A Geometric Introduction to Mathematical Induction · PDF file · 2015-09-09A Geometric Introduction to Mathematical Induction Problems ... Nicholls State University Math 584 Fall

Running head: A GEOMETRIC INTRODUCTION 1

A Geometric Introduction to Mathematical Induction Problems using the Sums of Consecutive Natural Numbers, the Sums of

Squares, and the Sums of Cubes of Natural Numbers

Nicholls State University

Math 584 Fall 2014

Galina Landa

A GEOMETRIC INTRODUCTION 2

Abstract

The Principle of Mathematical Induction (PMI) is a powerful method of proof. The illustrated

examples often come from algebra rather than geometry. This paper explores mathematical

induction in geometry. The traditional algebraic examples of the PMI involve proving formulas

for the sums of consecutive natural numbers, the sums of squares of natural numbers, the sums of

cubes, etc. However, it is not always clear how one would arrive at these formulas or in what

circumstances they would be useful. This research paper presents several geometry problems to

give context to such abstract formulas. It also demonstrates how one can use the PMI to prove

important theorems in geometry. The problems considered in this research involve finding the

number of lines through n points in a plane; finding the number of diagonals in convex polygons

with n sides; and finding the number of squares of all sizes enclosed by a two-dimensional

checkerboard or the number of cubes of all sizes enclosed by a three-dimensional checkerboard

with each side length equal to n units.


Introduction

A powerful method of proof, the Principle of Mathematical Induction (PMI) allows us to verify a

particular property for all consecutive integers greater than some smallest one. It works like this:

we start by checking if our conjecture is true for the first few initial values. Then, after assuming

that the conjecture is true for the given element, we check whether we can show that it is also true

for the next element. If we manage to complete both steps successfully, then the property is true

for all elements.

Mathematical induction is used to prove various sum formulas. However, it is not always clear

how one would arrive at these formulas, why they would be legitimate candidates for verification,

and in what circumstances they would be useful. In this paper, we will “discover” these formulas

in the contexts of several geometry problems and then prove them using mathematical induction.

Hopefully, by combining inductive investigation of geometry concepts with inductive proofs of

the resulting sums, we will gain a new perspective on the meaning of these formulas and a better

understanding of the PMI.

The process of mathematical discovery is quite exciting, but a proper mathematical investigation

can be time consuming. Thus, in this paper, we will limit our discussion to the following three

sums:

the sum of consecutive natural numbers

the sum of squares of consecutive natural numbers

the sum of cubes of consecutive natural numbers

These three sums will give names to the three major sections of this paper, which constitute the

body of related and original research. This body of research is called The Sums.


The Sums

The Sum of Consecutive Natural Numbers: 1 2 3 ... n

We will consider this first sum in the context of the following geometry problem.

Problem 1: Lines and Points

Find “the number of lines determined by a given number of points in a plane, no three of

which are collinear” (Wiscamb, p. 402).

Solution:

We are interested in the finite number of lines, so we will disregard a case of infinitely many lines

through a single point. We will start with natural n>1. Let us consider the first few cases for the

number of points n=2, n=3, n=4, n=5, and n=6 (see Table 1). Notice that while 2 arbitrary points

determine a single line, 3 non-collinear points determine 3 lines, 4 points determine 6 lines, 5

points determine 10 lines, and 6 points determine 15 lines.

Table 1. Lines and Points

n, points

nl ,

lines

1 2 3 ... ( 1)nl n 1

1( )

2n nn p pl

2 1 1=1

3 3 3=1+2 3(

23

12)

(The table continues on the next page)


4 6 6=1+2+3 6(

24

13)

5 10 10=1+2+3+4 5 4(

21

1) 0

6 15 15=1+2+3+4+5 6 5(

21

1) 5

Note: Diagrams constructed with The Geometer's Sketchpad® software.

If we look for patterns in the table, we may notice two interesting details applicable to each case.

1. The number of lines determined by n points is equal to the sum of natural numbers from 1

to n-1:

1 2 3 ... ( 1)nl n

2. The number of lines at each step is equal to half the product of the number of points at the

current step and the number of points at the preceding step:

1

1( )

2n n nl p p


Based on these observations, we can make the following conjecture.

Conjecture 1: For natural numbers greater than 1,

11 2 3 ... ( 1) ( 1)

2n n n

Proof:

We can prove this conjecture by the PMI.

Let P(n-1) be

11 2 3 ... ( 1) ( 1)

2n n n , n>1

1. P(2) is true:

12 1 2 (2 1) 1

2

2. Suppose P(n-1) is true for n>1.

3. We will show that P(n) is true. (We want to show that,

( 1)1 2 3 ...

