Because you learned to lineup the units, or ones, long before youlearned to line up the decimal points, youreally did not need to memorize that rule.

Look what happens to 4.37 + 9.2 +10.586 if we forget to line up the ones.The exercise could end up looking like

in which the decimals are ig-

nored and the sum does not make sense.If we merely line up the ones, then the restof the place values take care of them-

selves as is shown here,

The situation can be perplexing if .78 isan addend. All that is necessary is to write.78 as 0.78. When the exercise is 4.37 +9.2 + 10.586 + 0.78, and we write 0.78 in-stead of .78, the lining up the ones rule isfunctional and that zero eliminates a huge

Expanded Notation

As we convert from denominate numbersto expanded notation, the focus shifts toa different way of expressing place value.The addends 45.67 + 3.19 + 2.08 are indi-

vidually expanded to become 40 + 5 + 0.6+ 0.07, 3 + 0.1 + 0.09, and 2 + 0.0 + 0.08,respectively. At this point, we are dealingwith addition problems we have alreadyhandled. Even regrouping is familiar terri-tory. The problem 45.67 + 3.19 + 2.08 isexpressed as

Your Turn

1. Do the following problems showingany scratch work you generate:

Conclusions

We have provided a review that shouldhelp you tackle adding decimal numbersin any problem. We have shown you howto start concretely and relate back towhole number addition if necessary. Dili-gence from you is all that is required now.Don't forget to plan when and how youare going to use technology.

SUBTRACTION OF DECIMALS

FOCAL POINTS

. Concrete Beginnings

. Subtracting With the Same Numberof Places

• Zeros at the End

Much of the groundwork for decimal sub-traction has already been established. Es-

90 CHAPTER 2

sentially you are going to encounter onlyone new idea in this section.

Concrete Beginnings

Base 10 blocks are a nice and effectivetool to use to show what is going on withdecimal subtraction. If the problem doesnot involve regrouping, for example, 2.6 -2.5, Fig. 2.45 shows the subtraction con-

FIG. 2.45.

cretely. In this model, the Flat is the unitand a Long is the tenth. The top row ofblocks represents the sum and the bot-tom row represents the given addend.The double-headed arrows indicate thematching that eliminates the pairings,leaving one Long as the missing addend.Because the Long is designated as atenth, the missing addend would be 0.1and 2.6 - 2.5 = 0.1. Vertically, the problem

would be shown as: Either way is ac-

ceptable, however the vertical alignmentpoints out the need to line up the ones.

FIG. 2.46.

Figure 2.46 shows how a unit (Long inthis case) can be traded for 10 tenths tocomplete a subtraction exercise such as3.1 - 0.6. Scene 1 shows one Long indashed segments to indicate that it is go-ing to be traded for 10 tenths; Scene 2shows that 6 tenths are removed. The re-maining two units and 5 tenths representthe answer. Written horizontally, 3.1 - 0.6

= 2.5, or vertically, the missing ad-

dend is the same.

Subtracting With the SameNumber of Places

Here are a few examples to help you clar-ify any questions you might have

NUMBER AND OPERATIONS 91

All of the regrouping procedures are thesame as those discussed when weworked with whole number subtraction.Essentially, you can ignore the decimalpoints and proceed as usual when doingthe problem—as long as you line up theones digits in the sum, addend, and miss-ing addend.

decimal places a valid, as well as

helpful, alignment for the regrouping thatwill be required as this exercise is com-pleted.

Your Turn

1. Do each of these exercises usingonly pencil and paper:

MULTIPLICATION OF DECIMALS

FOCAL POINTS

• Concrete Beginnings. Relating to Fractions• Zeros in Multiplication

We will present a carefully sequenced setof ideas as you broaden your skills withand understanding of multiplication. Youshould continue to see the cumulative ef-fect of the underpinnings of mathematicalknowledge.

Concrete Beginnings

Any multiplication problem is like findingthe area of a rectangle. We will expandthe set of examples to include positivedecimals. When the Long Base 10 blockis the unit, the little Cube represents atenth. Figure 2.47 shows 2 times 0.3 andhow 0.6 is the answer. You have two setsthat are 0.3 big and end up with one setthat is derived from 2 x 0.3 = 0.6 as an ex-pression of the situation. Examining theanswer, the exercise looks a lot like 2 x 3 ,

Zeros at the End

Zeros can cause some problems in sub-traction, and even more so in decimalsubtraction. Actually, the absence of ze-ros is where the trouble starts. Supposethe problem involves a ragged alignment

as in . The problem is that there is

nothing from which to subtract the 0.001or the 0.05. There is something there tosubtract from, but it has not been written.The discussion about putting zeros afterthe last written decimal digit applies here.An equivalent expression for 34.8 is34.800, which allows the exercise to be

written as making the operation

more convenient.There is another ragged alignment

problem that can cause you to pause for

reflection. Suppose the problem is

Because there is an understood decimalpoint at the right of any whole number, 34can be written as 34., 34.0, 34.00, 34.000,or with as many zeros as we find conven-ient. For this problem, 34.000 is helpful inrewriting the exercise so that the sum andgiven addend have the same number of

92 CHAPTER 2

FIG. 2.47.

but a decimal point appears both in onefactor and in the product.

Figure 2.48 shows how 4 x 0.3 would

FIG. 2.48.

be done, again with the Long defined asthe unit. This time you have four sets thatare 0.3 big and when you combine all theelements, you get enough little Cubes totrade 10 of them for a Long and you stillhave 2 little Cubes left, as shown in Fig.2.49. The Long is a unit, so we have a unit

FIG. 2.49.

and two tenths, or 1.2. Thus, 4 x 0.3 = 1.2.Again, the problem looks a lot like 4 x 3 ,but decimal points are still present.

Using the same mentality, 17 x 0.9 =15.3 and this is an important problem inthe discussion. We are headed towardsome conclusions here. First, you shouldbe saying that if there is one decimal pointin the problem, then there will be one dec-imal point in the product. The location of

that decimal point is an issue, however.The 17 x 0.9 = 15.3 problem provides theneeded clue. Do the multiplication as if itwas 17 x 9 as far as the work is con-cerned. After you have done the multipli-cation, count from the right toward theleft, one place to locate the decimal pointin the product. This will work even withproblems like 25 x 0.4, where the productis 10. Doing the problem without worryingabout the decimals would give 25 x 4,which is 100. However, considering thedecimal points, the answer would be10.0, which is generally written as 10.

Moving Beyond the Concrete

It is important that you realize we couldcontinue to show various configurationsof Base 10 blocks to confirm the resultswe will discuss. If the problem is 3 x 0.97,the product is 2.91, and the beginnings ofa generalization are established. Examinethe following examples:

You should come up with a generalizationthat says you count the number of deci-mal places in the decimal factor, and youwill have that many decimal places in theproduct, COUNTING FROM THE RIGHTin both instances.

The problem 0.4 x 0.3 can be inter-preted as taking three tenths of fourtenths and is shown in Fig. 2.50. Thesquare is divided into tenths in both direc-tions. Because this figure is two dimen-sional, a complete column or a completerow of smaller squares would be the

NUMBER AND OPERATIONS 93

FIG. 2.50.

same as a Long in the Base 10 blocks.Similarly, the big square would be like aBase 10 block Flat and a little squarewould be like a Base 10 block little Cube.If we considered the big square as theunit, then either a column or a row wouldrepresent one tenth and one of the littlesquares would represent one one hun-dredth. The top four rows are cross-hatched with diagonals going from upperleft to lower right. That set of four rowsrepresents 0.4 in the problem. Similarly,the leftmost three columns are cross-hatched from upper right to lower left,representing 0.3 in the problem. The smallrectangle in the upper left corner of thebig square represents three tenths of fourtenths. But that rectangle is made up of12 little squares, each of which repre-sents one one hundredth. So, that threetenths of four tenths of the big square isequivalent to 12 one hundredths. Con-densing the notation, 0.4 x 0.3 = 0.12, asshown in Fig. 2.50.

Study the following exercises andcome up with a generalization aboutplacement of decimals in products, whenboth factors are decimals:

You should conclude that the total num-ber of decimal places in the factors is thenumber of decimal places in the product,when counting from the right. That is, be-cause there are four decimal places in thefactor, 2.3217, and one more in the fac-tor, 3.5, there must be five decimal placesin the product, 8.12595, and they arecounted from the right. The multiplicationis done as if the decimals were not thereand then the count of places locates thedecimal point in the product, countingfrom the right.

One more topic and you should have allyou need to do any decimal multiplicationproblem. Figure 2.51 shows 0.2 x 0.3 and

FIG. 2.51.

it helps explain a difficult situation. As inFig. 2.50, each little square representsone one hundredth. This time the rectan-gle in the top left of Fig. 2.51 only has sixlittle squares double crosshatched, whichrepresents six one hundredths. So, thecompleted problem is 0.2 x 0.3 = 0.06,and a new situation occurs. Up until now,we have been counting the number ofdecimal places in the factors and thencounting that many places from the rightto locate the decimal point in the product.Here, if we do the multiplication without

94 CHAPTER 2

considering the decimals in the factors,the product is six. In order to consider thedecimals in the factors and also do ourcounting from the right in the product, weneed to insert a zero to the left of the 6 inthe product to get the proper number ofplaces. Thus, 0.06 becomes the ex-pressed factor. Consider the following ex-amples and verify that the generalizationworks, if zeros are inserted in the productbetween the first significant digit andwhere the decimal point needs to be lo-cated. As you do this, mental images likeFig. 2.51 could prove helpful:

Multiplication works the same as before,except for the need to insert zeros in theproduct to get the right number of deci-mal places.

Relating to Old Ways

How does decimal multiplication connectwith earlier concepts? Recall that multipli-cation is a short way of adding. The sum0.4 + 0.4 + 0.4 is 1.2, as you know. Thisproblem could be interpreted as three setsof 0.4 or three 0.4s, which is really just3 x 0.4 and we have expressed the addi-tion problem via multiplication. This sce-nario could be repeated with any productof a whole number times a decimal.

When the problem involves two deci-mal factors, repeated addition is aban-doned in favor of prior work with frac-tions. Suppose 0.3 x 0.4 is the problem.Expressed as fractions, 0.3 x 0.4 be-

comes or 0.12 as a deci-

mal. Again, the decimal point location

generalization holds true. In a similarmanner, any repeating or terminatingdecimal multiplication could be convertedto fractions and worked out. Of course,somewhere along the line, you mightwant to shift to a calculator. Our assump-tion is that before you grab that calcula-tor, however, you do understand how toperform decimal multiplication.

One final thing that might help you in-volves the distributive property of multi-plication over addition on the set of realnumbers. A problem like 3 x 2.4 could berewritten as 3(2 + 0.4), which can be ex-pressed as (3 x 2) + (3 x 0.4) and this is fa-miliar material. In this case, the sum of 6and 1.2 would be 7.2 and, again, it lookslike multiplication is done as usual (tem-porarily ignoring the decimal point in thefactor) and then placing the decimal pointin the product.

Your Turn

1. Do each of the following problems topractice your multiplication skills whenone or both of the factors involve deci-mals. Although we are proponents of cal-culator use, you should not use calcula-tors here. You need to practice the skilland be sure you have a handle on decimalmultiplication:

Conclusions

It is important that you realize multiplica-tion of decimals is done just like withwhole numbers. All of the hurdles andchallenges encountered there exist heretoo, with one more: the decimal point

NUMBER AND OPERATIONS 95

placement. We have provided a logicaldevelopment and connections that shouldhelp you make sense of counting from theright in the product until you have thesame number of decimal places as wereexpressed in the factor(s), and put thedecimal point there.

DIVISION OF DECIMALS

FOCAL POINTS

• Concrete Beginnings• Whole Number Divided by a Whole

Number• Decimal Divided by a Whole Number• Whole Number Divided by a Decimal• Decimal Divided by a Decimal

You need to be able to do division prob-lems by hand, however, there is debateover how much is enough. It is reasonableto assume that you are comfortable doingsomething like mentally, regard-less of the format in which the exercise is

p r e s e n t e d M a n y authorities

in the field of education think that it wouldbe unreasonable to expect you to do43.81025 -r 796.38624 either mentally orby hand. We think that, somewhere be-tween those two exercises, a calculatorbecomes an acceptable tool for findingthe missing factor. You, your instructors,and the problem situations in which youare involved will determine when the useof technology becomes appropriate.There is simply no fine line between whenpencil and paper work is needed and theuse of a calculator is acceptable.

Prior to starting division with decimals,you need to be well founded in the basicsof division, place value, equivalent frac-tions, rounding, approximation, estima-tion, multiplication, and subtraction. Theassumption is that you are functioning

well beyond introductory levels with eachof these concepts. Building on equivalentfractions, the four categories of decimal

division are: , and

You should realize that each of

these four problem types could be ex-

pressed as using equivalent frac-

tions. For example, if you are given

you might multiply both the numera-tor and denominator by 100 to get

and eliminate decimal

numbers from the exercise. A quick reviewof place value will remind you that thevalue you use to create equivalent frac-tions without decimals will always be anappropriate power of ten. Actually, you maysoon realize that the only real necessity isto use an equivalent fraction in which thedenominator is a whole number.

Concrete Beginnings

If you and I have 1 Flat and 4 Longs froma set of Base 10 blocks to be divided be-tween us, we will have to trade the Flat for10 Longs. Once that is accomplished, thetask will be to share the pieces equallybetween us. If we do a one for you andone for me routine until all the Longs aredispersed, then each of us will have 7Longs. But, what will the Longs be worth?If we are dealing with money, we will tradethe dollar (the Flat) for dimes (the Longs)and then each of us will have 7 dimes, or70 cents. You have seen 70 cents writtenas $0.70, which is the decimal represen-tation. If the problem statement does notinvolve money, we would use 0.7 to de-scribe the 7 Longs. Figure 2.52 showshow the situation would be depicted with

96 CHAPTER 2

FIG. 2.52.

Base 10 blocks. You start with a Flat and4 Longs, make the trade and then dividethe set of 14 Longs into two sets of 7each. This is going to be the essence ofthe discussion that follows. You shouldcreate mental images or sketches of theBase 10 blocks as we go through eachexplanation.

Whole Number Dividedby a Whole Number

Initially, the product is a multiple of thefactor, so the missing whole number fac-tor represents the exact missing informa-tion. When that does not happen, wehave a means of expressing the situation.

We could say , which would suf-

fice sometimes. However, in the realworld, remainders frequently give way toother ways of telling the story. Certainly

could be expressed as which

could be expressed as As you

probably have observed over time, thereal world usually deals with decimal

numbers rather than common fractions.So, the task becomes one of expressing

in terms of decimals. Consider

which c a n b e written a s W e

know that and, from our addition

of decimals discussions, we know 1 + 0.5= 1.5. Thus, we have generated an equiv-

alent decimal expression for and we

can say that

Now the challenge is to make the

configuration match with which we

know to be the answer. If we are dealing

with the problem looks like

The only apparent differences between

is that there is a zero in

the units place in with no decimal

point, whereas has the decimal

point but no zero. But wait a minute! Weknow there is always a decimal point atthe end of any whole number, so 12.0 and12 must be equivalent. On top of that, weknow that we can put as many zeros tothe right of the decimal point as we needwithout changing the value of the number.In other words, 12 = 12.0 is a true state-ment. Similarly, if we look at the problemwith the answer expressed as a remain-

der, we have . But that 1 in the

missing factor is a whole number, so ithas a decimal point to its right. That samedecimal point would be to the left of the 5in 0.5, and voila! The whole thing is com-

NUMBER AND OPERATIONS 97

ing together. Insert the appropriate miss-ing zero and decimal points and the exer-

cise becomes —everything we have

been seeing and discussing matches up.Let's summarize what happened. We

started with but opted to express it

as nstead. We then rationalized

that the answer had to be 1.5, so we had

The decimal point in the missing

factor is directly above the decimal pointthat was inserted into the product. Youmay recognize that as a rule you learnedlong ago, put the decimal point at the endof the product and insert another decimalpoint directly above it in the missing fac-tor. Look at a few examples in which wehave shown all the steps:

Your Turn

1. Do these problems without a calcula-tor. You should check your work by multi-plying the factor times the missing factor:

Decimal Divided by a WholeNumber

Even something like does not/present much of a challenge. Althoughdecimals in fractions are not common, wecan express the problem as a fraction,

Use your knowledge of equivalent

fractions to multiply both the numeratorand denominator of the fraction by 100,

giving . We have

discussed this type of problem, becauseit is a whole number divided by a wholenumber. Your experience multiplying deci-mals by powers of 10 and knowledge ofplace value should clarify why we selected100 as the factor by which to multiply boththe numerator and denominator. We seethat 100 is the smallest power of 10 thateliminates decimals from the exercise.

You have learned that the decimal pointin the product determines where the deci-mal point is located in the missing factor,so this isn't a new type of problem andthere is no need to go through the maneu-ver of multiplying the numerator and de-nominator by some power of 10. Instead,with the exercise in the format

apply the technique of aligning the deci-mal point in the missing factor directlyabove the decimal point in the product.

Your Turn

2. Complete the following exercisesbeing careful to align the decimal point inthe missing factor:

98 CHAPTER 2

Whole Number Dividedby a Decimal

An exercise such as may ap-

pear to present a little bit more of a chal-lenge. Again, equivalent fractions makeeven this into a relatively simple task. Theproblem can be restated as:

Once again, an exercise involving a deci-mal factor is restated as a whole numberdivision exercise. The only hurdle is to de-termine by what power of 10 to multiplythe numerator and denominator of thefraction so that both are whole numbers.Restating the product and given factor aswhole numbers does not mean that themissing factor will be a whole number.The decimal point might need to be in-serted after the units digit of the product.More than likely, you were taught a rule thatsaid something about moving the decimalpoint to the right end of the factor and thenmoving the same number of places to theright in the product. Multiplying the numer-ator and denominator of the related frac-tion by the same power of 10 explains whythat works and is appropriate.

Your Turn

3. Tell the smallest power of 10 neededto make the factor a whole number:

Decimal Divided by a Decimal

The only type of problem remaining in-volving division and decimals is one suchas Once again, the hurdle is to

decide by what power of 10 to multiplythe numerator and denominator to elimi-nate the decimal in the factor. Becauseyou already know how to deal with divi-sion when the factor is a whole number,it seems logical to use the smallestpower of 10 that will do so. In this case,that is 10. So, would become

a problem type

you already know how to do. If the origi-nal problem had been then

we would have multiplied both the fac-tor and product by 100 and the exer-cise would be transformed into

Your Turn

4. Tell the smallest power of 10 neededto make the factor a whole number:

Conclusions

So there you have it. The hurdles pre-sented by decimal division are taken careof by looking at the problems in a fractionformat. Given that understanding, the pro-cedures used for division involving deci-mals should make sense now. The discus-

NUMBER AND OPERATIONS 99

sion of decimals and fractions in the lasteight sections should have given you botha review and an insight into the workingsof the mathematics behind these two fun-damental ideas in numeration. We will re-visit these ideas one more time in a dis-cussion about ratios and proportionalreasoning. Are you grounded well enoughto reach for your calculator now?

ADDITION OF INTEGERS

FOCAL POINTS

• Models• Rules for Adding Integers

Addition of integers provides an interest-ing extension of the operation of addition.You have already worked with addingwholes, fractions, and decimals. In all ofthose you used the operation sign, plus(+). Now we extend the operation to in-clude adding signed numbers (positive ornegative), and use a plus sign to indicatepositive numbers and a minus sign to indi-cate negative numbers. This extension in-troduces additional terminology consider-ations. The "+" can mean add or positiveand the"-" can mean subtract or negative.Thus, context becomes important.

With Integers, 4 + 3 = 7 is rewritten as+4 + +3 = +7, which you should read as,"Positive four plus positive three equalspositive seven." This provides reinforce-ment and practice to help you begin deal-ing with addition involving negative num-bers. Eventually, we will say there is noneed to worry about saying, "Positive fourplus positive three equals positive seven."It is assumed that numbers written with-out signs are positive, so we can simplysay, "Four plus three is seven."

Models

Initially, the existence of negative num-bers has to be established. Traditionally,this is done with discussions about tem-perature (people in Florida have troublethinking in terms of temperatures belowzero, because water actually freezes at32°F), below sea level (people in Coloradohave trouble identifying with this), debt(too many of us understand this concept),or below ground level. At least one ofthese real-world examples should giveyou a feel for the existence of negativenumbers. With this notion, we can intro-duce the other half of the number line bydiscussing numbers that are mirror im-ages of the whole numbers, with zeroserving as the mirror point. This impliesthat ~2 is in the opposite direction of +2and that these two numbers are equallydistant from 0. Figure 2.53 models the

FIG. 2.53.

number line with several integers indi-cated. It is important that number lines berepresented as never ending, with uni-form divisions.

One way of showing integer addition in-volves combining sets of colored chips.Figure 2.54 models integer addition using

FIG. 2.54.

black chips for positive numbers. Thismay remind you adding whole numbers,where we did not care about differences

100 CHAPTER 2

FIG. 2.55.

in color. Similarly, Fig. 2.55 models add-ing a negative four and a negative threewith white chips to represent negativevalues. To add integers that have differentsigns, consider the real-life concept ofmoney. If you get a dollar from someone,then spend a dollar, your net result is$0.00. This is the dollar version of +1 + ~1= 0. In this model, combining a red chipand a black chip results in no chips, or 0chips. A one-to-one correspondence be-tween a black chip and a white chip coun-terbalances them and you can discardthem both. Figure 2.56 models the exer-

FIG. 2.56.

cise +4 + ~3 = +1 because three of theblack chips in the first set can be put intoone-to-one correspondence with three ofthe white chips in the second set, leavingthe single black chip.

The number line is an effective way ofdeveloping rules for adding integers. Fig-ure 2.57 shows addition on the number

FIG. 2.58.

line using whole numbers. Figure 2.58shows the same problem using integers.As you compare the two figures, youshould notice that the process for doingthe problem on the number line with inte-gers is exactly the same as with wholenumbers. Figure 2.59 shows ~7 + ~6 = "13

FIG. 2.59.

on the number line. This should makesense in view of your prior experiencewith adding whole numbers on the num-ber line. In this model, the vectors arepointed in the negative direction, meaningthat our ability to model ~7 + ~6 = "13 isdependent on the inclusion of the nega-tive part of the number line.

Your Turn

1 . Do each of the following problemson a number line and record both theproblem and the answer in a separate list.State a rule for adding integers when theaddends have the same sign:

FIG. 2.57.

NUMBER AND OPERATIONS 101

So far, only number line examples inwhich the addends have the same signhave been modeled. Next we explore howto model addition on the number linewhen the signs of the addends are differ-ent. Figure 2.60 shows +7 + ~3 = +4 on the

FIG. 2.60.

number line. Notice that the first addendis +7 (7 units long in the positive direction)and the second one is a ~3 (3 units long inthe negative direction). Tracing along thefirst addend and then the second addendwill lead you to a final position at +4.

The only thing left is to determine howto get from where you start (0) to whereyou end up (+4). In this case, your netchange is four units in the positive direc-tion, yielding a sum of +4. This is shownusing a dashed vector.

A companion problem to +7 + "3 = +4 is~7 + +3 = ~4, which is modeled on thenumber line in Fig. 2.61. Here, the first ad-

FIG. 2.61.

dend is "7 and the second one is +3, re-sulting in "4, the sum. You should con-clude that if you start at 0 and go four

units to the left, or in the negative direc-tion, you arrive at the sum, ~4, shown as adashed vector.

Your Turn

2. Do some integer addition problemson the number line to ensure you can getthe sums. Each of your problems shouldbe similar to those in Figs. 2.60 and 2.61.

Rules for Adding Integers

All arithmetic involves performing binaryoperations, which means that we operateusing two values at a time. Although morethan two addends can be considered, in-teger addition is binary because, no mat-ter how capable you are, you can onlydeal with two addends at a time. Once thesum of two addends is determined, thatsum becomes an addend and the nextaddend is considered. For example, tocomplete +4 + +3 + +6, you begin by add-ing two of the addends, +4 + +3 = +7, andthen use the result as an addend, +7 + +6= +13. You might prefer to do M + +6 = +10first, but still, only two addends are dealtwith at any one time.

The calculator can be an ally in addingwhen several integers are involved. Doseveral integer addition problems wherethe addends have the same sign, whilewriting the problem and the answer. Asyou look for patterns, you should general-ize the rule for adding integers with thesame signs. As you refine the skill of look-ing for patterns and generalizing, the ruleto add the numbers as you normallywould and give the answer the commonsign, should develop rather quickly.

There are four problem types involvedin the addition of integers:

102 CHAPTER 2

Earlier, using the number line, we con-sidered problems where the addends hadthe same sign. One way to generalizeyour ideas about adding integers of thesame sign is to add the absolute values ofthe numbers and then to give the answertheir common sign. The definition of ab-solute value involves the idea of "distancefrom zero," which seems especially rele-vant as we discuss integers. It means thatwe can consider two addends having thesame sign as two distances that we aregoing to combine. When you look at thenumber line, you can tell that ~3 is not asfar from zero as ~5. When you deal withabsolute value, you are focusing only onhow far the number is from zero, or howlong the segment between the numberand zero is. You do not care about the di-rection from zero if the numbers have thesame sign. That should add some mean-ing to the generalization that, when add-ing numbers that have the same sign, addthe absolute value of the numbers andgive the common sign to the sum.

After completing work where the signsof the addends are the same, we intro-duced problems where the addends hadopposite signs. We modeled problemswhere the absolute values of the addendswere relatively small so there was littleopportunity for distraction. As you do ex-ercises like +5 + ~3 = +2, focus on thenumbers involved and ask how a two canbe the sum of a five and a three. Thisseems to involve two distances from zero,but in opposite directions. You mightconclude that the problem could be re-stated as 5 - 3 = 2, or the difference be-tween the distances. Then you might sur-mise that 5 - 3 = 2 and +5 + ~3 = +2 are twodifferent ways of doing the same problem.Focus on how one can be changed to the

other. You will deduce that the sign of thesecond number is changed, and the oper-ation is changed from addition to subtrac-tion.

Finally, you must deal with problemssuch as ~5 + +3 = ~2. Here, again, youmight view this as related to 5 - 3 = 2, butthe operation is addition and the signs areall scrambled around. Returning to ourdiscussion of absolute value, the sumgets the sign of the number with the largerabsolute value. Whereas this may notbe as immediately apparent as the lastexample, the same concept is in play-two distances from zero that are in oppo-site directions. Focus your thoughts on+5 + ~3 = +2 and ~5 + +3 = ~2, and how, ineach case, the two different ways of do-ing the problems can be interchanged.Ultimately, you will conclude that, whenadding numbers that have oppositesigns, you subtract the smaller absolutevalue from the larger absolute value andgive the sum the sign of the number withthe larger absolute value.

Conclusions

Addition of integers is familiar, particularlywhen the numbers have the same sign.One potential hurdle involves establishingan adequate background about integersand what they are. The number line pro-vides an excellent tool as you work on theconcepts of whole numbers and their op-posites. The second hurdle centers on theidea that, to add integers with oppositesigns, you might intuitively subtract; in onesense, you can use the inverse operationto complete the addition of numbers withopposite signs. Careful development ofthe addition operation using manipula-tives—a calculator, a number line, or allthree of these tools—relieves much of thepressure associated with an in-depth un-

NUMBER AND OPERATIONS 103

derstanding of integers. Additional prac-tice is necessary if you are not yet com-fortable with these two generalizationsthat deal with adding integers:

1. When adding integers with likesigns, add the absolute values and givethe sum the common sign.

2. When adding integers with unlikesigns, subtract the smaller absolute valuefrom the larger absolute value, and givethe sum the sign of the number with thelarger absolute value.

SUBTRACTION OF INTEGERS

FOCAL POINTS

• Models

As was the case with addition of integers,much of the groundwork for subtraction ofintegers has already been established. Wehave to make a few adjustments to yourknowledge base and you will be adept atsubtracting integers. One common inter-pretation of a rule for subtracting integerssays that you change the sign of the sec-ond number and then follow the rules foraddition. A common rule for adding inte-gers with unlike signs says you subtract.So, to subtract, you think add, which reallyinvolves subtracting some of the time.Confused? This section will help you makesense of those statements and developyour own enlightened interpretation.

Models

+6 - +2 = +4, using black chips to repre-sent the positive integers, as shown inFig. 2.62. A negative number take away a

FIG. 2.62.

negative number, such as 6 - 2 = 4, isshown in Fig. 2.63. Take away subtraction

FIG. 2.63.

is straightforward if the sum and addendhave the same sign. There are three casesthat are not as straightforward and theseare the cases that make this model useful.

Suppose the exercise is ~4 - ~6, asshown in Fig. 2.64. The given set has toofew white chips to allow us to take away 6white chips. The additive identity comesto our rescue! We can increase the num-

You can construct a good beginning forthe development of subtraction algo-rithms using white and black chips andthe subtraction idea of take away. First,consider the case of a positive num-ber take away a positive number, such as FIG. 2.64.

104 CHAPTER 2

ber of chips in the set as long as we bal-ance each white chip with a black chip.This is true because "1 + +1 = 0, which canbe extended to ~2 + +2 = 0 or ~n + +n = 0,for any number and its additive opposite.If we increase our set by 2 white chipsbalanced by 2 black chips, then we havenot changed the value of the set, but areable to take away 6 white chips as re-quired by the example.

Take a look at ~6 - +4 = ?, which asks usto take 4 black chips away from a set thathas only white chips. We can solve thisdilemma using the additive identity, insert-ing 4 black chips balanced with 4 whitechips, as shown in Fig. 2.65. Take away

FIG. 2.65.

the 4 black chips to show ~6 - +4 = ~10.The final case is very similar and uses theadditive identity again, as shown in Fig.2.66. It shows how we can take whitechips from a set that contains only black

chips. We express zero by inserting 5 whitechips balanced by 5 black chips, and thentake away the 5 white chips, which com-pletes the exercise, +6 - ~5 = +11.

The number line is a powerful tool forshowing how to determine the missingaddend in the subtraction of integers. Thesame format for number line subtractionas was developed in whole number sub-traction will be used here. The problem +3- +8 = ? is rewritten as +8 + ? = +3. Figure2.67 shows what this would look like onthe number line. The first addend is +8.

FIG. 2.67.

The sum is +3 and tells you where youwant to end. Your number line task is tofind the missing addend. In this case, youmove five spaces to the left or ~5, repre-sented by a dashed vector of length 5 thatpoints in the negative direction.

All number line subtraction exercisescould be done in a similar fashion and ex-plained in a similar manner. You have anaddend and a sum (solid vectors) and arelooking for the missing addend (dashedvector). Your task is to summarize themovement needed to get from the ad-dend to the sum in terms of how manyspaces are used and the direction of themotion. Figure 2.68 shows +8 - ~3 = +11,whereas Fig. 2.69 shows ~8 - +3 = "11.

FIG. 2.66.

FIG. 2.68.

FIG. 2.69.

NUMBER AND OPERATIONS 105

You need to do several problems of eachtype on the number line and then general-ize a rule from those examples.

Your Turn

1. Do each of the following problemson a number line and record both theproblem and the answer. State a rule forsubtracting integers.

Conclusions

As we moved through chips and numberlines toward abstract concepts, we es-tablished the idea that, when subtractingintegers, you change the sign of the sec-ond number and follow the rules for theaddition of integers. Within that setting,four problem types are encountered:

You are now an authority on subtract-ing integers. Your understanding of thisconcept has been developed through thediscussion of take away using black andwhite chips, followed by work on thenumber line, and finally arriving at theconclusion that, when subtracting inte-gers, you may change the sign of the ad-dend and then follow the rules for addingintegers. Before you continue, take a mo-ment to assure yourself that you under-stand what is happening in all types ofsubtraction exercises involving integers.

MULTIPLICATION OF INTEGERS

FOCAL POINTS

• Multiplying Integers

It is helpful, as we try to understand whathappens when we operate with numbers,to use our observations to write general-izations—statements that discuss num-bers and operations without using spe-cific numbers. The two generalizationsyou are looking for in this section shoulddevelop relatively quickly, given yourbackground with adding integers. Somewould argue that we should just tell youthe rules, because these generalizationsare well established. However, thatmakes you dependent on an outsidesource to tell you what to do, limiting theamount of thinking you do on your own.Our goal is to teach you to think throughthings and developing generalizationsfrom your observations is one element ofthis type of critical thinking. We don't askyou to reinvent the entire process, butlearning to think like an inventor is a valu-able skill for a mathematician.

Multiplying Integers

You have already dealt with multiplicationas repeated addition, thus a calculator ornumber line could be used to build theideas that follow. You have completed ex-ercises such as 4 x 3. In integer opera-tions, that exercise is +4 x +3. Figure 2.70shows +4 x +3 on the number line, showing4 vectors each of length +3. The observa-tion that the product of two positive fac-tors is positive should be familiar territory.

A positive factor times a negative factoris shown on the number line in Fig. 2.71and is very similar to the product of twopositive factors. Here, we show a graphic

106 CHAPTER 2

FIG. 2.70.

FIG. 2.71.

example of +4 x 3 and the figure is essen-tially Fig. 2.70 mirrored at zero; 4 vectorsof length 3 in the negative direction. Youshould generalize that a positive times anegative yields a negative product becauseof your previous experiences with re-peated addition using negative addends.

Multiplication of integers is a binary op-eration. If more than two factors are in-volved, then the product of two factorsmust be found first and then that productbecomes one of two factors that generatea new product, until all factors have been

used. For example, when finding theproduct of +2 x +4 x +5, you might find theproduct +2 x +4, which is +8. Then you findthe product of +8 and +5, which is +40.Given this and your ability to multiply twofactors, you will be able to handle morethan two factors in a systematic manner.With that assumption in mind, we willconsider multiplication of integers usingtwo factors, which requires considerationof four problem types: positive times pos-itive, positive times negative, negativetimes positive, and negative times nega-tive, taken in that order.

We have already dealt with two of thosefour possible problem types; a positivefactor times a positive factor yields a pos-itive product, and a positive factor times anegative factor yields a negative product.The commutative property of multiplica-tion on whole numbers extends to multi-plication of integers. With that in mind,~5 x +6 can be commuted into +6 x ~5.Therefore, the third problem type can begeneralized to a negative times a positiveyields a negative product. It should benoted that the factors would have to becommuted before the example could bemodeled with a number line. That is, youcould not have a negative five vectorsthat are a positive six long. Once the fac-tors are commuted, the situation wouldbe similar to that shown in Fig. 2.71.

The fourth problem type, which in-volves two negative factors, presents alittle more of a challenge to explain. Oneway to deal with the situation involves us-ing a calculator and exercise sets, allow-ing us to generalize from several specificanswers known to be correct. Becausethis type of reasoning leads only to a con-jecture, not a proof, it can be useful in sit-uations such as this. Calculators do differ,so care must be taken to insure that thesigns of the factors are properly enteredand considered as the exercises are

NUMBER AND OPERATIONS 107

done. Working through exercises such as

these with your calculator,

will lead you to conclude that, when twofactors are negative, the product is posi-tive. Again, although this is not a proof,the conjecture is, in fact, true.

Patterning is another way to develop theidea that a negative factor times a negativefactor yields a positive product. Assuminga firm understanding of the other three in-teger multiplication problem types at thispoint, we will begin with a product gener-ated using a negative factor with a positivefactor. You should read aloud the commentbeside each product as it is completed.Each of these comments compares theexercise with the one above it:

Comparing this line with theprevious line you say: "1stfactor stays the same, ("6);2nd factor decreases by 1 ,(from +6 to +5), product in-creases by 6, (from "36 to

"1st factor stays the same;2nd factor decreases by 1,product increases by 6"

"1st factor stays the same;2nd factor decreases by 1,product increases by 6""1st factor stays the same;2nd factor decreases by 1,product increases by 6""1st factor stays the same;2nd factor decreases by 1,product increases by 6""1st factor stays the same;2nd factor decreases by 1,product increases by 6"

"1st factor stays the same;2nd factor decreases by 1,product increases by 6"

BINGO! Following the pattern, "1st factorstays the same; 2nd factor decreases by1, product increases by 6," you just got apositive product out of two negative fac-tors. Using a number pattern such as this,developed from a comfortable concept, canbe a powerful way of supporting under-standings of abstract concepts until theybecome a part of our knowledge base.

Your Turn

1. Now that all four types of multiplica-tion problems involving integers have beenconsidered, state two generalizations thatdeal with multiplying signed numbers.

2. Develop a generalization relating tomultiplying more than two integer factorsbased on the number of negative factorsthat are involved.

3. State a generalization relating tomultiplying two or more integer factorswhen one of them is zero.

Conclusions

This whole section boils down to two rulesto remember when multiplying integers,assuming none of the factors is zero. Wehave asked you to develop these general-izations and we hope you have, eventhough we realize you could look them upin the solution section. Rememberingthose rules is important when doing multi-plication exercises involving integers asfactors. If you have not digested the proc-ess, then the result becomes a merememorization of disconnected rules pro-vided by someone else. That is not a wiseroute to the development of your mathe-matical skills and understandings.

DIVISION OF INTEGERS

FOCAL POINTS

• Division Using Signed Numbers• Signed Numbers in Inverse Operations

108 CHAPTER 2

Building on the concepts of whole num-ber division, we now divide using signednumbers. Building on the concepts of in-teger multiplication, we now discuss howto deal with signed numbers in the inverseoperation. Comfort with these two areasensures quick success with integer divi-sion. Remember that multiplication is de-fined as factor x factor = product and theinverse operation, division, is defined asproduct -=- factor = missing factor.

Division Using Signed Numbers

A calculator will be a handy tool for thisdiscussion. For a while, use the divisionfacts for whole numbers to avoid remain-ders or fractional parts; keep the prob-lems simple as you investigate what hap-pens with signed numbers. Create severalexercises in each of the following catego-ries: (a) Positive divided by Positive, (b)Negative divided by Negative, (c) Nega-tive divided by Positive, and (d) Positivedivided by Negative. Review your calcula-tor procedures for entering negative num-bers. You will see either a +/- or (-) keythat is to be used to enter the negativesign for a number. The "minus" key is dif-ferent from these keys and its use willcause a calculator error or an incorrect re-sponse. As you do your exercises, recordthe problem and answer, being sure to in-clude signs for all of the numbers in-volved. Look for patterns within the cate-gories and between the categories.

Look at two of the four possible configu-rations of division problems involving inte-

gers. The first one is

Here in the typical division format, thefactor (+4) is outside, the product (+20) isinside, and the missing factor (+5) is onthe top of the vinculum. You also see theproblem expressed as a fraction with the

This third problem type TOIIOWS the sameconfiguration as the other two, this timeusing ~4 as the factor or denominator and+20 as the product or numerator. Here themissing factor is now ~5. The fourth prob-

lem type, has +4 as

the factor or denominator and ~20 as theproduct or numerator. Again the missingfactor is "5.

Signed Numbers in InverseOperations

Compare these with the exercises youcreated earlier and do more of each type,writing the problem in fraction form, alongwith the missing factor. That is, record

them all like this: (fraction with

~20 over ~4 equals +5). How does this re-mind you of the section dealing with inte-ger multiplication?

Your Turn

1. Considering problems only of the

types for example

(positive divided by positive and negative

product as the numerator and the factoras the denominator, giving the missingfactor to the right of the equal sign. Thesecond problem type we consider here is

Both configurations

are the same, replacing the factor (+4) andproduct (+20) with a ~4 and ~20, respec-tively. Note that the missing factor (+5) forboth problem types remains the same.

One of the remaining two possible prob-

lem configurations is

NUMBER AND OPERATIONS 109

. Cross Products• Problem solving

Rational Numbers as Ratios

Rational numbers are fractions that can

be written as where a and b are inte-

gers and b is not zero. The vinculum thatwe use to define the numerator and de-nominator of a rational number has multi-ple meanings. In this section, we use it tohelp us compare two quantities.

Suppose we want to compare the num-ber of girls to the number of boys in a par-ticular classroom, as in Fig. 2.72. We

FIG. 2.72.

might satisfy our needs by stating thatthere are 12 girls and 14 boys. Using a ra-tional number, we could say that the ratioof girls to boys is 1? tn 14, which can be

written as 12:14, or Although this in-

formation might be all that we need, thereare several related ratios that might bejust as useful, depending on the situationor problem. We might change the orderand say that the ratio of boys to girls is 14

to 12, 14:12, or ; we could say:

divided by negative), describe a general-ization for working with dividing a productby a factor when their signs are the same.

2. Considering problems only of the

types and - for example

(positive divided by negative and negativedivided by positive), describe a generaliza-tion for working with dividing a product bya factor when their signs are not the same.

Conclusions

This section is relatively short because ofprevious work in division of whole num-bers and multiplication of integers. Theconnections with these two conceptshelped summarize the work with divisionof integers down to two rules that willcover all situations. Dividing when theproduct and factor have like signs gives apositive missing factor. Dividing when theproduct and factor have unlike signs yieldsa negative missing factor. No new meth-ods for finding missing factors were devel-oped here. We built integer division on thesame generalizations about signed num-bers that were developed in the sectionabout multiplication of integers. Remem-ber, you should always be looking for con-nections between and among concepts.This section is an excellent example of re-ducing the number of concepts thatneeded to be learned by scaffolding newknowledge onto previously obtained skills.

RATIOS AND PROPORTIONS

FOCAL POINTS

• Rational Numbers• Ratios• Percents• Equivalents• Proportions

there are 6 girls for every 7 boys,

there are 7 boys for every 6 girls, 7:6 or

there are 12 girls in the class of 26 stu-

dents, 12:26 or

110 CHAPTER 2

there are 6 girls out of each 13 stu-

dents, 6:13 or

there are 14 boys out of a total of 26

students, 14:26 or

there are 7 boys out of each 13 stu-

dents, 7:13 or

In each of these forms, there is an or-dered pair of numbers on which we canoperate using the properties of fractions.Although we can treat the numbers as ifthey were fractions, it is not necessarilytrue that equivalent ratios have the exactsame meaning as they do with equivalentfractions. For example, we say that thereare 12 girls to 14 boys, but we must saythat there are 6 girls for each 7 boys. No-

tice that are equivalent fractions,

but do not have the same meaning in the

context of our situation because not

only compares the number of girls to thenumber of boys, but also tells us exactlyhow many girls and boys are actuallypresent.

Ratios and proportions react differentlythan fractions when numbers change,too. If two new girls and one new boy jointhe class, then the total changes to 29children, 14 girls and 15 boys. The ratio of

new girls to new boys would be and the

ratio of girls to bovs in the class would

change from We could express

that as That is OK when

dealing with ratios. Think about it and youshould be able to come up with a lot ofexamples like that. Consider the basket-

ball player who made 3 out of 4 foul shotsduring the first game, 5 out of 9 foul shotsduring the second game, and 7 out of 7foul shots during the third game. Howwould you determine that player's foulshooting percentage? You would form aratio between the number of foul shotsmade and the number of foul shots at-

tempted over the three games, or But,

to get that ratio, you add the numeratorsof the individual ratios to get the overallfoul shots made during all three gamesand you add the denominators of the indi-vidual ratios to get the total number offoul shots attempted during all threegames. That is not how you operate withfractions. So, you see, ratios and frac-tions are treated differently when you op-erate with them.

Generally, the context will tell youwhich way to do it. For example, if youoperated on these three ratios as frac-tions, then you would have

, which is about 2.3. Now

there is a player you want on your team!Operating this way implies that every foulshot made is worth 2.3 points rather thanthe normal one point. An answer of 2.3does not make sense in basketball foulshooting.

Your Turn

1. Write ratios that compare the de-signs in Fig. 2.73.

FIG. 2.73.

NUMBER AND OPERATIONS 111

Percents

If we take the word "percent" apart, itmeans per hundred, implying that everynumber expressed as a percent is actu-ally a ratio with a denominator of 100. Ifwe find that a softball team wins 45% ofits games, then that means the team wins

45 out of 100 games, or . But does the

team have to play 100 games to earn thisstatistic? Suppose the team only plays 20games in the entire season; how could wesay they win 45% of the time? The key tothis issue is in the denominator. If theteam wins 9 out of the 20 games, then we

can set up a ratio of wins to games, If

we multiply this ratio by , then we will not

change its value because , the multi-

plicative identity. However, we will find

that the rewritten fraction

has a denominator of 100, and that is howthe 45% is established.

Suppose we are given the ratio and

asked to write it as a percent. We couldgrab a calculator and divide 6 by 15. Thiswould give us the decimal number 0.4 or,because we want to talk in terms of hun-dredths, 0.40. We often read a numbersuch as this as "zero point four zero," butif we read it as a decimal fraction we

would say "zero and forty-hundredths,"

which brings to mind the ratio or

40%. However, this procedure does nothelp us understand what is going on. Putaside the calculator for a moment and see

if we can get to by hand. First, sim-

plify the ratio to We would

like to change that denominator to 100,which can be done by multiplying boththe numerator and denominator by 20,

which is

Suppose you are charged a $36.00 re-stocking fee on a returned item that soldfor $250.00. What percentage of your re-fund does the merchandiser keep? The

ratio involved is and it might not be

obvious what the numerator would be ifthe denominator is forced to be 100.However, any power of 10 will allow us towrite the number as a percent. Multiplying

gives a product of

Because this denominator has one factorof 10 too many, we divide both thenumerator and denominator by 10 to get

, The merchandiser kept

14.4% of your refund. These examplesmay contain more steps than you wouldgenerally use to solve this type of prob-lem. The extra steps in this last examplehelp us make the point that a ratio doesnot always have to be a rational number

as you typically see them. The ratio is

a perfectly good and informative number,but it does not fit the classic definition of arational number, because 14.4 is not aninteger. However, multiplying both the nu-

112 CHAPTER 2

Your Turn

2. The following ratios can be changedso that they have denominators that arepowers of 10. Use the procedure shownin the examples to write these ratios aspercentages.

a) A bowler got 7 strikes out of the first20 frames at the bowling tourna-ment. What was the percentage ofstrikes during the first 20 frames?

b) Only 18 out of 250 people at thewedding selected chicken. Whatpercent of the people at the wed-ding selected chicken?

c) For 135 of the 2500 cars that passedthrough the tollbooth on Tuesday,an error in the amount of the toll paidwas indicated. What percentage oftoll errors does this imply?

3. The following ratios cannot bechanged so that they have denominatorsthat are powers of 10. Explain why this istrue. Use your calculator to write these ra-tios as percentages. Round to the nearesttenth of a percent.

a) 19 of 23 students earned passingscores on the first exam. What per-centage of the students earnedpassing scores?

b) 11 of 18 Matchbox™ cars representAmerican automobiles. What per-centage of Matchbox cars representsforeign automobiles?

Equivalents

One of the big ideas in this section is thenotion of equivalence between rationalnumbers, decimals, and percents. It is im-portant for you to realize that there areequivalencies between each of these top-ics, not just within each topic. That is, youneed to be comfortable with the notion

that the common fraction, is equivalent

to the decimal, 0.25, is equivalent to theratio, 1:4, is equivalent to 25%.

One way of showing the relations be-tween all of these different representa-tions is by using a chart such as Table2.1. Examine the relations among the col-

TABLE 2.1

umns of equivalents. Observing theseconnections is a key part of proportionalreasoning and you should review the sec-tions about fractions, decimals, andpercents if you are not comfortable withthis concept.

Proportions

When two ratios are set equal to one an-other, a proportion is established. Propor-tions offer us an elegant means of solvingmany types of problems, but first we mustlearn to use them with great care. Thereare 14 shaded squares in Fig. 2.74, which

merator and denominator of by 10

will give which does look like a typi-

cal rational number, before common fac-tors are divided out.

NUMBER AND OPERATIONS 113

FIG. 2.74.

has a total of 28 squares. We can set up aratio to compare the number of shadedsquares to the total number of squares,

. Another ratio describes how much

area of the figure is shaded, Whereas

the first ratio concerns how many individ-ual squares are shaded, the second ratioconcerns the amount of area that isshaded. Because these two fractions areequivalent to one another, we can write

When dealing with proportions, it is im-portant to remember that ordered pairsare involved. We could write the ratio of

total squares to shaded squares, , or

the ratio of shaded to unshaded squares,

, but neither of these ratios would be

equivalent to A common error people

make when working problems that in-volve proportions is to forget about theimportance of the ordered pairs of num-bers. We can demonstrate this using anordinary problem:

If a car can go 300 miles on a single 10-gallon tank of gas, then how far could it go ifit had a 20-gallon tank?

To solve this problem, set up a proportionwith one value missing—the distance the

car could travel if we doubled the capacityof the gas tank. If we are conscious of theimportance of the ordered pairs of num-bers, then we would set up the problem

as

where m represents miles traveled, thequantity for which we would like to solvethe proportion. In each ratio, the numberof miles is on top and the number of gal-lons is on bottom. We can solve this prob-lem by inspection because doubling thedenominator implies that we must doublethe numerator. The car can go 600 mileson 20 gallons of gas. If we had not paidcareful attention to the order in which thenumbers should appear, then we might

have set up the problem as .In

order to turn 300 into 20, we must divideby 15. If we divide 10 by 15, we find thatwe cannot go very far if we double thesize of the tank. The best way to avoidmaking this classic error is to write a ratio

with words, such as , and

then set up all of the numbers in ratios tomatch.

Your Turn

4. Write two equivalent ratios that rep-resent the quantities in the following state-ments. Use a letter to represent the miss-ing quantity and set the ratios equal to oneanother to find that missing quantity.

a) I would like to make 3 identicalshirts. My pattern for a single shirtrequires 2 yards of fabric. Howmuch fabric should I buy for my 3shirts?

b) There are 4 tables and 16 chairs inour classroom. If we continue thesame arrangement, then how manychairs would we need for 8 tables?

114 CHAPTER 2

c) A 4-pound turkey requires 1 hour toroast. Assuming all other things areequal, how much time would be re-quired to roast a 16-pound turkey?

Cross Products

In our discussion of ratios and propor-tions, we have been manipulating frac-tions. First we wrote fractions to representthe quantities we wished to compare.Then we manipulated fractions so that wecould have denominators that were pow-ers of 10 in order to compare using per-centages. In the last section, we usedequivalent fractions to find missing quan-tities. As you work with problems that in-volve ratios and proportions, the numbersmay become confusing to manipulatewithout an additional strategy—that ofusing cross products. As with all alge-braic strategies, this can be used in morethan one way. We can use the strategy ofcross products to prove or disprove thetruth of a proportional statement or wecan use the very same strategy to solve aproportion that has one quantity missing.

If a proportion is true, the product of thenumerator of the first ratio and denomina-tor of the second ratio will be equal to theproduct of the denominator of the first ra-tio and numerator of the second ratio.This is the long way of saying, "Cross

multiply." For example, is a true

proportion because 105 x 12 = 180 x 7.

However, because

8x91*17x43.If one quantity is missing in a propor-

tion that we consider to be true, we canuse cross products to determine themissing quantity. For example, in the pro-

portion the cross products are

18a and 441. If we set these two productsequal to one another and solve for a, wefind that a = 24.5.

Your Turn

5. Use cross products to determine ifeach of the following is a true proportion:

6. Solve the following proportions (ifnecessary, round to the nearest tenth):

Solving Problems UsingProportions

The father of one of your authors was amagic man. He could stand near a tree,look at his shadow, and tell the height ofthe tree. When the author was a littlechild, she only knew that her father couldsolve the problem without climbing thetree to measure it; she did not know thathe was using ratios and proportions! Be-cause the rays of the sun are virtually par-allel by the time they create shadows, theshadows of the tree and the man are di-rectly proportional to the heights of thetree and the man, as in Fig. 2.75. If theman's shadow is half as long as theshadow of the tree, then he must be halfas tall as the tree. Suppose the man is 6 fttall and casts a 10.75 ft shadow and the

NUMBER AND OPERATIONS 115

portions, such as

, which are

four of the many possible combinations.We have a great deal of freedom when weset up proportions, as long as we remem-ber that the two ratios represent orderedpairs and must be kept in the same orderon both sides of the equal sign.

Your Turn

FIG. 2.75.

tree casts a 21.5 ft shadow. We could use

the proportion to find that the

height of the tree is 12 ft. This is a directapplication of the properties of similarright triangles, which are often sketchedto solve such problems. If the right trian-gles are similar, then the same ratio canbe used to compare each set of corre-sponding sides. Assuming that the manand the tree stand perpendicular to theirshadows, imagine rays of sunlight tocomplete the hypotenuse (longest side ofa right triangle) of the right triangle asshown in Fig. 2.76. The hypotenuse wasnot needed for this problem, only theheights and shadows of the man andthe tree. In many problems of this type,the hypotenuse will be important.

Ratios and proportions are associatedwith similar triangles. Comparing the cor-responding sides of the similar trianglesin Fig. 2.76, we can set up several pro-

FIG. 2.76.

7. List at least four additional propor-tions for the triangles in Fig. 2.76.

8. Your recipe for Party Mix calls for 7cups of wheat cereal, 2 cups of mixednuts, 2 cups of pretzels, 1.5 sticks of but-ter, 1 tablespoon of special sauce, andseasoned salt to taste. You find that youhave only 1.5 cups of mixed nuts. Howmuch of each of the other ingredients doyou use if you want to keep the recipe inproportion?

9. For some incredibly interesting rea-son, you need to know the height of thetelephone pole in front of your home. Youknow that you are 5.5 feet tall and that youcast a shadow that is 11 feet long. Thetelephone pole casts a shadow that is 26feet long. How tall is the telephone pole?

10. Trace the triangles in Fig. 2.76 andlabel them as follows. Smaller triangle—the shorter leg is 9 cm and the longer legis 12 cm. Larger triangle—the hypotenuseis 37.5 cm and the shorter leg is 22.5 cm.What is the length of the hypotenuse ofthe smaller triangle?

11. As a teller in a bank that sells for-eign currency, you are responsible forhelping customers who are planning fortrips abroad. Because the rates changevery quickly, you must calculate eachtransaction separately. A customer asksyou how much it would cost to buy200.00 DM. You check the rate and find

116 CHAPTER 2

that $1.00 is worth 1.32 DM. How muchwill it cost your customer to get the deut-sche marks needed for the trip?

Conclusions

You have been introduced some of themyriad uses of ratios and proportions.With this introduction, you may find that

you use them intuitively in your dailylife—perhaps when shopping, cooking,working on your hobbies, enjoyingsports, or working on the monthly bud-get. As you go through your day, watchfor examples of ratios and proportions.You might be surprised to see how manythere are.

3Algebra

FOCAL POINTS

• Historic Underpinnings• When Does Algebra Begin?• Laying the Foundation for Algebra• Having Fun With Algebra• Integrating Algebra. Patterning• Representing Situations With Algebra• Using Models• Rate of Change• Sequences• Formulas

Algebra is often thought to be a specificsubject area studied in the middle schooland beyond. Components of algebra be-gin to appear early in the curriculum, if onlyin a problem like 2 + 3 = ?. Algebraicallythis might be expressed as 2 + 3 = x, butthe difference between the two symbols, ?and x, is not major. The study of subtrac-tion problems written in the form 7 + 4 = 7builds a background for an algebra-basedsolution involving subtracting 4 from bothsides of the equation to determine themissing addend. It is possible that alge-bra and related topics have been part ofyour world from the beginning of your for-mal education.

Consider the following problem, whichis rich in algebraic opportunities. Pete andRepeat are traveling the 96 miles fromPahokee to Holopow on their bikes. Petecan average 8 miles per hour (mph) andRepeat can average 12 mph. On the fol-

lowing graph, show how far each will havetraveled after 1, 2, and 3 hours.

Place the information about the travels ofPete and Repeat into the table.

If you think of questions like the following,you are beginning to think algebraically:

Describe a pattern for each row of thetable.

How far from Pahokee is each rider af-ter 3 hours?

How long will it take each rider to travel24 miles?

After Repeat gets to Holopaw, howlong will it take Pete to get there?

If h is the number of hours traveled, andm is the distance covered in miles, isthere a rule that shows the relation be-tween m and h for each rider?

Could I use the rule I just devised toshow how far each rider could go in anhour?

How far will each rider travel betweenthe end of hour 4 and the end of hour5?

If it is 96 miles from Pahokee to Holo-paw, how long will it take each rider tocomplete the trip?

117

118 CHAPTER 3

HISTORIC UNDERPINNINGS

Believe it or not, algebra did not simplyappear within the last 60 or 100 years orso. The Babylonians were using algebrabasics in 2000 BC. It has been discoveredthat they were solving second- and third-degree equations of the form x3 + x2 = b(Eves, 1990) around that time. The Baby-lonians did not use the letters x or b, nordid they use our current numeral, placevalue, or exponential systems, which hadnot been developed yet.

The first records dealing with adding orsubtracting the same magnitude on bothsides of an equation are found in theArabic writings of AI-Khowarizmi (Mo-hammed ben Musa, which means Ma-homet, the son of Moses) about 830 AD.Leonardo da Pisa (Fibonacci) introducedsome basics of algebra to Italy about1200 AD, and Robert Recorde introducedit to England in a 1557 AD publication.Even those introductions were not alge-bra as we know it today, but the ideas,as we know them, were beginning toform. Algebraic methods and notationshave been improved and revised throughthe centuries. Unlike arithmetic, where3 + 4 = 7 is constant, algebraic notationslike x + y = z take on different meaningsin different contexts.

Descartes (about 1637 AD) used sym-bolic notation to express algebraic calcu-lations. He also used letters at the begin-ning of the alphabet (a, b, c) to denoteknown quantities (constants) and lettersfrom the end of the alphabet (x, y, z), par-ticularly x, to indicate unknown quantities(variables). Does this terminology remindyou of your high school days?

Organizations such as NCTM suggestthat ALL students can learn mathematics.That means all students can learn algebraat the appropriate time in their develop-

ment. Everyone in our society does not ac-cept this philosophy. Traditionalists say,and many parents agree, "The mathemat-ics I learned and the way I learned it wasgood enough for me, so it is good enoughfor my child." Unfortunately, that statementis far from the truth. Limit your skills tocomputation and you limit your career op-tions. Technology can do the arithmetic af-ter the skills and notions behind the arith-metic are learned. Society needs thinkers,both inside and outside of the box. Today'sworld is much more mathematical thanyesterday's and mathematics is so muchmore than arithmetic. Productivity in to-day's world requires greater mathematicalabilities and more complex problem-solv-ing skills. Tomorrow's world will be evenmore mathematical than today's, thus wehave a responsibility to help future genera-tions gain insights into the mathematicsthat will be such a large part of their world.Consider the following example of a pri-mary school algebra problem: * + 4 = 9and * + + + * = 11. The first equation hasfive solutions in the set of whole numbers,one of which must satisfy the second equa-tion. You might list the addition facts for 9and try each number pair in the secondequation. Another strategy might involve aone-to-one matching of the parts of theequations to see what is duplicated. If one* and one + are subtracted from the leftside of each equation and 9 is subtractedfrom each right side, then the correct addi-tion fact can be determined without tryingall five pairs of addends. A third strategymight be to replace one * and one 4 in thesecond equation with 9. Easy you say?Think any second grader could solve thisproblem? This problem involves solving alinear system of equations, x + y = 9 and2x + y = 11. The three strategies discussedare guess and check, linear combinations,and substitution.

ALGEBRA 119

LAYING THE FOUNDATIONFOR ALGEBRA

Although algebra is an abstract way ofviewing mathematical concepts, manipu-latives can help us understand what ishappening. Suppose you have threeblocks and another student has seven, asshown in Fig. 3.1. How many blocks do

FIG. 3.1.

you need so that you have as many as theother student? In reality, you are asked tosolve a problem of the form 3 + A = 7, us-ing manipulatives. You are beginning thefoundation for algebra as you look for themissing addend. You could line up bothsets of blocks as shown in Fig. 3.2. This

Other Student

FIG. 3.2.

would allow you to look concretely for aone-to-one correspondence and find thenumber of blocks needed to have seven.You could add blocks until the number ofblocks in each set is the same.

Is there any way to solve 3 + A = 7 usinga calculator? What does this problemmean? Three is the known addend andseven is the known sum. The triangle rep-resents the missing addend, which canbe determined by subtracting the knownaddend from the sum. Subtraction using

the missing addend model is the begin-nings of algebraic thinking.

Let's kick this idea up a notch and findthe missing addend in 3x + = 7x. Oneway to do this is to use a model similar tothe one in Fig. 3.2. The bottom line is thatyou have three unknowns and you want toknow how many more unknowns you willneed to have a total of seven unknowns.

The Algebra FX-2.0 calculator, devel-oped by Casio, provides the power of acalculator with an algebraic tutorial. Youenter 3 + X = 7 and, with the push of abutton, the calculator supplies an on-screen step-by-step process for solvingthe problem. With 3 + X = 7, the first step,subtracting 3 from both sides, will show3 + X-3 = 7-3. Next, the calculator willcollect like terms on each side of theequation, and X = 7.

A computer program called MathXpert(www.mathxpert.com) will solve thisproblem in a similar symbolic manner. Inaddition, MathXpert provides a state-ment and reason for each step taken as aproblem is solved. Each of these pro-vides options for step-by-step generalhints, specific hints, or automatic com-plete solutions.

If you want to learn algebra, then tech-nology can be very helpful. Several calcu-lators, such as Casio's fx-9970 or TexasInstruments' TI-89 orTI-92, will solve alge-braic problems, but these will not show theintermediate steps involved in the process.

HAVING FUN WITH ALGEBRA

One way to increase enthusiasm formathematics is to use number tricks. Trythis one:

Pick a counting number.Multiply your number by two.

120 CHAPTER 3

Add four to your new product.Subtract 10 from your new sum.Add six to your new number.Now subtract your original number.What did you get?

Figure 3.3 shows a physical model of thisnumber trick.

There are many number tricks of thissort that can provide entertaining ways ofpracticing algebraic manipulations. Manyhave made the rounds on the Internet.One such trick claims that you will end upwith your age by following the directions,but that it will only work for the currentyear. The next time someone sends thistype of trick to you, analyze it algebra-ically and then fix the trick so that it willwork for any year. Algebra isn't mathe-matical magic, it just seems that way untilyou take a closer look. Number tricks arefun, but you won't be mystified for long ifyou look at them while wearing your alge-bra glasses.

Your Turn

1. Do this trick using a specific numberand then using x for the number picked.

FIG. 3.3.

Using algebra, we can demonstrate thisprocess using symbolic manipulation anda variable for the original number. Thismethod shows that, no matter what num-ber you pick, you will always end the trickwith that number:

Pick a number.Double it.Add 4.Divide by 2.Subtract your original number.

What do you get? If you begin the trickwith a different number, will you still getthat answer?

2. Do the trick in Exercise 1 using afraction. Is your answer still the same?

3. Do the trick in Exercise 1 using anegative number. Is your answer still thesame?

4. Do this trick using a specific numberand then using m for the number picked.

ALGEBRA 121

What do you get? Will this always work?Why or why not?

5. Do this trick using a specific numberand then using p for the number picked.

Pick a number.Triple it.Add 12.Divide by 3.Subtract your original number.

What do you get? Is this problem signifi-cantly different from the one given in Ex-ercise 1?

INTEGRATING ALGEBRA

Algebra should be infused throughout themathematics curriculum. As you haveseen, number operations provide founda-tional underpinnings. In beginning alge-bra, you learn to simplify expressionssuch as 4y + 5y by combining like terms.When dealing with 4y + 5y, you get 9y,just as you would get 9 pencils if you wereadding 4 pencils and 5 pencils. Think oftwo groups of identical blocks where onegroup contains four blocks and the otherfive blocks. Which problem is algebra?Which problem is arithmetic? Is there re-ally a difference?

As you learn to simplify more compli-cated expressions, you begin to realizethat you cannot combine unlike terms. Ifyou have three apples and four peaches,then can you combine them to give oneanswer? You could say you have sevenpieces of fruit, but the kind of fruit is lost.If you want to know what kind of fruit,then you have to stay with the three ap-ples and four peaches. You could write anexpression, 3a + 4p, where a stands forapple and p stands for peach, but youcan't simplify the situation beyond that.

Suppose you have three squares andfour triangles and a friend has fivesquares and two triangles. If you combinethe sets of figures and arrange the shapesinto like groups, how many would be ineach group? By writing the expression

., you have moved fromthe concrete situation to a symbolic rep-resentation of the problem. Combininglike figures will give >, which isas far as you can go. That is much likesimplifying 3x + 4y + 5x + 2y to 8x + 6y bycombining like terms.

In the section dealing with multiplyingwhole numbers, we discussed partialproducts and the connection with alge-bra. Suppose the exercise is:

Rather than doing the exercise as shown,we could use expanded notation andwrite the factors as 10 + 4 and 10 + 2. Theproblem would still be done the same waybut the formatting would be different.

As you can see, the rest of the problem isthe same but now the factors for eachpartial product are easier to trace. Figure3.4 shows this problem done with Base10 blocks, a reminder of our discussionabout multiplication as finding the area ofa rectangle.

122 CHAPTER 3

FIG. 3.4.

Now the stage is set for doing an alge-bra multiplication problem. Rather than10 + 4 and 10 + 2, suppose you had m + 4and m + 2 as factors. Using the same for-mat, the problem would be:

As you compare the algebra version andthe partial product version, things shouldlook similar. Figure 3.5 represents theproduct of m + 4 and m + 2 and it looks alot like Fig. 3.4. This problem easily transi-tions to the FOIL method for multiplying

FIG. 3.5.

FIG. 3.6.

binomials, as shown in Fig. 3.6. FOIL(Firsts, Outers, Inners, Lasts) is simply away of remembering to distribute eachnumber in the first binominal to eachnumber in the second binomial. There isnothing special about the order of the let-ters in FOIL; you could just as efficientlydistribute from the right and use LIOF.

PATTERNING

You have dealt with number patterns 2, 4,6, 8 , . . . and 1, 3, 5, 7 , . . . saying that theyare odd and even numbers. A pattern like1, 3, 6, 10, ... is not as easy to classifyand yet bowling pins are arranged usingthe fourth term of this pattern called trian-gular numbers.

Some patterns are not immediately ob-vious. Try your hand at this one:

Figure out the pattern before reading on.We're watching you! The question is,what is the next term in the pattern. Didyou try it? You should, because we areabout to tell you the answer and it isbetter for you to figure out the pattern be-fore we tell you. As you look at it, you

ALGEBRA 123

need to start at the top and read the firstline. What do you see? You saw one one,or 11, which is shown on the second line.Now, read the second line and what doyou see? You see two ones, or 21, whichis shown on the next line. Read that lineand you see one one and one two or1112. And so it goes. As you read eachline, you count the number of ones, thenthe number of twos, and so on. What youget becomes the next line. Continuing,the line after 1112 should be 3112 toshow that the line above contained threeones and one two.

Your Turn

6. What are the next two lines in thepattern 1, 11, 21, 1112, 3112, 211213,312213, 212223, ? ?

The pattern 1, 11,21, can be altered. Itwill start the same, but the fourth termand each one after that will be differentfrom the one used before. It is differentbecause a different pattern is being used:

Try to determine the next term in this onebefore reading on. As we said before, it isbetter for you to try to figure out the pat-tern before looking at the answer. In thiscase, we are reading left to right and re-porting what is seen. If the digits are thesame, then they are grouped together.Again, read the first line and you see oneone. Reading the second line, you seetwo ones or 21. Reading that third line,you see one two followed by one one, or

1211. Read the fourth line and you seeone one, one two, and two ones or111221.

Your Turn

7. What are the next two terms in thepattern 1, 11, 21, 1211, 111221, 312213,1311221113, ?, ?

There are a lot of number patterns. Wehave shown you only a few, but as youbegin to sort them out, you are beginningto strengthen your algebra skills.

REPRESENTING SITUATIONSWITH ALGEBRA

Algebraically, we try to express patternsby using a generic expression for anyterm. Consider the pattern created by theset of even numbers. If n is the ordinal, orpositional, number of the term (first, sec-ond, third, and so on), 2, 4, 6, 8, ... canbe expressed by 2n. If you need the 87thterm in this pattern, then you know it is(2)(87), or 174. Similarly, 2n - 1 can beused to express the pattern of odd count-ing numbers; if you wanted to know the87th odd counting number, it would be(2)(87) - 1, or 173. Triangular numbers canbe expressed in general by the formula

, which is not as simple to develop.

A pattern you have seen is 1, 4, 9, 16,25, 36, . . . . You may recognize this as abeginning of the list of square numbers.Another way of writing them would be 12,22, 32, 42, 52, 62 This format makes iteasier to see how the generic term is writ-ten. What should it be? If you determinedthat it should be n2, then give yourself apat on the back.

Sometimes we are asked to set up andsolve word problems. An important skill isthe ability to take the problem apart and

124 CHAPTER 3

use the information given. If the problem isnot immediately obvious, then it might bea good idea to look for a pattern. Supposeyou are asked to solve the following prob-lem: "In my home town there are two rentalcompanies. I need to rent a jackhammerand would like to spend as little as possi-ble. On Monday, I will need the equipmentfor 8 hours and on Tuesday I will need it for3 hours. At Rent It, the cost is a flat fee of$20.00 plus $3.00 per hour. At Get It Here,the cost is $8.00 per hour. What should Ido?" A table to compare the cost of theequipment at each rental company wouldbe a good way to look for a pattern.

The cost at Rent It is higher at the begin-ning of the table, but at 4 hours the costsare the same, and Rent It is the betterdeal after that. The solution to this prob-lem is to rent the equipment at Get ItHere for $24.00 on Tuesday and at RentIt for $44.00 on Monday. The more youpractice expressing patterns in generalterms, the better your algebraic founda-tions will be.

USING MODELS

Sometimes models make patterns easierto figure out. Look back at 1, 4, 9, 16, 25,36, . . . . Switching to exponents was oneway of doing it. Figure 3.7 shows how we

FIG. 3.7.

could have done it with a graphic. At thetop is a white unit square representing thefirst square number. Below that is ashaded V made up of three unit squareswhich, with the top center white unitsquare, forms a 2 by 2 square to repre-sent the second square number. The nextV is made up by five white unit squares,but if we include all the unit squaresabove, then the total is nine unit squares.The next V is made up by seven shadedunit squares, but if we include all the unitsquares above, then the total is 16 unitsquares. Continuing this procedure, youhave a 5 by 5 square made up of 25 unitsquares, a 6 by 6 square made up of 36unit squares, a 7 by 7 square made up of49 unit squares, and an 8 by 8 squaremade up of 64 unit squares. If we look atthe squares in terms of the number of unitsquares that make them up, then wewould have 1, 4, 9,16, 25, 36, 49, and 64.But, if we consider the dimensions in-stead, we have 12, 22, 32, 42, 52, 62, 72, 82,which connects to our abstract model.The graphic model could be extended toinclude as many unit squares as wewould like, but eventually the advantageof generalizing to the exponential form,and ultimately to n2, should be apparent.

ALGEBRA 125

Wait a minute. Look at Fig. 3.7 again. Ifyou take the number of unit squares mak-ing up each V (count the top center whitesquare as the first V), there is another wayto express the square numbers. The firstsquare would be one. That square plusthe three unit squares in the next V wouldbe 1 +3 = 4. Forming a sum for the next Vand all those above gives 1 +3 + 5 = 9.The whole figure could be described as:

1 = 11 + 3 = 4

1 + 3 + 5 = 91+3 + 5 + 7 = 16

1 +3 + 5 + 7 + 9 = 251 +3 + 5 + 7 + 9 + 11 =36

1+3 + 5 + 7 + 9 + 11 +13 = 491+3 + 5 + 7 + 9 +11 + 1 3 + 15 = 64

Check it out! The model gave a clue to an-other pattern that can be used to expresssquare numbers—adding consecutiveodd counting numbers starting with one.Models can come in handy.

Another type of model that leads tosome nice algebraic thinking comes fromsomething that we call figurate numbers.Figurate numbers are represented by pat-terns of dots to model geometric num-bers. Figure 3.8 is a model of the squarenumbers we described earlier.

FIG. 3.8.

For the square numbers, the patterncould be identified as Sn = n2, but Sn

could also be defined using the pattern1 + 3 + 5 + . . . + n. Sometimes one modelis clearer than another—like beauty, clar-ity is in the eye of the beholder.

Similarly, triangular numbers can beidentified using a pattern called

Tn = — -, found by generalizing the

model shown in Fig. 3.9. There are other

FIG. 3.9.

ways of looking at this pattern. For exam-ple, we might say that Tn = Tn _ 1 + n, thatis, any triangular number is just the previ-ous triangular number plus the ordinalnumber of the triangular number needed:T4 = T4 _ 1 + 4 = T3 + 4. This method re-quires that you to know T3 = 6. Anothernice pattern defined by the triangularnumbers is a little more manageable andhas a rich mathematical history. We willgive the history lesson and then let youmake the application to this model.

Carl F. Gauss (1777-1855) was a math-ematical prodigy. A story is told that his tu-tor assigned a considerable amount ofwork, certainly enough to keep him busyfor some time. The assignment was to findthe sum of all of the counting numbersthrough one hundred. Gauss thoughtabout it for a moment, then had the an-swer to the problem. The instructor wasamazed and unprepared for this astonish-ing outcome. How did Gauss add thenumbers so quickly? Certainly in the early1780s he did not use his electronic calcu-lator! Magic? No, models and patterning!

From his later work, we suspect thatGauss reasoned along these lines. Firsthe listed several counting numbers at thebeginning and at the end of the sequence,below that, he listed the same number inreverse order:

126 CHAPTER 3

This gave him 100 pairs of addends, eachwith a sum of 101. He quickly found thetotal to be 10100, but he realized that wastwice as much as he needed, because heused each number twice. To rectify thesituation, he divided it by two, getting5050. Is that cool or what?!

Now look at the triangular numbers-does Gauss' trick apply? The first triangu-lar number is one, the second is three, thethird is six, and so on. Do you see it? Canyou write this algebraically? You saw theanswer earlier.

Figurate numbers can be determined foralmost any geometric shape. The last ofthe figurate numbers we are going to modelfor you are the rectangular numbers. Thefirst rectangular number is 2, the second is6, the third is 12, the fourth is 20, and so onas seen in Fig. 3.10. The rectangular num-

FIG. 3.10.

bers are related to both the triangularnumbers and the square numbers. We willleave generalizing this pattern to you.

Your Turn

8. Make a model that shows howGauss' trick applies to the triangular num-bers.

9. Using the model of the rectangularnumbers, write a generalizing statementabout how to find any rectangular num-ber, Rn.

10. Rewrite the rectangular model toshow a relation between the rectangularnumbers and the square numbers.

11. Generalize the model you drewabout the relation between rectangularand square numbers.

12. How do the rectangular numbersrelate to the triangular numbers?

RATE OF CHANGE

Many topics in algebra depend on an un-derstanding of rate of change and how todetermine rate of change. A common ap-plication of rate of change is the distanceformula, d = rt in which the rate, r, can be

expressed as Geometrically, when we

think about rate of change, we look at rise(change in the y direction) divided by run(change in the x direction). The Greekdelta symbol is used to represent change,

so algebraically this would be

Ordered pairs identifying points on thex-y plane are written in the form (x, y).

Recall the trip Pete and Repeat took aswe began this chapter. Pete rode his bikeat a rate of 8 mph. At the end of 1 hour,Pete had ridden 8 miles. At the end of 2hours, Pete had ridden 16 miles, an addi-tional 8 miles, and he rode an additional 8miles each hour until he got to Hollopaw.If we plot Pete's rate representing dis-tance in miles on the x axis and time inhours on the y axis, it will look like thegraph in Fig. 3.11. We start at the first or-dered pair (0, 0), representing no milestraveled and no time elapsed. From there,we go up one unit and right eight units un-til we hit the second point at (8, 1). Foreach eight miles and 1 hour, we plot apoint by rising one and running eight fromthe previous point. What was the changein y? What was the change in x? Because

ALGEBRA 127

FIG. 3.11.

Pete was riding continuously at a con-

stant rate of hour per mile, we can

sketch a line segment that starts at (0, 0),goes through each of the plotted points,and ends at (96, 12).

The rate of change is also called the

slope, m and . Pete's rate of

change, or the slope of the line segmenttelling us how far Pete had traveled at any

given time, is What does this have

to do with algebra? The equation for aline can be written in what is known asthe slope-intercept form, y = mx + b,where m is the slope and b is the pointwhere the graph of the equation inter-sects the y axis. What if we have twopoints and we want to figure out the rateof change for a graph? We call the twopoints (x1, y1) and (x2, y2). Two distinctpoints from our discussion about Peteare (8, 1) and (16, 2), where y, = 1, x1 = 8,y2 = 2, and x2 = 16. We know that thechange in the y direction is 1, and by ex-amination we suppose thaty2 - y1 = 2 - 1 = 1. Likewise, we supposethat x2 - x1 = 16 - 8 = 8. As it turns out ourconjecture is correct and

The equation for the graph of Pete's trip is

; , which tells us that the slope is

and the graph intersects the y axis at (0,

0). In Fig. 3.11, we have shown a fewpoints on the graph.

Your Turn

13. Repeat traveled at a rate of 12 mph.Use the information you placed into thetable at the beginning of this chapter.

a) Make a continuous graph of Re-peat's rate of change using the xaxis for time and the y axis for dis-tance traveled.

riseb) Find m for Repeat by countingrun

on the line segment.c) Use the two points from the table for

how far Repeat had traveled at 2hours and at 5 hours and find m al-gebraically. Does your answer herematch your answer for part b?

14. In the word problem about rentinga jackhammer, we found a cost patternfor each of the rental companies. For RentIt, the pattern was 20 + 3h ($20.00 plus$3.00 per hour) and for Get It Here, thepattern was 8h ($8.00 per hour).

128 CHAPTER 3

a) Use the cost pattern for Rent It towrite a cost equation for renting theequipment. State the rate of changeand the y intercept.

b) Use the cost pattern for Get It Hereto write a cost equation for rentingthe equipment. State the rate ofchange and the y intercept.

c) At what point would the graphs ofthese two cost equations intersect?

15. Peggy, a member of this book'sdynamic author team, wants to paint herhouse a particular shade of blue such thatthe ratio of gallons of blue paint to whitepaint is 2:3.

a) How many gallons of blue paint andhow many gallons of white paint willbe needed to mix up 15 gallons?

b) What is the algebraic solution for theslope that represents this color mix-ture no matter how many gallons ofpaint are needed?

c) If Peggy would like to make the trimof her house a darker shade of thesame color, then how should shechange the ratio of the paint mixture?

d) Using the trim mixture you decidedon for Peggy, how many gallons ofblue paint and how many gallons ofwhite paint would be needed tomake 3.5 gallons of paint?

SEQUENCES

There are two interesting types of se-quences, which are quite common, thatalgebra can help explore. One of thesesequences is called arithmetic becausethe difference between any two succes-sive numbers is a constant. The mostcommon pattern is the counting numbers,1, 2, 3, 4, . . . . The difference between 1

and 2, 2 and 3, or 3 and 4, and so on isone, a constant or unchanging number. Ifthe difference between each number iscalled d, then to find the 50th countingnumber, multiply 50 times d to get 50, the50th counting number.

Starting at 5, 6, 7, 8, . . . , d would stillbe 1 , but now you use the number youstarted with to get to the right term in thesequence. If you call the 50th number a50,then the first number would be called a1,The 50th number in the sequence startingwith 5 is four more than the 50th numberin the sequence that started with one, soa50 = 54. Subtract a1 from 54 to get 49,which is one less than the nth numberneeded. Let n equal the nth number in thesequence. Then an = a1 + (n - 1)d is theformula for the nth term in an arithmeticsequence.

Another common sequence is the evennumbers: 2, 4, 6, 8, . . . . The differencebetween any two consecutive numbers istwo, so d = 2. Try the formula to see if itworks. What is the 137th even number?

Knowing how to find the nth term in anarithmetic sequence, the next step is tobe able to find the sum of a certain num-ber of elements in the sequence. What isthe sum of the first 10 counting numbers?When n is even the sum can be found by

using the following formula: i,

where a1 is the first number in the se-quence, an is the last number in the se-quence, and n is the number of terms to

ALGEBRA 129

be added together in the sequence. Thenit follows that:

When n is odd, the formula changes toavoid getting a fractional answer:

Try adding the

first 11 counting numbers:

The sum of the first 11 counting numbersis 66.

Your Turn

16. What is the 121 st number in the fol-lowing sequence: 4, 8, 12, . . . ?

17. What is the 326th number in thefollowing sequence: 9, 20, 31, . . . ?

18. What is the sum of the first 84 evennumbers: 2, 4, 6, . . . ?

19. What is the sum of the 9th throughthe 101st numbers in Exercise 16?

A sequence is geometric if each suc-cessive term is a constant multiple of theprevious term. Another way to think about

this is to say that the ratio of successiveterms is a constant; that is, a2 divided bya1 is equal to a3 divided by a2, and so on.An example of a geometric sequence is:1, 3, 9, 27, . . . . To find the nth term in ageometric sequence, multiply the firstterm by the ratio constant raised to then - 1 power:

Geometric sequences get very large veryquickly. For the sake of this discussion,use a small number of terms to avoid cal-culator overflow. What is the 20th term inthe following sequence: 1 ,3 ,9 , . . .? Firstdecide the ratio of the second to the firstterm, which is 3:1 or 3. Next plug this in-formation into the formula:

Wow! The 20th term is already over a bil-lion!

If the ratio constant is less than one,then a geometric sequence gets smallerand smaller like in: 27, 9, 3, . . . . What isthe ratio? What is the 20th term in this se-

quence? The ratio is and the 20th term

is . Wow, again!

The 27 gets really small really fast and isalmost what number? The 326th term inthis sequence would definitely be almostwhat number? How is this going to affectthe sums calculated next?

130 CHAPTER 3

The sum of the first 326 terms of 27, 9, 3,... is:

Huh, is that what you expected? Whatdoes this lead you to believe about thistype of geometric sequence? Mathe-matically, this sum has a limit as n ap-proaches infinity because the contribu-tions made by the ratio raised to the nthpower become insignificant to the sum.This is powerful stuff—adding an infinitenumber of elements from a decreasinggeometric sequence and getting a finiteanswer!

Your Turn

20. What is the 15th number in the fol-lowing sequence: 5, 10, 20, . . . ?

21. What is the 19th number in the fol-lowing sequence: 99, 33, 11, . .. ?

22. What is the sum of the first 20 num-bers in the sequence: 5, 10, 20, . . . ?

23. What is the sum of the first 500numbers in the sequence: 99,33,11,... ?

FORMULAS

Mathematical formulas are algebrabased. When you write the general termfor a number pattern, you have developeda formula. Some patterns provide infor-

One formula summarizes any geomet-

ric sequence: . Notice for

this formula that there is no need to firstdetermine the nth term. Once the ratio,the first term, and how many terms tosum are known, all the necessary infor-mation is present. Find the sum of the first20 terms in the sequence 1, 3, 9, 27,. . . :

The sum of the first 20 terms of 27, 9, 3,... is:

ALGEBRA 131

mation that is so useful that their formulasbecome standardized. How many formu-las have you been asked to memorize inyour educational experience? A betterquestion might be to ask you how manyformulas you remember.

Think about some formulas you re-member. Of all the formulas you haveseen, why do you suppose you remem-bered those specific formulas? Ofcourse, there are many factors that haveeffects on why we remember particularformulas. The real question is notwhether you memorized a formula, it iswhether or not you can apply it as yousolve a problem. Did you remembersome formulas but have no inkling aboutwhat they do? Why do you think you re-member a formula without rememberingits purpose? The following is a list ofsome common formulas. Did you re-member any of these? Did you remem-ber the application of each?

Meansum of data points

number of data points

VolumeRectangular Prism (box)

Iwh (I is length, w is width,h is height)

Sphere4 -5— nr (n » 3.14, r is radius)O

Right circular cylindernn

nrh (n » —, r is radius,

h is height)Perimeter of a rectangle

2(l + w) (I is length, w is width)Pythagorean Theorem

a2 + b2 = c2 (a and b are legs,c is hypotenuse ofa right triangle)

Formulas come in many differentshapes, sizes, and purposes. Think aboutthe pros and cons of memorizing a multi-tude of formulas. Generally, you are ex-pected to know where to find needed for-mulas and, above all else, know when andhow to effectively utilize the formula.

CONCLUSIONS

Algebra accounts for a large proportion ofthe Pre-K-16 curriculum. It seems that weare always laying the foundation for alge-bra concepts, learning algebra concepts,or applying algebra concepts. The repre-sentations of problems spiral from con-crete to semi-concrete to semi-abstractto abstract—and back again. We needthe concrete and semi-concrete modelsto aid in development and comprehen-sion of the concepts; we move to semi-abstract and abstract models becausethey are more efficient as problems be-

Meansum of data points

number of data points

VolumeRectangular Prism (box)

Iwh (I is length, w is width,h is height)

Sphere4 -5— nr (n » 3.14, r is radius)O

Right circular cylindernn

nrh (n » —, r is radius,

h is height)Perimeter of a rectangle

2(l + w) (I is length, w is width)Pythagorean Theorem

a2 + b2 = c2 (a and b are legs,c is hypotenuse ofa right triangle)

132 CHAPTERS

come more complex. With concrete be- REFERENCESginnings and a solid understanding of ba-sic algebraic concepts, the whole subject EveS, H (1990) An introduction to the history of

becomes more familiar, easier to grasp, mathematics (6th ed.). Fort Worth, TX:and ultimately, to understand. Saunders.

4Geometry

FOCAL POINTS

• Undefined Terms• Angles• Simple Closed Curves, Regions,

and Polygons• Circles. Constructions. Third Dimension• Coordinate Geometry• Transformations and Symmetry

You might be surprised about how manyreal-life concepts are included in thestudy of geometry. Young children exper-iment with ideas such as over versus un-der, first versus last, right versus left, andbetween, without realizing that they arestudying important mathematics con-cepts. Additionally, early attempts atlogic, even the common everyone elsegets to do it arguments that are so popu-lar with children, are geometry topics. Acube is a three-dimensional object withsix congruent faces and eight vertices—often called a block. You may also use theterm block to identify a prism that is not acube, but, although you may not think of itvery often, you can probably identify thedifferences between a cube and a prismthat is not a cube, as shown in Fig. 4.1.

In this chapter, you will review, refine,and perhaps, extend your understandingof geometry. When Euclid completed aseries of 13 books called the Elements in300 BC, he provided a logical develop-ment of geometry that is unequaled in our

history and is the foundation of our mod-ern geometry study. Geometry is adynamic, growing, and changing body ofintuitive knowledge. We will let you ex-plore conjectures and provide opportuni-ties for you to create informal definitions.There will be some reliance on terms andprevious knowledge, especially when weget to standard formulas.

UNDEFINED TERMS

Some fundamental concepts in geometrydefy definition. If we try to define point,space, line, and plane, then we find our-selves engaging in circular (flawed) logic.The best we can do is accept these fun-damental concepts as building blocksand try to explain them.

A fixed location is called a point, whichis a geometric abstraction that has no di-mension, only position. We often use atiny round dot as a representation of apoint. As the series of dots in Fig. 4.2 getsmaller and smaller, we observe that thedimensions are diminishing—but any dotthat we can see has some dimension,

FIG. 4.1.

133

134 CHAPTER 4

FIG. 4.2.

even the period at the end of this sen-tence. The fact that a point has no physi-cal existence does not limit its usefulness,either in geometry or everyday activities.Although a dot covers an infinite numberof points, it represents the approximatelocation of a distinct point so well that weforget the difference and freely identifythe dot as a pinpointed location. In math-ematics, we label the point representedby a dot with a printed capital letter, forexample, the points in Fig. 4.3. The set

FIG. 4.3.

of all possible fixed locations is calledspace. We have no method of represent-ing space, but we use the concept inmathematics and in real life.

The most direct (straight) path betweentwo points is called a line segment and, asshown in Fig. 4.4, the two points, P and

FIG. 4.4.

Q, are called the endpoints of the seg-ment. Because the most direct path mustbe the shortest distance, we say that thesegment joining two points must bestraight. Unlike a point, a line segmenthas a dimension because there exists adistance between the points. However, aline segment has only a single dimension,length. Just as we use a dot to representa point, we trace along the path of a seg-ment to create a physical representation,call it the line segment, and use the end-points to name it. A tiny line segmentdrawn above the two capital letters tellsus that the figure is a line segment.

A line is a geometric abstraction that in-finitely extends a line segment in both di-rections; it is a straight array of points thathas no endpoint. Figure 4.5 shows sev-eral models of lines. Sometimes it is con-venient to show dots to represent someof the points on the line, but this is notnecessary. If a few of the points areshown, then any two can be used toname the line. For example, the line thathas points A, B. C. D. and E could becalled , and so on. Alowercase letter could also be used toname the line. It is important to use arrow-heads on the model to indicate that theline continues, without end, in oppositedirections. The distance you choose toplace between the arrowheads is unim-portant, because the arrowheads them-selves mean that the line has no definedlength. When you name a line using twoof its points, you must also put arrow-heads on the ends of the tiny line seg-ment above the capital letters. Any twopoints on a line defines a line segment.So, line _ contains Similarly, <contains / and so on.

Sometimes we have a starting pointfrom which we proceed in a single direc-tion. You might think of a beam of light

GD o

o

F

GEOMETRY 135

FIG. 4.5.

shining out across an ocean from a light-house. Because there is a starting point, itdoesn't model a line, and because it con-tinues in a single direction without end, itcan't model a line segment. This subsetof the points on a line that starts at somepoint and continues, without end, in a sin-gle direction, is a ray. To sketch a ray, wefirst identify the endpoint and then an-other point in the desired direction, asshown in Fig. 4.6. As you might expect,we use a tiny ray to name the ray and in-dicate its direction, like . The ar-row points left over and right overBoth of these notations identify the sameray and we read the K first, and then the J.

The beginning point of the ray is placedover the letter that is its representativeand then another point on the ray isnamed. In the case of ra y we couldname the same ray with be-cause each of them starts with A and B,C, D, and E are all points along the path ofthe ray. Ray AE is different from EA be-cause starts at E and passes throughA, whereas starts at A and passesthrough E. Did you notice that AE andEA share and

The models look similar, but you shouldnot confuse the concepts of rays andvectors. A ray has one endpoint and a di-rection, but no length. Vector t, as shown

This ray is KJ or JK

It begins at point K and continues,

without end, in the direction of J.

AB, AC, AD, and AE all name the same ray

B "*^^ C

AB and BA name the same ray

FIG. 4.6.

__ _AB and BA name two rays that havedifferent starting points and go inopposite directions

136 CHAPTER 4

0 11

XY is a vector that extends from 0 to 11 on a number line

FIG. 4.7.

in Fig. 4.7, has a specific direction and aspecific length; in one sense, a vector hastwo endpoints, the initial endpoint and theterminal endpoint represented by an ar-rowhead. To help you tell the difference,you must refer to the context of the prob-lem under discussion.

Perhaps the most difficult of the unde-fined terms in geometry is the idea of aplane. Physical models of geometricplanes are everywhere, for example, thesurface of a table, wall, or piece of paper.One way to visualize a plane is to use awindow glass, because you can seethrough the surface of the glass, it ap-pears to have no thickness. However, justas a line continues, without end, in oppo-site directions, a plane continues, withoutend, in every direction within any two di-mensions forms a plane. Just as geomet-ric points, segments, and lines can berepresented only by physical models, ageometric plane can be represented onlyby real-world flat surfaces. You see, ageometric plane is nothing other than asurface; it has absolutely no depth andtherefore no visible presence in the physi-cal world. A plane is usually representedby a parallelogram, named with a capitalscripted letter, as shown in Fig. 4.8.

Your Turn

1. For each of the following, provide alabeled sketch and explain in your ownwords:

a) Point Qb) Line segment ST

c) Line kd) Plane Re) Ray UVf) Vector u of length 3 and heading to

the rightg) Which of these can be measured?

2. Answer each of the following ques-tions and provide a sketch:

a) How many different lines could yousketch through a single point?

b) How many different lines could yousketch through two points?

c) Is it always possible to sketch oneline through three points?

ANGLES

If two rays that go in different directionsare joined so they share their endpoint,the result is an angle. The common end-

FIG. 4.8.