2

n nn

)

By the induction hypothesis,

11 2 3 ... ( 1) ( 1)

2n n n

Let us add n to both sides:

( 1)( 1)

11 2 3 ... ( 1) ( 1)

2P n

P n

n n n n n


( 1) 2

2

n n n

( 1 2)

2

n n

( 1)

2

n n

Thus, P(n) is true. Then by the PMI, P(n-1) is true for all natural numbers n>1.

Therefore, we have proved the formula:

11 2 3 ... ( 1) ( 1)

2n n n , for natural n>1

From now on, we can use it to find the number of lines determined by a given number of points in

a plane, no three of which are collinear.

For this particular task, it was convenient to use the formula for the sum of natural numbers from

1 to n-1. We can also rewrite this formula for the sum of natural numbers from 1 to n (like we did

in our proof above for P(n)). Since we have already proven this result, we no longer call it a

conjecture, but a theorem.

Theorem 1 (The Sum of Consecutive Natural Numbers):

For natural numbers greater or equal to 1,

( 1)1 2 3 ...

2

n nn

,n

We can now use this result to discover some other mathematical relationships. For example,

Avital and Hansen (1976) posed the following question.


Problem 1a: Diagonals of n-gon

“What is the number of diagonals one can draw in a convex polygon of n vertices?” (Avital

& Hansen, p. 404)

Solution:

Clearly, a triangle has no diagonals; we will consider n > 3 (see Table 1a.)

Table 1a. Diagonals of n-gon

Polygon n, vertices d, diagonals

Triangle

3

0

Quadrilateral

4

2=2

Pentagon

5

5=2+3

Hexagon

6

9=2+3+4

Heptagon

7

14=2+3+4+5

Note: Diagrams constructed with The Geometer's Sketchpad® software.


After considering polygons with 4, 5, 6, and 7 sides, we find that the number of diagonals is 2, 5,

9, and 14, respectively. The last four numbers can be written as 2, 2+3, 2+3+4, and 2+3+4+5.

Thus, the number of diagonals in each of these cases is the sum of natural numbers from 2 to n-2,

where n is the number of vertices. Let us denote the number of diagonals d; then for an n-gon,

2 3 4 ... ( 2)d n

This resembles the sum from Theorem 1, minus 1, minus (n-1), and minus n. Then we can use the

result from Theorem 1 to find the expression for d:

2 3 4 ... ( 2)d n 1

1 2 3 ... 1 ( 1)

Theorem

n n n

1

( 1)1 1

2Theorem

n nn n

( 1) 4

2

n n n

( 1 4)

2

n n

( 3)

2

n n

Thus,

( 3)2 3 4 ... ( 2) , 3

2

n nn n

From now on, we can use this result to find the number of diagonals in a convex n-gon. We have

demonstrated that this result follows from Theorem 1, so we will call it Corollary 1.

Corollary 1: For natural numbers greater than 3,

( 3)2 3 4 ... ( 2)

2

n nn

,n


This result can be verified by the PMI for n>3; however, we will omit the proof here because it is

very similar to the proof of Theorem 1.

The Sum of Squares: 2 2 2 21 2 3 ... n

Wiscamb (1970) suggested introducing the formula for the sum of squares in connection with the

following problem.

Problem 2: Checkerboard in 2D

“Given an n-x-n checkerboard with n2 small squares each measuring 1 square unit, how

many squares of all sizes does it contain?” (Wiscomb, p. 403)

Solution:

Let us consider several cases for n=1, n=2, n=3, and n=4 (see Table 2).

Table 2. Checkerboard in 2D

Squares: 1x1 2x2 3x3 4x4 Total Number of Squares

n=1 1 21 1

n=2

4 1 2 21 2 5

n=3

9 4 1 2 2 21 2 3 14

n=4

16 9 4 1 2 2 2 21 2 3 4 30

Note: Square images above were formatted with MS Word Picture Tools.

We can see that that for n=1, there is only one square. For n=2, there are four 1x1 squares and one

2x2 square. For n=3, there are nine 1x1 squares, four 2x2 squares, and one 3x3 square. Finally, for


n=4, there are sixteen 1x1 squares, nine 2x2 squares, four 3x3 squares, and one 4x4 square. The

total number of squares in each case is the sum of squares of natural numbers from 1 to n (see

Table 2).

Continuing in this manner, we can conjecture that an nxn checkerboard will contain

2 2 2 21 2 3 ... n

squares of all sizes, but what is this sum equal to? How can we guess the closed-form expression

for the sum of squares of natural numbers?