GEOMETRY 137

point of the two rays is called the vertex ofthe angle. The rays are called the sides, orlegs. Sometimes, for the sake of conve-nience, line segments are used as sidesof an angle, but you should understandthat any line segment is a part of a ray.The best way to name an angle is to usethree points in order— first a point on oneleg, then the vertex point, and then a pointon the other leg (the vertex letter is alwaysthe middle of the three points defining theangle). The symbol for an angle looks likea tiny angle (Z) and sometimes it has atiny arc drawn across it (4)-

The sides of an angle are rays, so weneed not worry about sketching them anyspecific length. This has implications forour use of the word "congruent" and thesymbol that means congruent (=). Con-gruent is a very strong word that meansfigures must be exactly the same shapeand exactly the same size. The sides of anangle are rays, so they extend forever nomatter how long we make them appearto be. This means that we need onlymeasure the rotations of two angles todecide if they are congruent; in Fig. 4.9,

set of all points that are outside the legs.This may not be obvious when you look atan angle, because we draw only a tinypart of each ray. Figure 4.10 shows inte-

D DEF means the angle with a vertex at point E,with points D and F on its legsThe vertex of D HIJ is point I

FIG. 4.9.

ZDEF = ZHIJ because they have thesame measure, even though we have notsketched the rays to look the same.

Because the legs of an angle are rays,extending forever, an angle divides theplane into three distinct parts, the set ofall points that are inside the legs, the setof all points that are on the legs, and the

FIG. 4.10.

rior points, like P, are between the rays,exterior points, like Q, are outside thesides, and angle points, like D, E, and F,are on the legs of the angle.

Angles are commonly described by theirdegree measures. The most common an-gle found in buildings is the right angle,which has a degree measure of exactly 90°and is formed by two lines, rays, or linesegments that are perpendicular (JL) to oneanother. Some other angles are describedby their relation to the right angle. Angleswith degree measures greater than 0° butless than 90° are called acute. Angles withdegree measures greater than 90° but lessthan 180° are called obtuse. If two raysare joined at their endpoint and go in ex-actly opposite directions, then we say thatthey form a straight angle, which measures180°. See Fig. 4.11 for examples ofstraight, obtuse, right, and acute angles.Angles can have more than 180°, but themost commonly used ones in daily life areacute, right, obtuse, and straight.

Two of the angles in Fig. 4.11, ZJMIand ZHMI, can be combined to form aright angle. When this happens, we saythat the two angles complement one an-other. Taken one at a time, these anglesare acute; however, if you ignore theircommon leg, then you can see the rightangle, ZJMH. Another special relation ex-ists when two angles can be combined to

138 CHAPTER 4

FIG. 4.11.

form a straight angle. In Fig. 4.11, ignorethe common leg between ZKMI andZJMI, and you can see the straight angle,ZKMJ. We say that these two angles sup-plement one another.

Vertical angles apply some of the thingswe are have discussed so far. Think of anX and you have two pairs of vertical an-gles. As you look at that X, the top angleand the bottom angle make one pair ofvertical angles. The left and right pair ofangles formed by the X make another pairof vertical angles. Gee, that soundsstrange: Two horizontal angles are calledvertical angles. Do not dismiss that state-ment too quickly, because the emphasisfor vertical angles is how they are formed.Their orientation as far as being vertical orhorizontal has nothing to do with how ver-tical angles are defined. Look at the topand bottom angles in the X. They are bothacute and they have the same measure.The same is true for the left and right an-gles in that X. They are both obtuse andtheir measures are the same. Could verti-cal angles be right angles? If you lookaround as you read this sentence, thenyou should be able to see examples thatwould confirm your conjecture.

Your Turn

3. Use a straightedge to sketch an an-gle. Label a few interior points, exterior

points, and points on your angle. Shadethe part of the interior that is within therays you drew. Would this interior shadingcontinue to expand if you continued therays?

4. Given that name every an-gle in Fig. 4.11; then identify each angleas right, acute, obtuse, or straight. (Thereare at least 10 angles in the figure.)

5. In Fig. 4.11, explain why there is atiny box drawn in one of the angles at theintersection labeled with an M.

6. Explain why we need to use threeletters to identify any angle in Fig. 4.11.

7. In Fig. 4.11, we say that ZHMI andZHMK are adjacent angles. Write a defini-tion of adjacent angles.

8. Write an informal definition for theterm complementary angles.

9. Write an informal definition for theterm supplementary angles.

10. Do you think it is possible for twoangles to be complementary or supple-mentary without being adjacent? If youranswer is yes, then provide sketches thatindicate your thinking.

SIMPLE CLOSED CURVES,REGIONS, AND POLYGONS

A line segment is a straight array of all thepoints between two given points. Youcould choose a path that is not straight,but it would not represent the most directpath between the two points. Any paththat does not pass through any of itspoints more than once is called a simplecurve. As you can see in Fig. 4.12, evensegments can be considered simplecurves, because part of the path fromPoint E to Point F is a segment. Curve CDis not simple because it crosses itself atPoints H and I.

The simple closed curve is a basic ideafor many concepts in geometry. Any

GEOMETRY 139

The path from A to B is simple; the path from E to F is simple;the path from C to D is not simple

FIG. 4.12.

curve that returns to its starting point isclosed, but a simple closed curve returnsto its starting point without crossing anyof its points more than once. It does notmatter whether we use straight segmentsor wavy curves; the important idea is thata region on a plane can only be describedusing a simple closed curve. The pointsbetween the legs of an angle are called in-terior points, but they do not define a fi-nite region because an angle is not aclosed figure. Identifying a finite region isimportant for discussing concepts suchas area, which requires two dimensions(usually length and width). A simpleclosed curve divides the plane into threesubsets; the points that are inside thecurve, the points that are outside thecurve, and the points that are on the curve.In Fig. 4.13, we can identify the exterior

FIG. 4.13.

and interior points for the figures in thetop row, but it would be impossible to dothe same for figures in the second row.

POLYGONS

Simple closed curves created by con-necting straight segments at their end-

points are called polygons. Table 4.1 liststhe names for several polygons. The pre-fix poly- means "many" and the suffix -gonmeans "side," so we could call thesemany-sided figures polygons, or n-gons,no matter how many sides each has.When working with polygons, it is conve-nient to use more descriptive names toavoid confusion. Whereas some namesare familiar, others may not be easy to re-member. It is acceptable to use numbernames for polygons, such as 7-gon and11-gon. In some cases, you may want touse the general number name, n-gon.

Your Turn

11. Explain why Table 4.1 begins witha three-sided polygon.

12. There are several common poly-gons, called quadrilaterals, that have foursides. Use the diagram of the quadrilat-eral family in Fig. 4.14 to answer the fol-lowing questions:

a) What are the similarities and differ-ences among the special quadrilat-erals in Fig. 4.14?

140 CHAPTER 4

FIG. 4.14.

b) What does it mean if there is not anarrow connecting two quadrilater-als?

c) Is every square a rectangle?d) Is every rectangle a square?e) Where does a four-sided figure

shaped like a child's kite fit in this di-agram?

13. Using a piece of nonelastic stringto form a closed loop that is about 2 feetlong, make a large model of each polygonin Table 4.1. What do all your modelshave in common? How do your modelsdiffer as you add sides to create newpolygons?

Naming Polygons

We use letter names for individual n-gonsby listing the vertices. Although it doesn'tmatter where you start, you must moveeither clockwise or counterclockwisearound the figure to name it. Among themany ways to name the two polygons inFig. 4.15, you might call the four-sided

Concave QuadrilateralEKAJ

Regular PentagonEIRAM

FIG. 4.15.

figure EKAJ, JEKA, or AKEJ, and the five-sided figure EIRAM, RIEMA, or AMEIR.

The 4-gon in Fig. 4.15 seems to havebeen poked in at Point A. You couldsketch a line segment connecting VertexJ to Vertex K and it would be completelyoutside the figure. Although concavepolygons such as quadrilateral EKAJ arecommon, we will confine this informal dis-cussion to convex polygons such as pen-tagon EIRAM in Fig. 4.15, for which nostraight segment joining two vertices ex-tends outside the figure. Most of the com-mon algorithms used with polygons as-sume that they are convex. A regularpolygon, like pentagon EIRAM, has all an-gles the same measure and all sides thesame length.

Your Turn

14. Most of our sketches were createdusing Geometer's Sketchpad® and weencourage the use of a dynamic softwareapplication as you do this exercise. Youmay complete the sketches using astraightedge and protractor, but it will bea lot more work. Draw and label polygonswith the given characteristics. Name thepolygons based on the letters at their ver-tices, and based on their characteristics.Choose from these terms, using eachterm only once: acute triangle, equilateraltriangle, isosceles triangle, obtuse trian-gle, parallelogram, quadrilateral, rectan-

GEOMETRY 141

gle, rhombus, right triangle, scalene trian-gle, and trapezoid.

a) Three sides with as many right an-gles as possible.

b) Three sides with as many obtuse an-gles as possible.

c) Three sides with as many acute an-gles as possible.

d) Three sides with no sides the samelength.

e) Three sides with two sides the samelength.

f) Three sides with all three sides thesame length.

g) Four sides with no sides the samelength.

h) Four sides with two opposite sidesparallel and the other two oppositesides not parallel.

i) Four sides with as many right anglesas possible.

j) Four sides with all sides the samelength and as many obtuse anglesas possible.

k) Four sides with opposite sides par-allel and as many acute angles aspossible.

In addition to sides and vertices, poly-gons have other important parts. In Fig.4.16, one diagonal of parallelogram DCBAis shown as a dashed segment. The re-quirements for a quadrilateral to be a par-

allelogram are that opposite sides areequal in l nnth. denoted as 'and , and opposite sides areparallel, denoted byThere are special parallelograms, asshown by the quadrilateral family tree inFig. 4.14, but DCBA is a general parallelo-gram. Trace parallelogram DCBA on apiece of paper and cut it out. Cut along thediagonal from A to C and compare the twopolygons you have made.

If you repeat this experiment with a fewdifferent parallelograms—including rec-tangles, rhombi, and squares—you canform a conjecture about the relations be-tween parallelograms and triangles. Youmight trace DCBA again and repeat theexperiment using diagonal BD. Althoughyou do not need a formal definition of di-agonal of a parallelogram to conduct andunderstand this experiment, you definitelymust know what a diagonal is.

Another important polygonal part is thealtitude, or height, of a figure. In a poly-gon, an altitude can be drawn from anyvertex as long as it is perpendicular to theopposite side. In Fig. 4.17, you can see

FIG. 4.16. FIG. 4.17.

142 CHAPTER 4

that it is not necessary for an altitude tobe inside a figure. Side DA was extendedso Altitude BF from Vertex B could beshown and Side BA was extended soHeight DH from Vertex D could be shown.Sketches that include several auxiliarysegments can be visually confusing. InFig. 4.17, the parallelogram is shown withthick segments, the altitudes as thin seg-ments, and the side extensions as dashedrays. If you find it more comfortable toview an altitude as a vertical segment,then you may want to rotate the page asyou examine the four altitudes of ABCD.Notice that a new point is created whereeach altitude meets the opposite side (orextended side); for example, the heightfrom Vertex B is perpendicular to DA at F.The sides of ABCD are congruent in pairs,thus you will see that the height of thepolygon depends on your orientation asyou view the sketch.

Your Turn

15. Write an informal definition for eachterm.

a) Side of a polygon—How many sidesdoes a 12-gon have?

b) Vertices of a polygon —How manyvertices does a 7-gon have?

c) Diagonal of a polygon —How manydiagonals does a 5-gon have?

d) Altitude of a polygon—How manyaltitudes does a 3-gon have?

CIRCLES

Some people consider a circle to be aspecial polygon—with an infinite numberof infinitely short sides. Others prefer toput the circle in a special category all itsown. Because a circle is a simple closedcurve, it divides the plane into three setsof points (inside the circle, outside the cir-cle, and on the circle). When you used apiece of string to model the polygons inTable 4.1, did your model start to look abit like a circle as you added more andmore sides without changing the length ofthe string? It is easy to see the connectionwith polygons. However, as you look atcircles, such as the one shown in Fig.4.18, you do not see infinitely short sides,only smooth and perfectly roundedcurves, making it easy to argue that cir-cles should have their own category.

FIG. 4.18.

GEOMETRY 143

Sometimes people refer to circularpieces of material as circles—perhaps acoin, poker chip, or other disk. In fact,these items are right circular cylinders;they may be very short cylinders, but theyare not figures in a plane and they havemore dimensions than do circles. A circleis a point and every member of the set ofpoints is the exact same distance fromthe center in the plane. The circular arrayof points is the circumference of the cir-cle. The circle is the ring. The set of pointsinside the ring is not the circle; it is the in-terior of the circle. Similarly, the pointsoutside the ring are not the circle; theymake up the exterior of the circle. A circlehas no substance, so the best way tomodel a circle is with a wire ring. Ofcourse, the wire of the ring has thickness,but it represents a much better model of acircle than a solid plastic disk.

There are several important terms as-sociated with circles, some of which areshown in Fig. 4.18. The diameter is a seg-ment that joins two points on the circleand passes through the center. A radius ishalf a diameter and goes from the centerto any point on the circumference. Achord is a line segment that joins any twopoints on a circle. The diameter is the lon-gest chord of a circle. A few additionalterms are shown in Fig. 4.19. A sector of a

FIG. 4.19.

circle is a pie-wedge region bounded bytwo radii and an arc. A segment is a re-gion bounded by a chord and an arc.

Your Turn

16. Use the information in Fig. 4.18 andFig. 4.19 to write your own informal defini-tions for the following terms:

a) Centerb) Chordc) Circled) Circumferencee) Diameterf) Radiusg) Sectorh) Segment

CONSTRUCTIONS

Classic constructions, completed withnothing more than a compass andstraightedge (not a ruler), are elegant andbeautiful in form. However, the basic con-cepts of construction can be introducedusing paper folding. For this, you need astraightedge, a pencil, and some paper.Waxed paper, tracing paper, or meatpatty paper are best, because these arethin and show the creases of the con-struction well. We have selected a fewstandard geometric constructions for youto try using paper folding. Many more canbe accomplished using this method, al-though the more complex constructionsare actually easier to complete using acompass and straightedge. An introduc-tion to construction using paper foldingwill strengthen your understanding andimprove your level of confidence. Geome-ter's Sketchpad® or another dynamicsoftware application could be used togenerate additional conjectures once youhave mastered the basics.

144 CHAPTER 4

Your Turn

17. Use a straightedge and a sharppencil to score a line segment on your pa-per, labeling the endpoints A and B. Foldthe paper so that A and B coincide andcrease it. What geometric figure have youmade? Place a few points at random po-sitions on the crease and label them C, D,E, and F (be sure that one of the points isat the intersection of the crease with AB).From each of these points, one by one,compare the distance to A with the dis-tance to B. What might you conjecturebased on this construction?

18. Start on a new piece of paper withAB. Place a point C on the segment, notvery near the center. Fold the paperthrough point C so that the parts of AB oneither side of C coincide and crease it.How does this figure compare with theone in Exercise 16? How does it differ?

19. Start on a new piece of paper withAB. Place a C somewhere off AB_and foldthe paper through C so parts of AB lie ontop of each other and crease it. How doesthis figure compare with the ones in Exer-cises 16 and 17? How does it differ?

20. Start on a new piece of paper withAB. Place points C and D on the segment.Using first C and then D, follow the in-structions in Exercise 19. How does eachof these two creases meet AB? Whatconjecture can you make about these twocreases?

21. On a new piece of paper, score anangle on your paper; label the vertex Band the rays t and s. Fold the paperthrough B so Rays t and s coincide andcrease it. What construction have youcompleted? If you used an acute angle,try this again with an obtuse angle—if youused an obtuse angle, try it again with anacute angle. Finally, try it again with aright angle.

22. Using only a compass and straight-edge, construct a triangle and the altitudeto one side of the triangle. Does this applyany of the paper folding work you justdid? (You might want to use Geometer'sSketchpad® or some other dynamic soft-ware for this one.)

THIRD DIMENSION

All the figures we have discussed sofar are planar. They range from a point,which has no dimension, to a seg-ment, which has one dimension, to arectangle, which has two dimensions. Thethird dimension is a vital part of geometry,opening the door to real-world conceptslike capacity. An initial step into the thirddimension is learning to represent thingsthat have length, width, and depth in atwo-dimensional medium—your paper.Following a few procedures won't makeyou an artist, but it will help you createuseful sketches. If you already havemethods for sketching that are morecomfortable, then feel free to use them ortry the procedures presented here.

To sketch a sphere, start with a circleand imagine how the sphere would sinkinto and protrude from the paper. Sketcha solid curve to represent the bulge and asimilar dashed curve to represent the un-seen part of the sphere. The sphere in Fig.4.20 seems to bulge down toward theviewer.

To sketch a right circular cylinder asshown in Fig. 4.21, start with an oval or

FIG. 4.20.

GEOMETRY 145

FIG. 4.21.

football shape and a copy of that shapefor the top and bottom. Sketch line seg-ments forming the left edge and rightedge. When the upper and lower ovalsare vertically aligned, the figure is a rightcircular cylinder. Dash the arc represent-ing the unseen part. If the ovals are notvertically aligned, then the cylinder is nota right circular cylinder. If the segmentsrepresenting the vertical edges aren'tstraight, then the figure seems to wiggle.Right circular cylinders are common ingeometry classes, but all sorts of cylin-ders exist in the world.

To sketch a cone, start with an oval orfootball shape and a point directly overthe center of the shape. Sketch segmentsconnecting the apex (top) with the left-most and right-most points of the footballand dash the unseen curve, as shown inFig. 4.22. If the apex is not directly over

FIG. 4.22.

the center or if the figure on the bottomdoes not represent a circle, then it will notrepresent a right circular cone.

To sketch a rectangular prism, startwith a rectangle. Make a copy of the rec-

tangle a little to the right and a little abovethe original. Sketch line segments to con-nect the corners of the two rectangles,top left to top left, top right to top right,bottom left to bottom left, and bottomright to bottom right. Make the edges thatwould not be seen into dashed segments,as shown in Fig. 4.23.

FIG. 4.23.

Perhaps the most challenging three-dimensional figure to sketch is the pyra-mid. Consider a right hexagonal-basedpyramid. Start with a hexagon that ap-pears regular, but is in perspective, and apoint directly over its center. Sketch a linesegment to connect each vertex of thehexagon with the apex. Dash segmentsthat represent unseen edges, as shown inFig. 4.24.

FIG. 4.24.

Once you gain confidence withsketches such as these, you will be ableto produce other figures. Sketchingthese shapes will help you understandthe connections between polygons andthree-dimensional figures. With practice,you will develop additional techniques

146 CHAPTER 4

for creating even better and more usefulsketches.

Your Turn

23. All of the figures in this discussionwere sketched as if we were viewing themfrom above and in the front. Sketch eachfigure as if you were looking at it from be-low and in the front.

COORDINATE GEOMETRY

In algebra, coordinate geometry is calledgraphing. It is unfortunate that topics inmathematics sometimes are taught in iso-lation, divorced from natural connections.Perhaps coordinate geometry is one areaof subject matter integration you have rec-ognized. You knew about the coordinateplane before your geometry course—youjust called it the x-y system, rectangularplane, or Cartesian plane. We place twonumber lines perpendicular to one another

at their zero points and call the intersectionthe origin of the axis system, as shown inFig. 4.25. This system divides the planeinto four quadrants. Both the x and y coor-dinates of a point are positive in the firstquadrant, the x coordinate is negativewhereas the y coordinate is positive in thesecond quadrant, both coordinates arenegative in the third quadrant, and the xcoordinate is positive whereas the y coor-dinate is negative in the fourth quadrant.Because the coordinates are always listedas an ordered pair, with the x coordinatefirst, every address on the coordinateplane can be clearly provided. The nameof the x coordinate is abscissa and thename of the y coordinate is ordinate; al-though these names aren't used very of-ten, they provide another way to think ofthe ordered pair in alphabetical order.Other concepts, such as slope and inter-cepts, are clarified with the visually orien-tated geometric approach. In algebra welearn about slope by first counting the rise

FIG. 4.25.

GEOMETRY 147

and the run and then learning the slopeformula. We learn about intercepts bylooking for the place where a line crossesthe y axis or the x axis.

Coordinate geometry can be useful inother areas of study, such as statistics.Did you ever think about a histogram as ageometric figure? You need a coordinateplane with a vertical axis and a horizontalaxis to construct a histogram, bar chart,or even a pictogram. You also need a co-ordinate plane if you want a scatter plot.

TRANSFORMATIONSAND SYMMETRY

We will consider rigid transformations,called isometries, because they preservesize and shape. We will not discusstransmography, although these amazingtransformations are the basis for many ofthe special effects for movies, television,and video games. Special effect expertschange the figure point by point to trans-form an image from one appearance toanother. Parts of the image are changedby sliding, rotating, flipping, stretching, orshrinking. Because there is no set factorfor all the transformations, it is often im-possible to recognize any part of the origi-nal image after it has been transformed.

Translations, rotations, reflections, andglide reflections are transformations thatpreserve shape and size and can createimages that have symmetry. As a begin-ning, compare your left hand with yourright hand. Did you find that they are verysimilar, but have opposite orientation?This is an example of symmetry createdby reflection. For a completely differenttype of symmetry, consider a wallpaperpattern that repeats. If you cut out a pieceof the wallpaper and slide it up or down,you will find it matches a new section ex-actly. This is an example of symmetry cre-

ated by translation or sliding. A third typeof symmetry can be seen on fancy carwheels. As a wheel turns, the same imageis seen over and over. This is an exampleof symmetry created by rotation or turn-ing. We will not expect you to create pat-terns or tessellations such as M. C.Escher did, but we do recommend thatyou read about this artist/mathematicianand review some of his famous images,many of which were created using trans-formations. Information can be found athttp://www.mcescher.com.

Translations preserve not only size andshape, but also orientation. When wetranslate (or slide) a triangle with verticesat (3, 4), (4, 1), and (1, 2) four steps to theleft and two steps down, we have a trian-gle that appears identical, except that ithas vertices at f 1, 2), (0, ~1), and (~3, 0), asshown in Fig. 4.26. The two triangles, the

FIG. 4.26.

original and the copy, or image under thetransformation, demonstrate translation-al symmetry. To make the translationclearer, the image is shaded.

Rotations preserve size and shape, butthe orientation changes as we turn the fig-ure about a given point called the center

148 CHAPTER 4

FIG. 4.27.

of rotation. The triangle in Fig. 4.27 hasbeen rotated 130° clockwise. The twodashed figures indicate the rotation. Theimage, or copy, has been shaded. Thecenter of rotation may be near or far fromthe figure; it may even be on or inside thefigure as it is in the quadrilateral in Fig.4.27, which shows a rotation of 80° coun-terclockwise about a vertex of the figure(the image has been shaded).

An axis of symmetry, reflection line, ormirror is required to reflect a figure. Aswith the point of rotation, this axis may befar from the figure, near the figure, on thefigure, or even inside the figure. In Fig.4.28, the mirror is indicated by a thick line

Line of reflection

FIG. 4.28.

segment, over which the figure on the lefthas been flipped to create a congruent,but reversed image on the right. You cansee that, whereas the orientation of the

shaded image is different from that of theoriginal figure, size and shape have beenpreserved. Additionally, the original hexa-gon and its image are equally distant fromthe line of reflection.

The last transformation, and perhapsthe most fun one, is the glide reflection. InFig. 4.29, a pattern of footprints was cre-

FIG. 4.29.

GEOMETRY 149

ated from a single original sketch. Foreach image, the original is either trans-lated or both translated and reflected.Can you figure out which is the originalsketch? Do you think each of the imagesis congruent to the original? Can you findthe mirror line?

Your Turn

24. Consider the following capital let-ters of our alphabet:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

a) Determine which have reflectionalsymmetry about a vertical mirrorwithin the letter.

b) Determine which have reflectionalsymmetry about a horizontal mirrorwithin the letter.

c) Determine which have rotationalsymmetry within the letter.

d) Do any of the letters have more thanone type of symmetry within the let-ter?

e) Do any letters have all three types ofsymmetry within the letter?

f) Do any have no symmetry within theletter?

CONCLUSIONS

We have highlighted some important top-ics in geometry. It will be up to you to de-termine which topics in geometry are likecomfortable old clothes and which topicsare more like scratchy new wool sweat-ers. You need to work on the scratchytopics until you soften them up.

This page intentionally left blank

5Measurement

FOCAL POINTS

• Terminology. Attributes• Systems. Units, Tools (instruments), and Preci-

sionLinear MeasureAreaVolumeCapacityWeightTimeMoneyTemperatureAngle MeasureDimensional Analysis

What is the first thing that comes to yourmind when you hear the word "measure-ment"? Do you think about how manygallons of gasoline you had to buy yester-day, how many miles you must travel toget to class, how many minutes you willbe in class, how many dollars will be inyour next paycheck, how many ounces ofpotato chips are in a bag, how warm it willbe today, or even how many square yardsof carpet you had to buy to cover yourbedroom floor? Perhaps you consideredone or more of these without thinking ofthe word "measurement."

Measures are important in our world andwe are so comfortable with the conceptsthat we often accept and understand theinformation without much thought. Have

you ever considered how much the gaso-line in your tank weighs, how many timesyour left front tire revolves as you drive toclass, how many seconds you spend inclass, what your net salary per hour is, orhow temperature is related to humidity?These are all valid measures, you justdon't think of them as relevant to yourdaily life.

The measurement concepts that are sofamiliar to us have developed over manyyears. Primitive people used rudimentarymethods to measure distances for theirforaging or hunting boundaries, but therewere no standard units. Figure 5.1 repre-

FIG. 5.1.

sents ancient Egyptians, who lived alongthe Nile River and used standard linearmeasures and elaborate survey methodsto mark and remark their crop boundariesas the river flooded and receded, takingand restoring cropland. The Babylonianslooked to the stars to measure the lengthof a year; their calculations were surpris-ingly accurate and their methods of astro-nomical observations are the basis forcalculations today.

151

152 CHAPTER 5

TERMINOLOGY

Many early measurement techniques werebased on human anatomy, as shown inFig. 5.2, making standardization problem-

FIG. 5.2.

atic. For example, the cubit (length from el-bow to longest finger tip) was a linear mea-sure common to several cultures, but eachdefined it differently. The Egyptian RoyalCubit, used to build the pyramids, was20.63 inches, whereas the Greek OlympicCubit was 18.93 inches.

An argument for the inherent fairness ofa standard system of measures can bemade using a tale of two tailors. Tailorsonce measured cloth using the distancefrom the tip of one outstretched arm tothe center of the chest as a unit. Whowould want to pay more for a suit just be-cause a tailor, like the one in Fig. 5.3, had

FIG. 5.3.

short arms? In 1130 AD, with theregulation of trade in mind, King Henry I ofEngland decreed the distance from thepoint of his nose to the end of his thumbto be the lawful yard. Later, King Henry VIIcommissioned a bronze bar, 3 feet long,and ruled that to be the standard yard.

As we shift our attention from the ev-eryday world to the mathematics contentin Pre-K-16 classrooms, we must ac-knowledge that the terminology associ-ated with even basic measurement con-cepts is often approached differently inthe classroom. Dichotomous terminology,such as biggest and smallest, can bedoubly confusing, for certainly the biggestkitten is not as large as the smallest ele-phant. Comparison can be either quanti-tative or qualitative and a brief review ofsuch concepts and the related terminol-ogy is necessary at this time.

Your Turn

1. Make a list of all the measurementwords you can think of in 30 seconds.Share your list with classmates.

2. Make a list of all the dichotomousmeasurement words you can think of in30 seconds. Compare your list with thoseof your classmates.

3. Find a subtle example of confusingdichotomous terminology using the termslarger and smaller.

4. When can the smallest be largerthan the biggest? When can the biggestbe smaller than the smallest?

ATTRIBUTES

Consider the different attributes pos-sessed by a measurable object. When wefind the dimensions of an object, wechoose a unit of measure, but we alsomake a qualitative decision about the at-tribute we want to consider. A pound of

MEASUREMENT 153

feathers weighs the same as a pound ofchocolate, but they do not have the samevolume. A backyard might be completelyenclosed using 250 feet of fencing, butthe cost will differ according to the heightand style of fencing that is selected. Thechoices we make, including those relatedto measurement, are based on experi-ences and prior knowledge.

Your Turn

5. When you say that a beach ball isbigger than a bowling ball, what attributesare you ignoring?

6. What attributes, other than the num-ber of square yards required, influencethe cost of carpeting?

7. Attribute Blocks™ help childrenlearn the terminology and concepts of dif-ferences. The 60-piece set includessquares, rectangles, circles, triangles,and hexagons, in two sizes, three colors,and two thicknesses. List at least 10 dif-ferent ways by which Attribute Blocks™could be sorted.

SYSTEMS

Some metric units, such as kilowatt ofelectricity, appear regularly in the UnitedStates. The system used in general com-munication has many names: inch/foot/pound, U.S. Standard, U.S. Common,and English Customary. Ironically, we en-counter both systems and get around us-ing either name by saying things like acarton of milk, which can be taken to be aliter or a quart. On top of that, measures inboth units are listed on many containers.For example, a bottle of water shows 16.9FL OZ, 1.06 PT, and 500 mL Althoughboth the metric and inch/foot/pound wereadopted as official measurement systemsin the United States long ago, manyAmerican adults prefer to use the inch/

foot/pound system. President Fordsigned a voluntary Metric Conversion Actin 1975. Soon after that, metric speedlimit signs appeared on our interstatehighways, but they have long since disap-peared. Yet, the dual system lives on. Me-chanics must still maintain two completesets of tools in order to repair cars. Manycars have speedometers that list bothsystems. Digital gauge systems in manycars allow selection of systems. Meas-uring cups are dual-scaled, with ouncesand milliliters.

Few examples of the problems we ex-perience with this dual system are ascostly as the crash of the Mars ClimateOrbiter in 1999. Findings by NASA's inter-nal review indicated that the spacecraftwas not placed in the proper orbit be-cause one team of scientists was usingthe inch/foot/pound system and the otherteam was using the metric system. Thisfailure of communication between thespacecraft team in Colorado and the mis-sion navigation team in California resultedin faulty data for critical maneuvers, de-stroying an opportunity for scientific gainand costing American taxpayers millionsof dollars. Such problems are likely tocontinue until we all learn to be cognizantof the units being used, no matter the sys-tem. This is significant because theUnited States is the only industrialized na-tion, and one of a few nations in theworld, that is not using the metric systemof measure.

The inch/foot/pound system of meas-urement was the result of attempts tostandardize measures. In the 14th century,King Edward decreed that three kernels ofbarleycorn, taken from the center of theear, would be an inch or one twelfth of afoot. If we accept the Greek legend thatHercules' foot was the original basis forthis measure, it must have measured 36kernels of barleycorn. Learning to use the

154 CHAPTER 5

inch/foot/pound system is no easy task!We use words such as ounce or quart forboth liquid and dry measures, and thereare confusing relations between and withinthe units used for length, weight, and ca-pacity as shown in Table 5.1.

TABLE 5.1Commonly Used Inch/Foot/Pound Abbreviations

Quality

length

capacity

weight

Measure

inchfootyardmileteaspoontablespooncuppintquartgallonouncepoundton

Symbol

inftydmitTcptqtgalozIbT

Comparisons

12 in = 1 ft3 ft = 1 yd1760 yd = 1 mi5280 ft = 1 mi3 t = 1 T16 T= 1 c2 c = 1 pt2 pt = 1 qt4 qt = 1 gal

16 oz = 1 Ib2000 Ib = 1 T

The metric system grew out of a needfor more coherent measurement and theInternational Bureau of Weights andMeasures was established in 1875. Dur-ing the French Revolution, French scien-tists led the way and the system we call SI(le systeme international d'unites) is theresult of many decades of effort to estab-lish a system that uses Base 10. All metricmeasures within a unit concept are re-lated using powers of 10, so this is thesystem preferred by scientists around theworld. Once you master the units for lin-

ear measures, mass, and capacity theprefixes will begin to make sense. We useGreek prefixes, such as tera (1012), giga(109), mega (106), kilo (103), hecto (102),and deca (101) for measures that aregreater than the unit and Latin prefixes,such as deci (10~1), centi (10~2), milli (10~3),micro (10-6), nano (10~9), and pico (10~12),for measures that are smaller than theunit. Of course, there are many orders ofmagnitude greater and smaller thanthese, but we commonly use only the fewshown in Table 5.2.

The scientists developing the metricsystem had the advantage of startingfrom scratch. In 1791, the French Acad-emy of Sciences defined the meter as oneten millionth of the length of the meridianthrough Paris from the north pole to theequator. Whereas the original prototypemissed this target by 0.2 of a millimeter(flattening of the earth due to its rotationcaused a miscalculation), this length wasaccepted as the standard. The definitionof a meter has changed several times andin 1983 the definition was standardized tobe the distance traveled by light through

a vacuum during second.299,792,458

Note that only the definition of a meterchanged, not the length. This intercon-necting system has helpful relationsamong the measures for length, mass,and capacity. At sea level, one milliliter ofpure water at 4° Celsius weighs one gramand has a volume of one cubic centime-

TABLE 5.2Commonly Used Metric System Prefixes

Prefix

kilohectodekaunitdecicentimilli

Value

thousandshundredstensonetenthshundredthsthousandths

Numerical

103 = 1000102 = 100101 = 1010° = 110-1 = 0.110-2 = 0.0110-3 = 0.00

Linear

kmhmdammdmcmmm

Capacity

kLhLdaLLdLcLmL

Weight

kghgdaggdgcgmg

MEASUREMENT 155

ter. Because our ability to precisely meas-ure mass has improved over the decades,we know that this relation is not perfect,but rather an incredibly close approxima-tion.

UNITS, TOOLS (INSTRUMENTS),AND PRECISION

Linear Measure

Almost anything can be used as a linearmeasuring unit. You could measure thelength of this book in paperclips or eras-ers. A standardization problem would re-sult if you used large paperclips or newerasers and your classmate used smallpaperclips or partly used erasers. Meas-uring with nonstandard units is an inter-esting activity and makes the point thatstandardization is important. It also re-minds us that it is critical to always pro-vide both parts of the measure—howmany and what unit.

Once you have moved beyond non-standard measuring devices and pickedup a ruler, precision becomes an issue.Whereas it may seem natural to beginmeasuring at zero, it isn't necessary. Youcan place the length you want to measureanywhere along the rules. The smallerreading subtracted from the larger read-ing will be the length. Of course, if youplace one end of the length to be meas-ured at the origin, the smaller reading willbe zero (this process is called predeter-mining the measure). Be careful, however,because on some rulers, the origin is atthe edge, whereas on others it is indentedslightly. The first issue of precision is toknow your tool. After that, the degree ofprecision required will be determined orassigned by the task at hand. For sometasks, you may need to be accurate towithin one half or one fourth of an inch,

but such estimates might be consideredsloppy for others. Many rulers also pro-vide graduated marks for eighths and six-teenths of an inch, with halves indicatedby the longest mark and sixteenths indi-cated by the shortest. Some tools are cal-ibrated to even smaller parts of an inch.We have used inch/foot/pound in this dis-cussion, but similar comments could bemade about metric and we encourageyou to learn both systems.

Your Turn

8. Measure the height of your coffeecup using anything except a ruler. Wouldyou want to use a new pencil as your unit?

9. Measure the length of a room inpaces. (The Romans figured 1000 paceswere a mile and defined a pace as twosteps or about 5 feet.) Would it have madesense to use a new pencil as the unit?

10. Trace your hand and wrist. Howmany measurements can you discuss inreference to your tracing? Measure thetrace in inches, then in centimeters.Which unit was easier to use? What is thefinest degree of precision you can reachwith your ruler?

11. Measure the line segments in Fig.5.4 to the finest degree of precision avail-

FIG. 5.4.

able to you in the inch/foot/pound systemand the metric system. Measure the dis-tances AB, AC, AD, AE, BC, BD, BE, CD,CE, and DE.

12. What is the total length of a cornershelf that is 5 yd 1 ft 9 in along one walland 4 yd 2 ft 7 in along the other wall? Becareful as you regroup the measures inyour solution.

156 CHAPTER 5

FIG. 5.5.

13. What is the total length of a cornershelf that is 5 m 10 cm 5 mm along onewall and 4 m 44 cm 5 mm along the otherwall?

14. Use a piece of string to measurethe distance an ant would walk aroundthe outer edge of the leaf in Fig. 5.5. Com-pare your string measure to a ruler to findthe distance.

Area

A concrete way to review the concept ofsurface area is to make square tiles for

measuring area. You should make squaretiles using different colors of paper forsquare inch tiles and square centimetertiles. A square inch is much larger than asquare centimeter, so you need fewersquare inch tiles to cover a surface.Making both types of tiles will highlightthe idea that you have to decide what unityou are going to use as you measure, re-inforcing the idea that the units are just asimportant as the numbers in measure-ments. You may not be able to accuratelytile irregular shapes, so you will need toestablish a few estimating procedures.

A similar way to measure area involvesprinting square grids on transparencysheets. When you place the grid on a sur-face of an object, an outline can be tracedand the squares—and parts of squares—inside the outline can be counted to esti-mate the area of the object. The leaf inFig. 5.6 has an approximate area of 10square cm or 1.25 square in.

The right angles so evident in squaretiles or grids remind us that the heightsand bases used in algorithms must beperpendicular to one another. You may

FIG. 5.6.

MEASUREMENT 157

find that the familiar area algorithms aremore obvious after you have explored theconcept of area in this way.

Your Turn

15. Make a set of unit square tiles formeasuring area using any unit you like.Find an everyday item with a flat surfaceand estimate the area of the flat surfaceby tiling. You may need to use partial tiles.How do you know this is a good esti-mate?

16. Create a one foot square showing144 square inches. Create a 1 yardsquare using 9 square foot tiles.

17. Find the perimeter of each region inFig. 5.7 using your string and a centimeterruler; then find the area of each by count-ing and estimating the number of squarescovered by each.

18. At some point, the string and gridpaper are replaced by measures that leadto formulas for perimeter and area. Fromyour previous experience with algorithms,identify these common formulas and tellthe meaning of each letter and symbol.

FIG. 5.7.

158 CHAPTER 5

19. Use the appropriate algorithms inExercise 18 to check your measures andestimates in Exercise 17.

20. Which algorithms in Exercise 18were not helpful in completing Exercise17?

21. Why is there no algorithm in Exer-cise 18 for finding the perimeter of thetrapezoid or the triangle? Do you thinkperimeter algorithms are needed?

As you were completing Exercises 17and 18, did you notice that, in Fig. 5.7, thetriangular region has exactly one half asmuch area as the rectangular region? Acloser look at the area algorithms forthese two figures confirms this relation,because the terms length and base orwidth and height are used interchange-ably. You might have noticed that there isnot a separate algorithm for the area en-closed by a parallelogram. What happensif you use the slant height (not the heightbecause it is the slanted side) of the par-allelogram to find the area? You could re-arrange the squares enclosed by the par-allelogram to form a rectangle (which, is aparallelogram with right angles). For theparallelogram in Fig. 5.8, think of makinga cut through Vertex A that is perpendicu-lar to Side DC. The resultant right triangleis moved to the other end, creating a rect-angle. You don't have to chop up a regiondefined by a parallelogram to find its area.However, it is handy to use the termsbase and height instead of length andwidth. Whereas either of the two algo-rithms A = I • w and A = b • h can be used,

FIG. 5.8.

the second reminds us not to use theslant height when dealing with area.

The algorithm for finding the area of atriangular region works even if the baseand height (or altitude) of the triangle aredetermined by a parallelogram that is nota rectangle, such as the triangular regionshown in Fig. 5.9. For triangular region

GHJ, the base is determined by JH andthe altitude is determined by the dottedperpendicular from G to K. Even thoughthis altitude is completely outside of theregion defined by triangle GHJ, it is theheight of the triangle.

Consider the triangular region in Fig.5.10, in which the altitude from Vertex Nto Side MP is provided. The decision touse NT as the height and MP as the base

MEASUREMENT 159

FIG. 5.10.

is arbitrary. Any one of the three sidescould be used as the base, but you wouldneed to use the altitude to that side as theheight. The region in Fig. 5.10 is on a gridso that you can count the squares to esti-mate the area, then use a centimeter rulerto measure the two perpendicular seg-ments (base and height), and use the for-mula to confirm your estimate.

We included a circular region in Fig.5.7, even though the formulas for circum-ference and area are difficult to discoverbased on measuring with string or count-ing squares. Additionally, the two circlealgorithms look very much the same,each has two constants, n and 2, and thevariable r. The only difference in appear-ance is that the 2 is a factor in the circum-ference formula and an exponent of thevariable in the area formula. Just remem-ber that area is measured in squares and

that r2 will help you remember which for-mula is for area.

22Did you use 3.14 or — for 71 in Exercise

17? Either approximation is adequate formost Pre-K-16 applications. New com-puters are often used to determine n tomore and more decimal places. As wewrite this, 51.5 billion decimal places forTI have been determined, by the time youread this it will be many more. Still, thedecimal number for n does not terminateor repeat. This continues to confirm that7i is an irrational number. If you used acalculator as you completed the exer-cises, then perhaps you used the n but-ton. Did you notice how many decimalplaces of TI were involved? Some calcu-lators provide only a few, and others pro-vide 8 or 12 decimal places in this ap-proximation.

160 CHAPTER 5

A short experiment will help you under-stand 7i. Cut a piece of string to measurethe diameter of the circle in Fig. 5.11. How

FIG. 5.11.

many times must you use this measure totrace the circumference of the circle? Itisn't easy to be accurate with a piece ofstring, but this experiment will strengthenthe idea that n is the ratio of the circum-ference of a circle to the diameter of thecircle. Why not take the string around andacross the top of a can?

The area of any polygonal region canbe determined. We have standard formu-las for parallelograms, trapezoids, and tri-angles, and any polygonal region can besectioned into parts for which we can usethese standard formulas. The pentagon inFig. 5.12 has been sectioned into trian-

FIG. 5.12.

gles ABD, BCD, and DEA. If the polygonis not regular, we would want to use asectioning procedure to determine thearea of the region. Of course, we aren'tlimited to using only triangles, but they

provided a convenient sectioning of pen-tagon ABODE.

The area formula from Exercise 18,

, is an extension of the idea of di-

viding polygons to find their area. If apolygon is regular, such as the hexagon inFig. 5.13, then it can be sectioned intocongruent triangles that have a commonvertex at the center of the polygon. Thebase of each triangle is a side of the poly-gon and, because the polygon is regular,the bases have the same measure. Thecommon vertex is at the center of thepolygon, thus the altitudes of the triangleshave the same measure. We call thiscommon measure the apothem of thepolygon and it is shown as a dashed seg-ment in Fig. 5.13. Two of the three vari-

FIG. 5.13.

ables in the formula have been explained,a stands for the length of the apothem (al-titude of the triangles) and s stands for thelength of a side (base of the triangle). Thethird variable, n, tells the number of trian-gles that are needed, in this case 6.

Your Turn

22. Use the formula, A = -ans, to find

the area of a region defined by a regularoctagon with sides of length 4 cm and anapothem of length 4.828 cm.

MEASUREMENT 161

Volume

Just as area is directly measured using atwo-dimensional model, volume can bemeasured using a three-dimensionalmodel. For example, how many rolls ofquarters will it take to exactly fill your sockdrawer? You might decide that you needto break your unit roll into 40 subunits ordisks in order to completely fill the spaceand obtain a reasonably close approxi-mation of the volume. Perhaps you wouldlike a more convenient unit, such as acube. How many sugar cubes will it taketo fill your coffee mug? If your mug is acylinder rather than a right, rectangular-based prism, then you will find that youmust develop (or recall) some estimationstrategies. As you stuff the cubes in thecup, some of them will be out of sight.

In our discussion about area, we talkedabout covering a region with unit squares.For volume, we could use a unit cubewhose face is the same size as the unitsquare that was used to discuss area.Suppose a rectangle is 8 units long and 3units wide. From the area work, the rec-tangle would be covered by exactly 32 unitsquares. We could place a unit cube oneach of those squares, as shown in Fig.5.14, and now a wondrous thing has hap-

FIG. 5.14.

pened. That rectangle we used to discussarea is still visible (the tops of the cubes),but there is now a depth factor as well.Counting the cubes, rather than the topface of each cube, tells us that the figurehas a volume of 32 unit cubes. We have alength of 8 units, a width of 4 units, and a

height of 1 unit, giving a total of 32 unitcubes. If a second layer of cubes is placedon top of the first, then the length is still 8units, the width is still 4 units, but theheight is now 2 units, and now we haveused 64 unit cubes. A pattern is emerging:

32 = 8 x 4 x 1

64 = 8 x 4 x 2

The model could be made taller by fol-lowing this pattern of adding a layer of 32unit cubes for each unit increase inheight. The prism in Fig. 5.15 has a

FIG. 5.15.

height of 3 units and a volume of 96 unitcubes. The top faces of the models inFig. 5.14 and Fig. 5.15 have areas of 32square units, but the volume in Fig. 5.15is three times the volume in Fig. 5.14.The pattern defined by this layeringmeans that the volume is found by multi-plying the area of the top face by thenumber of layers. A familiar formula maycome to mind: V = I x w x h, where thelength times the width is the area of thetop face. The terms length, width, andheight can be replaced with other terms,such as base, height, and depth, but theidea of three dimensions is the critical is-sue. As a matter of fact, we can glue thecubes together and stand the prism on adifferent face, as shown in Fig. 5.16. Nowthe area of the top face is 3 x 4 = 12 unitsquares and the prism is 8 layers tall, butthe volume is still 96 unit cubes. Movingthe prism around does not change the

162 CHAPTER 5

FIG. 5.16.

measures, so it doesn't matter what wecall the three measures.

Figure 5.17 is reminiscent of the figureswe used while discussing the formula for

FIG. 5.17.

the area of a triangle. In this case, the areaof the top face of the prism is 32 unitsquares and the area of the triangleformed by slicing straight down throughtwo opposite corners is 16 unit squares.You might suspect that there is a similarconnection between the volumes of therectangular prism and the triangularprism—that the volume is halved whenthat cut is made. If we use the idea of mul-tiplying the area of the top face by thenumber of layers, the suspicion is con-firmed. Because the area of the top face ishalved, the volume is halved. Does thismean that we need to write a new formula

each time we use a prism with a differenttop face? Well, each formula will requireonly the area of the top face and theheight. Consider the skew trapezoidalprism in Fiq. 5.18 and think about how we

FIG. 5.18.

can find the volume. The top face is a trap-ezoid, so we need that area formula andthe height:

Your Turn

23. Build a cube using 27 unit cubes.Use any unit cube, even a sugar cube.

a) What are the linear measurementsassociated with this cube?

b) What are the area measurementsassociated with this cube?

MEASUREMENT 163

24. Use 24 cubes to answer the follow-ing:

a) How many different rectangularprisms can you form using the 24unit cubes?

b) Do all of the prisms have the samevolume?

c) Use your prisms to justify the vol-ume formula, V = Iwh, for rectangu-lar prisms.

d) Add the areas of the six faces to de-termine the total surface area foreach of your prisms. Do they all havethe same surface area?

set, we have provided several formulaswith sketches. As you work through theexercises, see if you find connectionsamong the formulas. We have limited ourdiscussion to some degree, because theexercises involve figures with regularbases and perpendicular sides. We knowthere are volumes where the bases willnot be regular polygons and the sides willnot be perpendicular to the base, as thesketches in Fig. 5.19 show. However, we

25. Consider a solid cube that is 6 unitcubes in height, width, and depth. If youpaint the six faces of this large cube, eachunit cube may have no, one, two, or threepainted faces, depending on its locationin the cube.

a) How many of the unit cubes willhave paint on at least one face?

b) How many of the unit cubes willhave only one face painted?

c) How many of the unit cubes willhave two faces painted?

d) How many of the unit cubes willhave three faces painted?

e) How many of the unit cubes willhave 4, 5, or 6 faces painted?

f) How many of the unit cubes willhave no faces painted?

Rectangular prisms are handy for ex-ploring the concept of volume because, ifyou stay with unit measures for thelength, width, and height, as we havedone, you can build convenient models.In the real world, volume won't always bethat convenient. However, if you remem-ber that the height must be perpendicularto the base, the formulas should not betoo confusing. In the following exercise

FIG. 5.19.

feel that the exercises we have providedare sufficient and that you will be able tolocate formulas for nonregular volumeswhen the need arises.

Your Turn

26. The formula for the volume of aright circular cylinder is V = Tii h, or thearea of the circular top face times theheight. Use the information in Fig. 5.20 tofind the volume of the cylinder.

FIG. 5.20.

164 CHAPTER 5

27. The formula for the volume of a

right circular cone is .Use the in-

formation in Fig. 5.21 to find the volume ofthe cone.

FIG. 5.21.

28. How do the volume formulas in Ex-ercises 26 and 27 compare?

29. To find the volume of a regular rightpolygonal prism, find the area of the topface and multiply by the height. For this

exercise, use the formula V = -asnh to

find the volume of the right hexagonalprism in Fig. 5.22.

FIG. 5.22.

30. The volume of a pyramid also de-pends on the number of edges on the

base. Use V = -s2h and the information in3

Fig. 5.23 to find the volume of the rightsquare pyramid.

Capacity

Volume, or capacity, is not always meas-ured in cubic units. When was the last time

FIG. 5.23.

you stopped by a service station to buy acubic yard of gasoline? The word "gallon"was derived from an Anglo French word,galon, and was standardized by the Eng-lish in 1215 AD for measuring grain andwine. A gallon represents 213 cubic inchesof liquid. When you must consider capac-ity using the inch/foot/pound system, de-nominate numbers are useful in dealingwith inch/foot/pound units. Consider thisexample of an addition problem that in-volves regrouping cups to pints, pints toquarts, and quarts to gallons:

Several regroupings are required tocomplete this exercise. In the first line be-low the vinculum, you see the sum, 14gallons, 5 quarts, 2 pints, and 2 cups. Inthe second line, the 5 quarts have beenregrouped to 1 gallon 1 quart, the 2 pintshave been regrouped to 1 quart, and the 2cups have been regrouped to 1 pint. Inthe third line, the 1 gallon from the quartscolumn has been added to the 14 gallons,the 1 quart from the pints column hasbeen added to the 1 quart, and the 1 pintfrom the cups column has been moved tothe pint column. The result of all this re-grouping is an answer of 15 gallons, 2quarts, and 1 pint.

The capacity unit for the metric systemis Liter. Perhaps the most well-known

MEASUREMENT 165

metric measure in the United States is 2Liters, which is a little more than half agallon. Many household measuring cupshave inch/foot/pound on one side andmetric measures on the other. On thehandy 4-cup measure, you will find thatthe 1 L or 1000 ml_ mark is a bit higherthan the 4-cup mark, 750 mL is a bit morethan 3 cups, 500 mL is a bit more than 2cups, and 250 mL is a bit more than 1cup. As any cook will tell you, the differ-ences in these measures have little effecton a finished meal. You just need to prac-tice using the other side of the measuringcup to become comfortable with eithercapacity measure. Any measure is only asaccurate as the measurer.

Weight

Do you know why we speak of weightwhen using the inch/foot/pound and masswhen we use the metric system? In phys-ics, mass is a measure of the inertia of abody and weight is mass multiplied bygravity. This implies that a body has noweight without gravity, whereas the massof a body remains constant—with or with-out gravity. Because we do most of ourweighing here on earth, we often use theterms mass and weight interchangeably.The best experience we can provide in thisarea is to suggest that you weigh lots ofthings using a simple balance. If you haveaccess to a precision instrument and amass set, then it is fun to measure exactlyhow many grams that special gold chain orbracelet actually masses. Again, denomi-nate numbers can be used to simplifyarithmetic operations that involve regroup-ing weight units in the inch/foot/poundsystem. For example, find the differencebetween the weight of a one ton, six hun-dred pound, fourteen ounce car and a twoton, one hundred pound, nine ounce car:

1 ton 2099 pounds99 pounds 25 ounces

2 tons 100 pounds 0 ounces-1 ton 600 pounds 14 ounces

1499 pounds 11 ounces

To complete the subtraction, 100 poundsneeds to be regrouped to 99 pounds and16 ounces so we can subtract 14 ouncesfrom the 9 ounces plus the 16 ounces. Ina similar manner, 2 tons needs to be re-grouped to 1 ton and 2000 pounds so wecan subtract 600 pounds from the 99pounds plus the 2000 pounds.

Your Turn

31. Name everyday items that youcould use to approximate each of the fol-lowing: 1 g, 1 kg, 1 oz, 1 Ib, 1 ton

32. What is the sum of 8 tons 1700pounds 13 oz and 5 tons 980 Ib 6 oz?

The metric unit for mass is the gram. Abar mass set may contain bars for 1 g, 2g, 5 g, 10 g, 50 g, 100 g, 500 g, and 1 kg.A gram is also defined in terms of one cu-bic centimeter of pure water at sea level inideal conditions at 4° Celsius, so you canimagine that it represents a very smallmass. When buying groceries, such ascoffee, it is handy to know 500 g is ap-proximately 1 pound. The 500 g of coffeewould be about 2 cups.

Time

Do you have an intuitive understanding oftime? How long is a minute when some-one says, "Wait a minute." What does asalesperson mean when asking, "May Ihave a minute of your time?" Using awatch or clock that shows the passage ofseconds, close your eyes and see if youcan tell when exactly one minute haspassed—no fair counting or peeking! Try

166 CHAPTER 5

this little experiment with your friends ofdifferent ages.

Some people seem to struggle withlearning to tell time using an analog clock.You may argue that it isn't important totell time with an analog clock: After all,who has bothered to learn to use a sun-dial recently? The skills used in learning totell time with an analog clock will be use-ful. A clock face with 60 divisions helpswith the concept of elapsed time. Con-sider a delicate surgical situation, whereeven seconds count. You might need todetermine how much time elapses be-tween 11:42:51 AM and 2:07:08 PM. Aconvenient way to complete this as asubtraction problem is to use the military24-hour clock, which keeps counting upfrom 1200 hours, or noon, to 2400 hours,or midnight. In our example, 11:00 in themorning is just 1100 hours, but 2:00 in theafternoon is 1400 hours:

Using denominate numbers reminds usthat regrouping 1 hour results in 60 min-utes and regrouping 1 minute results in60 seconds. In this example, 1 minutewas regrouped to allow the subtractionof 51 seconds, then 1 hour was re-grouped to allow the subtraction of 42minutes.

For everyday problems involvingelapsed time, we usually count or esti-mate. To determine the elapsed timefrom 11:45 AM to 2:15 PM, you couldcount the 15 minutes until noon, then the2 hours and 15 minutes after noon, anddetermine that 2 hours and 30 minuteselapsed.

Your Turn

33. Explain why you would know that aclock with one hand pointed directly atthe 9 and the other hand pointed directlyat the 3 was broken.

34. What other measure uses Base60? Is there a connection between thesetwo measures?

35. Use denominate numbers to deter-mine how much time elapses between9:24:13 AM and 1:15:08 PM? (Don't for-get that, on the 24-hour clock, you canuse 13:00 for 2:00 PM.)

Money

How did you learn about money? Do youeven remember? Most people seem tohave picked up the concepts of our mon-etary system before entering school.Some confusion is possible when chil-dren or adults first encounter our systemof coins. Shel Silverstein's poem, shownin Fig. 5.24, provides a humorous re-minder of this confusion.

A few children may prefer a shiny nickelover a dull old quarter, but we hope thatmost will prefer the new golden dollar toany other coin. Even with our variety ofsize and value in coins, it can be seen thatour monetary system is an application ofBase 10: penny (10°), dime (101), dollar(102) ten dollar (103), hundred dollar (104).

Your Turn

36. List as many different ways as youcan think of to use U.S. coins to make adollar.

37. How much change should you re-ceive if you have a $10 bill and the amountyou must pay is $4.15? How many differ-ent ways might you receive the change?

MEASUREMENT 167

(Please remember that 585 pennies mightnot be reasonable, but is possible.)

Temperature

We know when it is too warm or too chilly.There are two commonly used scales tohelp us describe temperature. The Fahren-heit scale is probably the one with whichyou are most familiar, but the other is theCelsius scale. Another scale, often usedby scientists, is called Kelvin. Anders Cel-sius, the Swedish astronomer, suggestedusing a Base 10 scale comparing twocommon fixed points, the melting point ofthawing snow and the boiling point of wa-ter. He wanted to use 100° for the freezingpoint and 0° for the boiling point. His scale,turned upside down, is what we use today,and the scale is named after him.

Figure 5.25 shows the two scales sideby side, with the boiling and freezingpoints lined up. There are formulas for

converting between Fahrenheit and Cel-sius, but many thermometers have bothscales. Although using the conversionformulas is good practice for algebra, youmay want to just read the scale you needand skip the conversion.

Your Turn

38. Using Fig. 5.25, find the boilingpoint and the freezing point of water atsea level in degrees Fahrenheit and de-grees Celsius.

39. Using Fig. 5.25, determine thecomfort range for humans in degreesFahrenheit and degrees Celsius.

Angle Measure

About 2000 years ago, the astronomerClaudius Ptolemy divided a full rotationinto 360 equal parts, which we now calldegrees. The protractors you have usedrepresent only half a rotation and therefore

FIG. 5.24. From "Where the sidewalk Ends"(p.35) by S. Siverstein, 1974, San Francisco: EvilEye Music, INc.

168 CHAPTER 5

FIG. 5.25.

indicate 180°, which is sufficient for begin-ning geometry. Later in the study of math-ematics, there are discussions of angleswith degree measures greater than 180°and two other units of measure for rotationare introduced, radians and gradients. Aswith any measurement concepts, the unitsare part of our answer. When we say thatan angle measures 45°, we are using de-nominate numbers in a practical way.

To use a protractor, such as the onerepresented in Fig. 5.26, place the origin

FIG. 5.26.

of the protractor (the point of the wedge inthe very center on the straight edge) onthe vertex of the angle, lining up one legof the angle with the straight edge of theprotractor. The other leg of the angle indi-cates the number of degrees of rotation

the angle represents. The angle superim-posed on the protractor in Fig. 5.26measures 30°. Most protractors have 10°between each number so that we canmeasure angles to within 1 °.

Can you name the angle that occursmost often in our daily lives? Can you ex-press the same concept two more ways?We are intrigued when we see a cornerthat is not a right angle. Carpenters use aninstrument called a square to verify rightangles and we can express the same con-cept by using the terms perpendicular and90°. The definitions of acute and obtusedepend on our knowledge of right angles.A common error is to read the wrong scalewhen using a protractor. To overcome thistendency, always decide whether an angleappears to be acute, right, obtuse, orstraight before you measure. The angle inFig. 5.26 appears to be an acute angle, butthe ray goes through two rules, 150° aswell as 30°, and we must make a decisionabout which scale to use. In this case, ameasure of 150° does not make sense forour acute angle. Notice that the correctscale is the one that counts up as we fol-low the rotation from the first leg, at 0°, to-ward the second leg, at 30°.

Your Turn

40. How many degrees are in one fullrotation?

41. Do the lengths of the legs of an an-gle have any meaning in degree meas-ure?

42. Is there a special name for an angleof 180°?

43. Identify each named angle in Fig.5.27 as less than, exactly, or greater than90°, then measure the angles using a pro-tractor.

Angles are created if two lines, rays, orsegments intersect. An application of an-

MEASUREMENT 169

FIG. 5.27.

gle measure occurs when parallel linesare cut by a transversal (a line in the sameplane as the parallel lines that intersectseach of the lines). Figure 5.28 shows that

sponding angles and they have the samemeasure. That accounts for four of theeight angles. For the other four, noticethat Angles 1 and 2 are supplementary, ifAngle 1 measures 35°, then Angle 2 mustmeasure 145° (recall that the measures ofsupplementary angles add up to 180°). IfAngle 2 measures 145°, so do Angles 3,6, and 7. There you are! Eight angle meas-ures determined by measuring only one.

Another application of angle measureoccurs when we examine the interior an-gles of polygons. There are several waysto demonstrate that the three interior an-gle measures in any triangle total 180°.You might use your protractor to measureand total the interior angles of many,many triangles. You might cut out many,many triangles and tear off the vertices ofeach, lining them up to form a straight an-gle as shown in Fig. 5.29. A quick and

FIG. 5.28.

eight angles are formed when a trans-versal cuts through two parallel lines. Youcan determine the measure of each ofthose eight angles after making only onemeasurement. Suppose you measure An-gle 1 and find that it is 35°, then Angles 4,5, and 8 also measure 35°. Angles 1 and 4are formed by the same pair of lines,share a vertex, and are directly oppositeone another across that vertex—they arecalled vertical angles and they have thesame measure. Angles 1 and 5 have thesame orientation, each is above one ofthe parallel lines and has the transversalas the second leg—they are called corre-

FIG. 5.29.

convincing demonstration is to use dy-namic geometry software such as Geom-eter's Sketchpad®.

The sum of the measures of the interiorangles of any polygon can be determinedby sectioning the polygon into triangles,as was done in Fig. 5.12. The measures ofthe interior angles of each triangle total180°, thus you only need to know howmany triangles are involved. For the penta-gon in Fig. 5.12, the sum of the measuresof the interior angles is 3 x 180 = 540°,whereas for the parallelogram in Fig. 5.9,the sum of the measure of the interior an-gles is 2 x 180 = 360°.

170 CHAPTER 5

Your Turn

44. Given that Angle 1 measures 48°,Angle 10 measures 90°, Angle 20 meas-ures 115°, and Line m is parallel to Line n,use your knowledge of perpendicularlines, parallel lines, supplementary an-gles, complementary angles, and trian-gles to determine the measure of eachangle in Fig. 5.30.

FIG. 5.30.

Pythagoras is credited with discoveringa special relation among the sides of righttriangles. The Pythagorean Theorem is anapplication of the concept of perpendicu-lar lines or segments. The PythagoreanTheorem says that the sum of the squaresof the measures of the two legs of a trian-gle is equal to the square of the measureof the hypotenuse if and only if the trian-gle is a right triangle. This means that wecan apply the Pythagorean Theorem toany right triangle. It also means any trian-gle with measures that satisfy the Pythag-orean Theorem is a right triangle. Fig. 5.31shows one way of understanding the the-orem. It also shows the geometric modelof a number to the second power, be-cause the squares on the three sides rep-resent a2 + b2 = c2. In a right triangle, thehypotenuse is the longest side and thesquare drawn using the hypotenuse as aside has the greatest area. Its area is equal

FIG. 5.31.

to the sum of the areas of the squaresdrawn using the two sides that are legs ofthe right angle. The implication of the the-orem is that the length of the third side ofa right triangle is determined as soon asthe lengths of any two sides are known.

Pythagorean Triples represent a classof right triangles that have whole num-ber measures for all three sides. A com-mon Pythagorean Triple is 3-4-5, because32 + 42 = 52. In other words, this set ofmeasures satisfies the Pythagorean The-orem, determining a right triangle. Asshown in Fig. 5.32, any multiple of 3-4-5determines another right triangle.

FIG. 5.32.

Your Turn

45. Find and confirm five more Pythag-orean Triples.

For generations, students in algebraclasses have been required to memorizethe distance formula:

But did you

think you would find it in the measure-ment chapter? Unless a teacher

MEASUREMENT 171

helped you connect this formula to thePythagorean Theorem, you probably arewondering if we are getting forgetful. Ex-amine the individual parts of the formulaand you will see the connection!

First, consider a triangle with legs thatare 6 units and 3 units. Substitutingthese values into the Pythagorean Theo-rem will determine the length of the hy-potenuse.

Now, place that same problem on acoordinate plane, as shown in Fig. 5.33,

FIG. 5.33.

and the question might become, "Whatis the distance between (2, 3) and (8, 6)?"To use the distance formula, you stillmust begin by substituting values intothe formula:

Compare the last 4 lines using thePythagorean Theorem with the last 4lines using the distance formula. Theylook a little different, but the arithmetic isthe same. The horizontal displacement,(x2 - x1), is one leg of a right triangle andthe vertical displacement, (y2 - y1), is theother leg. The distance between the twogiven points is the hypotenuse of the righttriangle. Use a couple of PythagoreanTriples from Exercise 45 to check yourunderstanding of this connection.

Dimensional Analysis

Sometimes it isn't possible to measureusing the units desired for reporting. In anexperiment using a matchbox car and aramp, it would not be reasonable to mea-sure speed in miles per hour (mph) be-cause neither the distance traveled northe time period involved would be greatenough. However, you can determine theaverage speed of the little car in mph bymeasuring the distance in feet and thetime in seconds. Suppose you start thestopwatch when the car is released at thetop of the ramp and find that it covers 14feet of ramp in 0.2 seconds. Using dimen-sional analysis, you can figure out the av-erage rate of change for your matchboxcar in mph. The beauty of dimensionalanalysis is that it sounds so serious andyou already have the skill you need to useit—it is a straightforward application offraction multiplication and ratios. You willalso need to know, or look up, unit com-parisons in order to complete the proce-dure. You can find the length comparisonneeded for this example in Table 5.1 and

172 CHAPTER 5

Conclusions

As you read this chapter, you may havefelt that you were reading about history,science, arithmetic, geometry, algebra, oreven trigonometry. The concepts of meas-urement are absolutely vital to our studyin many areas of mathematics and sci-ence. Close your book for a moment andtry to list all of the areas of life outside theclassroom that are somehow impactedby the concepts of measurement. Whatare some of the roles that measurementplays in, for example: construction, medi-cine, commerce, manufacturing, account-ing, computers, transportation, landscap-ing, real estate, art, music, cooking,parenting, entertainment, and taxes—orany other activities you listed? We hopethat you will always think about measure-ment in terms of uses and never think of itas a stand-alone concept or skill.

you will need to remember that

and We will show the matchbox

car problem in two formats that are incommon usage:

Your Turn

46. How do the two formats demon-strated compare? How do they differ?

47. How did the units go from feet persecond in the first column to miles perhour in the last column?

48. If it is reported that a matchbox carhad an average speed of 30 mph, howmany feet did it travel in 1 second?

6Data Analysis and Probability

DATA COLLECTION ANDREPRESENTATIONS

FOCAL POINTS

• Data and Where to Get It• Representations of Data• Venn Diagrams. Percentages• Circle Graph. Line Plot• Bar Graph• Line Graph• Histogram• Frequency Polygon• Box and Whisker Plot• Scatter Plot• Stem and Leaf Plot

As citizens of the 21 st century, we live in adata rich world. It would be hard to findsomeone who has not heard of or, morelikely, participated in at least some sort ofopinion poll. Do you like this candidate orthat one? Do you like this cola or thatone? Do you like this restaurant or thatone? On and on, we are inundated withquestions, percentages, circle graphs,bar graphs, and a myriad of other databits when we watch the news on televi-sion, listen to the radio, or even talk withfriends. You do a course evaluation sur-vey at the end of each class; you areasked for input about the course's useful-ness, knowledge gained, and your in-structor's presentation clarity, availability,and concern for you. The data center

people analyze the responses and pro-vide a summary to your instructor, whothen uses the information to improve thecourse for future students.

Data and Where to Get It

In order to perform any type of statisticalanalysis, you need data. Data collectionprovides the information to be analyzedso questions can be answered aboutsome situation. Statisticians follow strictrules for the proper collection of data sofinal analyses and generalizations arevalid. If data is collected or used improp-erly, then conclusions are not valid. Forour purposes, we will discuss the attri-butes of random samples and conven-ience samples. Data are not only col-lected and used by statisticians for formalstudies. For example, do you check theinventory of canned goods in your cup-boards to make a shopping list before youleave for the grocery store? If so, you col-lect and analyze data! Is your inventory arandom sample or a convenience sam-ple? What is the proper use of your data?

If you wanted to know how many 45-year-old people in the United States re-sponded to the 2000 U.S. Census, youwould go to the Census and count howmany people were 45 years old on thatdate. If we wanted to know how manypeople in the United States have greentelephones in their kitchens, we wouldneed to figure out another way to get the

173

174 CHAPTER 6

information. Random sample data collec-tion is a powerful tool because it allows arepresentative sample of a population tobe used to infer or generalize about theentire population.

Perhaps you are thinking to yourself,"So what?" or "Why is that important?"Two quick reasons are time and money.Only governments can afford to do a cen-sus of the entire population, especiallyone as large as that of the United States.A small random sample is both affordableand takes far less time than asking mil-lions of people the same question. Fur-thermore, if the sample is taken properly,the results can be generalized or used tomake predictions about the entire popula-tion. At times, you might be asked to con-sider your classmates as a population.You will see that collecting data from theentire class provides more robust infor-mation than just collecting a sample fromthis population. Collecting the sample willbe faster and easier for you and will giveyou a feel for differentiating the ideas ofsample versus Census.

Random samples require careful plan-ning and design. The classroom data wesuggest you use for this fundamental ex-cursion into statistical analysis is not a ran-dom sample; any grouping of your class-mates will represent a conveniencesample. Convenience sampling can beused to explore fundamental ideas, butshould not be used for generalization pur-poses. Data gathered from your class-mates will be convenient and handy for ouruse, but we caution you to NOT generalizethe data you collect from your classmatesto other sections of students taking thiscourse, other courses, or to your school asa whole. Because random sampling tech-niques are beyond our reach, there will betimes when we will say, "Suppose thesedata WERE randomly collected, what in-ferences could you make about the popu-

lation?" It is important that you know thedifference between randomly collecteddata and convenience samples.

The most data rich source in any class-room is the students! Heights, weights,birth months, birth dates, favorite colors,preferred games, or desirable foods all pro-vide rich, meaningful, data. We recommendthat you be particularly sensitive about thetypes of data you collect; some types ofinformation may be too personal for com-fort. Think about your current classmates,what kind of data will you be able to gatherjust by observation? What kind of data willnot be available without communicatingwith individual classmates?

Your Turn

Collect the following convenience data,by observation only, from your class-mates in this course, your classroom orbuilding for this course, and from the en-vironment in which you currently live:

1. Gather three types of data aboutyour classmates, for example: How manyof your classmates are wearing glasses?How many of your classmates are wear-ing shoes?

2. Gather three types of data about theroom or building in which you are takingthis class, for example: How many desks/tables are in the room? How many ceilingtiles are there?

3. Gather three types of data about theenvironment in which you currently live(dorm, house, apartment, etc.), for exam-ple: How many books? How many chairs?

Representations of Data

Statistical data can be represented in avariety of ways. Data appears in severalformats. You can show a plethora of infor-

DATA ANALYSIS AND PROBABILITY 175

mation in Venn diagrams. Look in yournewspaper. You will see budget items re-lating to headline stories, car prices, storeadvertisements listing sales, discounts,bargains, and so on. But, it does not stopthere. Go to the financial section and youwill see charts, tables, line graphs, circlegraphs, and bar charts. Check out thesports section. Box scores, batting aver-ages, foul shooting percentages, goodfishing locations, and more are there.Data! It is almost anywhere you look. Youjust need to learn to spot it.

Venn Diagrams

Raw data, or data that have been col-lected, but not analyzed, are seldom use-ful. Venn diagrams provide a handymethod of organizing information so thatanalysis is possible. This brings up theidea of sets. Recall the information youstudied earlier about the different types ofproperties that sets have and how theycan be depicted in Venn diagrams. We willuse some of those same ideas to enrichyour understanding of statistics. Perhaps,when responding to the question for dataabout your classmates, you noted that 15people wore glasses and 15 did not, 20were female and 10 were male, and 30people wore shoes. You could representyour data using a Venn diagram like theone in Fig. 6.1.

Percentages

Another way to represent data is to useproportional reasoning, such as percent-ages and circle graphs. In order to find thepercent of information in each category ofcollected data, divide the number of ob-servations of interest (people who wearglasses, gender, or footwear for this exam-ple) by the total number of subjects for thisexample. To convert the decimal numberto a percentage, multiply by 100. We ob-tained the percentage in the top row of Ta-ble 6.1 by writing a ratio with 15 as the nu-

TABLE 6.1

Category

Glasses: YesGender: FemaleShoes: Yes

Observations

152030

Percentage

50%66.67%

100%

FIG. 6.1.

merator (the number of people who wearglasses) and 30 as the denominator (thetotal number of people in the class). An

equivalent form is which means 50

per 100, or 50%. For the second^row, we

used the data to write or ap-

proximately 66.67%. For the third row, wedid not find an equivalent fraction, be-cause 100% of the people were observedwearing shoes. Organizing data in a tablewill be a helpful first step as we exploreother ways of representing data.

Circle Graph

Figure 6.2 shows another way to repre-sent our classmate data. Half of the circlegraph is shaded to show that 50% of theclassmates wear glasses. You can thinkabout why we shaded half of the circle inat least two ways. We wrote a ratio,

176 CHAPTER 6

FIG. 6.2.

, and used the equiva-

lent ratio, which is 50

per 100, or 50%. If you think of the ratioas a fraction and remove the common

factors, we have -, exactly the part of the

circle we shaded. Another way to thinkabout 50% is to use its decimal equiva-

lent, and again exactly half the

circle is shaded.

Line Plot

A line plot is often used to efficiently rep-resent data. We will use a set of data toanswer the question, "How many keys doyou have with you today?" The followingdata set, ranging from 1 key to 8 keys,was obtained from a convenience sampleof 30 people:

To represent this data using a line plot, ahorizontal line is marked in even stepsfrom 0 to 10, as shown in Fig. 6.3. Theneach data point (the number of keys re-ported) is plotted using an "x" or othermark (e.g., balloons, stars, diamonds) torepresent each time a number of keys oc-

FIG. 6.3.

curs. The resultant plot allows us to seesummary information about the data eas-ily. What number of keys occurs most of-ten? What number occurs least often?Did anyone in our sample have no keys?

Bar Graph

What if we extended our key survey toseveral groups of people? A line plotwould not be very convenient if the sam-ple included a hundred or more people. Abar graph is similar to a line plot and ismuch more practical for large data sets(more than 40 data points). Instead ofcounting the number of data points ineach column as in the line plot, you readthe total for each column using the scaleprovided in the bar graph. In Fig. 6.4, nine

FIG. 6.4.

DATA ANALYSIS AND PROBABILITY 177

people carry two keys, whereas only oneperson carries six keys.

Occasionally, bar graphs are con-structed with horizontal bars instead ofvertical ones. Rotate your book so thatFig. 6.4 is turned 90° counterclockwise tosee how a horizontal bar graph mightlook. The figure's text would need to bereoriented if we decided to actually makeFig. 6.4 into a horizontal bar graph. Addi-tionally, we could elect to have the barsextend to the right instead of the left.

Your Turn

Use a tally technique to collect the follow-ing data about you, your classmates, andyour professor: wear eyeglasses, gender,wear shoes, have cats or dogs, number ofkeys. Condense your data into a tablesimilar to Table 6.1:

4. Using proportional reasoning, whatpercentage of the people in your class:wear glasses; are male; wear shoes; havecats; have dogs; have 3 keys?

5. Construct circle graphs to representthe data you collected about eyeglasses,shoes, and pets.

6. Construct a line graph to representthe number of keys each person in theclass reported.

7. Construct a bar graph to representthe number of keys each person in theclass reported.

Line Graph

A line graph is constructed using thesame axis system as a bar graph, a hori-zontal and a vertical axis. Look at the bargraph used for the key data collected inFig. 6.4. We will construct a new graph byplacing a point at the top middle of eachbar. After marking these points, connectthem with dashed line segments in orderfrom left to right as seen in Fig. 6.5.

When our data represents discreterather than continuous information, weconnect the data points with dashed linesegments. Discrete data includes only in-formation that is counted with whole num-bers. Think of a digital clock. It flips minuteto minute (assuming the absence of a sec-ond display), not showing any point alongthe continuum between minutes. An ana-log clock models continuous data, be-

FIG. 6.5.

178 CHAPTER 6

cause the minute hand sweeps from min-ute to minute, passing through every pointon the continuum. We use solid line seg-ments when the graph contains continu-ous information. A line graph provides agood feel for when data has a sharp turnupward or downward. For this reason, it isfrequently used to show economic out-comes and forecasts for stocks andbonds, where anticipating dramatic upsand downs is important. Because you areprobably not in this course to become astock analyst, you will most likely use thisinformation in your future career to graphstudent test results and behavior data.

Histogram

One popular way to represent statisticaldata is by using a histogram, which issimilar to a bar graph. For the histogram,one axis shows range, class, or categoryand the other axis gives the frequency ofeach category. If you are thinking, "I couldhave made a histogram out of the class-mate data by labeling all of the catego-ries," then you are absolutely right! Thedata are graphed vertically and horizon-tally in histograms in Fig. 6.6.

Frequency Polygon

Just as the bar graph lead into a linegraph, a histogram leads into a frequency

FIG. 6.7.

polygon. Figure 6.7 is a frequency poly-gon of the classmate data, which comesfrom the frequency table found in Table6.2. Whereas this graph looks a bit like a

quadrilateral, an actual polygon, not manyfrequency polygons actually look likeclosed figures.

Consider how you go about creating afrequency polygon for the data about howmany keys people report. You will need a

FIG. 6.6.

DATA ANALYSIS AND PROBABILITY 179

histogram as your basis. One way to sim-plify the task is to group the individual fre-quencies into equal ranges like 1-3, 4-6,7-9. Notice that these ranges are distinctand do not overlap one another. It is im-portant that no piece of data fit into morethan one range of the histogram. Table6.3 shows the frequency distribution forthe key data tabulated according to thenew ranges.

Class

1-34-67-9

TABLE 6.3Key Data

Frequency

1983

Figure 6.8 is a frequency polygon of thedata in Table 6.3. It is a good representa-

FIG. 6.8.

tion of how many keys a classmate islikely to have. You might have elected touse different ranges, such as 1-2, 3-4,5-6, and 7-8 to create an equally effectivefrequency polygon. The frequency distri-bution for this set of ranges is shown inTable 6.4.

Class Frequency

1-23-45-67-8

111153

Your Turn

Use the tables of classmate data you cre-ated earlier:

8. Construct a line graph from the datayou collected about how many keys yourclassmates had with them.

9. Construct a horizontal histogram forthe data you collected about eyeglasses,gender, shoes, and pets.

10. Construct a frequency polygon forthe data you collected about eyeglasses,gender, shoes, and pets.

11. Construct a table for your key datawith equal classes and tabulate the fre-quency of each. From the table constructa frequency polygon for the key data.Compare and contrast the information inthe frequency polygon and the line graphyou made earlier. Which graph do you be-lieve summarizes the data best, and whydid you make that selection?

Box and Whisker Plot

A box and whisker plot is based on therange and quartiles of the data. The rangeof the data is plotted as the whisker, a linesegment that begins at the lowest valueand ends at the highest value in the data.A rectangular box is constructed on top ofthe whisker and spans the middle 50% ofthe data. The left side of the box is posi-tioned at the value of the first quartile andthe right side of the box is located at thethird quartile. The second quartile, themedian, or middle value is marked with adot on the whisker and a line segmentacross the box. Figure 6.9 is the box and

FIG. 6.9.

180 CHAPTER 6

whisker plot of the key data. The dataranges from 1 key to 8 keys, with a me-dian value of 3 keys. The middle 50% ofthe data is from 2 keys to 5 keys. We willdiscuss, in detail, how to obtain each ofthese points and give you a turn at con-structing a box and whisker plot in thenext section.

Scatter Plot

Suppose we want to investigate a relationbetween one type of data and another.For your classmates, is there a relationbetween the year their cars were manu-factured and how many keys they report?Additional information is required, so youmust re-poll the class and organize thedata as shown in Table 6.5. You can'tsimply ask for the last two digits of theirmodel year, because you can't match themodel year with the key data for any indi-vidual.

Now that we have the data in an appro-priate table, we can plot it using Car Yearon a horizontal axis and Number of Keyson a vertical axis as shown in Fig. 6.10. Is

TABLE 6.5Car Year Versus Keys Data

Classmate

123456789101112131415161718192021222324252627282930

Car Year

828283838585868787898981909195969798979897969596979599999899

Keys

112222222223333333344455556778

FIG. 6.10.

there a relation between someone's carmodel year and the number of keys theycarry? If you said to yourself, "Accordingto the scatter plot, it seems the newer thecar a person has the more keys they will

carry," give yourself a pat on the back!Way to go!

You might suspect that the data waspurposely contrived to support that con-clusion. A trend where both values in-crease is a positive relation. How wouldthe scatter plot look if students carriedfewer keys as they obtained newer cars?A trend where one value increases whilethe other decreases is a negative relation.How would the scatter plot look if therewere no relation between the age of a stu-dent's car and the number of keys car-ried? Work with a partner or group to fig-ure out how these two plots might look.Chances are you will be asked to createsuch a plot when it is your turn.

DATA ANALYSIS AND PROBABILITY 181

Stem and Leaf Plot

Stem and leaf plots are similar to lineplots, because you can reclaim all of theoriginal data points after organizing them.A stem and leaf plot is also handy as afirst step in creating a box and whiskerplot. A stem and leaf plot organizes datain two parts: a stem, which is one or moredigits, and a leaf, which is always com-posed of single digits. It is necessary toinclude a key with this plot so that it canbe interpreted properly.

One thousand rainbow-colored centi-meter cubes were brought into the class-room. Each class member took one hand-ful of centimeter cubes and countedthem. The number of cubes they col-lected is shown in Table 6.6. The "Key 115

TABLE 6.6Cm Cube Data

Number of Cubes

402031232540243727402721312835

Number of Cubes

254120322539263028322222384229

= 15" is used to interpret the plot. Thestem represents digits in the tens placeand the leaf represents digits in the unitsplace. Because all of the centimeter cubedata can be represented with two digits,this is sufficient. If the stem and leaf plotwere about the altitudes of the RockyMountains, it would look the same. Butthe key would look like this: Key 1|5 =

15000, where the stem part representsdigits in the ten thousands place and theleaf part symbolizes digits in the onethousands place.

A stem and leaf plot always starts witha decision about how many digits thestem will require. An examination of therainbow centimeter cube data from Table6.6 gives us the information. Which digitsare needed in the stem? Why don't weneed digits other than 2, 3, and 4 for thestem? For the leaf part of the plot, we listthe units digit in the row with its respec-tive tens digit, without comas betweendata points, as shown in Fig. 6.11. Why isthere no need for punctuation to separatethe data points?

The data in a stem and leaf plot are gen-erally organized so that the leaf entries gofrom smallest to largest. The plot in Fig.6.11 is called an unordered stem and leaf

FIG. 6.11.

plot, an intermediate step often necessaryfor large data sets. After getting all the leafdigits into their correct row, order them toget the final stem and leaf plot as shown inFig. 6.12. Although each class member's

FIG. 6.12.

182 CHAPTER 6

handful of centimeter cubes is shown inthe stem and leaf plot, it isn't possible toreconstruct Table 6.6 because we didn'tkeep track of what count went where. Inthis hypothetical situation it is unimpor-tant, but if your professor lost the gradebook and were trying to recreate it from astem and leaf plot, even though your gradeis listed, your professor would have noidea what grade belongs to you!

Your Turn

Re-poll your classmates and keep trackof the model year of each person's caralong with how many keys they eachhave. Place the data into an appropriatetable and do the exercises that follow:

12. Construct a scatter plot for yourdata using car year versus keys.

13. What, if any, relation do you see be-tween the model years of your classmates'cars and the number of keys they had?

14. How would a scatter plot look ifthere were a positive relation between themodel years of your classmates' cars andthe number of keys they had?

15. How would a scatter plot look ifthere were a negative relation betweenthe model years of your classmates' carsand the number of keys they had?

Conclusions

There are an unlimited number of datatypes that can be collected from your datarich classmates and classrooms. None-theless, you should not limit yourself tothese convenience samples. If you haveInternet access, then you can locate andanalyze random samples from the NationalEducation Longitudinal Study (NELS) da-tabase, Nielson television program ratings,or other Web sites. Although informationfound on the Internet should not be trusted

without regard to the source, even ana-lyzing invalid information can be informa-tive.

Although we had you plot or graph ev-erything by hand in this section, technol-ogy may assist you. Most of the graphsand all of the tables in this section weremade using Microsoft Excel®. When andwhere will you use technology in makinggraphs and tables. It is important to re-member that even if you use technologyfor assignments, you may be asked totake tests without it.

Data can be represented in a wide vari-ety of ways. Some data is better orga-nized in a line plot or stem and leaf plot,whereas other material is better summa-rized using bar graphs, histograms, andso on. Sometimes the choice of represen-tation is not clear and you will have tomake judgments based on your experi-ence and understanding. Welcome to thewonderful and robust world of statistics!

DATA ANALYSIS AND STATISTICS

FOCAL POINTS

• Measures of Central Tendency andScatter

MeanMedianModeRangeVarianceStandard deviationQuartilesStatisticsAssumptionsGeneralizations

Collecting data and summarizing datawith a variety of representation tools wasfun, but the fun doesn't stop there! Nowyou need to analyze the data to makesense of it and draw some conclusions.

DATA ANALYSIS AND PROBABILITY 183

As you explore the measures of centraltendency and scatter, it is important thatconclusions are supported by the data.We will discuss statistics, random sam-ples, assumptions, and generalizability.

Measures of Central Tendencyand Scatter

Measures of central tendency help us de-termine where the central part of the datais located. Statisticians recognize that un-der different circumstances, differentmeasures of the central part of the dataare more helpful or more meaningful thanothers. Sometimes the mean is the bestmeasure of central tendency; other times itis the median or the mode. As you gothrough this section, think about the differ-ent measures and decide when eachwould be more beneficial than the other ina given situation.

The type of data we have can also influ-ence which measure of central tendencyis the most appropriate measure to beused. Data comes in a variety of forms.Some data is nominal or categorical, suchas the color of each student's car in yourclass or the different types of candy col-lected in a bag of Halloween treats. Othertypes of data are seen as being ordinal innature. Ordinal data is data that can beplaced into a logical order like low, me-dium, and high or infant, toddler, child,teenager, adult. The last type of data weroutinely gather is interval ratio data. In-terval ratio data is data that comes in theform of numbers, such as those associ-ated with measurements of time, dis-tance, area, or volume and test scores.With nominal data, we are limited to onlyfinding the mode of the data, with ordinaldata we can determine the mode and themedian. Interval ratio data is the only typeof data we are able to determine all three(mean, median, and mode) measures ofcentral tendency.

Scatter is a term used to describe howthe data points are spread out within thedata set. One measure of scatter is therange; others are variance and standarddeviation. Quartiles also give you a feelfor how spread out the data is in a set.Just as with measures of central ten-dency, the data type determines whetheror not measures of scatter are appropri-ate. Nominal data usually do not havemeasures of scatter, unless they are alsoordinal. Ordinal data can have a range de-termined and quartiles, but not varianceor standard deviations. Only interval ratiodata meets the requirements for having allof the scatter values calculated. Under-standing these concepts along with themeasures of central tendency will helpyou better understand, analyze, and inter-pret statistical information.

Mean. Mathematics and statistics of-ten define words differently from how theyare used in everyday language. Before youcan talk the talk of statistics, you must ex-amine some everyday words that are de-fined differently in mathematics andstatistics. The first is mean, which hasnothing to do with kicking someone in theleg or providing meaning for an unclearcommunication. This is an instance wheremathematics is like a foreign language anda new definition for an everyday word mustbe added to your vocabulary.

The mean is probably the most com-monly used measure of central tendency.Many people are aware of the term andthink of it as the arithmetic average of agroup or set of numbers. For example,your grade may be the mean of your testscores; if there are five tests during acourse, add your five scores and divide byfive. A mean (u) can be taken of an entirepopulation, but is far too costly and timeconsuming in most instances, or a mean(x) can be calculated from a sample. Youwill stick to finding x (read "x bar").

184 CHAPTER 6

Students like to compare grades withthe class average to get some sense ofhow they are doing in a course. Although itmight be interesting to look at your class-mates' grades while we explore this topic,confidentiality issues outweigh whateverbenefit might be derived from such an ex-ercise. Instead, we will take another look atthe data gathered earlier when we askedhow many keys people reported:

The key data here represents a sampleof the number of keys students at theschool possess. How do we calculate_themean number of keys? To calculate x forthis sample, we find the sum of all ofthe data points and divide by the numberof data points we added,

. In the numera-

tor, x1 represents the first data point to beadded, x2 the second, and so on until thelast term, xn. The sum is divided by thenumber of terms, n. This can be statedmore concisely using a summation sym-

bol: where the 2 means to add

all the data points and xi represents eachvalue, starting with x1 for the first and go-ing on to xn for the last.

If we read the data points from left toright across each row, then x1 = 1; x2 = 3;x3 = 5; . . . ; xn = 4; and n = 30. Using

we have:

The mean number of keys is approxi-mately 3.47. But we ask, does this answermake sense? We used the formula cor-rectly, but can you have 3.47 keys? In thereal world, you have three keys, or youhave four keys. Do you think anyone car-ries 0.47 of a key?

Sometimes it makes sense to give deci-mal answers for weights or temperatures,but it does not when we are talking aboutkeys, people, or any other discrete object.What does two thirds of a person looklike? When we are calculating the meanfor items that must be thought of in termsof whole numbers, it is appropriate andnecessary to always round to the nextwhole number. This type of estimating isdifferent from the regular rule for round-ing, which you probably learned as rounddown if the decimal is less than 0.5 andround up if the decimal is 0.5 or more. Themean rounding rule says always round tothe next whole number for discrete data,so that 3.1 rounds to 4 the same as 3.9rounds to 4. That being said, what is themean number of keys someone had intheir pocket?

Median. The next measure of centraltendency, median, also has a differentmeaning when we use it mathematically.Median does not describe the grassy partdividing a highway that you are routinelycautioned to keep off. Median, as a meas-ure of central tendency, refers to the mid-dle value after the data points have beenput in order from highest to lowest or low-est to highest. Sometimes we call the me-dian the positional average, because it isthe data value located at the mean dis-tance from one end of the data to the

DATA ANALYSIS AND PROBABILITY 185

other. When you have an odd number ofterms, the median is the data point in theexact center of the ordered data. If there isan even number of data points, then themedian is found by taking the average ofthe two middle terms in the ordered data.

One clever way of making sure that youget the correct median, after putting thenumbers in order, is to slash out datapoints on each end, in pairs, working to-ward the center. If you are careful when or-dering the data and slashing out terms,then this method gets you what you arelooking for every time. Consider the fol-lowing data about keys reported by peoplein two different sample groups: Group A:1,2,3,1,2,1,2,3,5,7,8,5,8,7,4,4,5,8,9 and Group B: 1, 2, 3, 1, 2,1, 2, 3, 5, 7, 8,5, 8, 7, 4, 4, 5, 8, 9, 9. Using this methodyou can find the median for Group A:

The median for Group A is 4 keys.Typically, when using this procedure, onlyone the last row would be visible. Thismodel was created to show the work thatwas done, starting from the top and work-ing down.

When finding the median for Group B,there are two terms in the middle of theordered data. The mean of these twoterms is the median for Group B. The laststep would look like this:

The median for Group B is 4.5. Notice that4.5 is not a value listed in the original dataand that you will need to apply the round-ing scheme to conclude that the medianfor Group B is 5 keys. The median will notalways be equivalent to a data point, itjust happened to be true this time.

The median is not affected by extremevalues the way the mean is and often pro-vides a more useful measure of the cen-tral tendency of the data. The mean forthe data in Group A is 4.47 or 5 keys andthe mean for Group B is 4.70 or 5 keys. Ifyou replace the last 9 in each group with1000, the medians will remain the same,but the means will shift dramatically to56.63 or 57 keys and 54.25 or 55 keys, re-spectively. Similarly, one really expensivehome in a neighborhood can skew or shiftthe mean value, whereas the medianvalue for a home in a neighborhood re-mains constant. This is something youshould think about when you are shop-ping for a house. Ask for the median priceof homes in a neighborhood, not just themean or average price.

Mode. Mode is not a mode of trans-portation—or "a la mode," the ice creamfor your desert—when we are speakingmathematically. Mode is another measureof central tendency. Mode refers to datathat appear most often in the set. A dataset has exactly one mean and one median.But, there may be zero, one, two (bimodal),or more modes in any given set of data.

There are several ways to find the mode.Which column has the most data points inFig. 6.3? Which bar in Fig. 6.4 representsthe most data points? Which peak in Fig.6.5 represents the most data points? Twokeys? Then two keys is the mode for thatkey data. Which leaf entry occurs most of-ten in Fig. 6.12? The centimeter cube datahas two modes, 25 cubes and 40 cubes; itis a bimodal data set.