To answer this question, we will try to use an analogy with the formula for the sum of natural

numbers from Theorem 1:

( 1)1 2 3 ...

2

n nn

This formula has a polynomial of the second degree on the right side. Thus, we will assume that

the sum of squares of natural numbers should add up to a polynomial of the third degree, something

like 3 2an bn cn d , for real coefficients , , ,a b c d . Our guess is that

2 2 2 2 3 21 2 3 ... ,n an bn cn d (Eq.1)

Where,

, , ,a b c d

However, what are these real coefficients , , ,a b c d ?

If we rewrite Equation 1 four times for n=1, n=2, n=3, and n=4, respectively, we will get a

system of four linear equations in four variables:


1:

2 :

3 :

4 :

n

n

n

n

3 2

3 2

3 2

1

2 2 2 5

3 3 3 14

4 4 4 30

a b c d

a b c d

a b c d

a b c d

1

8 4

27 9

64 16

2 5

3 14

4 30

a b c d

a b c d

a b c d

a b c d

We can solve this system by hand (using a method like Cramer’s rule, Gaussian elimination, or

several others), or we can use a calculator to save time and paper space.

The solutions are:

1 1 1, , , 0

3 2 6a b c d

Then,

3 2an bn cn d 3 2

3 2

2

1 1 1

3 2 6

2 3

6

(2 3 1)

6

.( 1)(2 1)

6

n n n

n n n

n n n

n n n

Then we can make the following conjecture.

Conjecture 2: For natural numbers greater or equal to 1,

2 2 2 2 ( 1)(2 1)1 2 3 ...

6

n n nn

,n

Proof:


Let P(n) be,

2 2 2 2 ( 1)(2 1)1 2 3 ...

6

n n nn

,n


1. P(1) is true:

21

1 (1 1)(2 1 1)1

6

2. Suppose P(n) is true.

3. We will show that P(n+1) is true. (We want to show that,

2 2 2 2 2 ( 1)1 2 3 ...

6

( 1) 1 (2( 1) 1)( 1)

nn

n nn

)


2 2 2 2 ( 1)(2 1)1 2 3 ...

6

n n nn

Let us add 2( 1)n to both sides. Then,

2 2 2 2 2

( )

1 2 3 ... ( 1)P n

n n 2

( )

( 1)(2 1)( 1)

6P n

n n nn

2

( 1)(2 1) 6( 1)

6

n n n n

( 1)

6

(2 1) 6( 1)n n n n

2( 1)

6

2 6 6n n n n

2( 1)

6

2 7 6n n n

( 1)

6

2 (2 3)n n n

( 1)

6

( 1) 1 (2( 1) 1)n n n

Thus, P(n+1) is true. Then by the PMI, P(n) is true for all natural numbers.


Therefore, we have proved the formula for the sum of squares of natural numbers,

2 2 2 2 ( 1)(2 1)1 2 3 ...

6

n n nn

,n

From now on, we can use it to find the number of squares of all sizes contained in a two-

dimensional checkerboard. Since we have already proved this result, we will no longer call it a


Theorem 2 (The Sum of Squares of Consecutive Natural Numbers):


2 2 2 2 ( 1)(2 1)1 2 3 ...

6

n n nn

,n

We will now consider a similar problem for a three-dimensional checkerboard. We will use this

problem to introduce the formula for the sum of cubes of natural numbers.

The Sum of Cubes: 33 3 31 2 3 ... n

Problem 3: Checkerboard in 3D

“How many cubes of all sizes are there in a n-x-n-x-n three-dimensional checkerboard?”

(Wiscomb, p. 404)

Solution:

Let us consider several cases for n=1, n=2, and n=3 (see Table 3.1).


Table 3.1 Checkerboard in 3D

Cubes: 1x1x1 2x2x2 3x3x3 Total Number of Cubes

n=1

1 31 1

n=2

8 1 3 31 2 9

n=3

27 8 1 3 3 31 2 3 36

Note: Cube images above were formatted with MS Word Picture Tools.

We can see that that for n=1, there is only one cube. For n=2, there are eight 1x1 cubes and one

2x2 cube. For n=3, there are twenty-seven 1x1 cubes, eight 2x2 cubes, and one 3x3 cube. The total

number of cubes in each case is the sum of cubes of natural numbers from 1 to n.

Continuing in this manner, we can conjecture that an nxnxn checkerboard will contain

33 3 31 2 3 ... n

cubes of all sizes, but what is this sum equal to? How can we guess the closed-form expression for

the sum of cubes of natural numbers?

Let us consider the sum of cubes when n=1, n=2, n=3, and n=4. The numbers that we get in each

case are perfect squares: 1, 9, 36, and 100, respectively (see Table 3.2, left column).