186 CHAPTER 6

Consider the data we used to find themedians for Groups A and B. There arefour modes present in each data set. Theone, two, five, and eight occur three timeseach, making them the modes for thedata sets. Modes come in handy whenyou need information that might be ob-scured by the mean or median. Supposeyou surveyed the students at your schoolto find the location of the best tastingpizza. Which measure of central tendencyshould you use to decide which pizza isthe best? Do you want to eat at the mean,the median, or the mode restaurant intown? You can meet me at the mode be-cause that is where the most people goand that indicates it is the best!

Range. The range is the most com-mon measure of dispersion, rather thancentral tendency. The range measureshow the data is spread and is calculatedby subtracting the smallest data pointfrom the largest data point. Unfortunately,like the mean, it is adversely affected, orskewed, by extreme values. Going back tothe Group A data, the largest value is 9 andthe smallest is 1, and the range is 9 -1 =8.In our discussion about median, we intro-duced the idea of changing the last 9 in thedata set to 1000. If we do that, the range is1000 -1, or 999, which shows how one ex-treme value can significantly affect thismeasure of dispersion.

Your Turn

1. Find the mean, median, mode, andrange of the key data you collected fromyour classmates.

2. Find the mean, median, mode, andrange of the rainbow centimeter cubedata.

Variance. Variance is a much bettermeasure of spread or dispersion for a setof data, because variance is not affected

as strongly as the range by extremevalues. We will be asking you to find thesample variance (s2) rather than the popu-lation variance (a2). We talked earlier abouthow only governments and large corpora-tions have the money necessary to collectdata on entire populations. Although youwill not be asked to compute or work withthe population variance (a2), we want to re-mind you that it is different from the sam-ple variance (s2). You will not generally findthe variance mentioned in statistical re-ports, but it is an important intermediatestep in finding the standard deviation.

To find the variance for a given set ofdata, first determine the mean of the data.Look at the key data reported by studentsin Group C: 1, 2, 3, 1, 2, 1, 2, 3, 5, 7,where x = 2.7 or 3 keys. For the purposesof calculating the variance, we need touse the mean as calculated, even thoughno one had 0.7 of a key. The variance isbased on the actual arithmetic mean, notthe practical mean, so we must use x =2.7 to calculate variance.

One of the best ways to manage thecalculations needed to find variance is toorganize the data in a spreadsheet, eitherby hand, using a calculator, or computerapplication. Subtract the mean from eachdata point as shown in Table 6.7. Noticethat the sum of all the Xi - x values is zero,because x represents the arithmetic aver-age. This implies a need to do somethingelse to the information in the Xi - x columnin order to obtain meaningful information.This is accomplished by squaring eachvalue in the Xi - x column before findingthe sum.

For small data sets like Group C, thecalculations are not too troublesome todo by hand; calculate each of the (xi - x)2

values for Table 6.7 and compare yourresults with Table 6.8. Next sum all of the(Xi - x)2 values. The final step, which isalso shown in Table 6.8, is to divide the

DATA ANALYSIS AND PROBABILITY 187

TABLE 6.7Computing Variance

Sum =Mean =

Keys

1231212357

272.7

TABLE 6.8

x, - x

-1.7-0.70.3

-1.7-0.7-1.7-0.70.32.34.3

0

Computing Variance

Keys

1231212357

Sum = 27Mean = 2.7

xi - x

-1.7-0.7

0.3-1.7-0.7-1.7-0.7

0.32.34.3

0Variance =

Of, ~ xf

2.890.490.092.890.492.890.490.095.29

18.49

34.13.41

sum in the (Xi - x)2 column by the numberof data points, 10 for this data set, yield-ing a variance of 3.41, or s2 = 3.41. Acommonly used formula for this calcula-

tion is: -, which is exactly

what is shown in Table 6.8. Now that weknow the variance of the data for GroupC, you might be asking, "Isn't there aneasier way of getting this answer?" Thereis another way, but it is up to you to de-cide which you think is easier.

The second way of calculating the vari-ance is one you might have discoveredfor yourself after doing many of these cal-

culations by hand. First find the mean ofthe data as before. Then square each ofthe data points, and add those squares,as shown in Table 6.9. Next, multiply thesquare of the mean by the number of datapoints which, for this example, would be10(2.7)2. Subtract this product (72.9) fromthe sum of the squares (107) and dividethe difference by the number of datapoints (10). What did you get? This for-

mula is:

AlternateTABLE 6.9

Method for Computing Variance

Keys xi

Sum =Mean =

*1nx2

variance

1231212357

272.77.29

72.93.41

14914149

2549

107

Standard Deviation. Standard devia-tion is directly related to variance. Thestandard deviation for the population (a) iscalculated in the same way as for a sample(s). Comparing the symbols for variance ofa sample (s2) and standard deviation of asample (s) should make you sigh with re-lief. Finding the standard deviation fromthe variance is not much work. We calcu-lated the variance for Group C to be 3.41.The standard deviation for Group C is thesquare root of 3.41 or s = 1.85. The formulawe use for standard deviation isCombining this with the variance formula

188 CHAPTER 6

The standard deviation is an importantmeasure of dispersion. As a matter offact, we say that the mean and the stan-dard deviation, together, are necessaryand sufficient statistics to describe orsummarize an entire data set. This is truebecause, in Fig. 6.13 you can see that68% of the data in a set are within plus orminus one standard deviation (±1s) fromthe mean; approximately 95% of the dataare within plus or minus two standard de-viations (±2s) from the mean; and nearly100% of all data in a set are containedwithin plus or minus 3 standard devia-tions (±3s) of the mean. Figure 6.13 is

FIG. 6.13.

known as a normal curve. When a randomset of data is normally distributed, youcan make assumptions and predictions.

A data point falling outside two stan-dard deviations (x ± 2s) is considered tobe an outlier or an unexpected input.When we were discussing the median val-ues and ranges for Groups A and B, wepurposely introduced an outlier, 1000keys, to show what affect an outlier canhave on the mean and the range while notaffecting the median or mode. Sometimesit is easy to recognize an outlier. In our ex-ample, 1000 keys is a pretty extreme data

point. However, if the data point is onlyone or two values outside the expectedrange, it may be difficult to spot. Look atthe key data for Group C. How many ofthe data points are within one, two, orthree standard deviations of the mean?Are there any outliers? Are any of the datapoints farther than 1.85, 3.70, or 5.55from the mean of 2.7 keys?

Your Turn

Use the first 10 key data points reportedby your classmates to answer the follow-ing:

3. Find the variance of the sample us-ing both formulas. Calculate each byhand and then verify your results using asoftware application.

4. Find the standard deviation for thissample of your data.

5. Look back at your full data set forthe key data. Are there any outliers basedon the standard deviation of the sample?

6. Does this sample accurately repre-sent the whole class? Why or why not?

Quartiles. Quartiles? What are quartilesand why do I have to know about them?You need to be comfortable with quartilesso you will be able to adequately and com-fortably discuss standardized test scores.Quartiles are directly related to percentiles.So naturally the next question is, "What's apercentile? Is it like a percent?"

Percentiles should only be used whenthere is a very large data set, like scoresfrom a whole state or at least scores fromseveral school districts. You may havetaken a standardized test on which youscored in the 85th percentile. Did youthink that meant you made an 85% on theexam? Nope, you might have gotten 90%or 60% of the problems correct, but youdid better than 85% of the other people

DATA ANALYSIS AND PROBABILITY 189

who took the exam. Another way to thinkabout it is that only 15% of the peopletaking the exam got more right answersthan you did.

The 50th percentile is approximatelyequal to the median score. A box andwhisker plot, such as the one in Fig. 6.14,is often used when discussing the resultsof standardized tests. Remember, for thebox and whisker plot we needed fivethings: the lowest score, the highest score,the second quartile (the median score), thefirst quartile (the median of the scores be-low the median), and the third quartile (themedian of the scores above the median). Ifthe median or second quartile is equivalentto the 50th percentile, then you mightguess that the first quartile is the 25th per-centile and the third quartile is the 75thpercentile. This connection between per-centiles and quartiles points out the use-fulness of quartiles as well as box andwhisker plots.

Data points falling outside of two stan-dard deviations are considered outliers.But what if you only have a box and whis-ker plot? Aha! We can determine outli-ers by using the interquartile range (IQR).The IQR is found by subtracting the firstquartile (Q1) from the third quartile (Q3) orQ3 - Q1 = IQR. The expected range is cal-culated by subtracting 1.5 times IQR fromthe first quartile, and adding 1.5 timesIQR to the third quartile [q1 - 1.5IQR,Q3 + 1.5IQR]. Any data point outsidethese values is considered an outlier. Fig-ure 6.14 is the box and whisker plot de-veloped for the key data.

Key Data

0 1 2 3 4 5 6 7 8 9Number of Keys

FIG. 6.14.

Without peeking back, use Fig. 6.14 todetermine what value represented thefirst quartile. What value represented thesecond quartile or median? What valuerepresented the third quartile? Youshould have said 2 keys, 3 keys, and 5keys, respectively. Using this information,Q1 = 2 and Q3 = 5, the IQR is 5 - 2 or 3.Because the product of the IQR and 1.5 is4.5, Q1 - 1.5 (IQR) = -2.5, and Q3 + 1.5(IQR) = 9.5 for the expected range. Whatwere the fewest and most keys presentaccording to the Fig. 6.14? The outlierswould be outside the interval [-2.5, 9.5].Did anyone have fewer or more keys thanexpected? The left side of the interval is anegative number; what meaning does thishave for the data set? What is the fewestnumber of keys that would be consideredan outlier?

Your Turn

Use your key data for the following prob-lems:

7. Find the first, second, and thirdquartiles.

8. Find the IQR.9. Create a box and whisker plot.10. Determine the expected range us-

ing the IQR and identify any outliers.

Statistics

Statistics are based on samples frompopulations and if the samples are col-lected properly, they allow us to makegeneralizations about the entire popula-tion fairly accurately. Using samples isboth a cost effective and efficient way offinding information about an entire popu-lation. When you make generalizationsabout the population, however, certainassumptions go along with this process.

190 CHAPTER 6

It is important for you to be aware of theseassumptions and when it is and is not ap-propriate for generalizations to be madeabout a population.

Assumptions. Assumptions play animportant role in the statistics you create,no matter what procedure you use. Noth-ing invalidates research and wastes morehard work faster than breaking a statisticalassumption. If these assumptions are socritical, then why did we wait until now totalk about them? We needed to developyour statistical vocabulary to providecommon ground on which to build. As-sumptions must be verified or qualified fora given data set. The data representationsyou have been learning can be examinedto determine whether assumptions areholding or failing.

Whereas it is beyond the scope of thiscourse to talk about all the assumptionsmade in statistical analysis, some of themare critical to daily survival in a data richsociety. When you find the mean, vari-ance, and standard deviation for a givenset of data you make some assumptions.First you assume that the data is intervalratio data or that the spaces between thenumbers are fixed and the numbers havemeaning. This meaning is not arbitrarilyassigned, such as blue equals one, redequals two, but meaning that developsfrom the data collected, such as testscores or numbers of keys.

Second, you assume the data was de-rived from a random sample. Data col-lected in your classroom represents aconvenience sample. To get a truly ran-dom sample of students is a rather com-plicated process. It is okay to use thedata and make calculations based onnonrandom samples in order to learn theprocesses. It is not okay to break the as-sumption outright or ignore it for conve-nience sake when doing research.

The last assumption you make whencalculating a mean, variance, and stan-dard deviation is that the data is evenly ornormally distributed and there are no ex-treme values or outliers falsely skewing orshifting the mean. Although we found noevidence of outliers in the key data, thebox and whisker plot showed that the keydata are skewed, indicating that the dataare not normally distributed. Rememberwhen we replaced the last 9 in the GroupsA and B data with 1000? Certainly, thatoutlier had a profound impact on themean value. Sometimes we are allowedto assume a data set is normally distrib-uted if the data set is large enough. Ingeneral, a data set must have a minimumof 30 random data entries in order to beassumed normally distributed.

There are other assumptions in statis-tics, some of which seem quite obvious.For example we assume that every circlegraph represents 100%. Not 99% or101 %, but exactly 100%. When you use apie chart or circle graph in which the sec-tor values total more or less than 100%,you violate the assumption for a circlegraph.

Generalizations. Your classmate datawas a convenience sample, as opposedto a random sample. The data is usefulfor learning procedures, but this viola-tion of the second required assumptionimplies that we should not attempt to gen-eralize the information to any population.All of the assumptions must be met in or-der to have any generalizability for a givenset of data. Whereas most people in thesample we provided had two keys, wecannot say that most people, in general,carry two keys.

As you read and study research, it isimportant for you to be a critical thinker.Tainted generalizations, due to statisticalassumption violations, will make the re-

DATA ANALYSIS AND PROBABILITY 191

searcher look pretty silly. It may not beobvious during casual reading, but impor-tant assumptions are often ignored or de-liberately broken in published articles, in-validating the conclusions drawn by theresearchers or authors. We want you torecognize this fact, so that you can makedecisions based only on reliable andsound data analysis.

Conclusions

You should understand that a statistic is arandom sample of data that can be gener-alized to a population by looking at meas-ures of central tendency and dispersion,as long as assumptions have not been vio-lated. Some measures of central tendencyand dispersion, like the mean and range,are adversely affected by extreme values.Values like median, mode, variance, andstandard deviation are not as affected byoutliers. Percentiles, quartiles, and medi-ans are related to one another and arehelpful in describing and interpreting data.Sometimes data is normally distributedand sometimes it is skewed. Recognizingthese foundational concepts as you readdata analyses will help you be a discrimi-nating consumer of research.

COUNTING AND PROBABILITY

FOCAL POINTS

• Factorial• Permutations• Combinations• Independent• Dependent• Conditional Probability• Lottery• Odds

Many people think that the words "proba-bility" and "statistics" are synonymous.They are related, but they are also very dif-ferent. Statistics are numerical data thathas been tabulated and organized to pro-vide reports or information. When we talkabout probability, we discuss the out-comes of experiments and how those out-comes might be used to make predictions.This often involves counting how many dif-ferent ways there are of combining events.Some fundamental ideas about independ-ence or dependence of data will be pre-sented. After we have explored this vocab-ulary and worked some problems, we willexplore lotteries. You can decide for your-self if playing is a good idea.

Factorial

Factorials provide a convenient and shortway to write products of several relatedfactors. The mathematical notation usedto indicate this operation is an exclama-tion point placed to the right and adja-cent to a number, 5! is shorthand for writ-ing 5 • 4 • 3 • 2 • 1 = 120. We say thatn! = n(n - 1)(n - 2)(n - 3 ) . . . 3 2 1. Theeasiest way to remember this is to think ofa countdown and then multiply all thenumbers together.

Many of the problems we do involve di-viding a factorial by another factorial ortwo factorials. These can be done byhand or with a calculator; this is an in-stance where you are smarter than thecalculator because you have the ability todo higher order thinking. The calculatordoes the exercise by finding 10!, then 7!,and then 3628800 -f 5040. As you dosome of these by hand, you will gain anunderstanding of what is happening. We

begin this exploration by looking at —'-.

192 CHAPTER 6

The computation of 10! -r 7! shows howyou are smarter than a calculator. Youcan divide out common factors—the cal-culator cannot. This becomes significantwhen you encounter something like 500!-^ 499! The calculator gives up, even somebig fancy ones overload and quit. But,with the previous process, you can getthe answer, 500, because all factors arecommon except the 500.

Next look at an exercise that has two

factorials in the denominator: —-4!-6!

There is one more example that will helpyou gain a good understanding of how thisoperation works. After the common fac-

tors in are divided, the result will be a

fraction instead of a whole number.

A few more things about factorials be-fore we move on. First, 0! = 1 by defini-tion. Second, 1! = 1 because in 1!, 1 is theonly factor. Third, for your work in thisbook, ~5! is written as ~1 (5!) = ~1 (5 • 4 • 3 • 2

1), then -1(120) = "120. As mentionedearlier, factorials get really big really fastand calculators are handy when dealingwith them. Some calculators have a fac-torial key; for others, you have to enter theindividual factors. Some calculators canhandle 50! or even 150!, but it may take awhile to come up with the number be-cause it is so huge. Some calculators willexpress such huge results in scientific no-tation and may use front-end estimation.For example, 200! = 7.88658E374 or 7.89x 10374 to two significant digits. Again, theexciting part is that you are smarter thanyour calculator!

What?! Yes! That is because you cando many problems that your calculatorcannot! How can this be so? Revisit 500!

divided by 499! or Examine what we

are saying:

DATA ANALYSIS AND PROBABILITY 193

You don't have to write out all of thefactors if you see a relation between thefactors in the numerator and denomina-tor. You can complete many outrageousproblems by hand, or with the aid of yourcalculator after making a few adjustments.

For , think of 500! as 500 499 498 •

497 496 495!, because you can divideout the common factor of 495!. Then useyour calculator to find 500 499 498 497

496 = 30629362512000. All right!

Your Turn

Simplify each of the following by hand.Check your answers using a calculator orsome other form of technology. It will behelpful to you if you become comfortabledoing these types of problems using bothmethods.

Permutations

How many different ways can you orderthree colored blocks that are red (R), pur-ple (P), and blue (B)? There are only threeobjects, so you can make a list to figureout all the different ways to order R, P,and B. Try it. You should get: RPB, RBP,PRB, PBR, BRP, and BPR. There are ex-actly six ways to arrange the three blocks.Add a green (G) block to the R, P, and Bones. Now how many arrangements?Wow! There are many more than six dif-ferent ways to order four blocks. Isn'tthere an easier way to determine howmany arrangements there are? Yes!

Finding the total number of differentways to order a given set of objects iscalled a permutation. The notation for thisoperation can be written in two differentways, P(n, n) or nPn. Here comes the coolarithmetic to all of this. P(n, n) = n!, wheren is the number of objects in the set to bearranged. So, instead of using trial and er-ror, and a lot of time, to figure out howmany different ways to order R, P, B, andG, just do the arithmetic: P(4, 4) = 4! = 24.There are 24 different ways to order R, P,B, and G blocks.

Of course, sometimes we only want tosee how many different ways there are torearrange part of a set at a time. Howmany different ways could 5 students bechosen from a class of 30 students? Ournotation changes to P(n, r), or nPr, where nis the total number of objects in the setand r is the number of things to be ar-ranged. (Note that n is always at least aslarge as r.) You could say that you arelooking for how many different ways to ar-range n things taken r at a time. This for-

mula involves a fraction:

Speaking like a statistician, read, "The per-mutation of n choose r, is n factorial di-vided by n minus r the quantity factorial."

Suppose you want to know how manydistinct ways you can order subgroups of5 out of a total group of 30. You would belooking for P(30, 5):

194 CHAPTER 6

Did you notice that this is the procedureyou learned for division problems with fac-torials?

Combinations

When you looked for all the different waysto order the R, P, and B blocks, the orderyou placed them in was important. That isRPB was different from BPR. When youlook at forming combinations, order is notimportant, because rearranging thepieces doesn't increase your count. Theletters A through G are used to name mu-sical notes, the seven major tones. Forthis exercise, define a chord as havingthree notes and CEG sounds exactly thesame as GEC. How many different chordsor combinations of three different notescan be formed (not taking into accountthat some will sound awful) using theseven major tones? We could try to findthem all by trial and error as we initiallydid for finding permutations, but youshould guess that there is a formula tohelp us with the arithmetic.

Combinations are expressed as C(n, r)i= :—. Combinations are related to

r!(n - r)!permutations. Multiplying this formula byr! returns you to the formula for permuta-tions. This also indicates that there will befewer combinations than permutations ifwe are using the same numbers. Whenworking with combinations, n must belarger than r. To see why, try workingthrough the formula when n = r. Did youget 1 ? Well, there is only one way to com-bine the entire set, all other arrangementsare duplicates.

Back to our musical note problem, howmany combinations of three notes out ofseven notes are there?

3!-4!

C(7 3) = 35

There are 35 different combinations ofchords that can be made using the majortones A through G.

Your Turn

For each of the following problems:

a) Determine whether the problem re-quires a permutation or a combina-tion for its solution.

b) Give your reasoning for choosingthat function.

c) Find the solution to the problem.

10. Suppose the rainbow colors couldcome in any order. How many differ-ent patterns of colors of the rainbow(ROYGBIV) could there be?

11. A certain guitar has 12 strings. Inhow many different orders can a guitaristpluck any 5 strings on this guitar?

12. How many different pairs of colorscan be made using colors of the rainbow(ROYGBIV)? (We understand that youwould not wear or be seen in some ofthese combinations, but we do want youto consider and count them all.)

Independent

We are ready to discuss probability. Weare a society of risk takers and the ideasof chance and the percent chance of be-ing a winner or a loser permeates our cul-ture. We say things like "fat chance,""sure thing," or "take a chance" on a reg-

DATA ANALYSIS AND PROBABILITY 195

ular basis. The weatherperson reportsthere is a slight chance of rain or a strongpossibility of thunderstorms in the area.But how much do we really know aboutprobability and what it really means?

All probabilities lie between zero andone inclusive. If something has no chanceof occurring, or is impossible, then we sayit has a probability of zero. If somethingwill always happen, or if it must occur,then it has a probability of one. All of theother probabilities can be expressed ascommon or decimal fractions betweenzero and one. Another way to think aboutthis is that all of the probabilities are be-tween 0% and 100%, inclusive.

Okay, now we know a little bit aboutprobabilities, but how do we figure out theprobability that something will occur? Weare going to take a look at experimentsand their outcomes, and then figure outthe probability of a particular outcome.We will assume that the possible out-comes are uniformly likely to occur. Thismay seem to be a fairly sophisticatedstatement and not all experiments havethis quality, but for now it is enough to getyour feet wet.

One popular experiment with equallylikely outcomes is a fair coin toss, wherethe outcomes are either a head (H) or atail (T). Each coin toss is independent (itdoes not rely on the result of any othercoin toss). The set containing all of thepossible outcomes is called the samplespace. The sample space for a fair cointoss is {H, T}. The cardinality of this sam-ple space is two. Because both outcomesare equally likely to occur, the probabilityof getting a head is one out of two, onehalf, 0.50, or 50%, and the probability ofgetting a tail is one out of two, one half,0.50, or 50%. Notice that the sum of theoutcomes is one, or 100%, as it alwayswill be.

Another popular experiment is rolling afair die. Each roll is independent of an-other roll and each face is equally likely tobe on top. The sample space of the out-comes for rolling a fair die is {1, 2, 3, 4, 5,6} and the probability of rolling a 1 is oneout of six, one sixth, approximately 0.167,or approximately 16.7%. The probabilityof rolling either a 2, 3, 4, 5, or 6 is also oneout of six. Notice that the fractions giveexact answers, whereas the decimalequivalents and percents require round-ing. If you like using decimal equivalentsor percents, then you will have to usegreat care to be sure that the total of allthe probabilities is 100%.

Another common experiment involvesdrawing a card from a fair standard deckof playing cards. A standard deck hasfour suits, hearts (H), diamonds (D), clubs(C), and spades (S), and 13 cards in eachsuit, ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, jack(J), queen (Q), and king (K). Thus, thesample space could be written {AH, 2H,3H 10H, JH, QH, KH, AD, 2D, 3D,. . . , 10D, JD, QD, KD, AC, 2C, 3C, . . . ,10C, JC, QC, KC, AS, 2S, 3S 10S,JS, QS, KS}. The sample space has 52

cards, with each card having chance

of being selected. But what are thechances of drawing an ace (A) if you don'tcare about the suit?

In order to figure out this problem, welook back to the original sample space tofind the sample space for a success. Thesample space for a success is {AH, AD,AC, AS} and the cardinality of the successis four. To find the probability of a suc-cess, you divide the cardinality of a suc-cess (n) by the cardinality of the entire

sample space (s) or -. The probability ofs

drawing an ace, from a fair standard deck

of playing cards is r.

196 CHAPTER 6

In general, the probability of a successin any experiment is found by dividing thecardinality of the success of the event bythe cardinality of the entire sample space

or where n represents a success,

or the cardinality of a success samplespace, and s represents the cardinality ofthe entire sample space. We can use thatinformation to go back and explore somemore complicated ideas about rolling afair die. How do we determine the proba-bility of rolling a single fair die where anevent is a roll of less than 3?

What happens if the success samplesets overlap? Finding the probability ofdrawing a king (K) or a heart (H) will re-quire us to use some of the ideas aboutintersections of sets and unions of sets:

What do you think would happen if ourexperiments had multiple parts? Insteadof tossing a coin once, suppose we tosstwo coins once. What are all the possible

outcomes for the sample space? Listthem and see if you get four possible out-comes: {HH, HT, TH, TT}. Four is the car-dinality for the sample space. If youtossed three fair coins, what is the samplespace? Watch for a pattern so you candetermine the cardinality of a sample setfor tossing any given number of coins.The sample space for tossing three faircoins is {HHH, HHT, HTH, THH, HTT,THT, TTH, TTT}. What is the pattern? Canyou fill in the question marks in Table6.10? Generalize the pattern and deter-

TABLE 6.10

Number ofCoins Tossed

123

n

Cardinality of theSample Space

248

?

Pattern

22 x 2

?

?

mine the cardinality for the sample spacefor tossing six fair coins.

A tree diagram could be used to deter-mine the cardinality of a sample space. Aword of caution on tree diagrams—if yournotation is sloppy or if you write very large,then you are likely to make errors usingthis method. Although a tree diagram be-comes cumbersome if you flip more thanthree coins, many people favor thismethod for its ease of use. As long as youare careful, it will serve you well. Figure6.15 shows how the tree diagram works.To determine the probability of a givenoutcome, multiply the probability of eachtoss along a direct path to the desired out-come. For example, to find the probabilityof HT, consider the tree diagram for toss-ing two coins; the path to the first H has aprobability of 0.5 and the path to the T thatfollows (down and to the right of the H) hasa probability of 0.5. Multiply 0.5 x 0.5 =0.25, and there you have it, the probability

DATA ANALYSIS AND PROBABILITY 197

FIG. 6.15.

of tossing HT in a two-coin toss experi-ment is 0.25, or one fourth.

We now have a convenient rule. Whenthe probabilities of individual outcomesare independent (one does not rely on an-other), multiply the respective probabili-ties together to get the final probability ofa success. That may not sound like a verybig deal, but mathematically and for easeof obtaining probabilities it is huge! Con-sider the following scenario and you willsee. There are 10 marbles in a bag. Five ofthe marbles are red (R), 3 are Violet (V),and 2 are blue (B). The probability ofdrawing a R out of the bag is P(R) =

or 0.5; the probability of drawing a

V out of the bag is or 0.3; and

or 0.2 is the probability of

drawing a B. What is the probability ofdrawing RRWB, in any order, as long aseach time a marble is drawn it is placedback into the bag before the next marbleis drawn? Because no draw depends onthe results of another draw, the events areindependent. P(RRWB) = 0.5 x 0.5 x 0.3x 0.3 x 0.2, or 0.0045. The probability of

drawing 2 red, 2 violet, and 1 blue, in anyorder, is 45 out of 10000. Doesn't seemvery likely, does it?

A hidden idea we just explored with in-dependence is randomness. When we flipa coin, roll a die, pick a card (with replace-ment), or draw a marble (with replace-ment), we do so at random. Randomnessis a statistical assumption. As we are find-ing probabilities, it is assumed that theoutcomes occur randomly. If a coin istossed 10 times, resulting in four H andsix T, does that mean the coin wasn'ta fair coin? No, it means that, in order toapproach the theoretical probability,P(H) = 0.5 and P(T) = 0.5, a large numberof coin tosses, like 1000, or more, areneeded. Even with 1000 tosses, there isno guarantee that you will get 500 H and500 T. Theory and practice are not neces-sarily the same.

Your Turn

13. What is the cardinality of the sam-ple space for each of the following?

a) an experiment tossing 5 coinsb) an experiment rolling 3 dice

14. What is the probability of obtainingthe following outcomes from the experi-ments in Exercise 13?

a) HHTHTb) 352

15. There are R = 3 marbles, O = 4marbles, Y = 4 marbles, G = 2 marbles,B = 1, I = 3 marbles, and V = 3 marbles ina bag. What is the probability of drawingBORG from the given bag of rainbowmarbles when there is replacement?

Dependent

We will explore one aspect of dependenceusing nonreplacement. Our focus will beon experiments such as drawing cards from

198 CHAPTER 6

a standard deck or drawing marbles from abag without replacing the item before draw-ing again. When drawing without replace-ment, the total number of cards or marblesin the bag decreases with each draw. Howdoes this affect the probabilities?

We start by exploring 10 marbles in abag. Five of the marbles are red (R), 3 areviolet (V), and 2 are blue (B). The probabil-ity of drawing a R out of the bag the first

time is or 0.5, the probabil-

ity of drawinq a V out of the bag the first

time is and the probabil-

ity of drawing a B out of the bag the first

time is . What is the

probability of drawing RRWB, in any or-der, when a marble is drawn it is notplaced back into the bag? We are still go-ing to multiply the probabilities of each in-dividual outcome together, but this timewe have to adjust the probability of eachsuccessive draw to reflect its depend-ence on the previous draw.

, just as before, but P(R2)

because now a red marble has been

removed, changing the number of totalmarbles and the number of red marblesleft in the bag. The probability of drawing

a V is affected, so which affects

the next draw, so I Our last

draw is a B, but now there are only 6marbles left in the bag and 2 are blue, so

Now we have all the informa-

tion we need and we multiply each proba-bility toaether, P(RRWB) =

placement, P(RRVVB) = 0.0045. Withwhich method are you more likely toachieve a desired outcome? This is justsomething for you to ponder.

For a dependency example involv-ing drawing three cards without replace-

ment, for the

first card drawn. For the second draw,

no H or A was drawn and

if a H or A was pulled

on the first draw. For the third draw, the

pattern continues if

neither the first nor the second draw was

aHorA, if only one of

the other draws was a H or A, and finally

if both the first and

ocounu uiawo wcie H or A. For thesake of simplicity, stick to the ideathat each draw was H or A. Then |

= 0.0253. Card ex-

periments get complicated when there isno replacement, but you now have thebackground to handle different cases.

What would the P(AH, AC, or 3H) be ifcards were not replaced, assuming eachdraw was a success? P1(AH, AC. 3H) =

and

3H) = —. Now P(AH, AC, 3H) =

Although the probability of drawing threecards, any ace or heart, without replace-ment is approximately 0.0253, the proba-bility of drawing the AH, AC, or 3H in anyorder is much smaller. Things are not aslikely to happen as you might think.

DATA ANALYSIS AND PROBABILITY 199

Your Turn

16. What is the probability of drawingBORG from the given bag of rainbowmarbles when there is no replacement?There are R = 3 marbles, O = 4 marbles, Y= 4 marbles, G = 2 marbles, B = 1, I = 3marbles, and V = 3 marbles in the bag.

17. What is the probability of drawingan AD through 10D, any J, or a 3S from afair standard deck of 52 playing cardswhen there is no replacement and fivecards are drawn? Consider that eachdraw was a success.

Conditional Probability

Sometimes we are given information orknow things that influence how we calcu-late the outcome of an experiment. Hav-ing advanced information limits or refinesthe sample space. Suppose that there is a20% probability of rain today. Given thatit is already cloudy and thundering in-creases the probability that it is going torain, compared with not having this infor-mation. Knowing some conditions prior tocalculating the probability of an occur-rence increases the probability of a suc-cess. When you are asked the probabilityof some event, given some information, itis written P(A|B), which is read "the prob-ability of A given B."

We start with the sample space for athree-coin toss experiment: {HHH, HHT,HTH, THH, TTH, THT, HTT, TTT}. What isthe probability of obtaining two heads andone tail, in any order? There are threesuch possibilities: HHT, HTH, THH, thusP(two heads and one tail) = 0.375. What isthe probability of two heads and one tail,given the first toss was H? There are onlytwo elements of the sample space that fitthe criteria: HHT and HTH. Knowing that

the first toss was H eliminates any ele-ment for which H was not listed first. Thenour new sample space is {HHH, HHT,HTH, HTT}. Thus, P(two heads and onetail|H first toss) = 0.5 for this conditionalstatement.

Your Turn

18. What10H card,drawn?

19. Whaton either ofof the dice

20. Whatred K card,drawn?

is the probability of drawing agiven that a red card was

is the probability of rolling a 2two dice, given that the sumis 7?is the probability of drawing agiven that a black card was

Lottery

Suppose there is a lottery for which youpick 6 out of 55 numbers. This lottery in-volves 55 individually numbered, identi-cally weighted, ping-pong balls in sealedbin. They mix the balls using compressedair and then capture 6 balls, one at a time.You win the lottery if you have picked, inany order, all 6 captured numbers. Thelottery sponsor is only going to chargeyou a dollar for a chance to win. Such adeal!

Or is it?

P(winner) = — x — x — x — x — x —55 54 53 52 51 50

1 - 0.000000000048.20872566000

Up until now we have only looked at theprobability of a success, but this is asgood a time as any to take a look at thelikelihood of a failure. In this case,P(losing)= 54 X 5 3 X 5 2 X 5 1 X 50 x 48 = 48

55 54 53 52 51 50 55

200 CHAPTER 6

Hey, wait a minute, that is the probabilityof not getting a single one of the numbers.We shouldn't go wait in line just yet.

We need to explore this lottery proba-bility a bit further. This particular lotterywill also pay small amounts of money forthree, four, or five numbers correctlypicked. The probability of getting threenumbers is

The probability of not winning any ofsmaller prizes is calculated by subtractingthe probability of winning from one. Sothe probability of not getting any moneyat all for any lottery ticket is the comple-ment of gtting three numbers or

We will let you figure out the likelihood oflosing for the other amounts. We will alsolet you decide if you want to wait in line fora lottery ticket.

Odds

What are the odds that we would wait untilthe end to discuss odds? Asking about theodds of something occurring is fairly com-mon in our culture. The thing to rememberis that, although probability and odds arerelated, thev are not eouivalent. This is be-

cause the odds in favor of something

occurring is where m equals the cardi-

nality of a success and m' is the comple-ment of m or the cardinality of a failure. Inother words, m + m' = n, where n is thecardinality of the sample space. The prob-ability of a favorable odds ratio is found by

The odds against something

occurring are and the probability of the

odds against something happening is

One important difference be-

tween odds and probability is that theodds ratio can be larger than one, whereasthe probability ratio is always betweenzero and one, inclusive.

What are the odds in favor of rolling afive using a fair die? We know the proba-

bility is We know that a fair die has only

one face with five dots, so m = 1. Thecomplement of m is m', which can befound by subtracting m from the cardinal-ity of the sample space. Here n = 6, so m'= n -morm ' = 6-1 =5. Thus, the oddsin favor of rolling a five using a fair die are1:5. The odds against rolling a five are 5:1.You notice we use different notation here.We include this to help you become com-fortable with seeing different notation.The probability of the odds for rolling a

five is and the prob-

ability of the odds against rolling a five is

Notice again thatithe odds against rolling a five are greaterthan one, and the probability of the oddsagainst rolling a five remain less than one.

Your Turn

For each of the following find:

DATA ANALYSIS AND PROBABILITY 201

A. The odds in favor of the eventB. The odds against the eventC. The probability of the odds in favor

of the eventD. The probability of the odds against

the event

21. Drawing a black card from a fairstandard deck of 52 playing cards?

22. Flipping two coins and obtainingHT?

23. Drawing BORG from the given bagof marbles when there is replacement?There are R = 3 marbles, O = 4 marbles, Y= 4 marbles, G = 2 marbles, B = 1, I = 3marbles, and V = 3 marbles in the bag.

Conclusions

You have just been through several math-ematical ideas as they relate to countingand probability. We have endeavored toopen your eyes and increase your mathe-matical prowess in these two areas. Youknow the difference between a permuta-tion and a combination and when to usethem. You have explored independentand dependent probability. You haveeven looked at how the lottery works andthe difference between an odds ratio anda probability ratio. It has been said thatthe lottery is a tax on the mathematicallyilliterate. Is the lottery worth it?

This page intentionally left blank

7Problem Solving

FOCAL POINTS

• What Makes a Problem a Problem?• Polya's Steps• Strategies• Collection of Problems

Ever have any problems? How did yousolve them? As you ponder that for a mo-ment, think about the steps you took toresolve whatever situation you were in.Problem solving in mathematics usestechniques similar to those you apply toreal-life problems. Problem solving is oneof the core ideas in mathematics and themain thing that mathematicians do. Be-cause we all experience problems of onenature or another, learning how to be agood problem solver is important for ev-eryone.

Problem solving is central to all disci-plines, and one reason for includingmathematics in every Pre-K-20+ curricu-lum. According to NCTM, "in everyday lifeand in the workplace, being a good prob-lem solver can lead to great advantages"(2000, p. 52). Do all situations in mathe-matics require problem solving? No. Mostof the exercises you completed in thisbook represented guided practice as youwalked through the process of figuringout how to use unfamiliar algorithms orreflected on the processes behind familiarones. We did that to raise your comfortlevel in doing mathematics. In this chap-ter, however, to guide you too muchwould be a disservice to you and the

topic. We will guide you through defini-tions and some general guidelines, butyou will have to use your own thinking tosolve the problems.

WHAT MAKES A PROBLEMA PROBLEM?

Not every situation we encounter requiresproblem solving. If we asked, "How muchis four plus three?", you would say seven.In fact, you might roll your eyes if we im-plied that this was something for whichyou needed problem-solving techniques.However, one person's firmly understoodfact might be another person's problem.Even the most difficult problem becomesjust another exercise once you unlock thesolution. Exercise and problem are ofteninterchanged, but we distinguish betweenthem by saying that an exercise is prac-ticing an algorithm or technique, whereasa problem has no immediate or obvioussolution. So what makes an exercise intoa problem? How do you know whensomething will require problem solving?What makes a good problem solver?

In order for a situation to be a problemfor you, first it has to be something youcannot answer by blindly applying a regu-lar algorithm to get a viable answer. Fourplus three could require problem solvingfor a child learning addition, but for you itis not problem solving. This criteria is im-portant and means that, if you alreadyhave done a particular type of problem,

203

204 CHAPTER 7

even if you had to do problem solving toget the answer, it is no longer a problemfor you. It might still be a problem for oth-ers who have not seen it or resolved it.

The second requirement is that a prob-lem needs to engage you or get your at-tention. If you are not engaged, or donot accept the challenge the problem pre-sents, then you will not attempt it. Anyonecan be turned off if the problems pre-sented are too difficult or if the learnerdoes not have the necessary backgroundto attempt a solution. Problems must bechallenging, appropriate, and engaging forthe learner.

The third requirement for problem solv-ing is tenacity—you must work on theproblem long enough to come up with asolution. Making one or two attempts at anew problem is rarely sufficient for findinga viable solution. It is important to makeconjectures (guesses that can be tested)about a problem as you work through it.Conjectures, even when they are wrong,often lead to a solution as they may elimi-nate false paths or provide insights toother possibilities. Sometimes we call thisattribute stick-to-itiveness.

POLYA'S STEPS

George Polya (1898-1972) is creditedwith outlining steps for solving a problem.Polya stated there are four phases re-quired for a person to solve a problem, asshown in Fig. 7.1. The first step is to un-derstand the problem. That is, the learnermust read and correctly interpret theproblem. You are probably saying toyourself, "Well, of course you have to un-derstand the question before you can be-gin to solve the problem. How could youeven start if you didn't understand?"

Polya's second step is to make a plan,which will help you solve the problem. We

Polya's 4 Steps for Problem Solving

FIG. 7.1.

will discuss strategies that can be used inproblem solving later. Any plan you de-velop will be based on your prior knowl-edge and experiences and may combineyour ideas in new or different ways. None-theless, you may be saying, "Of course thenext step is to make a plan. What elsewould you do next?"

Polya's third step is to carry out theplan. Your plan may require revision orperhaps you will scrap it and develop anew plan after attempting to carry outyour original one. You may even go backto make sure you understood the problemcorrectly, make an entirely new plan, andthen try carrying it out. You are probablysaying to yourself, "Duh? How else wouldyou ever get a solution?"

We distinguish between the words an-swer and solution. The answer to fourplus three is seven. When you solve aproblem, you come up with a solution.There may be more than one solution fora problem. A solution also implies thatthere is some sort of rationale or reason-ing behind your method.

Polya's fourth step is to look back andmake sure that your understanding, plan,and execution of the plan resulted in a so-lution that makes sense and resolves the

PROBLEM SOLVING 205

original situation. This is the time to lookfor errors. You should also examine yoursolution for faults in your plan or difficul-ties in the execution of your plan. Are yousaying, "I know I should always check myanswers, but sometimes I forget or get ina rush?"

Polya's four steps for problem solvingmake good sense. Understand the prob-lem, make a plan to solve the problem,carry out the plan, and look back andcheck your answer. It seems like just plaincommon sense. Then why do we strugglewith problem solving? Take a look at Fig.7.2, which indicates the interconnected-

FIG. 7.2.

ness of the four problem-solving stepsand you will begin to see that, althoughthe steps themselves seem simplistic,their interconnectedness and interde-pendence veil a complexity the stepsthemselves do not necessarily evoke.

STRATEGIES

Now that we have discussed what makesa problem a problem and the steps youtake to find a solution to a problem, youare ready to explore different strategiesfor solving problems. Will we give everypossible strategy? No. The strategies weare about to discuss, just like Polya's

problem-solving steps, will seem naturaland, for the most part, straightforward.These, coupled with your continuedgrowth in problem solving, should providethe tools you will need to become an ef-fective problem solver.

Probably one of the most popular strat-egies used by students is guess, test, andrevise. After reading, your plan for solvingthe problem is to guess, test the solution,and then revise your guess according tothe results of testing. For example, trysolving Riddle Me This #1: "I am a numberbetween one and twenty. I can be writtenas a base raised to a power to obtain my-self. Exchange the base and the powerand you still get me. The base is not equalto the power. What number am I?"

You can start by guessing any number,say four, our first conjecture. We test fourand find that four is two to the secondpower. The power and the base are equal.Looking back, four is not the solution.You could guess nine as your next con-jecture. Nine is three to the secondpower, but two to the third power is eight,not nine, so looking back, nine doesn'twork either. If you guess six, well there isno easy way to figure out what power youwould raise a number to for six, so thatone gets eliminated in the testing phase.Continue to guess, test, and revise untilyou solve the riddle.

How'd you do? You probably startedthinking about using another strategy tohelp keep track, like making a list. Did youmake a list of the numbers between 1 and20? Another handy strategy for this prob-lem is eliminating impossible answers. Didyou cross out one and the prime numbers?That would leave 4, 6, 8, 9, 10, 12, 14, 15,16, 18, and 20. What could you do next?

Giving you the answer will not assistyou in building problem-solving skills, butwithout the answer, how can you checkyour work? We think the simplest way

206 CHAPTER 7

around this is to put the solutions for eachof the text discussions in a section called"Riddle Me This." That way you can havesomeone else check the solution to see ifyou are right. Of course, after followingPolya's four steps, you already know yoursolution is right!

We are trying to help you think aboutyour thinking, a process called metacog-nition. Thinking about your thinking and be-ing able to communicate about your think-ing is one of the NCTM process standards.Many standardized tests now require ex-planations. It would be wise for you to artic-ulate your thinking process to your partnersin class. Verbalizing your thinking helpsclarify it. When working in a group, it helpsin two ways: first, by letting you know oth-ers were thinking similar things and, sec-ond, knowing what others are thinking re-duces overlap in the group effort.

Making a chart or a table can be helpfulin solving some problem types. Riddle MeThis #2: "Suppose you are in charge of a128-team single elimination softball tour-nament for which all tie games will go intoextra innings until a winner is determined.If no team forfeits a game, how manygames will have to be played before thewinner is declared?" What categories willyou need for your chart or table? Wouldsoftware be helpful?

You may have played in tournamentsand recall the diagrams used to pairteams. If that is part of your previousknowledge base, then you might try solv-ing the 128-team tournament problem bydrawing a picture or diagram. In manyways, your past experiences and comfortlevel with a particular type of strategy de-termines whether or not you use it. Weencourage you to draw pictures as astrategy, because they often provide in-sight that makes a difficult appearingproblem easier.

One of the key strategies used for anyproblem is looking for a pattern. Most

mathematics is based on one type of pat-tern or another. Before you see thepattern, solving the problem seems im-probable, if not impossible. Once you seethe pattern, the problem may becometrivial. Looking for a pattern is a subgoalof almost every problem-solving strategy.You might think of the tournaments inwhich you have played and use them asexamples for the 128-team problem, evenif they involved far fewer than 128 teams.Using a simpler version of a problem canprovide the needed insight for solving acomplicated or complex problem.

Some problems lend themselves to so-lutions involving algebraic models. Re-stating the problem to use an algebraicstrategy more effectively sometimeshelps. Take a look at Riddle Me This #3:"If the length of the Loch Ness monster is20 meters long plus half of its own length,how long is Nessie?" Your understandingof the English language is important insolving problems such as this. Certainlythis problem begs for an algebraic model,but would restating the problem help youavoid an error during the process oftranslating the English into the algebraicmodel? Whereas phrases add descriptiveinformation and clarity to English sen-tences, they are sometimes movable orremovable in mathematical situations.

Some people shy away from doingstory problems because they strugglewith forming algebraic models. Problemsin the real world are not stated in mathe-matical terms. Understanding the Englishstructure of problems goes a long way to-ward helping you become more proficientat solving problems, because it helps youtranslate situations into algebraic terms.

Have you ever wondered what purposethe answers in the back of your bookmight have—other than to see if you areright? One reason we provide a solutionmanual (not just answers, but how we got

PROBLEM SOLVING 207

them) is to give you the opportunity towork backward. Sometimes, if you knowthe answer, then you can work your wayback to the question.

Riddle Me This #4: "You have been con-tracted to use gold to guild the page num-bers in a reproduction of an ancient manu-script, starting with page 1. Because of theexpense and time required for this proc-ess, you will be paid by the number of dig-its you guild. If you guild a total of 642 dig-its, how many pages did you number in themanuscript?" Try your hand at using theworking backward strategy to come upwith a solution for this problem.

Did you figure it out? Perhaps the veryfirst strategy for solving any problem is touse reasoning to determine how to pro-ceed. Few problem-solving strategieswork well in isolation. One of the impor-tant aspects of the reasoning strategy isthe estimation of a solution and checkingthe reasonableness of a solution. An esti-mate of 500 pages for Riddle Me This #4should seem unreasonable to you. Why?Did you use Polya's step number four,looking back, before you looked up theanswer? If so, you put the reasoningstrategy to good use. Every time youcheck back you are using this strategy!

Riddle Me This #5: "In the early days ofmovie making, a villain might do thingsthat cannot be done in real life. In today'smovie making, directors often use consul-tants to avoid this type of blunder. A direc-tor has asked you to determine if it is rea-sonable for the villain to grab a $1,000,000ransom, in one-dollar bills, and run. Youknow a dollar bill weighs about one gram.Is the scene reasonable?"

COLLECTION OF PROBLEMS

The difficulty level of any particular prob-lem is dependent on the backgroundknowledge of the problem solver. Difficult

for one person may be easy for another,not because one is smarter but becausethey have had experiences with a type ofproblem or have the background andtools to solve the problem. Another influ-encing factor, and a big key to solving al-most any problem, is how much tenacitya problem solver possesses. One or twoattempts at a problem are often insuffi-cient. Take a look at all of the problemshere. Anyone can do "easy" problems.These may not be so easy. Take this op-portunity to challenge yourself to do theseproblems, even the ones you perceive as"difficult." Please note they may not bearranged in ascending degree of difficultyfor you:

1. Take an ordinary sheet of paper andfold it in half. Fold it in half a second time.Fold it in half a third time. If you couldcontinue folding it in half 50 times, howhigh will the stack of paper be?

2. A farmer had 26 cows. All but 9 died.How many lived?

3. A uniform log can be cut into threepieces in 12 seconds. Assuming the samerate of cutting, how long will it take for asimilar log to be cut into four pieces?

4. How many different ways can youadd four odd counting numbers to get asum of 10?

5. What is the sum of the first 100 con-secutive counting numbers?

6. How many cubic inches of dirt arethere in a hole that is 1 foot deep, 2 feetwide, and 6 feet long?

7. How many squares are there in a 5by 5 square grid?

8. A little green frog is sitting at the bot-tom of the stairs. She wants to get to the10th step, so she leaps up 2 steps andthen slides back 1. How many leaps willshe have to take if she follows this patternuntil she reaches the 10th step?

208 CHAPTER 7

9. If there are 7 months that have 31days in them and 11 months that have 30days in them, how many months have 28days in them?

10. There are exactly 11 people in aroom and each person shakes hands withevery other person in the room. When Ashakes with B, B is also shaking with A;that counts as ONE handshake. Howmany handshakes will there be when ev-eryone is finished?

11. What number does thisrepresent?

12. There are 9 stalls in a barn. Eachstall fits only 1 horse. If there are 10horses and only 9 stalls, then how can allthe horses fit into the 9 stalls withoutplacing more than one horse in eachstall?

13. You are given 5 beans and 4 bowls.Place an odd number of beans in eachbowl. Use all beans.

14. You are to take a pill every halfhour. You have 18 pills to take. How longwill you be taking pills?

15. If you got a 40% discount on a$150.00 pair of sport shoes and 20% off a$200 set of roller blades, what was thepercent discount on the total purchase(assuming no taxes are involved)?

16. Where should the Z be placed andwhy?

17. Estimate how old will you be inyears if you live 1,000,000 hours?

18. A child has $3.15 in U.S. coins, butonly has dimes and quarters. There aremore quarters than dimes. How many ofeach coin does the child have?

19. There are three children in a family.The oldest is 15. The average of theirages is 11. The median age is 10. How oldis the youngest child?

20. A famous mathematician was bornon March 14, which could be written 3.14.This date is the start of a representationfor pi. It is interesting that this mathemati-cian was born on "pi day." Give his name.

CONCLUSIONS

Polya provided the general four-step proc-ess for solving problems. We added agroup of common strategies and insightsinto what makes a situation a problem ver-sus an exercise. We also gave you oppor-tunities to practice Polya's four-step proc-ess and use a variety of strategies to solveproblems as they were presented to you.As you solve a greater variety of problemsand use different strategies, your problem-solving abilities will grow. It is up to you tobecome a good problem solver, but likeany other worthwhile endeavor becominga good problem solver requires interestand practice on your part.

BIBLIOGRAPHY

NCTM. (2000). Principles and standards for schoolmathematics. Reston, VA: Author.

Polya, G. (1957). How to solve it (2nd ed.). GardenCity, NY: Doubleday Anchor.

8Reasoning and Proof

FOCAL POINTS

. And

. Or• If, Then• Negations• Tautologies• Informal Proofs• Beyond the Informal Proofs. Two-column Proof• Paragraph Proof• Indirect Proof

In a regular conversation, how manytimes have you heard, "That seems logi-cal,"—once, twice, a lot of times? Try torecall why the statement was made. Mostlikely it was made because the conversa-tion followed some sort of procedure oroutlined a step-by-step process thatseemed to make sense. In mathematics,we use truth tables to determine if state-ments are true or false. In some ways,your intuition about logic will serve youwell as you explore the ideas behind rea-soning and proof. However, you may findthat, when dealing with compound state-ments, the rules of formal logic differ fromeveryday conversation.

AND

"And" statements in logic are called con-junctions and they are linked to the ideaof intersection. The symbol logicians usefor "and" statements is A, which makessense, as it closely resembles n, the in-

tersection symbol for sets. Whereas anyvariables can be used to make logicalstatements, you will find that logicians arefond of the letters p and q when they talkabout ideas in general. Using the letters pand q to represent sentences or parts ofsentences shortens our work. For exam-ple, use "Today is Friday" for p and "It issunny" for q. Our first logical expression pA q is read "p and q". But what does p A qmean? Inserting the sentences into thesituation helps sort out the meaning,based on what is said in English.

We can figure out whether p A q is atrue statement or a false statement by as-signing truth values for p and q. A truth ta-ble is the tool of choice for this procedure.To construct a truth table, you assign oneof two values to p and to q—true (T) orfalse (F). It is possible to say whether thestatements, "Today is Friday," and "It issunny," are true or false. Because eitherstatement might be true or false while theother is true or false, there are severalpossible combinations. Table 8.1 orga-nizes every combination of true and falsefor p and q. The third column helps us in-terpret the meaning of the compoundstatement p A q. We can generalize fromthis table that a conjunction is true only

TABLE 8.1

P

TTFF

9

TFTF

P ^Q

TFFF

209

210 CHAPTER 8

when both p and q are true. In otherwords, everything must be true or thestatement using "and" is false. If you arereading this on a sunny Friday, the con-junction is true. Otherwise, one or both ofthe statements is false and the conjunc-tion is false.

You have already explored the commu-tative property of addition on differentsets of numbers. Do you thinkp A q = q A p? Why do you think so? Doesit seem logical to you? What would thetruth table look like? Does your hypothe-sis match Table 8.2?

TABLE 8.2

Q

TFTF

P

TTFF

q ^P

TFFF

Will the idea of the associative propertyof addition on different sets of numbers,also hold true as conjunctions are consid-ered? We will use r as the third letter forthis property. In other words, is p A (q ^ r)equivalent to (p ^ q) ^ r? A truth table forthree variables is shown in Table 8.3.

TABLE 8.3

PTTTTFFFF

QTTFFTTFF

r

TFTFTFTF

Your Turn

1. Using p, q, r, and the truth table inTable 8.3, determine if the associativeproperty is true for conjunctions. In other

words, does the following statement hold:p A (q ^ r) = (p ^ q) ^ r. (Hint: You will needcolumns for both q ^ r and p ^ q.)

2. Using what you have learned aboutprobability, what is the probability of get-ting:

a) a true "and" statement (p A q) withtwo variables?

b) a true "and" statement with threevariables?

c) a false "and" statement with twovariables?

d) a false "and" statement with threevariables?

OR

Just as conjunctions, which use "and" asthe connecting word, are related to inter-sections of sets, statements using "or" asthe linking word are related to unions ofsets. Logicians call "or" compound state-ments disjunctions and use the symbol vwhich is similar to the symbol for union(u) used when operating on sets. Take alook at Table 8.4. How is the result of p v

TABLE 8.4

P

TTFF

qTFTF

P v q

TTTF

q, read p or q, different from p ^ q (p andq)? You might find it helpful to compareand contrast Table 8.4 with Table 8.1 asyou think this through. We can generalizefrom Table 8.4 that a statement using "or"as the linking word is true as long as atleast one of the variables is true. In otherwords, the only time a disjunction is falseis when all of the variables are false, justas the only time the union of two sets isempty is when both sets are empty. Does

REASONING AND PROOF 211

Table 8.4 make sense based on the Eng-lish using "It is Friday" for p and "It issunny" for q?

IF, THEN

Compound statements connected using"if" and "then" are called conditionalstatements and are written p -> q, read,"If p, then q." English sentences aid un-derstanding these, also. This time we willuse "You live in North America" for p and"You live in the United States" for q. Fourdifferent possibilities exist:

If you live in North America (true), thenyou live in the United States (true).

If you live in North America (true), thenyou live in the United States (false).

If you live in North America (false), thenyou live in the United States (true).

If you live in North America (false), thenyou live in the United States (false).

The conditional statement for all four ofthese is: If you live in North America, thenyou live in the United States. The "if" por-tion of the statement is called the ante-cedent, or hypothesis, of the conditionand the "then" part is called the conse-quent, or conclusion, of the condition.The first of the four possibilities, "If youlive in North America (true), then you livein the United States (true)," would makesense, because both the antecedent andthe consequent of the condition are true,the conditional statement is true.

Looking at the second of the four pos-sibilities, "If you live in North America(true), then you live in the United States(false)," implies that you live in NorthAmerica, but do not live in the UnitedStates. In other words, if you live in Can-ada, the original compound statement isfalse. This may seem counterintuitive at

first glance. Remember that we are deal-ing with a very specific case here. Wehave to stick strictly with what is written inthe original compound statement andmust not reword, make leaps of faith, orassume Canada into the statement. Inthis case, the statement is that you live inNorth America and also in the UnitedStates. The conditional statement is falsebecause the antecedent is true and theconsequent is false.

The third statement is "If you live inNorth America (false), then you live in theUnited States (true)." You might think thatthis has to be false because, if you do notlive in North America, then how could youlive in the United States? Well, maybe youlive in Hawaii! Logically, when we look at aconditional statement, if the antecedent isfalse, and the consequent is true, then thestatement is true.

Finally, the fourth statement, "If you livein North America (false), then you live inthe United States (false)", is a true state-ment because if you do not live in NorthAmerica, most likely you do not live in theUnited States. You could live in Hawaii,but, when the antecedent is false and theconsequent is false, the conditional state-ment is logically true. Thus, if the ante-cedent is false, then the conditional state-ment will be true regardless of whether ornot the consequent is true or false. All fourof these situations are summarized in atruth table in Table 8.5. Did you noticethat the only time a conditional statementis false is when the antecedent is true andthe consequent is false?

TABLE 8.5

PTTFF

q

TFTF

p ->gTFTT

212 CHAPTER 8

NEGATIONS

Logicians have devised a concise way forreversing the meaning of a statement. Ne-gation is used to convey the messagethat the opposite of what was said is whatwas meant. That is, if you say, "It issunny," then the negation is, "It is notsunny." Similarly, if you say, "It is notsunny," the negation is, "It is sunny." Wesay that negation changes the truth valueof a variable or statement. Although thesymbols ~p, p', -p, and can be used toshow negation, the most common one is~p. Table 8.6 shows the truth table for thenegation of a variable.

TABLE 8.6

PTF

~P

FT

You can think of the negation as actinglike the minus sign in front of a number.You know that -(-p) = p is true, so it fol-lows then that ~(~p) = p. How about~(~(~(~(~p))))? Because there are an oddnumber of ~ signs (5 in all), then~(~(~(~(p)))) = ~p.

TAUTOLOGIES

Take a look at Table 8.7. Each entry in thelast column is True. This special case iscalled a tautology, which has to be trueall the time. In Table 8.7, the column with(p A q) (p v q) has all true values. Com-

TABLE 8.7

pTTFF

qTFTF

P A q

TFFF

P vq

TTTF

(p A q) (p

TTTT

vq)

paring the first and second columns withthe last column, the statement (p A q)(p v q) is true no matter what truth valuesare assigned to statements p and q. Tau-tologies represent basic laws of logic.

The commutative and associative rulesfor conjunctions you explored earlier arealways true, and form two tautologies,(p A q) (q A p) and p A (q A r) (p ^ q) A r,respectively. The symbol is read, "Ifand only if," and is similar to the equalsymbol used with ~(~p) = p. Another tau-tology is implied in the discussion aboutnegation, ~(~p) p. Were you able to un-derstand the uses of the two symbols (im-plies [ ] and logically equivalent [ ]) thatwe have been using?

LOGICAL EQUIVALENCE

Compare Table 8.8 with Table 8.9. Noticethat the truth values for the p v q column

TABLE 8.8

pTTFF

P

TTFF

qTFTF

q

TFTF

TABLE

p v q

TTTF

8.9

~p ~p –> q

F TF TT TT F

and the ~p –> q columns are the same.When this happens, the two compoundstatements are said to be logically equiv-alent, symbolized by +*. Putting two logi-cally equivalent statements in a truth tablewill always generate a tautology, such asthe one in Table 8.10. This allows the re-placement of a statement with a logically

REASONING AND PROOF 213

TABLE 8.10

pTTFF

9

TFTF

p v q

TTTF

~P —> 9

TTTF

(p v q) (~p –>q)

TTTT

equivalent statement without affectingthe truth value of the original compoundstatement. As you move through higherstages of proof, this capability will be-come a very handy tool.

Your Turn

3. Experiment with the different typesof logical expressions that have been dis-cussed. Find at least two different tautol-ogies that have not already been dis-cussed. Make a truth table for each toprove it is a tautology.

INFORMAL PROOFS

Proof can be a strange word, at times. Itcarries various assumptions and mean-ings with it, and sometimes it is difficult todiscern the exact meaning. There is aneed for proof in the study of mathemat-ics. Otherwise, we arrive at incorrect con-clusions. Mathematical proofs come in avariety of formats, the most fundamentalof which is an informal proof. With an in-formal proof, we might see compellingevidence that something is so but, at thislevel, it is possible that an exception ex-ists somewhere.

Consider the medians of a triangle asshown in Fig. 8.1. A median is a segment

FIG. 8.1.

that connects a vertex with the midpointof the side opposite the vertex. It looks asif the three medians in each of the trian-gles are concurrent (meet at a commonpoint). If you use a dynamic geometryprogram and move the vertices of a trian-gle around, then the medians will appearto be concurrent. As compelling as thismight be, it is not a proof. We can onlyprove by example if we do every possibleexample and, even with a dynamic geom-etry program, you can't examine all possi-ble triangles. Zoom in really close to asuspected point of concurrency—couldthere be a tiny triangle formed where thepoint of concurrency appears to be? Inthis particular example, we know that willnot happen, because a formal proof hasbeen done to show that the medians ofany triangle are concurrent. Although theformal proof is beyond the scope of thisdiscussion, you need to realize the signifi-cance of that last statement.

Whereas you are unable to prove thatany particular statement is true by usingany number of examples (unless you doevery possible one), you are able to dis-prove any statement with one single coun-terexample. This is not only true for an in-formal proof, but for any formal proof aswell. Proof and disproof by counterex-ample are powerful devices in a mathe-matician's tool belt. You can show thatthe set of digits is not closed for multipli-cation using a counterexample such as3 x 4 = 12. A mathematician might writethis proof: Suppose the operation of mul-tiplication is closed on the set of digits.Arbitrarily choose 3 and 4 to multiply to-gether, yielding 3x4 = 12. Because 12 isnot a digit, we have a contradiction. Thus,the operation of multiplication is notclosed on the set of digits. (QED, quid er-rata demonstratum—the proof is finished,which was to be shown, quite easilydone, quit exit desist.)

214 CHAPTER 8

Your Turn

Informally prove or disprove the followingstatements:

4. The operation of addition is closedon the set of whole numbers.

5. The operation of addition is closedon the set of digits.

BEYOND THE INFORMALPROOFS

Formal proofs can explain many things.Do what is asked in each of the followingsteps:

Pick a number.

Triple it.Add 12 to the product.Divide the sum by 3.

Subtract your original number.What is the result?

If you followed the instructions, and did allarithmetic correctly, then you should haveended up with 4. Many people, young andold, are fascinated by number tricks. Of-ten, after expressing amazement, theywill ask if it always comes out that way.That is a paraphrasing of a request for aformal proof. In this particular case, alge-bra is helpful to see how and why the trickworks:

This trick will work no matter what num-ber you pick for x. Can you explain why?

The next time you hear a number trick,you should ask yourself why the numbertrick works. We have used number tricksas examples several times. It would begood practice for you to see if you can fig-ure out how they all work. The more skillyou develop in determining how thingswork, the more successful you will be inyour pursuit of mastering the wonderfulworld of mathematics. As you investigatehow things are put together, you are be-ginning to formalize proofs.

Without proof, sometimes you arrive atfalse conclusions. For example, lookcarefully at each of the following prob-lems, trying to find some things that aretrue for each example:

In each case, the denominator of the sumis the product of the denominators of theaddends. Also, in each case, the numera-tor of the sum is the sum of the denomi-nators of the addends. Will this alwayswork? You might wonder why no one hasexplained fraction addition this way be-fore because this looks a lot easier thanthe set of rules you learned. We kind ofset you up for that one—in these specialproblems, only unit fractions are being

added. If the problem had been

then the answer would have been and

in this more general case, the proceduredoes not work. We set you up to thinksomething would be true and thenshowed you it was not to amplify the needfor proof. Remember, it takes only one

REASONING AND PROOF 215

example where something is false toprove that it is untrue.

Here is another example in which wemove beyond an informal proof. Given theset of numbers in Fig. 8.2, select any one

FIG. 8.2.

of them and eliminate all the other num-bers in the row and column occupied bythat number. There will still be some num-bers left, so select one of those and elimi-nate all the numbers in the row and col-umn in which that number appears. Youwill still have four numbers that have notbeen crossed out. Select one of them andcross out all the numbers in its row andcolumn. Now you have one number left.Select it, too. The sum of the four num-bers you selected is 100. Are you wonder-ing how that worked? We will show you(prove to you) why the answer for thisproblem will always be 100.

Look at Fig. 8.3. You see that 33 hasbeen selected as the first number and all

FIG. 8.3.

the other values in its row and column ofthe original table have been crossed out.But Fig. 8.3 has some additional numbersthat were not shown in Fig. 8.2. That isbecause this is an addition table. Thenumbers in the row and column that were

hidden in Fig. 8.2 are the addends, butthe shown numbers are the sums of theaddends in any given row and column.When the row and column for 33 are elim-inated, so are the addends (the shaded 31and 2 in this case). Those addends willnot be used again in this application. Se-lecting another number and eliminating itsrow and column will eliminate two more ofthe initial addends. When all four numbersare selected, all eight of the original ad-dends will have been accounted for. Weknew the answer had to be 100 (barringarithmetic errors). It does not matterwhich numbers you pick, because17 + 2 + 16 + 9 + 4 + 13 + 31+8 = 100.

Your Turn

6. Make up a 5 by 5 addition table likethe one show in Fig. 8.2 (be sure the ad-dends are hidden) and give it to someonewho is not in this class to do, following thedirections given for Fig. 8.2. Record theirreaction to the trick.

Two-Column Proof

After discussing informal proofs, wemoved to a new level where we startedshowing why things work. The next stepin that progression is the formal, two-column proof. In this approach, a state-ment is given, accompanied by a reason,for each step that is taken in a logical ar-gument, taking you from a beginningpoint to a conclusion. All forms of proofs,including two-column proofs, have strictrules that must be followed. Statementsthat are given to you as a part of the prob-lem are accepted as true. Next, you havea set of rules (axioms, theorems, and pos-tulates) that you are permitted to use.These will be statements that you haveproven earlier or know to be true. Thisframework is used to build an argumentto support your proof.

216 CHAPTER 8

For example, suppose that you wantedto prove that, for all real numbers, a, b,and c, if a = b, then it must be the case

that as long as c * 0. Before begin-

ning a formal proof, you need to deter-mine what is given. In this case, you havethat a, b, and c are real numbers, a = b,and c 0. Those facts, along with priorwork, are all you have to use as you startyour formal, two-column proof:

You might say it all makes sense and youknew that. Well and good! However, aswe build a formal system, we need to gobeyond what makes sense and is intuitiveto establish a carefully developed set ofideas and rules that build an unshakeablefoundation. It should be noted that someproofs would have combined the twogivens into one step. That decision is leftto personal preference.

You might wonder why we opted togive an algebra proof as our first exampleof a formal proof. We did this becausemost people have not seen two-columnproofs unless they took high school ge-ometry. The exposure to two-columnproofs in high school geometry coursescan be both intimidating and formidable.If, on the other hand, students are ex-posed to proofs, both informal and for-mal, before getting to high school geome-try, then two-column proofs can be muchmore palatable.

Reconsider the trick where you pickeda number, tripled it, added 12, divided thesum by 3, and subtracted your original

number. We replicate the explanation ofthat trick below, but replace the x with atriangle, which might make the whole ap-proach appropriate for elementary schoolstudents. The point is, proofs should bestarted much earlier in the curriculumthan students currently experience. Earlyexperience with proofs, at appropriatelevels, would be a huge asset for studentstaking subsequent classes that empha-size proof.

A proof of this sort, shown after studentshave been intrigued by the trick, couldspark the beginning of something won-derful.

Pick any two odd counting numbers andmultiply them. Repeat the process withdifferent pairs of numbers until you arriveat a conclusion about the respective prod-ucts. The conclusion you just generated issomething that appears to be true, basedon your observations of several similarproblems. Suppose, however, that youwanted to prove your conjecture.

REASONING AND PROOF 217

It is important that you start provingthings you know to be true so you canbuild the skills that will help you provethings you do not know to be true. It is likeyou are training for some athletic event—you need to get the right muscles tunedup before you take on the challenge of theevent.

Paragraph Proof

The two-column proof is not the only wayto prove things. They are handy whenthings are long and complex. Sometimes aparagraph can be written as a means ofpresenting the logic involved. Supposeyou are given Fig. 8.4, in which ACB is

FIG. 8.4.

a straight line and there are the 4 anglesshown as a, b, b, and a. A paragraph proofthat could be developed.

Given that ACB is a straight line,

Like we said, paragraph proofs are ac-ceptable and a matter of preference. Oneargument against them is that the reasonsfor the steps are not presented. They canbe inserted with each part, but then theproof is really a two-column proof, justnot written in columns.

Indirect Proof

You may have been using a form of indi-rect proof for some time without realizingit. Consider the multiple choice question,"When was the classmate you just metborn?" The four options for your answerare: (a) 1492, (b) 1776, (c) 1983, (d) 2054.You can arrive at the conclusion indi-rectly. It is highly unlikely that individualsin your class are over 100 years old, andthat rules out answers a and b. Similarly, itis highly unlikely that one of your class-mates would have been born in 2054, be-cause that would imply the individual is atime traveler. Thus, answer d is not an op-tion. This process of elimination leaves cas the only logical answer and you wouldsay your new classmate was born in1984. In an indirect proof, all the possibili-ties under consideration are examined.Rule out each item that contradicts somefact or rule. If all options except one canbe eliminated, then it must be acceptedthat the one remaining option must betrue, and the proof is complete.

CONCLUSIONS

The basic ideas of proof have been estab-lished. You need to start asking whythings work out as they do in the world ofmathematics. Then, after you ask, youneed to begin a course of investigationthat will logically answer your question.You will find that, with repeated effort, theactivity will become easier and easier foryou. Like any routine, you learn to excelthrough practice. Get ready. Get set. Go!

This page intentionally left blank

9Communication

"I know what I want to say but can't findthe words." Have you ever said that toanyone? How frustrating it is to be unableto put knowledge into a concise set ofwords. We do not use only words to com-municate. We use gestures, body lan-guage, objects we hold and point to,things we refer to, and a host of othermeans, all designed to get our messageacross.

OK! If you checked the Table of Con-tents, you found that this chapter is theshortest in the whole book. Please do notassume this chapter is less important be-cause it is shorter than the others. Mathe-matics as communication has been dis-cussed implicitly on almost every page sofar! We include this chapter because wewant you to think explicitly about mathe-matics as communication.

Most of us have learned mathematicsby observing teacher demonstrations.Perhaps that is why things like Cuisenairerods, Base 10 blocks, dynamic software,and models seem foreign to you. Thereare places where teacher telling is an ap-propriate method of instruction. But, it isnot the only way.

How do teachers communicate? Thefollowing vignette will help make a point.A teacher is explaining something to aclass. After the presentation is com-pleted, the teacher asks for questions.The dialogue between the teacher and astudent is:

S: "I don't get it."

T: (The teacher is quite patient with allsuch questions. The explanation isrepeated, almost verbatim from thefirst one.)

S: "I still don't get it."T: (The explanation is repeated again,

still calmly and patiently, again, al-most verbatim.)

S: "I still don't get it."T: (Exasperation is beginning to show

but patience persists and again, thematerial is covered, almost verba-tim.)

S: "I still don't get it."T: (Finally, pushed beyond the limit)

"Haven't you been listening?"

The teacher is failing miserably when itcomes to communication. Perhaps a ver-batim explanation is in order the first timethe student asks for clarification. How-ever, after that, it should be clear that acommunication gap is present. One solu-tion is to explain the material a differentway. That mandates that you understandmathematics. Do you see now why wehave been presenting so many differ-ent ways of approaching topics? May-be this student needs to see the conceptat a concrete, semi-concrete, or semi-ab-stract stage. Something different needsto be done. It is the teacher's responsi-bility to portray knowledge in a mannerthat is understandable. Teacher tellingis not necessarily equivalent to studentlearning.

219

220 CHAPTER 9

We were mighty hard on the teacher aswe went through the discussion. The stu-dent bears some responsibility to com-municate, too. "I don't get it," is not veryhelpful or specific. We can develop bettercommunication by trying to elicit morespecific comments. Ask things like,"Where did you begin to get confused?"or "Tell me what you do understand."These questions begin to force thinkingabout what is going on and to expressideas more specifically.

You probably have never consideredmathematics as communication.Learning to do mathematics, without un-derstanding how and why it works, ismerely learning to complete a mechanicaltask; any computer can do the mathemat-ical manipulations quickly and accurately,but only a human can understand and ex-plain the how and why. Mathematics is anessential way of expressing ideas and weall, at one time or another, communicatequantitative and qualitative ideas, argu-ments, concepts, or requirements. Evenas you discuss the fact that you must takesome college courses dealing with math-ematics in order to earn your elementaryschool teaching degree, you are usingmathematics to communicate. Studentstell us many things as we try to lead themto view mathematics as a way of commu-nicating. "No one ever asked me to ex-plain how I got the answer," "I never gotto discuss my answer with someone elsebefore," or "I know what I want to say butcan't find the words" are commentsheard repeatedly. Mathematical commu-nications are embedded in each of thefollowing:

Negotiate with a bank teller about howyou want your change

Explain how to navigate through abuilding to a given location

Brag about how quickly you traveled

Boast about what great gas mileageyour car gets

Sketch a map

How can we promote mathematicalcommunication? Begin by asking for ex-planations of the mathematics beingdone. Promoting communication must bean ongoing process. It cannot be a one-time shot. Continually build verbal andwritten explanations of concepts. Learnhow to ask questions.

Earlier we mentioned Gauss' procedurefor finding the sum of the first 100 con-secutive counting numbers. We provideda discussion of how Gauss communi-cated with himself to solve the problem.He knew what questions to ask and howto answer them. That is communication!

Discovery has a place in learning howto communicate mathematically. That isone reason why we have talked so muchabout investigation of patterns. Manythings can be extracted from a set ofnumbers when a pattern is spotted,sometimes even going to the point ofgenerating a formula that works for allsimilar situations. That is really one of thethings mathematics is all about—reduc-ing a set of experiences to a formula orrule that makes the solving of similarproblems simple. Then, when it is gener-alized, it is easily communicated.

An unwanted tradition is inherent in theworld of mathematics. Ask someone whatthey think of when you say mathematicsand most people will say numbers, arith-metic, or formulas. Think back over themathematical instruction you had in thePre-K-16 environment. Most of it proba-bly focused on numbers. How manytimes did you write an explanation of howto do a problem? Had you heard of a

COMMUNICATION 221

paragraph proof before the section onProof and Reasoning?

We could go on, but you get the point.Communication is everywhere: in mathe-matics, in your daily conversation, in thereal world, and in the list of responsibili-ties that you inherit as a future teacher.

Your Turn

1. Write a paragraph explaining howyou would add 842 + 136. Put the para-graph aside and don't look at it for 24hours. Is your paragraph organized and

coherent? Does it concisely explain whatyou did and why you did it?

2. Exchange paragraphs with a class-mate. Do you understand that person'sexplanation? Ask questions that mighthelp the person clarify their explanation.

3. Compare the two paragraphs, iden-tifying the strong points of each. Reasontogether logically about disputed points.

4. Using the strong points of each, col-laborate to write a paragraph that is betterthan either of you wrote alone. Ask some-one who is not in your class to use theparagraph to complete the addition prob-lem.

This page intentionally left blank

10Connections

FOCAL POINTS

• Independent or Interconnected Topics• How Are Things Connected?

People often see mathematics as being acollection of independent skills or activi-ties. Individuals who visualize mathemat-ics as a bunch of unrelated facts, skills,and procedures to be memorized cannotattain mathematical maturity. Mathemat-ics is not a collection of separated topics;it should be viewed as a comprehensiveand coherent science without which hu-man progress would be severely limited.

INDEPENDENT ORINTERCONNECTED TOPICS

Why do people have trouble thinking ofmathematics as interconnected and inter-dependent concepts? One culprit is thecurriculum that is taught in the UnitedStates. It is common for mathematics tobe taught as a collection of topics thathave no apparent connection. Do you re-member how you learned your multiplica-tion facts? Was there any discussionabout how 3 x 5 is a shortcut method ofadding 5 + 5 + 5? Even if the repeated ad-dition definition for multiplication is men-tioned early in the curriculum, the con-nection is made simply to help childrenwith the transition from addition to easymultiplication facts, it soon fades away.Although there are other ways to definemultiplication, the repeated addition

method strengthens the connection toprior work. One of the most effectiveways to learn any subject is to take a newtopic and relate it to previously masteredwork. The reference makes the new workseem familiar and, therefore, not so fright-ening. As the connections between addi-tion and multiplication are strengthened,the learner may discover the advantage ofusing multiplication as a shortcut, aban-doning the old method in favor of the newapproach. This may not happen immedi-ately. A learner may not see much advan-tage to abandoning 5 + 5 + 5 in favor of3 x 5 , but might not be overly thrilled to do9 x 47 in repeated addition form (47 + 47+ 47 + 47 + 47 + 47 + 47 + 47 + 47). If us-ing repeated addition to complete 9 x 47is preferred, then that is OK. It might beconsidered a less preferred method, but itis a valid strategy. Perhaps we can con-vince the learner to change routines bygiving problems with larger factors. Ulti-mately, everyone will be ready for an eas-ier way to do the problem. At that point,we are happy to provide the world's slick-est, quickest way to do the problem—multiplication. Mission accomplished andconnection made.

Stick with multiplication a little longer.The problem 3x5 could be solved usingan array as shown in Fig. 10.1. Countingthe unit squares is a reasonable way todetermine how many are needed. A dis-cussion could focus on 3 rows with 5squares in each row, 3 sets of 5, or threefives—each of which provides a total of

223

224 CHAPTER 10

FIG. 10.1.

15 unit squares. The example could alsobe viewed as 5 columns with 3 squares ineach column, 5 sets of 3, or five threes.Such a discussion provides a verbal tran-sition that solidifies the connection be-tween addition and multiplication. Haveyou noticed the connections that havebeen made throughout this text? We havebeen building them with you throughoutour discussions. We purposely waited un-til now to call this to your attention. Wewanted to show you how easily connec-tions can be inserted into the curriculum.

HOW ARE THINGS CONNECTED?

We used Fig. 10.1 as a means of generat-ing a multiplication fact, but it also modelsthe area of a rectangle that has a base of 5units and a height of 3 units (or is that alength of 3 and a width of 5?) and an areaof 15 square units. How is it that the pic-ture is the same and yet we are movingfrom the concept of multiplication to thearea of a rectangle? Surprise! Connec-tions! Hold on, we are not done. The prod-uct of almost any 2-factor multiplicationexercise can be expressed as the area of arectangle (would ~3 x ~4 give the area of arectangle?). Suppose the dimensions are3.5 units and 6 units. Figure 10.2 modelsthis situation. Sure, there are partialsquares, but two of the 6 partial squarescan be put together to make anothersquare because each partial square is a

FIG. 10.2.

half of an original square. Do that threetimes, add the 18 squares that are clearlywhole, and you end up with a total of 21squares. What is the product of 6 and 3.5?It better be 21. By the way, the results are

the same if you use I as the factor

to be multiplied by 6. The numbers canbe made to look more complex but thestory is the same—you end up with thearea of a rectangle when you do a two-factor multiplication problem. Now youhave another connection—from repeatedaddition through multiplication to areas ofrectangles. During your work in this text,you connected between decimals, im-proper fractions, and mixed numbers inmultiplication. These simple, as well asmuch more complex, connections aboundin mathematics. You only have to learn torecognize them. Few mathematical con-cepts can be understood in total isolation.

Another connection grows from the ar-ray definition used for multiplication andarea of a rectangle. We have said that thedimensions in Fig. 10.2 were 3.5 units by6 units. We counted the unit squares, butwe did not extend the discussion of di-mensions to the product of 21 unitsquares. An important part of learningabout area involves understanding whatis happening with the units. This can beestablished by remembering what hap-

COMMUNICATION 225

pened when you learned addition facts bycombining 3 blocks and 2 more blocks toget 5 blocks. That verbiage is fairly naturaland many students will say it reflexivelyas they do the problem or answer thequestion. Extend that to the idea of find-ing the sum of 3 inches and 2 inches. Forsome reason, there is a temptation toomit the inches and just give 5 as thesum. Recording the answer as 5 inches issignificant and must not be overlooked.The connection can be strengthenedwhen working with something like Fig.10.3. A key element of the discussion isthe length of the resultant vector, which isthe sum of the two shorter ones. Whenthe answer is given as 5, an immediatequestion should be, "5 what?" so that theunits are emphasized.

This discussion can be extended toarea problems (grid paper is handy here).The readiness skill for dealing with unitsgrows from experiences such as our dis-cussion about Fig. 10.3. In the two-

FIG. 10.3.

dimensional concept of area, the unit is asquare, so area is expressed in terms ofhow many little squares it takes to coverthe shape. The central issue involves notonly how many little squares are needed,but also the length of a side of a littlesquare. If the length of the side is an inch,then the area discussion can be builtaround square inches. You have con-nected the little square to square inches,the unit for area discussions. If the lengthof the side of a unit square is changed toa foot, the discussion must be builtaround square feet. It takes 144 inch

squares to cover a foot square, thus im-portant information is contained in theunit provided. Depending on the problem,the length of the side of a unit squaremight be a meter, centimeter, yard, mile,or paperclip. Sometimes we use the wordunit as a placeholder. Initially we write thearea for the rectangle in Fig. 10.2 as 21square inches or 21 square units. Even-tually, the notation is shortened to 21units2 or even 21 u2. This shorter notationshould be allowed only after the idea ofincluding units in every answer is firmlyentrenched and the concept of exponentshas been developed. Did you notice aconnection to the concept of denominatenumbers during this discussion?

How can unit squares be used to ex-press the area of a figure other than a rect-angle? Connecting the area of a triangle tothe area of a rectangle helps. Start with asheet of rectangular paper and fold the pa-per along the diagonal, as shown in Fig.10.4. Cutting along the fold and rotating

FIG. 10.4.

one piece shows that the two triangleshave the same area. One of the triangleshas an area that is half of the area ofthe original rectangle and the formula

is established. Un-

derstanding this connection does notcome as quickly as the time you took toread those last two sentences. Some stu-dents will struggle to see how the littlesquares can fit in a triangle. A different de-

226 CHAPTER 10

FIG. 10.5.

velopment of this, as shown in Fig. 10.5,might help. Scene 1 shows a rectangle,which happens to be a square with an areathat equals 16 u2. Scene 2 shows a diago-nal along which we might fold the rectan-gle in half. Scene 3 shows the little trian-gles, which are half unit squares, shaded.Half of the rectangle is removed in Scene 4to make it easier to see the triangle. Men-tally visualize the little triangles beingpaired together to make squares. The areaof the triangle can be counted—and is 8square units, exactly half the area of therectangle. Scenes 5 and 6 might helpsome people with the visualization, butmight confuse others because the trianglehas been rearranged into a rectangle.Scene 6 shows that none of the pieces arechanged in size. Again, the area is exactlyhalf of the original rectangle.

If repeating decimals are explored, thena rule is provided, and the discussion isterminated long before the connectionsbetween topics are understood. Often ashortcut is provided for convertinq re-peating decimals, such as tocommon fractions. The rule states thatyou count the number of digits involvedin the repeat and write a fraction withthat number of 9s in the denominator andthe repeated value as the numerator. So

set of three examples, each of the frac-tions has common factors. Both thenumerator and denominator divided by 3for starters (use the divisibility rule for 3 asa quick check of that statement).

Regretfully, using the rule stops short ofshowing connections. Why does the rulework? How did they come up with usingso many 9s as a denominator? Algebraanswers those questions. For example,take a look at

Let which we will call equation1.

If we multiply both sides of Equation 1by 10, we get Equation 2.

If we multiply both sides of Equation 1by 100, we get Equation3.

Although we could multiply by any value,Equations 2 and 3 provide the directionwe need. Using some algebraic skills,consider the following:

This process explains the derivation ofthe 9s rule. If Equation 2 had been usedrather than Equation 3, we would havehad

which presents a prob-lem. When Equation 3 is used, the valuesto the right of the decimal place in bothequations line up under each other. Thatis, there is a 5 subtracted from a 5 or a 7subtracted from a 7, regardless of theplace value and that assurance exists nomatter how far the repeat is extended.The subtraction is easy because despite

In this

COMMUNICATION 227

the place where the subtraction is started,the missing addend is zero. On the otherhand, if Equation 2 is used, the decimalplaces do not line up properly for the sub-traction.

That gives a part of the whole picturebut there are other problem types thatneed to be discussed. Suppose the taskis to convert to a fraction. With ex-panded notation, becomes

and when the rule is applied,

This does not address the all-importantquestion of how that works. The otherway to do this problem is use the processof multiplying x = 4.86 by some power of10. The question of which power to use isanswered by the need to align the 8s and6s to the right of the decimal point so thesubtraction will yield zeros only to theright of the decimal point:

The important distinction between thetwo methods is that the rule did not showwhy the answer came out as it did.

Although these examples do not ex-haust the discussion on converting a re-peating decimal to a fraction, it hasshown how one topic within mathematicscan be used to develop understanding inanother. We encourage you to explorethis topic in greater depth. As we learnmathematics today, these explanationsare critical to overall understanding ofwhat is happening. And, at the same time,connections are amplified.

CONCLUSIONS

In the multiplication of whole numberssection of this text, we discussed partialproducts, a connection to algebra. Wediscussed patterns in several places andended with extensions, rules, or formulas.More connections. They are everywhereand all you have to do is notice them. Youmight find that you will actually get ex-cited about some of the connections yousee and the mathematics you are learningand discovering in the world around you.Aaaahhhhhh, connections, they really areeverywhere!

This page intentionally left blank

11Representation

FOCAL POINTS

. Different Ways of Saying the SameThing

I am one of the authors of this book. Myname is Douglas Kent Brumbaugh. I havebeen addressed as Doug most of my life.In my younger days, if I was not behavingas I should have been, I heard Douglas. IfI was more out of line, I heard DouglasKent. If I was pushing the envelope to thelimit, I heard Douglas Kent Brumbaugh.My dad and my grandfather wanted toname me Mike, but they lost that discus-sion. Undaunted, they called me Mike. Myfriends awarded several different nick-names to me; many of my high schoolstudents called me Coach and some ofmy college students address me as DocB. Over the years, I have answered to allof them. I am still me, no matter what I amcalled.

DIFFERENT WAYS OF SAYINGTHE SAME THING

The notion of representation is somethinglike all my names discussed earlier. "Theterm representation refers both to proc-ess and product—in other words, to theact of capturing a mathematical conceptor relationship in some form and to theform itself. . . . Moreover, the term appliesto processes and products that are ob-servable externally as well as to those thatoccur 'internally,' in the minds of people

doing mathematics" (NCTM, 2000, p. 67).Figure 11.1 shows how a child showed

FIG. 11.1. Principles and standards for mathe-matics (p. 67), by National Council of Teachersof Mathematics, 2000, Reston, VA: Author.

her age to be 5 and one half. It appears asif the girl has begun to establish a feel forthe idea of a half by how she writes thesecond part of her age. Her style seemsstrange and yet, apparently, she knowsthat she is 5 and some more, and that themore is a half. She seems to be attempt-ing to convey that message by writing halfof the numeral 5. It makes sense, doesn'tit? And yet we adults would write 5.5 or

and might not have understood her

representation without an explanation.Representation consistently moves

from concrete to abstract as new con-cepts are encountered and understood.The ultimate goal is to have you functionabstractly with your mathematical under-standings because that is the most con-venient and accepted way to communi-cate about such things. On the otherhand, the main idea is to communicatethe idea under consideration. As your un-derstanding of a mathematical concept

229

230 CHAPTER 11

increases, you progress to a more so-phisticated or acceptable representationof the idea.

Young children represent numbers ofobjects using sets of items. They moveon to using pictures of the sets, tallymarks to represent the pictured objects,and finally to a numeral, which is a short-cut way of showing the tally marks—which represent the pictured objects,which represent the objects themselves.The last sentence describes parallel de-velopments that might take years to ac-complish, depending on the complexityof the concept being considered. Thinkback through the things covered in thistext. You should be able to conjure upideas that show the development of thedifferent representations.

Technology comes into play as ameans of representing ideas too. Con-sider fractions. Throughout this text, wehave been careful to represent fractionswith horizontal vinculums. But many cal-culators express three fourths as 3/4.Some calculators will show three fourthsas and a few show the pretty print

form of Many will shift the representa-

tion to 0.75 with the touch of a button.You become comfortable with one way,but easily gravitate to another becauseyou understand that all those forms aredifferent ways of saying the same thing.Those are only some of the ways to repre-

sent

Throughout this text, we have depictedor modeled different mathematical situa-

tions. Those have all been representations.We gave you examples of representationsand did not tell you we were doing it. Now,as you look back over the semester, youshould be able to see a variety of repre-sentations of different concepts. What isthe best way to resolve questions aboutrepresentation? Ask, "What did you meanwhen you wrote. .. ?" Sometimes the rep-resentation used is actually the goal of anactivity, but many times we just need tounderstand the representation withoutjudging it or requiring that it fit neatly withinour preconceived notion of how it shouldlook.

CONCLUSIONS

Howard Eves, a friend and world-classgeometer and mathematical historian,has said, "If you have algebra without ge-ometry you have answers to questionsnobody would ask, and if you have geom-etry without algebra you have questionsyou cannot answer." You might be askingwhy we would mention algebra and ge-ometry here, as you reach the end of thistext about elementary school mathemat-ics. This, however, is not the end of yourlearning about mathematics or how tocommunicate about it. With that in mindwe leave you with our hope for you . . .

Let your adventure begin.MaryE, Peggy, & Doug/Mike

BIBLIOGRAPHY

Eves, H. (January 26, 2002).

National Council of Teachers of Mathematics.(2000). Principles and standards for schoolmathematics. Reston, VA: Author.