They can be written as 2 2 2 21 ,3 ,6 ,10 . Consider the sequence 1, 3, 6, 10. Notice that we can rewrite

the terms of this sequence as 1, 1+2, 1+2+3, 1+2+3+4 (see Table 3.2, right column).


Table 3.2 Checkerboard in 3D

Total Number of Cubes 2ik

n=1 31 1 2

1 1 n=2 3 31 92 2 23 ( )9 1 2

n=3 3 3 3 62 31 3 2 26 (1 236 3)

n=4 3 3 3 41 2 3 4 100 2 210 (1 2 3 4)100

Note: Total Number of Cubes expressed in two different ways.

Notice that in each case, the sum of cubes can be written as the square of the sum of consecutive

natural numbers from 1 to n:

33 3 3 21 2 3 ... (1 2 3 ... )n n

The right side of this expression contains the sum 1 2 3 ... n , for which we already know the

formula (from Theorem 1):

( 1)1 2 3 ...

2

n nn

Then, 2 2

2 ( 1)

41 2 3 ...

n nn

and therefore, 3 3 3 3

2 2( 1)

1 2 3 ...4

n nn

Then we can make the following conjecture.

Conjecture 3: For natural numbers greater or equal to 1,

3 3 3 3

2 2( 1)

1 2 3 ...4

n nn

,n

Proof:



Let P(n) be

3 3 3 3

2 2( 1)

1 2 3 ...4

n nn

1. P(1) is true:

3

2 21 (1 1)

14

1

2. Suppose P(n) is true.

3. We are going to show that P(n+1) is true. (We want to show that

3 3 3 3

2 23

1 2 3 ...( 1) (( 1) 1)

( 1)4

nn n

n

)


3 3 3 3

2 2( 1)

1 2 3 ...4

n nn

Let us add 3( 1)n to both sides. Then,

2 2

33 3 3 3 3

( )

( )

( 1)1 2 3 ... ( 1) 1

4P n

P n

n nn n n

2 2 3( 1) 4( 1)

4

n n n

2 2( 1) ( 4 4)

4

n n n

2 2( 1) ( 2)

4

n n

2( 1)( 2)

2

n n

2( 1)(( 1) 1)

2

n n

2 2( 1) (( 1) 1)

4

n n


Thus, P(n+1) is true. Then by the PMI, P(n) is true for all natural numbers.

Therefore, we have proved the formula for the sum of cubes of natural numbers:

3 3 3 3

2 2( 1)

1 2 3 ...4

n nn

,n

From now on, we can use it to find the number of cubes of all sizes contained in a three-

dimensional checkerboard. Since we have already proved this result, we will no longer call it a


Theorem 3 (The Sum of Cubes of Consecutive Natural Numbers):


2 233 3 3 ( 1)

1 2 3 ...4

n nn

,n

Recommendations for Further Research

There are additional geometry problems that could provide interesting contexts to various formulas

used to illustrate the Principle of Mathematical Induction. There are also cases when the PMI

proves indispensable to verifying important results in geometry. This mutually beneficial

relationship between geometry and the Principle of Mathematical Induction deserves further

consideration. Unfortunately, the limited scope of this paper does not allow us to explore these

topics in more detail; however, below are some sample questions that could be investigated in

future research.


Questions about n lines in the plane:

Given n concurrent lines, how many pairs of supplementary angles are formed? How many

pairs of vertical angles are formed? What is the maximum number of regions in the plane

determined by n lines? (Wiscamb, 1970)

Questions about polygons with n sides:

What are the sums of interior angles of various convex polygons? Into how many triangles

can a convex polygon be divided by non-intersecting diagonals? How many such non-

intersecting diagonals are needed to divide a convex polygon into these triangles? What

are the answers to these questions for a concave polygon?

It takes some time to completely work through questions like these in order to actively explore

geometry concepts, to find patterns, to formulate conjectures in terms of natural numbers, and to

write induction proofs. However, by not skipping any steps of the process, we can gain not only a

better understanding of mathematical induction, but also “a feeling for the manner in which

mathematics is being created” (Avital & Hansen, p. 411).


References

Avital, S., & Hansen, R. T. (1976). Mathematical induction in the classroom. Educational

Studies in Mathematics, 7(4), 399411.

Wiscamb, M. (1970). A geometric introduction to mathematical induction. The Mathematics

Teacher, 63(5), 402404.

Documents

A Geometric Introduction to Mathematical Induction · PDF file · 2015-09-09A Geometric Introduction to Mathematical Induction Problems ... Nicholls State University Math 584 Fall