Index

Algorithm, 158-159addition, 33division, 55-57, 60low stress, 39-40multiplication, 49-55standard, 34-36, 40, 44, 48, 51, 52-53, 58, 60-61subtraction, 4

Acute, 137, 168Addend(s), 27, 33-34, 88-91, 100-103, 105-106,

118-119concrete beginnings, 43denominator, 78-79, 81expanded notation, 44low stress addition, 39missing addend, 46-47, 77, 90-91, 104, 119proofs, 214-215terminology, 34, 41-42, 57

Addition, 39, 86, 224algorithm, 33, 35decimals, 86left to right, 37low stress, 39-40

Addition facts, 86Additive identity, 20, 104Adjacent, 138Algebra, 117-132AI-Khowarizmi, 118Altitude, 141-142And, 209-210Angle, 136-140, 167-169Answer, 4Antecedent, 211Any column first subtraction, 38Arabic, 118Arbitrary, 158-159Area, 22-24, 58, 131, 139, 156-162, 224-226Arithmetic, 2, 101

fundamental theorem, 24binary operations, 101mental, 81sequence, 128mean, 186

Arithmetic average, 186Arithmetic sequence, 128Array, 134, 223-224

Arrowhead, 134Associative property of addition on a set, 18-21Associative property of addition on integers, 18-21Associative property of multiplication on a set, 18-21Assumptions, 188-191Attributes, 10, 152

sample, 173Average, 185Axis, 146-148, 177-178

B

Babylonians, 118, 151Bar chart, 147Bar graph, 176-178Base 10 blocks, 4, 121-122Basic skills, 4-5Behavioral perspective, 4Binary, 35, 101

operations of integers, 106Borrow-pay back subtraction, 45Box and whisker plot, 179-181, 189-190Braces, 11

Calculations, 88, 190-191history of, 118, 151variance, 186-187

Calculator, 5-6, 192-193Capacity, 154, 164Cardinal number, 13-14, 194-196, 200Carry out the plan, 204Casio, 6, 119Categorical, 183Category, 178Celsius, 167Census, 173Chart, 24, 112Chord, 142-143Circle, 131, 142-144, 160Circumference, 142Circle graph, 175-176, 190Class, 178Closed curve, 138-139Closure property of addition on a set, 19

231

A

C

232 INDEX

Closure property of multiplication on a set, 19Collecting date, 173-174Combinations, 194

linear, 118Combining like terms, 121Common denominator, 72-81, 84-85Common factor, 22, 76

factorials, 192-193greatest, 28-32

Commutative property of addition on a set, 18Commutative property of addition on integers, 18-19Commutative property of multiplication on a set, 18Commutative property of multiplication on the set of

real numbers, 19Commutative property of multiplication on the whole,

19Complement, 15-16Complementary, 137-138Concave, 140Concrete models, 90Concurrent, 213Conditional, 211

probability, 199Cone, 145Congruent, 137Conjecture, 6, 127, 205-206, 216Conjunction, 209-212Connections, 28, 41, 112, 142, 189, 223-225Consequent, 211Contradiction, 213Convenience sample, 173-174, 190Convex, 140Coordinate plane, 146, 171Counting numbers, 11, 16.-17, 19-21, 24-25, 32,

123-125, 128-129Cross products, 114Cubit, 152Curriculum, 2 (see a/so NCTM Curriculum and Proc-

ess Standards)Curve, 138-139Cylinder, 131, 143-145

Data, 173-191Data collection, 173-174Data points, 131, 176, 183-189Decagon, 139Decimal,

addition, 86-87divided by, 98-99division by a whole number, 97-98division of, 95-96multiplication of, 93-95subtraction, 90-91whole numbers, 97-98

Decimal equivalents, 176Decimal numbers, 86-87, 175

Decimal places, 91-93connections, 95whole numbers, 96-98

Decimal point, 87-94Degrees, 167-168Denominate numbers, 36-27, 44, 87-91, 166Denominator, 63-64

addition of fraction, 72-77cross products, 114division of decimals, 95-99division of fractions, 81-86division using signed numbers, 108equivalent fractions, 64-66factorials, 192-193fraction and mixed number, 71mixed numbers, 80mixed numbers to improper fractions, 66-67proportions, 113percentages, 111-112proof, 214rational numbers, 109-110subtracting of fractions, 77-80two fractions, 69-70two mixed numbers, 71whole numbers and fractions, 68

Dependent, 100, 197-198Descartes, 118Diameter, 142-143, 160Dichotomous terminology, 152Difference, 41Digits, 16-17, 19-20, 34-35, 226

any column first, 38decimals, 86-88, 91divisibility, 26-28lattice multiplication, 51-53leaf, 181

Dimension, 22-24, 124, 133-134, 139, 143-144, 152Dimensional analysis, 171-172Disjoint sets, 15, 17Disjunction, 210Dispersion, 186, 191Distance, 131Distributive property of multiplication over addition on

a set, 19Distributive property of multiplication over addition on

the wholes, 19Divide out common factors, 70Dividend, 57-58Division, 22, 25, 55-58, 60-61, 63, 81-85, 95-99Divisor, 57-58Dodecagon, 139Drawing a picture or diagram, 206

Element, 10-12, 16-17, 19-21, 67, 99Eliminating impossible answers, 205Empty set, 13-14, 210

D

W

INDEX 233

Equal sets, 14Equation, 18Equivalent denominators, 64-66Equivalent fractions, 63, 65-67, 74, 76, 78-79,

84-85, 95, 97-98Estimation, 56, 192Euclid, 133Eves, H., 118-119, 132, 230Exclamation symbol, 25, 191Exercises, 53, 60, 69, 106-107

addition, 73column addition, 39decimal, 97decimals in products, 93division using signed numbers, 108expanded notation, 37fraction addition, 71fraction over fraction, 84integers, 105number line subtraction, 104partial product method of multiplication, 51partial sum and denominate numbers, 36scratch method, 38

Expanded notation, 26-28, 37, 44, 89, 121left to right subtraction, 46repeated subtraction division, 58

Expected range, 188-189Exponents, 27, 29-30Exterior, 137-139, 143

Factor, 22, 57-58, 66, 69-70, 76, 78-80, 96-98, 159,192-193, 224

decimal, 92-95greatest common, 28-32multiplication of integers, 106-107standard algorithm, 50

Factorial, 191-193Fahrenheit, 167Failure, 199-200False, 209-211

proofs, 214-215Fibonacci, 118Field Axioms, 18Finite set, 12, 14Flat, 43-44

addition of decimals, 86-89division of decimals, 95-96subtraction of decimals, 89-90, 93

FOIL, 48, 51, 53, 122Formative assessment, 5Formula, 123, 126, 128, 130-131, 160, 161-163,

170-171, 186-187, 193-194, 225Fraction(s), 63, 108-176, 192-193

addition of, 72-77division of decimals, 95-99division of, 81-85

cross products, 114equivalent and multiplication of, 63-72equivalents, 112-113percentages, 111-112rational numbers, 109-110relating to old ways, 94subtraction of, 77-81

Fraction bar, 72Frequency polygon, 178

Gauss, C., 125-126General term, 124, 130Generalization(s), 68, 88, 92-94, 102, 105, 189, 190Generalize, 19-20, 101-102, 106, 174Generic expression, 123Geometric sequence, 129Geometer's Sketchpad, 140, 143-144, 169Glide reflection, 148Group(s), 9-12, 18, 33-36

closed, 20Guess, test, and revise, 205

H

Heptagon, 130Hewlett Packard, 6Hexagon, 139, 145, 148, 160

Identity element for addition on a set, 18Identity element for multiplication on a set, 18If, then, 209-212Images, 99, 147Implicit multiplication, 21Improper fraction, 66-71, 73-74, 84Improper subset, 13-14Inclusive, 195Increasing, 30Independent, 194-197, 223Indirect proof, 217Infinite set, 12, 14, 17Integer, 105-108Integer subtraction, 47Interconnected, 205-223Interdependent, 205Interior, 137-139, 143Intersection, 15-16, 196, 209-210Inverse element for addition on a set, 18Inverse element for multiplication on a set, 18IQR-lnter Quartile Range, 189Irrational numbers, 16-17

G

H

F

I

234 INDEX

J

Juxtaposition, 21, 70

K

Kite, 140

Least common denominator, 75-76, 80Least common multiple, 22, 28, 31, 84-85Left to right subtraction, 46Legs, 131, 137, 139, 168-169Line, 87, 127, 133, 134-136, 168

digits, 90-91Line graph, 177-178Line plot, 176, 182Line segment, 49, 127, 135-135, 137-138, 143-145,

148, 177-179Linear measure, 152, 154Logic, 53Long, 43-44, 86-88, 90-93, 95-96

representation of, 58-59Look back, 204-205Looking for a pattern, 206Looping, 11Lottery, 199-201Low stress addition, 39-40

M

Magnitude, 118Make a list, 205Make a plan, 204-205Making a chart of a table, 206Manipulative(s), 4, 41, 72, 77, 86, 119Mass, 154, 165Math Anxiety, 7-8MathXpert, 119 (see a/so Technology)Mean, 131, 183-191Measure of central tendency, 182-191Median, 179-189

informal proofs, 213Metacognition, 206Methods,

addition, 32-40any column first, 38denominate numbers, 36expanded notation, 37horizontal and vertical writing, 36-37left to right, 37-38low stress, 39-40partial sum, 35-36scratch, 38standard algorithm, 34-35

subtraction, 40-48

any column first, 47borrow-pay back, 45-46concrete, 42-44denominate numbers, 44expanded notation, 44integer, 47left to right, 46scratch, 46standard algorithm, 44-45

Metric system, 153-155, 164-165Minuend, 41Missing addend, 41-42

foundation for algebra, 119model, 43number line, 104

Missing factor, 96-97, 108-109Mode, 183, 185-186Model(s), 99-101, 103, 120, 134-135, 140, 142, 161

185base 10, 90big block, 88graphic, 124missing addend, 119mixed number, 71number line, 42take away, 41

Multiple, 22, 24-25, 27-32, 35-36Multiplication, 15-16, 18-22, 48-55, 55-58, 63-71,

78, 91-95, 105-109lattice multiplication, 52-53left to right, 53

Multiplication facts, 50, 52-53Multiplicative identity, 48Multiplicative inverse, 82Mutually exclusive, 17

N

National Council of Teachers of Mathematics(NCTM), 1-8, 118, 203, 229

NCTM process standards, 1-7, 206Negation, 212Negative, 42

addition of integers, 99-102division of integers, 107-109integer subtraction, 47multiplication of integers, 107-109scatter plot, 180subtraction of integers, 103-105

Nonagon, 139Normal curve, 188Notation, 26, 191-200

connections, 225-227expanded, 27history of, 118permutations, 193

Number line, 33-34, 42, 49, 66, 75, 77, 99-102,104-107, 121

L

INDEX 235

Number tricks, 119-120, 214Numerator, 63-70, 72-81, 83, 85, 95-98, 108-114,

226

Obtuse, 137-138, 168Octagon, 139-160Odds, 200One-to-one correspondence, 43, 100, 119Operation(s), 15-16, 19-21, 34, 40-41, 48

addition of integers, 99factorials, 191-193permutations, 193

Operation of addition, 99Opinion poll, 173Or, 209-210Ordered pair, 110, 113, 126, 146Origin, 34, 42Outcome, 194-199Outlier, 188-190Overlapping sets, 15-17

Paragraph proof, 217Parallel, 27, 141, 169Parallelogram, 136, 141-142, 158-160Partial product, 50-53Partial sums, 35-36

left to right, 37-38low stress, 39-40

Partitioned set, 15Pattern, 11, 107, 130-131

patterning, 122-123representing situations with algebra, 123-125volume, 161

Pedagogical content knowledge, 1-3, 7Pentagon, 139-140Percentages, 175Percentile, 188-189Permutations, 193-194Perpendicular, 115, 137, 141-142, 156, 158-159, 163Pi, 131, 159-160Pictogram, 147Pie chart, 190Place value system, 37Plane, 133, 136, 142-143Plus, 67, 99Point, 133-136, 138-139

angles, 136-138circles, 142-143exclamation, 25identifying, 126-127line graph model, 177-178polygons, 139

Polya, G., 204-206, 208

Polygon, 138-142, 160, 169, 178-179Population, 12, 186Positional average, 184Precision, 88Prime factorization, 27, 29-31Prime numbers, 22-24, 66Prism, 131, 145, 133, 145, 161-163Probability, 191

conditional, 199dependent, 197-199independent, 194-197lottery, 199-200odds, 200

Problems, 78-79, 84, 91collection of, 207-208using proportions, 114-115with adding integers, 101-102with ordered pairs, 113with zeros, 91

Product, 48-58, 97concrete beginnings and beyond, 92-95connections, 224cross products, 114division of integers, 107-108fraction and mixed numbers, 70-71multiplication of integers, 105-107signed numbers in inverse operations, 108-109two fractions, 69-70two mixed numbers, 71whole numbers and a fraction, 68whole numbers and a mixed number, 68-69whole number divided by a decimal, 98

Proper subset, 13-14Property, 18-22, 25, 53, 70-71, 77-78, 82, 94, 106Proportion, 109-110, 113-115Protractor, 167, 169Pyramid, 60, 145Pythagorean theorem, 131, 170-171Pythagorean triple, 170

Q

QED, 213Quadrilateral, 139-141Quartile, 188-189Quartiles, 183, 191Quotient, 57-58

Radius, 131, 142-143Random sample, 173-174, 190-191Randomness, 197Range, 178-179, 183, 186, 189, 191Ratio, 129-130, 160, 175-176, 183, 190, 200-201Rational numbers, 16-17, 69-70, 73-74, 78, 82,

109-111

O

P

R

236 INDEX

Ray, 135-137, 168Real numbers, 16Recorde, R., 118Rectangle, 22-24, 58-59, 91-93, 121-122, 131, 140,

224-226Reduced, 65Reflection, 147-148Reflexive property of equality, 216Region, 138-139, 157-160Regrouped, 39, 43-44, 46, 49-50, 81Regrouping, 36, 43-44, 47, 80-81, 90-91, 166Regular polygon, 140Related denominators, 74, 77-79Relatively prime, 65-66, 75-76, 79Remainder, 61-62Repeated addition, 94, 223-224Repeated subtraction division, 58-62Representations of data, 174-175Reynolds, J., 7Rhombus, 140Right, 92-93, 131Right angle, 137-138, 141, 156, 158, 168Right circular cone, 145Right circular cylinder, 131, 144-145, 163Rotation, 137, 147-148, 167-168Rounding, 38, 56, 184-185Rule, 18, 80, 88, 107, 109

for adding integers, 99-103for divisibility, 25-27for same denominators, 77for subtraction of fractions, 77

Rule-based, 4Russian peasant multiplication, 54

Same denominators, 77-78Sample, 174Sample space, 195-196, 199-200Scale, 176Scatter plot, 180Scratch subtraction, 46Sector, 143Segment, 135, 139, 142, 143-145Sells, L, 8Set, 9-25, 135, 175, 186-188, 209-210Set multiplication, 15, 18Set product, 16Sharp, 6Shulman, L, 8SI (le système international d'unités), 154Signed numbers, 99, 107-109Signs, 99-102, 108-109Simple closed curve, 138-139, 142Sieves of Eratosthenes, 24-25Simplified, 65Simplify, 121Skew, 186

Solution, 204Space, 52, 134Sphere, 131, 144Spread, 183, 186Square, 22-23, 43, 52, 92-93, 122-113, 124-125,

131Standard algorithm (see Algorithm)Standard deviation, 186-191Standard formulas, 160Standard notation, 37Standard subtraction algorithm, 45 (see a/so Algo-

rithm)Statistics, 188-191Stem and leaf plot, 181-182Stick-to-itiveness, 204Straight, 138-139Straight angle, 137-138Strategies, 205-207Stuart, V., 8Subset, 12-15Subtrahend, 41, 58Success, 199Sum, 26-27, 32, 34, 35-46, 57, 72-74, 76-79,

101-102, 106, 186-187Summarizing data, 173-182Summation symbol, 184Summative assessment, 5Supplementary, 138Surface area, 136Symmetry, 147

Take away model of subtraction, 46Tautology, 212Teacher centered, 4Teacher dependent, 4Technology, 5, 6, 55, 61, 119, 230Tenths, 86-87, 90, 92Term(s), 26, 29, 32-34, 53, 57, 65, 74, 121Texas Instruments, 6, 119Theoretical probability, 197Tiling, 157Transformation, 147, 148Translation, 147-149Transversal, 169Trapezoid, 131, 140, 160Triangle, 15, 115, 131, 139, 158-160, 169-171, 213,

225-226Triangular number, 122-125True, 18-19, 94, 96, 104, 107, 110, 114

uUndecagon, 139Understand the problem, 204-205Union, 15-18, 196, 210

T

S

INDEX 237

Unit, 22-23, 34-35, 43-47, 49-51, 53, 58-60, 86-93,99-101, 124-125, 152, 156, 161-164, 223-226

Unit fractions, 73, 214Use reasoning, 207

Valid, 24, 42, 44, 69, 91, 151, 173Variance, 186-187, 190-191Vector, 33-34, 42, 72, 75, 100, 101, 104, 105-107,

135-136Venn diagrams, 16-17, 175Vertex, 137, 140-142, 160, 168-169, 213Vertical angles, 138, 169Vertices, 133, 140Vinculum, 56, 72, 108-109Volume, 131, 161-164

W

Weight, 154-165Well defined, 9, 101Whole number(s), 16-17, 19, 24, 73

concrete beginnings, 86-87division of decimals, 95-98factors, and multiples, 22mixed numbers, 68-69models, 99-100whole number addition, 32-40whole number divided by a fraction, 82whole number division, 55-63whole number multiplication, 48-55whole number subtraction, 40-48

Zero, 20, 25-26, 33, 37, 48, 50, 58, 88-89, 91, 94,102, 105-107, 186, 200

V

Z

This page intentionally left blank

Solutions Manual

2: NUMBER AND OPERATIONS

Sets

1. Based on the examples in Figs. 2.1 and2.2, write a definition of what looping the ele-ments of a set means.

Answers may vary. You are going to put afence around all of the elements of the set.When you are finished, it is clear what doesand does not belong in the set.

2. Define a finite set.

Answers may vary. When you count thenumber of elements in the set, you stopcounting at some point. If you can stopcounting, no matter how large the number is,the set is finite.

3. Define an infinite set.

Answers may vary. When you count thenumber of elements in the set, you can neverstop counting because there are always moreelements to be counted, no matter how largethe number becomes.

4. Define a subset.

Answers may vary. In each case, the firstset is completely contained in the second set,making it a subset of the original set. Noticethat a subset may be all or part of the originalset.

5. Define a proper subset.

Answers may vary. A proper subset is part,but not all, of a set.

6. Define an improper subset.

Answers may vary. The only improper sub-set is the set itself. The definition of a propersubset gives a clue as to why the set itself isimproper. Part, but not all, indicates that ei-ther some or no elements of the set must bein the subset.

7. Define cardinality.

Answers may vary. The cardinality, or car-dinal number of a set, tells the number of ele-ments in the set.

8. What is the cardinality of the empty set?

Zero. There are no elements in it.

9. What is the cardinality of

One. There is one element in the set. It justso happens that the element is 0, which isused to represent the empty set. However, inthis setting, that 0 becomes an element of aset and takes on a different meaning. Youcould think of it as one of the Greek lettersjust like you would say that {A} has a cardinal-ity of one.

10. What is the cardinality of {BMW}?

One. There are no commas between theletters so the three are considered one ele-ment, how does this compare to {P, e, g, g,y}? ©

11. Generalize the pattern established bythe total number of subsets, as the number ofelements in the set is increased one elementat a time.

{2, 4, 8, 16, . . .} can be written many ways,but one convenient one for this exercise is {21,22, 23, 24 . . . 2n}, where "n" represents the car-

SM-1

SM-2 SOLUTIONS MANUAL

dinality of the set. The generalization wouldthat when the number of elements in a set isknown, the number of subsets can be deter-mined. Once the number of subsets is known,you also know the number of improper sub-sets and the number of proper subsets (al-ways one less than the number of subsetsbecause there is always one improper subsetfor any set with cardinality of at least one).

12. A set with 11 elements will havesubsets, of which are improper, and

of which are proper.

A set with 11 elements will have 2048 sub-sets, 1 of which is improper, and 2047 ofwhich are proper.

13. Is there a case where the number ofproper and improper subsets will be equal? Isthere more than one case? Why or why not?

Yes. The set {A} has one proper subset, {},and one improper subset, {A}. Because no sethas more than one improper subset, there canbe no other case where the number of properand improper subsets will be equal. In ourBase 10 numbering system, only 1 + 1 = 2 .

14. Define equal sets and equivalent sets.

Equal sets have the same cardinality andthe exact same elements, although they maynot necessarily appear in the same order.Equivalent sets have the same cardinality, butthe elements are not all the same.

15. Which tells you more about two sets,equal or equivalent? Why?

Equal is the stronger statement because ittells you that both the cardinality and ele-ments are the same. Equivalent only assuresthat the cardinalities are the same.

16. Define overlapping sets.

Overlapping sets have some elements incommon. Generally, those with one set con-tained within another (like {1, 2, 3, 4} and (2,3}) are excluded from this definition.

17. Define disjoint sets.

Disjoint sets have no common elements.

18. Define partitioned sets.

Sets may be partitioned according to somedefinition. For example, the counting numberscan be partitioned into odd numbers and evennumbers. When a set is partitioned, each ele-ment is in one and only one of the subsets.

19. Define complements of sets.

The complement of a subset is everythingthat is in the original set except for the ele-ments of the subset itself. The complement ofa subset and the subset are disjoint and to-gether include every element of the set.

20. Define u (the union of sets).

When the union of two sets is formed, all ofthe elements in any set are listed in the unionset. If an element is in more than one of thesets being joined by the union, then it isshown only once in the resultant union set.

21. Define n (the intersection of sets).

The intersection of sets is all elementscommon to all of the sets being considered. Ifthe sets being considered are disjoint, thenthe union will be the empty set.

22. Define X (set multiplication or set prod-uct).

The set product of two sets is a set of or-dered pairs with the first element in the or-dered pair coming from the first set in theproduct and the second element of the or-dered pair coming from the second pair. Thatis why {1, 2} x (A, B} = {(1, A), (1, B), (2, A), (2,B)} and {A, B} x {1, 2} = {(A, 1), (B, 1), (B, 2), (A,2)} look different. Their respective elementsare ordered pairs, so (A, 1) is different from (1,A) because the order is different.

23. Define relative complement (set sub-traction).

The relative complement of one set isfound by subtracting the intersection of thatset from a second set. The relative comple-ment of A to B or the complement of A rela-tive to B is written B - A = B - (B n A).

SOLUTIONS MANUAL SM-3

24. State a generalization about the car-dinalities of sets and their union.

If the sets in the union are disjoint, the sumof the cardinalities of the sets will equal thecardinality of the union. For example, {1, 2, 3}u {A, %} would be 3 + 2 in terms ofcardinalities of the sets. The union would be{1, 2, 3, A, %}, which has a cardinal numberof five. We know that 3 + 2 = 5. If the sets inthe union are overlapping, then addition can-not be shown. {1, 2, A} u {A, %} = {1,2, A, %},but the cardinalities are 3, 2, and 4, respec-tively. We know that 3 + 2 * 4 .

25. State a generalization about the car-dinalities of sets and their set product.

The cardinality of a set product will be theproduct of the sets involved. For example, in

ithe cardinality of {A, 1, &} is 3,

the cardinality of \ is 2, and the cardinal-ity ofis 6 [remember, (A, V) is one element of theset] and 3x2 = 6.

26. Describe what the universal set mightbe for each of these examples given previ-ously of unions and then construct a Venn dia-gram that accurately depicts each statement:

27. Describe what the universal set mightbe for each of these examples given previ-ously of intersections and then construct aVenn diagram that accurately depicts eachstatement:

SM-4 SOLUTIONS MANUAL

28. Describe what the universal set mightbe for each of these examples given previ-

ously of set multiplication and then constructa Venn diagram that accurately depicts eachstatement:

29. Let the universal set for each of theseexamples be the set of digits and then con-struct a Venn diagram that accurately depictseach statement:

Let A = {1, 2, 3, 4}, B = {1, 2, 3}, C = {5, 6,7}, D = (3, 4, 5}, E = {4}, and F = {7, 8, 9}.

SOLUTIONS MANUAL SM-5

30. We know 2 + 2 = 2x2 as discussedwith the commutative property of addition onthe set of counting numbers. Are there anyother examples that will react similarly?

31. Is there a situation where commuta-tivity of subtraction on some set would exist?

Yes. As long as the elements are the same.7-7 = 7-7 = 0.

32. Is there a situation where commuta-tivity of division on some set would exist?

Yes. As long as the elements are the same.9 ÷ 9 = 1 = 9 ÷ 9 .

33. Generalize the idea of commutativity ofoperations for some set in words.

Answers may vary. Pick two elements andoperate on them. Reverse the order of the el-ements under the same operation and the an-swer will be the same.

34. We know 2 + 2 = 2x2 as discussedwith the commutative property of addition onthe set of counting numbers. Are there anyexamples that will work like that for associa-tivity?

Yes. Use zeros throughout.

35. Is there a situation where associativityof subtraction on some set would exist?

Yes. Use zeros throughout.

36. Is there a situation where associativityof division on some set would exist?

Yes. Use all ones.For each of 34 through 38, select True orFalse and explain the reason for your choice.

37. Generalize the idea of associativity ofoperations for some set in words.

Answers may vary. You can switch whichelements you operate with first, but you willstill get the same result.

38. The even counting numbers are closedfor addition.

True. Answers may vary on the why by giv-ing a lot of examples. It is assumed that thesestudents have completed at least high schoolalgebra, if not college algebra, and thus theycould do something like 2y and 2z are guar-anteed even for any y and z as elements of

SM-6 SOLUTIONS MANUAL

the counting numbers. Their sum, 2y + 2zcould be expressed 2(y + z), which is alsoguaranteed to be even.

39. The odd counting numbers are closedfor addition.

False. Answers may vary. 2y = 1 is guaran-teed odd as is 2z + 1. Adding gives 2y + 2z +2 = 2(y + z + 1), which must be even.

40. The even counting numbers are closedfor multiplication.

True. Given 2y and 2z as even, (2y)(2z) =4yz = 2(2yz), guaranteed even.

41. The odd counting numbers are closedfor multiplication.

True. Given 2y + 1 and 2z + 1 as guaran-teed odds, then (2y + 1)(2z + 1) becomes 4yz+ 2y + 2z + 1. Three of the terms are even butthat 1 at the end forces the sum to be odd.

42. Give a set that is closed for additionand describe why it is so.

Answers will vary.

43. Give a set that is closed for multiplica-tion and describe why it is so.

Answers will vary.

44. Would the commutative property foraddition have a negative impact on the situa-tion if we insisted that it hold true while wediscussed closure?

No. Actually, many mathematicians preferto define closure to include commutativity forthe operation because of advanced situationswhere the property may not hold.

45. Generalize the idea of closure of oper-ations for some set in words. Don't forgetabout division and subtraction.

Answers may vary. Take any two elementsof a set, operate on them and the result mustbe an element of the given set.

46. Generalize the idea of the identity ele-ment for an operation for some set in words.

Pick an element out of a set, operate on itwith the identity element and the result will bean element in the set.

47. Generalize the idea of the inverse ele-ment for an operation for some set in words.

Pick an element out of the set, operate on itwith its operative inverse and the result will bethe identity element for that operation in thatset.

48. In general, is there a left distributiveproperty of multiplication over subtraction inthe real numbers?

Yes. It reacts just like addition.

49. In general, is there a right distributiveproperty of multiplication over addition orsubtraction on the set of reals?

Yes. It reacts just like addition.

50. In general, is there a left distributiveproperty of division over addition in the reals?

No.

51. In general, is there a right distributiveproperty of division over addition in the set ofreal numbers?

No. It will work for something like (8 + 4) ÷ 2,which is 12 ÷ 2 or 6. Applying the distributiveproperty of multiplication over addition fromthe right, (8 + 4) ÷ 2 becomes (8 ÷ 2) + (4 ÷ 2)= 4 + 2 = 6.

52. Generalize the idea of the distributiveproperty for multiplication over addition forsome set in words.

Answers will vary. Take the element out-side the parentheses and operate with it oneach element inside the parentheses, addingthe resulting products.

53. Using the following examples, write adefinition of multiple and a definition of factor:

SOLUTIONS MANUAL SM-7

Multiples of 12 are 12, 24, 36, 48, ...Multiples of 5 are 5, 10, 15, 20, 25, ...Multiples of 7 are 7, 14, 21, 28, 35, ...Multiples of 2 are 2, 4, 8, 16, 32, 64, 128,

Factors of 12 are 1, 2, 3, 4, 6, 12Factors of 7 are 1, 7Factors of 100 are 1,2,4,5,10, 20, 25, 50,

100Factors of 36 are 1,2,3, 4, 6, 9,12,18, 36

Answers will vary. A multiple of a number isthat number times any counting number.Some say a multiple is that number times anywhole number, making zero be a multiple ofevery number, sometimes referred to as the"trivial case." Still others say that a multiple ofa number is that number times any integer.

A factor of a number is any counting num-ber that divides the given number. Here, "di-vides" means that when the division is done,the remainder is zero.

54. What conclusions can you draw fromyour table?

Answers will vary. Some numbers haveonly one way of making a rectangle and oth-ers have more than one way. Eventually, theone way numbers will be called primes andthe more than one way numbers will be calledcomposites.

55. Complete the Sieve of Eratosthenesfollowing the directions given in the text im-mediately before this exercise. Although thischart stops at 100, it could be continued toany desired value.

Yes.

56. What is the greatest number of primesin any given row?

There can never be more than three afterthe first row. Consider columns for this dis-cussion. The columns headed by an evennumber will always contain only even num-bers. Thus, those five columns after the firstrow cannot contain a prime, eliminating five

columns from discussion. Similarly, the col-umns headed by 5 and 10 will always containonly multiples of five, eliminating those col-umns. This leaves three columns that couldcontain a prime number.

57. Would the answer in Exercise 53change if the chart were extended indefinitelyto include more rows or more columns?

Yes. If you go far enough, like to 105!, it ispossible to have over 100 consecutive com-posite numbers, giving at least eight rows ofconsecutive composite numbers in a10-column sieve. Ranted, they are going tobe mighty big numbers.

58. Next consider the least number ofprimes in any given row?

That number could be zero, if you considersomething like 12!

59. Would the answer in Exercise 55change if the chart were extended indefinitelyby adding more rows or columns?

Answers will vary, but it is yes in bothcases. As columns are added, even headingswill be eliminated from consideration as hold-ers of prime numbers, as explained in Exer-cise 56, as would be the case for multiples offive. Still, there would be more columns thatcould contain a prime.

60. Create a sieve on a 6-column chart,then answer the questions that follow (thesame questions asked for the 10-columnsieve). Even though the same questions areasked, the answers will change, thus enhanc-ing understanding.

61. What is the greatest number of primesin any given row of the 6-column sieve?

After the first row, two. The second, fourth,and sixth columns will always hold values that

SM-8 SOLUTIONS MANUAL

are multiples of 2. The third column will al-ways hold values that are multiples of 3. Thatleaves, at most, two columns that could con-tain a prime.

62. Would the answer in Exercise 58change if the 6-column sieve were extendedindefinitely?

Yes. See Exercise 57.

63. Now, what is the least number ofprimes in any given row of the 6-columnsieve?

Zero. See Exercise 57.

64. Would the answer in Exercise 60change if the 6-column chart were extendedindefinitely?

Yes.

65. Develop an argument that shows howthe divisibility rule for 9 would work with a4-digit number wxyz.

Consider a 4-digit number, wxyz, where w,x, y, and z are any digit:

This shows that the divisibility depends onthe sum of the original digits. Only if the sum+ w + x + y + z is divisible by 9, will the originalnumber be divisible by 9.

66. Why does the 6 rule break into an even3 rule? Explain why a similar rule could orcould not be devised for divisibility by 15.

The prime factors of 6 are 2 and 3. Thus, ifboth are met, the number must also be divisi-ble by 6. A similar argument would be madefor 15.

67. Describe a divisibility rule for somenumber other than those discussed and showwhy it works.

Answers will vary.

68. Are divisibility rules limited to integers?

In the sense of our discussion, yes. How-ever, they are often employed in fractions, forexample, to determine if there is a commonfactor that exists between the numerator anddenominator.

Whole Number Addition

1. Use the grouping method described inthe previous button example to find the totalof two sets of elements as a little child might.Use buttons, pennies, chips, or any other ma-nipulative.

Answers may vary, but one possible repre-sentation is given with each part.

2. Using only addition, write out a simpleword problem that would be appropriate foran early childhood addition problem. Useyour manipulatives to solve the problem.

Answers may vary. The manipulative re-sponse should be similar to those demon-strated.

3. Use the number line method describedpreviously to show:

SOLUTIONS MANUAL SM-9

4. 4837 + 3519 = ?

5. 647 + 10254 + 9938 = A

6. 24751 + 608 + 93 + 3562 =

7. 4837 + 3519 = ?

8. 647 + 10254 + 9938 =

9. 24751 + 608 + 93 + 3562 =

10. 4837 + 3519 = ?

4000 + 800 + 30 + 7 + 3000 + 500 + 10 + 9 =4000 + 3000 + 800 + 500 + 30 + 10 + 7 + 9 =7000 + 1300 + 40 + 16 =8000 + 300 + 50 + 6

11. 647 + 10254 + 9938 =

12. 24751 + 608 + 93 + 3562 =

13. 4837 + 3519 = ?

SM-10 SOLUTIONS MANUAL

14. 647 + 10254 + 9938 = K

64710254+ 9938100009000170012019

20839

15. 24751 + 608 + 93 + 3562 = O

2475160893

+ 35622000070001800200

+. 1429014

16. 4837 + 3519 = ?

4837+ 3519?3468 5

17. 647 + 10254 + 9938 = n

64710254+ 9938W392083

18. 24751 + 608 + 93 + 3562 = y

20. 647 + 10254 + 9938 = (

Answers will vary.

21. 24751 + 608 + 93 + 3562 = #

Answers will vary.

Whole Number Subtraction

Do Problems 1,2, and 3 using both the takeaway model and number line.

1. Jo has 14 marbles and loses 8. Howmany are left?

2. Shawn owns 4 high value stamps, andSean owns 11 high value stamps. How manymore high value stamps does Sean own?"

3. Chris has 5 cards left but started with17. How many are gone?

15. 24751 + 608 + 93 + 3562 = O

16. 4837 + 3519 = ?

17. 647 + 10254 + 9938 = n

18. 24751 + 608 + 93 + 3562 = %

19. 4837 + 3519 = ?

SOLUTIONS MANUAL SM-11

4. Shawn scored 512 on a video game andReggie scored 178 on the same game. Howmuch higher was Shawn's score thanReggie's? Do this problem using each of thestages: concrete, denominate numbers, ex-panded notation, and standard algorithm.Write a concluding paragraph explaining howthe different steps in the various stages areconnected across the stages.

6. 703 - 164

Answers will vary on written paragraph.Do the following subtraction problems usingeach of the following methods: borrow payback, left to right, scratch, any column firstand integer. Show your regroupings in eachstyle in a manner that would show how youare doing your work.

5. 8314 - 2756

Note that the any column first method wouldlook a lot like the left to right.

7. How would a problem like 8152 - 1936impact each of borrow pay back, left to right,scratch, any column first and integer subtrac-tion?

Answers will vary. The process will be simi-lar except that if there is a case where thedigit in the sum is greater than the digit in theaddend, no regrouping is required for thatcolumn.

Whole Number Multiplication

Use the partial product method of multiplica-tion (in either the vertical or horizontal format)to find the products in the following exercises.

Note that the any column first method wouldlook a lot like the left to right.

SM-12 SOLUTIONS MANUAL

3. 803 x 745 =

Use lattice, left to right, and the distributiveproperty of multiplication over addition on theset of whole numbers methods to find theproducts in:

4. 23 x 47=

5. 519 x 68 =

6. 803 x 745 =

i

Use the Russian peasant method of multipli-cation to find the products in:

7. 23 x 47 =

SOLUTIONS MANUAL SM-13

8. 519 x 68 =

9. 803 x 745 =

Whole Number Division

1. 5184 ÷ 9 = 576 Answers will vary

3. Answers will vary

Equivalent Fractions andMultiplication of Fractions

1. Find an equivalent fraction to each ofthe following; be sure to show the steps thatassure your result is correct.

Answers will vary. Some examples are:

if all common factors are divided out

2. Convert the following mixed numbers tofractions. Please note we have not discussedany shortcuts you might have learned in thepast, so the expectation is that you will notapply them at this point. You need to practicethe skills of converting at a basic level so youwill gain a full understanding of the shortcutsyou might have learned previously.

SM-14 SOLUTIONS MANUAL

3. Convert the following improper frac-tions to mixed numbers. Once again, we askthat you practice the skills of converting at abasic level; leave the shortcuts for later.

4. Find the following products:

5. Find the following products first by con-verting the mixed number to an improperfraction and then by using the distributiveproperty of multiplication over addition on theset of rational numbers:

SOLUTIONS MANUAL SM-15

6. Find the following products:

c) Answers will vary.

7. Use both the partial product and distrib-utive methods to complete the following exer-cises.

We leave the rest because it is all materialthat has been covered before.

We leave the rest because it is all materialthat has been covered before.

Addition of Fractions

1. Find each of the following sums, show-ing the intermediate steps:

SM-16 SOLUTIONS MANUAL

2. Do the following problems, showing theintermediate steps you made to get eachsum:

3. Do the following problems, showing theintermediate steps you made to get eachsum:

4. Do the following problems, showingthe intermediate steps you made to get thesum:

SOLUTIONS MANUAL SM-17

4. Discuss what happens if a fraction sub-traction problem involves two unit fractionswith related denominators.

Answers will vary. The smaller denominatorshould be listed first or the answer will benegative. Beyond that, the answer will be aunit fraction if the larger denominator is twicethe smaller denominator. If the multiple of thelarger denominator is more than twice thesmaller denominator, then the missing ad-dend will not be a unit fraction.

5. The examples we have shown in thetext involved two fractions with relativelyprime denominators. If the problem involvedthree (or more) fractions and the only com-mon factor shared by the denominators isone, then how would you work the problem?

Answers will vary. The numerator and de-nominator of each fraction must be multipliedby each of the other denominators.

6. Do each of these problems, showingbasic intermediate steps:

Subtraction of Fractions

1. Do each of these problems, showingbasic intermediate steps:

2. Discuss what happens if a fractionsubtraction problem involves two unit frac-tions with the same denominator.

The answer will always be zero because

you would have

3. Do each of these exercises, showingbasic intermediate steps:

SM-18 SOLUTIONS MANUAL

7. Do each of these problems, showingbasic intermediate steps:

8. Do each of these problems, showingbasic intermediate steps:

Division of Fractions

1. Do each of these problems, showingbasic intermediate steps:

SOLUTIONS MANUAL

2. We have shown the reasoning behindinverting the second fraction and followingthe rules for multiplication in division prob-lems involving fractions. Do these exercisesusing the fraction over a fraction routine.

3. Do each of these problems, showingbasic intermediate steps:

4. Use the least common multiple methodto do the following problems:

SM-19

SM-20 SOLUTIONS MANUAL

Addition of Decimals

1. Do the following problems showing anyscratch work you generate:

Subtraction of Decimals

1. Do each of these exercises using onlypencil and paper:

Multiplication of Decimals

1. Do each of the following problems topractice your multiplication skills when one orboth of the factors involve decimals. Althoughwe are proponents of calculator use, youshould not use calculators here. You need topractice the skill and be sure you have a han-dle on decimal multiplication.

12.5 x 4.2 = 52.5 NOTE that the answercomputes to be 52.50 but that last zero isgenerally not written.

Division of Decimals

1. Do these problems without a calcula-tor. You should check your work by multiply-ing the factor times the missing factor.

The decision in the last example on how far togo before you show the repeat is left you andyour instructor.

2. Complete the following exercises be-ing careful to align the decimal point in themissing factor:

The degree of the precision of the answers isleft to you and your instructor.

3. Tell the smallest power of 10 needed tomake the factor a whole number:

4. Tell the smallest power of 10 needed tomake the factor a whole number:

SOLUTIONS MANUAL SM-21

Addition of Integers

1. Do each of the following problems on anumber line and record both the problem andthe answer in a separate list. State a rule foradding integers when the addends have thesame sign.

The conclusion is that adding like signednumbers is just like adding whole numberswithout the signs and the answer has thecommon sign.

2. Do some integer addition problems onthe number line to assure you can get thesums. Each of your problems should be simi-lar to those in Figs. 2.56 and 2.57.

Answers will vary.

Subtraction of Integers

1. Do each of the following problems on anumber line and record both the problem andthe answer in a different location. State a rulefor subtracting integers.

SM-22 SOLUTIONS MANUAL

Multiplication of Integers

1. Now that all four types of multiplicationproblems involving integers have been con-sidered, state two generalizations that dealwith multiplying signed numbers:

If the two factors have the same sign, theproduct is positive.

If the two factors have different signs, theproduct will be negative.

2. Develop a generalization relating tomultiplying more than two integers factorsbased on the number of negative factors thatare involved.

If the number of negative factors in a multi-plication problem is even, then the product ispositive. If the number of negative factors in amultiplication problem is odd, then the prod-uct is negative.

3. State a generalization relating to multi-plying two or more integers factors when oneof them is zero.

If one factor is zero, then the product willbe zero, no matter how many factors are in-volved.

Division of Integers

1. Considering problems only of the

types = +5, for example (posi-

tive divided by positive and negative dividedby negative), describe a generalization forworking with dividing a product by a factorwhen their signs are the same.

Dividing with like signs gives a positivemissing factor.

2. Considering problems only of the

types , for example (posi-

tive divided by negative and negative dividedby positive), describe a generalization forworking with dividing a product by a factorwhen their signs are not the same.

Dividing with unlike signs gives a negativemissing factor.

Ratios and Proportions

1. Write ratios that compare the numbersof designs in Fig. 2.73.

d) Number of figures that have onlystraight segments to number of figures thatcontain curves

SOLUTIONS MANUAL SM-23

Approximately 82.6% of the students werepassing after the first exam.

b) 11 of Kim's 18 matchbox cars representAmerican automobiles. What percentage ofKim's matchbox cars represents foreign au-tomobiles?

18 is not a factor of a power of 107 -r 18 0.389

Approximately 38.9% of Kim's cars representforeign automobiles.

4. Write the two equivalent ratios that rep-resent the quantities in the following state-ments. Use a letter to represent the missingquantity and set the ratios equal to one an-other and find the missing quantity by inspec-tion.

a) I would like to make 3 identical shirts.My pattern for a single shirt requires 2 yardsof fabric. How much fabric should I buy formy 3 shirts?

b) There are 4 tables and 16 chairs in ourclassroom. If we continue the same arrange-ment, then how many chairs would we needfor 8 tables?

c) My small, 4-pound turkey requires 1hour to roast. Assuming all other things areequal, how much time would be required toroast a 16-pound turkey?

a) y = 6

6 yards of fabric are needed for 3 skirts

b) c = 32

32 chairs are needed for 8 tables

2. The following ratios can be changed sothat they have denominators that are powersof 10. Use the procedure shown in the exam-ples to write these ratios as percentages.

a) Jose got 7 strikes out of his first 20frames at the bowling tournament. What washis percentage of strikes during the first 20frames?

Jose had 35% strikes in the first 20 frames ofthe bowling tournament.

b) Only 18 out of 250 people at the wed-ding selected chicken. What percent of thepeople at the wedding selected chicken?

7.2% of the people at the wedding selectedchicken.

c) For 135 of the 2500 cars that passedthrough the tollbooth on Tuesday, an error inthe amount of the toll paid was indicated.What percentage of toll errors does this im-ply?

There was an error in 5.4% of the tolls onTuesday.

3. The following ratios cannot be changedso that they have denominators that are pow-ers of 10. Explain why this is true. Use yourcalculator to write these ratios as percent-ages. Round to the nearest tenth of a percent.

a) 19 of 23 students earned passingscores on the first exam. What percentage ofthe students earned passing scores?

23 is not a factor of a power of 10.19^23 0.826

SM-24 SOLUTIONS MANUAL

7. List at least four additional proportionsfor the triangles in Fig. 2.76.

8. Your recipe for Party Mix calls for 7cups of wheat cereal, 2 cups of mixed nuts, 2cups of pretzels, 1.5 sticks of butter, 1 table-spoon of special sauce, and seasoned salt totaste. You find that you have only 1.5 cups ofmixed nuts. How much of each of the otheringredients do you use if you want to keep therecipe in proportion?

**Seasoned salt will still be added "totaste"

9. For some incredibly interesting reason,you need to know the height of the telephonepole in front of your home. You know that youare 5.5 feet tall and that you cast a shadowthat is 11 feet long. The telephone pole castsa shadow that is 26 feet long. How tall is thetelephone pole?

c) h = 4

4 hours will be required to roast a16-pound turkey

5. Use cross products to determine ifeach of the following is a true proportion.

5 • 111 = 555, 31 • 18 = 558, therefore, thisproportion is not true.

8 • 342 = 2736,19 • 144 = 2736, therefore, thisis a true proportion.

6. Solve the following proportions (if nec-essary, round to the nearest tenth):

SOLUTIONS MANUAL SM-25

2. Do the trick in Exercise 1 using a frac-tion. Is your answer still the same?

10. Trace the triangles in Fig. 2.76 and labelthem as follows. Smaller triangle—the shorterleg is 9 cm and the longer leg is 12 cm. Largertriangle—the hypotenuse is 37.5 cm and theshorter leg is 22.5 cm. What is the length of thehypotenuse of the smaller triangle?

One possible solution is:

h = 15 cm is the length of the hypote-nuse of the smaller triangle

11. As a teller in a bank that sales foreigncurrency, you are responsible for helping cus-tomers who are planning for trips abroad. Therates change very quickly, so you must calcu-late each transaction separately. A customerasks you how much it would cost to buy200.00 Deutch Marks (DM). You check therate and find that $1.00 is worth 1.32 DM.How much will it cost your customer to getthe Deutch Marks he needs for his trip to Ger-many?

3: ALGEBRA

1. Do this trick using a specific numberand then using x for the number picked.

What do you get? 2If you begin the trick with a different num-

ber, will you still get that answer? Yes

Yes, the answer is still the same.

3. Do the trick in Exercise 1 using a nega-tive number. Is your answer still the same?

4. Do this trick using a specific numberand then using m for the number picked.

What do you get? I got the number I startedwith.

Will this always work? Yes.Why or why not? The algebra shows why it

will always work.

5. Do this trick using a specific numberand then using p for the number picked.

Pick a number.Triple it.Add 12.Divide by 3.Subtract your original number.

SM-26 SOLUTIONS MANUAL

What do you get? 4.Is this problem significantly different from theone given in Exercise 1? No. The numbersand variables are the same, but the algebra issimilar and shows that you will always getyour original number.

6. What are the next two lines in the pat-tern 1, 11, 21, 1112, 3112, 211213, 312213,212223, ?, ?

The next term is 114213.The term after that is 31121314.

7. What are the next two terms in the pat-tern 1, 11, 21, 1211, 111221, 312213,1311221113, ?, ?

8. Make a model that shows how Gauss'trick applies to the triangular numbers.

Thus, adding all of the counting numbers upto and including the number n can find any tri-angular number Tn.

9. Using the model of the rectangularnumbers, write a generalizing statementabout how to find any rectangular number, Rn.

10. Rewrite the rectangular model to showa relation between the rectangular numbersand the square numbers.

11. Generalize the model you drew aboutthe relation between rectangular and squarenumbers.

12. How do the rectangular numbers re-late to the triangular numbers?

A rectangular number is twice as much asa triangular number: Rn = 2Tn

13. Repeat traveled at a rate of 12 mph.Use the information you placed into the tableat the beginning of this chapter.

a) Make a graph of Repeat's rate ofchange using the x axis for time and the y axisfor distance traveled.

b) Find m for Repeat by counting onthe line segment.

c) Use the two points from the table forhow far Repeat had traveled at 2 hours and at5 hours and find m algebraically. Does youranswer here match your answer for Part b?

14. In the word problem about renting ajackhammer, we found a cost pattern foreach of the rental companies. For Rent It, thepattern was 20 + 3h ($20.00 plus $3.00 perhour) and for Get It Here, the pattern was 8h($8.00 per hour).

a) Use the cost pattern for Rent It to writea cost equation for renting the equipment.State the rate of change and the y intercept.

SOLUTIONS MANUAL SM-27

b) Use the cost pattern for Get It Here towrite a cost equation for renting the equip-ment. State the rate of change and the y inter-cept.

C(h) = 8h or y = 8x; m = 8; b = 0 or (0, 0)

c) At what point would the graphs of thesetwo cost equations intersect?

(4, 32)

15. Peggy, a member of this book's dy-namic author team, wants to paint her housea particular shade of blue such that the ratioof gallons of blue paint to white paint is 2:3.

a) How many gallons of blue paint andhow many gallons of white paint will beneeded to mix up 15 gallons?

b) What is the algebraic solution for theslope of the ray that represents this color mix-ture no matter how many gallons of paint areneeded?

c) If Peggy would like to make the trim ofher house a darker shade of the same color,then how should she change the ratio of thepaint mixture?

Answers will vary: example—3 gallons ofblue to 4 gallons of white

d) Using the trim mixture you decided onfor Peggy, how many gallons of blue paintand how many gallons of white paint wouldbe needed to make 3.5 gallons of paint?

Answers will vary: example—

16. What is the 121st number in the follow-ing sequence: 4, 8, 12, . . . ?

17. What is the 326th number in the fol-lowing sequence: 9, 20, 31, . . . ?

18. What is the sum of the first 84 evennumbers: 2, 4, 6, . . . ?

19. What is the sum of the 9th through the101st numbers in Question 16?

SM-28 SOLUTIONS MANUAL

The sum of the 9th through the 101st num-bers in the sequence 4, 8,12,.. . is 20468.

20. What is the 15th number in the follow-ing sequence: 5, 10, 20, . .. ?

21. What is the 19th number in the follow-ing sequence: 99, 33, 11, .. . ?

22. What is the sum of the first 20 numbersin the sequence: 5, 10, 20, . . . ?

23. What is the sum of the first 500 numbers in the sequence: 99, 33, 11 , . . . ?

GEOMETRY SOLUTIONMANUAL

1. For each of the following, provide a la-beled sketch and explain in your own words:

a) Point Qb) Line segment STc) Line kd) Plane Re) Ray UVf) Vector u of length 3 and heading to the

rightg) Which of these can be measured?

SOLUTIONS MANUAL SM-29

2. Answer each of the following questionsand provide a sketch:

a) How many different lines could yousketch through a single point?

b) How many different lines could yousketch through two points?

c) Is it always possible to sketch one linethrough three points?

a) You can sketch many (an infinite num-ber) of lines, segments, or rays throughone point.

b) You can sketch only one distinct line orsegment given two given points. How-ever, two distinct rays can be defined(each beginning at one of the points andgoing through the other point).

c) Unless three points line up in a straightarray, a line cannot be sketched that in-cludes all three.

3. Use a straight edge to sketch an angle.Label a few interior points, exterior points, andpoints on your angle (remember to use printedcapital letters when you label points). Shade

SM-30 SOLUTIONS MANUAL

the part of the interior that is within the raysyou drew. If you continued the rays, would thisinterior shading continue to expand?

Points G, H, and I are on the angle. Points Aand C are exterior points. Point B is an interiorpoint. If the shaded area were extended, itwould get wider and wider and extend infi-nitely, like a wedge of a plane.

4. Given that HL1JK, name every angle inFigs. 4.2-4.11; then identify each angle asright, acute, obtuse, or straight. (There are atleast 10 angles in the figure.)

Right angles: ZJML, ZKML, ZHMK, ZHMJAcute angles: ZHMI, ZIMJObtuse angles: ZIMK, ZIMLStraight angles: ZHML, ZJMK

5. Explain why there is a tiny box drawn atthe intersection labeled with an M in Fig.4.11.

The tiny box means that HL _L JK (Line Seg-ment HL is perpendicular to Line SegmentJK), so that ZHMK and three of the other an-gles, ZJML, ZKML, and ZHMJ are right an-gles.

6. Explain why we need to use three lettersto identify any angle in Fig. 4.11.

We use three letters to make sure we iden-tify the correct angle. If we say ZM, we could

mean any one of the 10 angles in Fig. 4.11,but if we identify a point on each leg of an an-gle, as well as the vertex, there will be nodoubt about which angle is being named.

7. In Fig. 4.11, we say that ZHMI andZHMK are "adjacent angles." Write a defini-tion of adjacent angles.

The two angles, ZHMI and ZIMK, are adja-cent because they share a vertex and one leg,Ray Ml, and are side by side (not overlap-ping). Two angles are adjacent if they share avertex and one leg without overlapping.

8. Write an informal definition for the termcomplementary angles.

Two angles are complementary angles ifthey can be combined (without overlapping)to form a right angle.

Another way to look at this is to considertheir measures, because a right angle has ameasure of 90°.

mZJMI = 47°, mZHMI = 43° The sum oftheir measures is 90°.

Two angles are complementary angles iftheir measures add up to 90°.

9. Write an informal definition for the termsupplementary angles.

Two angles are supplementary angles ifthey can be combined (without overlapping)to form a straight angle.

Another way to look at this is to considerthe angle measures, because a straight anglehas a measure of 180°.

mZJMI = 47° and mZKMI= 133° The sumof their measures is 180°.

Two angles are supplementary angles iftheir measures add up to 180°.

10. Do you think it is possible for two an-gles to be complementary or supplementarywithout being adjacent? If your answer is yes,then provide sketches that indicate yourthinking.

The requirement is that the angles be com-bined to form a right angle or a straight angle.

HL1JK,

Right angles: ZJML, ZKML, ZHMK, ZHMJAcute angles: ZHMI, ZIMJObtuse angles: ZIMK, ZIMLStraight angles: ZHML, ZJMK

HLUJK

SOLUTIONS MANUAL SM-31

You could cut out two angles that aren't adja-cent and rearrange them to form the requiredangle. The definitions could mention thesums of the measures; neither would requirethat the angles be adjacent.

11. Explain why Table 4.1 begins with apolygon of three sides.

Because polygons are made up of line seg-ments, it takes at least three sides to make afigure that will close.

12. There are several common polygons,called quadrilaterals, that have four sides.Use the diagram of the quadrilateral family inFig. 4.14 to answer the following questions:

a) What are the similarities and differencesamong the special quadrilaterals in Fig. 4.14?

b) What does it mean if there is not an ar-row connecting two quadrilaterals?

c) Is every square a rectangle?d) Is every rectangle a square?e) Where does a four-sided figure shaped

like a child's kite fit in this diagram?

a) The general quadrilateral doesn't haveanything special. It just has four sides andfour angles. The sides may or may not be thesame length; the angles may or may not be ofany specific measure.

All of the special quadrilaterals shown inthe diagram are similar because they all havefour sides and at least one pair of parallelsides.

The family tree is built on differences asmuch as similarities. The trapezoid has onlyone pair of parallel sides, whereas the resthave two pairs of parallel sides. The parallelo-grams get more and more special as youmove down the chart. The parallelogram hastwo pairs of congruent, parallel sides. Therhombus has two pairs of parallel sides, butalso has all sides congruent. The rectanglehas two pairs of congruent, parallel sides, butalso has right angles. The square has twopairs of parallel sides, all sides congruent,and right angles.

b) The arrows connect the quadrilateralsthat have similarities. If there is no arrow con-necting two quadrilaterals, then there is atleast one attribute that is not shared.

SM-32 SOLUTIONS MANUAL

c) A square has all the attributes of a rect-angle (two pairs of congruent, parallel sides,and right angles), so it is a rectangle.

d) A rectangle that has all sides congruentis a square. Some rectangles are not squaresbecause they have different lengths for theirheight and width.

e) In Fig. 4.14, a child's kite might be arhombus if all four sides are the same length.However, such a kite would be difficult to fly.Start a new branch for quadrilaterals with ad-jacent sides congruent in pairs, as the follow-ing figure.

13. Use a piece of nonelastic string(thread, dental floss, etc.) to form a loop. Useyour closed loop to model the polygons in Ta-ble 4.1. What do all your models have in com-mon? How do your models differ as you addsides to create new polygons?

The models all have the same perimeter,because the string has a fixed length. Thelengths of the sides change, growing shorteras more and more sides are added. The an-gles also change, growing larger as more andmore sides are added. Because the sketchesin the figure represent regular polygons, theareas of the enclosed regions increase assides are added.

14. Most of our sketches were created us-ing Geometer's Sketchpad®. You may wantto use this or another dynamic software appli-cation as you complete this exercise. Youmay, of course, complete the sketches usinga straightedge and protractor. Draw and labelpolygons with the given characteristics. Namethe polygons (a) based on the letters at theirvertices, and (b) based on their characteristics(Choose from these terms, using each termonly once: acute triangle, equilateral triangle,isosceles triangle, obtuse triangle, parallelo-gram, quadrilateral, rectangle, rhombus, righttriangle, scalene triangle, and trapezoid.):

a) Three sides with as many right anglesas possible.

b) Three sides with as many obtuse anglesas possible.

c) Three sides with as many acute anglesas possible.

d) Three sides with no sides the samelength.

e) Three sides with two sides the samelength.

f) Three sides with all three sides thesame length.

g) Four sides with no sides the samelength.

h) Four sides with two opposite sides par-allel and the other two opposite sides not par-allel.

i) Four sides with as many right angles aspossible.

j) Four sides with all sides the same lengthand as many obtuse angles as possible.

k) Four sides with opposite sides paralleland as many acute angles as possible.

SOLUTIONS MANUAL SM-33

a) CBA is a right triangle; it has exactlyone right angle

b) FED is an obtuse triangle; it has ex-actly one obtuse angle

c) IHG is an acute triangle; all three an-gles are less than 90°.

d) JKL is a scalene triangle; it has nosides the same length. We indicate that thesides are not the same length by placing asingle mark on the shortest, two marks on thenext longer, and three marks on the longest.

e) RQP is an isosceles triangle; it has atleast two sides the same length (all threecould be the same length). Congruent sides,PQ and RQ, are indicated because they eachhave a single mark.

f) STU is an equilateral triangle; allthree sides are congruent. The congruentsides are indicated because they all have thesame number of marks.

g) General quadrilateral ADCB has nosides the same length, as indicated by the

marks. This time the number of marks doesnot indicate relative length, only that no twosides are the same length.

h) Trapezoid EFGH is a quadrilateral withtwo opposite sides parallel (as indicated bythe double arrowheads on sides EH and FG)and the other two opposite sides not parallel.

i) A parallelogram that is a rectangle (or asquare) has four right angles (it is only neces-sary to mark one of them).

j & k) A general parallelogram (or a rhom-bus) has four sides with opposite sides paralleland as many acute angles as possible. Paral-lelogram ONQP has two (opposite) acute an-gles, . and and two (opposite) obtuseangles,. and, Although Rhombus STUVhas all four sides the same length, it has two(opposite) acute angles and . , and two(opposite) obtuse angles and .

15. Write an informal definition for each ofthe following terms:

SM-34 SOLUTIONS MANUAL

a) Side of a polygon —How many sidesdoes a 12-gon have?

b) Vertices of a polygon —How many verti-ces does a 7-gon have?

c) Diagonal of a polygon—How many di-agonals does a 5-gon have?

d) Altitude of a polygon—How many alti-tudes does a 3-gon have?

a) A side of a polygon is a straight linesegment. The sides of a polygon are joined bytheir endpoints to create a region on a plane.Polygons can have three or more sides. A12-gon has 12 sides.

b) The vertices of a polygon are the cor-ners where the sides are joined at their end-points. Polygons have the same number ofvertices as sides. A 7-gon has 7 vertices.

c) The diagonal of a polygon is an auxil-iary line segment that joins nonconsecutivevertices. The number of diagonals in a polygonis determined by the number of vertices it has.In a pentagon, there are five diagonals, repre-senting two connections across the interiorfrom each vertex, as shown in the figure.

d) The altitude of a polygon is a line seg-ment that starts at a vertex and ends perpen-dicular to a side. It is the height of the poly-gon. The number of altitudes is determinedby the number of vertices a polygon has. In atriangle, an altitude can be drawn from eachof the three vertices, as shown in the figure.

16. Use the information in Fig. 4.18 andFig. 4.19 to write your own informal defini-tions for the following terms:

a) Centerb) Chordc) Circled) Circumference

e) Diameterf) Radiusg) Sectorh) Segment

a) The center of a circle is a given pointfrom which all the points on the circle are thesame distance.

b) A chord of a circle is a line segment thatconnects two points on the circle.

c) A circle is a given point, called the cen-ter, and a set of points that are equidistantfrom the center. It creates a region in theplane and divides the plane into three parts:the points inside the circle, the points outsidethe circle, and the points on the circle.

e) A diameter of a circle is a chord thatgoes through the center of the circle. It is thelongest possible chord.

f) A radius is a line segment that connectsthe center and a point on the circle. All the ra-dii of a circle are the same length.

g) A sector of a circle is a pie wedge regiondefined by two radii and an arc of the circle.

h) A segment of a circle is a region definedby a chord and an arc of the circle.

17. Use a straightedge and a sharp pencilto score a line segment on your paper and la-bel the endpoints A and B. Fold the paper sothat Points A and B coincide; crease the pa-per along the fold. What geometric figurehave you constructed? Place a few points atrandom positions on the fold line and labelthem C, D, E, and F (be sure that one of thepoints is at the intersection of the fold with theoriginal segment from A to B). From each ofthese points, one by one, compare the dis-tance to A with the distance to B. What mightyou conjecture based on this construction?

SOLUTIONS MANUAL SM-35

When the segment is created so that PointA and Point B coincide, the resulting crease isthe perpendicular bisector of Segment AB.From a point on the perpendicular bisector,the distance is the same to each of the end-points of the segment. The point where theperpendicular bisector intersects the seg-ment is the midpoint of the segment.

18. Start on a new piece of paper with a linesegment from A to B scored. Place a point, C,on the segment, not very near the center. Foldthe paper through Point C so that the parts ofthe line segment on either side of Point C coin-cide; crease the paper on this fold. How doesthis construction compare to the constructionin Exercise 17? How does it differ?

The crease is perpendicular to the seg-ment, but the point of intersection is not themidpoint of the segment because it wasplaced away from the center of the segment.

construction compare to the constructions inExercises 17 and 18? How does it differ?

The crease is perpendicular to the seg-ment, but the point of intersection is not themidpoint of the segment because it wasplaced away from the center of the segment.

20. Start on a new piece of paper with aline segment from A to B scored. Place twopoints, C and D, on the segment. Using first Cand then D, fold through the point so that theparts of the line segment on either side of thepoint coincide; crease along the fold. Howdoes each of these two creases meet the linesegment from A to B? What conjecture canyou make about these two creases?

Each of the creases is perpendicular to theline segment. The two creases are parallel toone another.

19. Start on a new piece of paper with aline segment from A to B scored. Place apoint C somewhere off the line segment. Foldthe paper through point C so that the parts ofthe line segment on either side of point C co-incide; crease along the fold. How does this

21. On a new piece of paper, score an an-gle on your paper; label the Vertex B and theRays t and s. Fold the paper through Point Bso that Rays t and s coincide; crease the pa-per along this fold. What construction haveyou completed? (If you used an acute angle,

SM-36 SOLUTIONS MANUAL

try this again with an obtuse angle—if youused an obtuse angle first, then try it againwith an acute angle and a right angle.)

The crease divides the angle into two equalsize angles. It is the angle bisector of the an-gle whether the angle is acute, right, or ob-tuse.

22. Using only a compass and straight-edge, construct a triangle and the altitude toone side of the triangle. For which of the pa-per folding constructions is this an applica-tion? (If you wish, you may complete this con-struction using paper folding or Geometer'sSketchpad® or other dynamic software.)

The altitude of a triangle is an application ofconstructing a perpendicular to a segmentfrom a point not on the segment.

23. All of the figures in this discussionwere sketched as if we were viewing themfrom above (indicated by the segments orcurves that are dashed). Sketch each figureas if you were looking at it from below.

Changing only the perspective from whichthe objects are viewed, the sketches wouldappear as shown in the figure.

24. Consider the following capital letters ofour alphabet:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

a) Determine which have reflectional sym-metry about a vertical mirror.

b) Determine which have reflectional sym-metry about a horizontal mirror.

c) Determine which have rotational sym-metry.

d) Do any of the letters have more thanone type of symmetry?

e) Do any letters have all three types ofsymmetry?

f) Do any have no symmetry?

a) A, H, I, M, O, T, U, V, W, X, Yb) B, C, D, E, H, I, O, Xc) H, I, O, S, X, Zd) Both vertical and horizontal mirror: H, I,

O, Xe) Mirror and rotational: H, I, O, Xf) No symmetry: F, G, J, K, L, P, Q, R

SOLUTIONS MANUAL SM-37

5: MEASUREMENT

1. Make a list of all the measurementwords you can think of in 2 minutes. Shareyour list with classmates.

A few words to get your list started are am-plitude, area, bushel, calendar, capacity, cen-timeter, circumference, cubic, cup, degrees,diameter, digit, dimensions, distance, estima-tion, fat-free, fathom, foot, furlong, gallon,hand, height, inch, length, light year, linear, li-ter, meter, mile, millimeter, nautical mile, palm,peck, perimeter, pint, quart, radians, radius,rod, ruler, scale drawing, seriation, span, sur-veying, tape measure, temperature, tiling,time, volume, weight, width, and yard ...

2. Make a list of all the dichotomous mea-surement words you can think of in two min-utes. Compare your list with those of yourclassmates.

A few pairs of words to get your list startedare shorter-longer, littler-bigger,smaller-larger, shorter-taller, less-more,few-many, fewer-more, lighter-heavier, shal-lower-deeper, lower-higher, and nar-rower-wider . . .

3. Find a subtle example of confusing di-chotomous terminology using the termslarger and smaller.

Consider the books that are usually foundin an elementary classroom. Could a "large"book have fewer words than a "small" book?

4. When can the "smallest" be larger thaneven the "biggest"? When can the "biggest"be smaller than even the "smallest"?

The smallest second-grade student mightbe just a bit larger than the largest first-gradestudent, but the biggest child in kindergartenmight be a great deal smaller than the small-est child in sixth grade.

5. When you say that a beach ball is "big-ger" than a bowling ball, what attributes areyou ignoring?

Mass, density, or weight might be ignored.

6. What attributes, other than the numberof square yards required, influence the cost ofcarpeting?

Thickness, height of pile, or weight of ma-terial might influence the cost of carpet.

7. Attribute Blocks™ help children learnthe terminology and concepts of "differ-ences." The 60-piece set includes squares,rectangles, circles, triangles, and hexagons,in two sizes, three colors, and two thick-nesses. List at least 10 different ways bywhich Attribute Blocks™ could be sorted.

A few sorting criteria to get your list startedare color; size; shape; thickness; color andsize; color and shape; color and thickness;size and shape; size and thickness; shapeand thickness; color, size, and shape . . .

8. Measure the height of your coffee cupusing anything except a ruler. Would youwant to use a new pencil as your unit?

My coffee mug is approximately as tall as3.5 times the diameter of a quarter. Few cof-fee mugs are as tall as a new pencil.

9. Measure the length of a room in paces.(A pace is two steps or about 5 feet. TheRomans figured that 1000 paces was a mile.)Would it have made sense to use a new pen-cil for this measurement?

ANSWERS WILL VARY. It would take toolong (and return meaningless data) to mea-sure a room in new pencils.

10. Trace your hand and wrist. How manymeasurements can you discuss about yourtracing? Measure in inches, then in centime-ters. Which was easier? What is the finest de-gree of precision you can reach with yourruler?

ANSWERS WILL VARY.

11. Measure the line segments in Fig. 5.4to the finest degree of precision available to

SM-38 SOLUTIONS MANUAL

you in the customary system and the metricsystem. Measure the distances AB, AC, AD,AE, BC, BD, BE, CD, CE, and DE.

12. What is the total length of a cornershelf that is 5 yd 1 ft 9 in along one wall and 4yd 2 ft 7 in along the other wall? Be careful asyou regroup the measures in your solution.

13. What is the total length of a cornershelf that is 5m 10 cm 5 mm along one walland 4 m 44 cm 5 mm along the other wall?

14. Use a piece of string to measure thedistance an ant would walk around the outeredge of the leaf in Fig. 5.5. Compare yourstring measure to a ruler to find the actual dis-tance.

ANSWERS WILL VARY.

15. Make a set of unit square tiles formeasuring area using any unit you like. Findan everyday item with a flat surface and esti-mate the area of the flat surface by tiling. Youmay need to use partial tiles. How can youmake your estimate as "good" as possible?

ANSWERS WILL VARY.

16. Create a one-foot square using 144square inch tiles. Create a one-yard squareusing 9 square foot tiles.

You will need 144 one-inch squares for aone-foot square. You will need 9 one-footsquares for a one-yard square.

17. Find the perimeter of each region inFig. 5.7 using your string and a centimeterruler; then find the area of each by countingand estimating the number of squares cov-ered by each.

18. At some point, we will put away ourstring and grid paper and begin to use theconcepts of perimeter and area abstractly.From your previous experience with algo-rithms, identify these common formulas andtell the meaning of each letter and symbol.

a) A = I • w Area equals length timeswidth—Area of a rectangle or square

b) P = 21 + 2w Perimeter equals twice thelength plus twice the width—Perimeter of anyparallelogram

c) A = s2 Area equals length of sidesquared—Area of a square

d) P = 4s Perimeter equals four times thelength of side—Perimeter of a square

e) h Area equals one half of the

length of the base times the length of theheight (altitude)—Area of a triangle

f) Area equals one half of

the height (altitude) times the sum of thelengths of the bases—Area of a trapezoid(Special Note: This formula can be used tofind the area of any regular polygon.)

SOLUTIONS MANUAL SM-39

g) P = ns Perimeter equals number ofsides times the length of a side—Perimeter ofa regular polygon

h) A = -ans Area equals half of the

apothem (a line segment from the center ofthe polygon perpendicular to a side) times thenumber of sides times the length of aside—Area of a regular polygon

i) A = Ttr2 Area equals Tir times the radiussquared—Area of a circle

j) C = 2nr Circumference equals two timesTI times the radius—Circumference (perime-ter) of a circle

19. Use the appropriate algorithms in Ex-ercise 18 to check your measures and esti-mates in Exercise 17.

See answers for Exercise 17.

20. Which algorithms in Exercise 18 werenot helpful in completing Exercise 17?

There were no figures in Exercise 17 thatrequired the algorithms (g) and (h).

21. Why is there no algorithm in Exercise18 for finding the perimeter of the trapezoid orthe triangle? Do you think perimeter algo-rithms are needed?

ANSWERS WILL VARY. All that is neededfor perimeter is to add the measures for allsides of the figure. Perhaps a generic formulafor perimeter is P = da (distance around).

22. Use the formula, A = -ans, to find the

area of a region defined by a regular octagonwith sides of length 4 cm and an apothem oflength 4.828 cm.

A = 1 • 4.828 - 8 - 2

A = 77.248 square centimeters

23. Build a cube using 27 unit cubes. Useany unit cube, even a sugar cube.

a) What are the linear measurements as-sociated with this cube?

The cube will be three units in height,width, and depth.

b) What are the area measurements asso-ciated with this cube?

The area of each face will be 9 squareunits. Because there are six faces, the to-tal surface area will be 54 square units.

24. Use 24 cubes to answer the following:

a) How many different rectangular prismscan you form using the 24 unit cubes?

1 x 1 x 24; 1 x 2 x 12; 1 x 3 x 8; 1 x 4x 6 ; 2 x 2 x 6 ; 2 x 3 x 4

b) Do all of the prisms have the same vol-ume?All have the same volume, 24 cubic units.

c) Use your prisms to justify the volumeformula, V = Iwh, for rectangular prisms.

Each of the products in Part 1 is 24.d) Add the areas of the six faces to deter-

mine the total surface area for each of yourprisms. Do they all have the same surfacearea?

The sum of the areas of the faces of the 1 x1 x 24 is 24 + 24 + 24 + 24 + 1 + 1 = 98square units. The sum of the areas of thefaces of the 1 x 2 x 12 is 12 + 12 + 24 + 24+ 2 + 2 = 76 square units. This is sufficientto say that the prisms do not all have thesame surface area.

25. Consider a solid cube that is 6 unitcubes in height, width, and depth. If you paintthe six faces of this large cube, each unitcube may have zero, one, two, or threepainted faces, depending on its location inthe cube.

a) How many of the unit cubes will havepaint on at least one face?

152 unit cubes will have paint on at leastone face.

b) How many of the unit cubes will haveonly one face painted?

96 will have one face painted.

SM-40 SOLUTIONS MANUAL

c) How many of the unit cubes will havetwo faces painted?

48 will have two faces painted.

d) How many of the unit cubes will havethree faces painted?

8 will have three faces painted.

e) How many of the unit cubes will have 4,5, or 6 faces painted?

None will have 4, 5, or 6 faces painted.

f) How many of the unit cubes will have nofaces painted?

64 will have no faces painted.

26. The formula for the volume of a rightcircular cylinder is V = mPh, or the area of thecircular top face times the height. Use the in-formation in Fig. 5.20 to find the volume of thecylinder.

V = 7i • 52 • 18

V = 450.71 D 1313 cubic centimeters

27. The formula for the volume of a right

circular cone is V = -rci h. Use the informationO

in Fig. 5.21 to find the volume of the cone.

V = 1548 cubic units

30. The volume of a pyramid also dependson the number of edges on the base. Use V =

-s2h and the information in Fig. 5.23 to findO

the volume of the right square pyramid.

V = £ - 9 • 18.2

V = 54.6 cubic centimeters

31. Name everyday items that you coulduse to approximate each of the following: 1gm, 1 kg, 1 oz, 1 Ib, 1 T

ANSWERS WILL VARY. Examples:paperclip, two pounds of coffee, a McDonaldtoy, box of powdered sugar, half a car

32. What is the sum of 8 T 1700 Ib 13 ozand 5 T 980 Ib 6 oz?

8 Tons+ 5 Tons

13 Tons

14 Tons

1700 pounds980 pounds

2680 poundc1 Ton 680 pounds681 pounds

13 ounces6 ounces

19 ouncos1 pound 3 ounces3 ounces

33. Explain why you would know that aclock with one hand pointed directly at the 9and the other hand pointed directly at the 3was broken.

The tip of the shorter "hour" hand movesslowly from one number to the next in onehour, whereas the tip of the longer "minute"moves all the way around the clock face inone hour. If it is 9:15, then the tip of theshorter hand should be one fourth of the wayfrom the 9 to the 10. If it is 15 minutes until 3,the tip of the shorter hand should be threequarters of the way from the 2 to the 3.

34. What other measure uses base 60? Isthere a connection between these two meas-ures?

Angles are measured in degrees, minutes,and seconds. There are 60 seconds in a min-ute, 60 minutes in a degree, and 360 degrees

28. How do the volume formulas in Exer-cises 26 and 27 compare?

If a cylinder and a cone have identicalbases and heights, then the cylinder will havethree times the volume of the cone.

29. To find the volume of a regular rightpolygonal prism, find the area of the top faceand multiply by the height. For this exercise,

use the formula V = -asnh to find the volume

of the right hexagonal prism in Fig. 5.22.

31. Name everyday items that you coulduse to approximate each of the following: 1gm, 1 kg, 1 oz, 1 Ib, 1 T

ANSWERS WILL VARY. Examples:paperclip, two pounds of coffee, a McDonaldtoy, box of powdered sugar, half a car

32. What is the sum of 8 T 1700 Ib 13 ozand 5 T 980 Ib 6 oz?

SOLUTIONS MANUAL SM-41

in a revolution. The measures used for timeand rotations both stem from earth measures.

35. Use denominate numbers to deter-mine how much time elapses between9:24:13 AM and 1:15:08 PM? (Don't forgetthat, on the 24-hour clock, you can use 13:00for 1:00 PM.)

36. List as many different ways as you canthink of to use U.S. coins to make a dollar.

A few of the combinations are 100 pennies;95 pennies and 1 nickel; 90 pennies and twonickels; 90 pennies and one dime; 85 penniesand three nickels; 85 pennies, one dime andone nickel; .. .

37. How much change should you receiveif you have a $10 bill and the amount youmust pay is $4.15? How many different waysmight you receive the change? (Please re-

member that 585 pennies might not be rea-sonable, but is possible.)

One way is 1 $5 bill, 1 half dollar, 1 quarter,1 dime, and 1 nickel

38. Using Fig. 5.25, find the boiling pointand the freezing point of water at sea level indegrees Fahrenheit and degrees Celsius.

212°F, 100°C is the boiling point of water;32°F, 0°C is the melting point of ice;

39. Using Fig. 5.25, determine the comfortrange for humans in degrees Fahrenheit anddegrees Celsius.

Answers will vary around 72°F, 20°C

40. How many degrees are in one full rota-tion?

360°

41. Do the lengths of the legs of an anglehave any meaning in degree measure?

No. The degree measure of an angle is ameasure of rotation.

42. Is there a special name for an angle of180°?

Straight angle

43. Identify each named angle in Fig. 5.27as "less than," "exactly," or "greater than" 90°,then measure the angles using a protractor.

Angles HGI and ABC are acute

&

44. Given that Angle 1 measures 48°, An-gle 10 measures 90°, Angle 20 measures115°, and Line m is parallel to Line n, use yourknowledge of perpendicular lines, parallellines, supplementary angles, complementaryangles, and triangles to determine the meas-ure of each angle in Fig. 5.30.

45. Find and confirm five or six more Py-thagorean Triples.

5, 12, 13; 7, 24, 25; 5n, 12n, 13n; 7n, 24n,25n; 3n; 4n; 5n

SM-42 SOLUTIONS MANUAL

46. How are the two formats demon-strated the same? How do they differ?

They are the same because both involvemultiplying and simplifying fractions. They aredifferent because the first explicitly shows themultiplication and the second implies it.

47. How did the units go from feet per sec-

ond in the first column to miles per hour in thelast column?

The units changed because the fractionswere simplified:The ft in the numerator of the first fractionwas divided out by the ft in the denominatorof the second fraction; the sec in the numer-ator of the third fraction was divided out bythe sec in the denominator of the first frac-tion; the min in the numerator of the fourthfraction was divided out by the min in the de-nominator of the third fraction. This simplifi-cation process left only mi in the numeratorand hr in the denominator.

48. If it is reported that a matchbox carhad an average speed of 30 mph, how manyfeet did it travel in 1 second?

6: DATA ANALYSISAND PROBABILITY

Data Collectionand Representations

1. Gather three types of data about yourclassmates, for example: How many of yourclassmates are wearing glasses? How manyof your classmates are wearing shoes?

Answers will vary. Example: 15 peoplewore glasses and 15 did not; 20 are womenand 10 are men; 30 people wore shoes.

2. Gather three types of data about theroom or building in which you are taking thisclass, for example: How many desks/tablesare in the room? How many ceiling tiles arethere?

Answers will vary. Example: 30 desks; 3white boards; 8 banks of fluorescent lights.

3. Gather three types of data about the en-vironment in which you currently live (dorm,house, apartment, etc.), for example: Howmany books? How many chairs?

Answers will vary. Example: 3 cats; 9chairs; 1 TV.

4. Using proportional reasoning, what per-centage of the people in your class: wearglasses; are male; wear shoes; have cats;have dogs; have 3 keys?

Answers will vary. Example:

5. Construct circle graphs to represent thedata you collected about eyeglasses, shoes,

and pets.

Answers will vary. Example:

SOLUTIONS MANUAL SM-43

6. Construct a line graph to represent thenumber of keys each person in the class re-ported.

Answers will vary. Example:

7. Construct a bar graph to represent thenumber of keys each person in the class re-ported.

Answers will vary. Example:

8. Construct a line graph from the data youcollected about how many keys your class-mates had with them.

Answers will vary. Example:

9. Construct a horizontal histogram for thedata you collected about eyeglasses, gender,shoes, and pets.

Answers will vary. Example:

10. Construct a frequency polygon for thedata you collected about eyeglasses, gender,shoes, and pets.

Answers will vary. Example:

SM-44 SOLUTIONS MANUAL

11. Construct a table for your key datawith equal classes and tabulate the frequencyof each.

Answers will vary. Example:

Key Data

From the table construct a frequency polygonfor the key data.

Answers will vary. Example:Re-poll your classmates and keep track of themodel year of each person's car along withhow many keys they each have. Place thedata into an appropriate table and do the ex-ercises that follow.

12. Construct a scatter plot for your datausing car year versus keys.

Answers will vary. Example:

13. What, if any, relation do you see be-tween the model years of your classmates'cars and the number of keys they had?

Answers will vary. The most likely responseis, "No relation is seen between my class-mates cars and the number of keys."

14. How would a scatter plot look if therewere a positive relation between the model

years of your classmates' cars and the num-ber of keys they had?

A scatter plot with a perfect positive rela-tion would show that for each new modelyear, they would gain another key, with allclassmates with the same age of car havingexactly the same number of keys. The oldertheir cars, the fewer keys they had.

15. How would a scatter plot look if therewere a negative relation between the modelyears of your classmates' cars and the num-ber of keys they had?

A scatter plot that showed a negative rela-tion between the ages of my classmate's carsand their number of keys would have pointsthat started in the top left of the plot withpoints gradually shifting down and to theright. The newer their cars, the fewer keysthey had.

Data Analysis and Statistics

1. Find the mean, median, mode, andrange of the key data you collected from yourclassmates.

Answers will vary. See data in the Line Plotsection

Mean = 3.467Median = 3Mode = 2Range = 7

2. Find the mean, median, mode, andrange of the rainbow centimeter cube data.

Answers will vary. See data in Table 6.5

Mean = 30Median = 28.5Mode = 25 and 40Range = 22

SOLUTIONS MANUAL SM-45

Use the first 10 key data points reported byyour classmates to answer the following:

3. Find the variance of the sample usingboth formulas. Calculate each by hand andthen verify your results using a software ap-plication.

Answers will vary. An example follows:

First 10 data points from key data:

4. Find the standard deviation for thissample of your data.

Answers will vary. An example follows:

Yes, 7 keys would be considered an outlier,as it is greater than 2 standard deviationsaway from the mean.

6. Does this sample accurately representthe whole class? Why or why not?

Answers will vary.No, the mean for the whole sample was

3.47 keys per person, whereas the mean forthe sample was only 2.9 keys per person.There would have been less than 95% (only27 keys or 90% of the keys) of the whole dataset within (2s of the mean (2.9 keys).

Use your key data for the following problems:

7. Find the first, second, and third quar-tiles.

Answers will vary. An example is: Q1 = 2;Q2 = 3; Q3 = 5.

8. Find the IQR.

Answers will vary. An example is:Q3 - Q1 = 5 - 2 = 3.

9. Create a box and whisker plot.

Answers will vary. An example is:

5. Look back at your full data set for thekey data. Are there any outliers based on thestandard deviation of the sample?

Answers will vary. An example follows:

10. Determine the expected range usingthe IQR and identify any outliers.

Answers will vary.An example is:

SM-46 SOLUTIONS MANUAL

Counting and Probability

10. Suppose the rainbow colors couldcome in any order. How many different pat-terns of colors of the rainbow (ROYGBIV)could there be?

a) Permutation: P(n, n) = n!b) The question wants to know how many

different patterns or arrangements of the rain-bow colors there are. Order is important.

c) P(7, 7) = 7! = 5040

11. A certain guitar has 12 strings. Howmany different orders can a guitarist pluckany 5 strings on this guitar?

12. How many different pairs of colors canbe made using colors of the rainbow(ROYGBIV)? (We understand that you wouldnot wear or be seen in some of these combi-nations, but we do want you to consider andcount them all.)

i

13. What is the cardinality of the samplespace for each of the following?

a) an experiment tossing 5 coins 25 = 32b) an experiment rolling 3 dice 63 = 216

14. What is the probability of obtaining thefollowing outcomes from the experiments inExercise 13?

15. There are R = 3 marbles, O = 4 mar-bles, Y = 4 marbles, G = 2 marbles, B = 1, I =3 marbles, and V = 3 marbles in a bag. Whatis the probability of drawing BORG from thegiven bag of rainbow marbles when there isreplacement?

16. What is the probability of drawingBORG from the given bag of rainbow marbleswhen there is no replacement? There are R =3 marbles, O = 4 marbles, Y = 4 marbles, G =2 marbles, B = 1, I = 3 marbles, and V = 3marbles in the bag.

17. What is the probability of drawing anAD through 10D, any J, or a 3S from a fairstandard deck of 52 playing cards when thereis no replacement and five cards are drawn?Consider that each draw was a success.

18. What is the probability of drawing a10H card, given that a red card was drawn?

P(10H|red) = 1:26

SOLUTIONS MANUAL SM-47

19. What is the probability of rolling a 2 oneither of two dice, given that the sum of thedice is 7?

P(2|sum = 7) = 2:6 or 1:3

20. What is the probability of drawing a redK card, given that a black card was drawn?

P(red K |black) = 0:26 or 0

21. Drawing a black card from a fair stan-dard deck of 52 playing cards?

a) m = 26; m' = 26; 26:26 = 1:1 (we call 1:1even odds)

b) m' = 26; m = 26; 26:26 = 1:1c) m = 26; n = 52; P(m) = 26:52 = 1:2d) m' = 26; n = 52; P(m') = 26:52 = 1:2

22. Flipping two coins and obtaining HT?

a) m = 1; m' = 3; 1:3b) m' = 3; m = 1; 3:1c) m = 1; n = 4; P(m) = 1:4d) m' = 3; n = 4; P(m') = 3:4

23. Drawing BORG from the given bag ofmarbles when there is replacement?

There are R = 3 marbles, O = 4 marbles, Y= 4 marbles, G = 2 marbles, B = 1, I = 3 mar-bles, and V = 3 marbles in the bag.

a) m = 3; m' = 19997; 3:19997 (use the an-swer from 14. C. and m' = n - m)

b) m' = 19997; m = 3; 19997:3 (use the an-swer from 14. C. and m' = n - m)

c) P(m) = 3:20000 (we knew this from Ex-ercise 14. C.)

d) m' = 19997; n = 20000; P(m') =19997:20000 (use the answer from Exercise14. C. and m' = n - m)

7: PROBLEM SOLVING

Riddle Me This

#1: List the numbers 1 through 20:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, 17, 18, 19, 20

Cross out one (remember one is not prime)and all the prime numbers

>

Cross out all the numbers that can't easilybe written as a power of another number

Show the remaining numbers using expo-nents

4 = 22 8 = 23 9 = 32 16 = 42

We eliminate 4 because the power (2) isequal to the base (2)

We see that 23 * 32, eliminating both 8 and9

We know that 42 = 16, we test 24 = 2 x 2 x 2x 2 = 16, so the solution to the riddle is16. 16 = 42 = 24

#2: Make a list with one column for the numberof teams, one for the number of games, andone for the number of winners. It was singleelimination with no ties allowed, so only thewinning team of any contest gets to move on:

If 128 teams played in a single eliminationtournament, then a total of 127 games wouldbe played. Another way of looking at it wouldbe to say that, if the tournament is single elim-ination, then all teams except the championlost, so out of all 128 teams, 127 lost. A lossoccurs each time a game is played. Thus, 127games were played.

#3: Original question—If the length of theLoch Ness monster is 20 meters long plushalf of its own length, how long is Nessie?

SM-48 SOLUTIONS MANUAL

Rewrite and translate —How long isNessie if the length is 20 meters plus halfof its own length?

How long is NessieL = Nessie's length

if the length is 20 meters plus half its ownlength?

combine like terms by sub-

tracting from both sides of the equa-

tion, then

multiply both sides of the equa-

tion by 2, thenL = 40 m

Nessie the Lock Ness monster is 40 meterslong if the length of the Loch Ness mon-ster is 20 meters long plus half of its ownlength.

#4: 642 digits were written

9 pages + 90 pages + 151 pages = 250 pages

#5: Using reasoning: If every dollar weighs agram, then $1,000,000 weighs 1,000,000grams. There are 1000 grams in a kilogram,thus there are 1000 kilograms in 1,000,000grams. You can either recall or look up thatone kilogram is about 2.2 pounds. Then 1000kilograms is approximately equivalent to 2200pounds or about one ton. You should advisethe producer that no ordinary person wouldbe able to pick up $1,000,000 in $1 bills be-cause they would weigh about one ton!

Collection of problems:

1. Take an ordinary sheet of paper andfold it in half. Fold it in half a second time.Fold it in half a third time. If you could con-tinue folding it in half 50 times, how highwould the stack of paper be?

0.375 in. « 87 pages then 1 page « 0.00431in.

250 = 1.126 x 1015

0.00431(1.126 x 1015) = 5.43186 x 1012 in.or 85,730,114 miles

2. A farmer had 26 cows. All but 9 died.How many lived?

17 cows died and 9 cows lived

3. A uniform log can be cut into threepieces in 12 seconds. Assuming the samerate of cutting, how long will it take for a simi-lar log to be cut into four pieces?

2 cuts divide the log into 3 pieces; then 3cuts divide the log into four pieces

2 cuts take 12 seconds; then 3 cuts willtake 18 seconds. Another way of lookingat this is to make one cut parallel to thelong sides of the log and then a secondcut perpendicular to it. That will give 4pieces. Granted, the length of the logcould be such that the time requiredwould differ, but you have to admit thatthis is a pretty creative solution, and itwas proposed by a third grader.

4. How many different ways can you addfour odd counting numbers to get a sum of10?

SOLUTIONS MANUAL SM-49

5. What is the sum of the first 100 consec-utive counting numbers?

This gives 100 pairs of addends, each with asum of 101, for a total of 10100. That is twiceas much as needed, because each number isused twice. Divide by two, getting 5050.

6. How many cubic inches of dirt are therein a hole that is 1 foot deep, 2 feet wide, and 6feet long?

The hole is 12 in x 24 in x 72 in; there are 0cubic inches of dirt in it or it wouldn't be ahole.

7. How many squares are there in a 5 by 5square grid?

1 —5 x 5 square; 4—4 x 4 squares; 9—3x 3 squares; 16—2 x 2 squares; 25—1x 1 squares

1 + 4 + 9 + 16 + 25 = 55

8. A little green frog is sitting at the bottomof the stairs. She wants to get to the 10thstep, so she leaps up 2 steps and then slidesback 1. How many leaps will she have to takeif she follows this pattern until she reachesthe 10th step?

9 leaps. Ground to 2 to 1 to 3 to 2 to 4 to 3to 5 to 4 to 6 to 5 to 7 to 6 to 8 to 7 to 9 to8 to 10. The bold indicates leaps up.

9. If there are 7 months that have 31 daysin them and 11 months that have 30 days inthem, then how many months have 28 days inthem?

All of the months have 28 days.

10. There are exactly 11 people in a roomand each person shakes hands with every

other person in the room. When A shakeswith B, B is also shaking with A; that countsas ONE handshake. How many handshakeswill there be when everyone is finished?

11. What number does this rep-resent?

sevenT-nine

12. There are 9 stalls in a barn. Each stallfits only 1 horse. If there are 10 horses andonly 9 stalls, how can all the horses fit into the9 stalls without placing more than one horsein each stall?

NINESTALLS = 10 letters; one on each stallfor each horse

13. You are given 5 beans and 4 bowls.Place an odd number of beans in each bowl.Use all beans.

Visualize 4 bowls arranged from smallest tolargest and stack the bowls one inside theother. You could place all 5 beans in thesmallest bowl and be done. There are othersolutions.

14. You are to take a pill every half hour.You have 18 pills to take. How long will yoube taking pills?

3 are taken the first hour, then 2 each hourafter—8.5 hours

15. If you got a 40% discount on a$150.00 pair of sport shoes and 20% off a

SM-50 SOLUTIONS MANUAL

interesting that this mathematician was bornon "pi day." Give his name.

Albert Einstein

8: REASONING AND PROOF

1. Using p, q, r, and the truth table in Table8.3, determine if the associative property istrue for conjunctions. In other words doesthe following statement hold: p A (q A r) =(p A q) A r. (Hint: You will need columns forboth q A r and p A q.)

The associative property holds for "and"statements.

2. Using what you have learned aboutprobability, what is the probability of getting:

a) a true "and" statement (PAQ) with twovariables?

b) a true "and" statement with three vari-ables?

c) a false "and" statement with two vari-ables?

d) a false "and" statement with three vari-ables?

3. Experiment with the different types oflogical expressions that have been dis-cussed. Find at least two different tautologiesthat have not already been discussed. Make atruth table for each to prove it is a tautology.

Answers will vary.Informally prove or disprove the followingstatements:

$200 set of roller blades, what was the per-cent discount on the total purchase (assum-ing no taxes are involved)?

16. Where should the Z be placed andwhy?

Z goes in the top because it doesn't have anycurved parts.

17. Estimate how old will you be in years ifyou live 1,000,000 hours?

18. A child has $3.15 in U.S. coins, butonly has dimes and quarters. There are morequarters than dimes. How many of each coindoes the child have?

$2.85 - $0.10 = $2.75 —$2.75 is a multipleof 25 so 11 quarters and 4 dimes is a solution.Continuing this thinking, $2.25 is the nextpossible answer for quarters but that is 9quarters and that demands 9 dimes.

19. There are three children in a family.The oldest is 15. The average of their ages is11. The median age is 10. How old is theyoungest child?

8,10, and 15 are the ages. You are given 2of the ages. One of them is directly shown as15. Median, because there are 3 kids, mustbe the middle age. So, 10 + 15 + n must equal33, if the average is 11. 33 - (10 + 15) = 8.

20. A famous mathematician was born onMarch 14, which could be written 3.14. Thisdate is the start of a representation for pi. It is

SOLUTIONS MANUAL SM-51

4. The operation of addition is closed onthe set of whole numbers.

Answers will vary. 1 + 1 = 2; 1 + 2 =3; 1 + 3=4; ...

5. The operation of addition is closed onthe set of digits.

Answers will vary.Suppose the operation of addition is

closed on the set of digits. Arbitrarily choose5 and 6 as addends. But, we know that 5 + 6= 11, which is not a digit, a contradiction.Thus, the operation of addition is not closedon the set of digits. QED

6. Make up a 5 by 5 addition table like theone show in Fig. 8.2 (be sure the addends arehidden) and give it to someone who is not inthis class to do, following the directions givenfor Fig. 8.2. Record their reaction to the trick.

Answers will vary.

9: COMMUNICATION

aside and don't look at it for 24 hours. Is yourparagraph organized and coherent? Does itconcisely explain what you did and why youdid it?

Answers will vary.

2. Exchange paragraphs with a classmate.Do you understand that person's explana-tion? Ask questions that might help the per-son clarify their explanation.

Answers will vary.

3. Compare the two paragraphs, identify-ing the strong points of each. Reason to-gether logically about disputed points.

Answers will vary.

4. Using the strong points of each, collab-orate to write a paragraph that is better thaneither of you wrote alone. Ask someone whois not in your class to use the paragraph tocomplete the addition problem.

Answers will vary.

1. Write a paragraph explaining how youwould add 842 + 136. Put the paragraph