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1Introduction to Abstract Mathematics
Chapter 4: Sequences and Mathematical Induction
Instructor: Hayk Melikya [email protected]
4.1- Sequences .
4.2 , 4.3 Mathematical Induction
4.4 Strong Mathematical induction and WOP
2Introduction to Abstract Mathematics
Sequences
A sequence (informally) is a collection of elements (objects or
numbers usually infinite number of) indexed by integers.
a1, a2, a3, …, an, …
1, 2, 3, 4, …
1/2, 2/3, 3/4, 4/5,…
1, -1, 1, -1, …
1, -1/4, 1/9, -1/16, …
Examples
General formula
Each individual element ak is called a term, where k is called an index
Sequences can be computed using an explicit formula: ak = k / (k + 1) for k > 0
Finding an explicit formula given initial terms of the sequence
6Introduction to Abstract Mathematics
Factorial defines a product:
Turn product into a sum taking logs:
ln(n!) = ln(1·2·3 ··· (n – 1)·n)
= ln 1 + ln 2 + ··· + ln(n – 1) + ln(n)n
i=1
ln(i)
Factorial
How to estimate n!?
7Introduction to Abstract Mathematics
Arithmetic Series
Given n numbers, a1, a2, …, an with common difference d, i.e. ai+1 - ai =d.
What is a simple closed form expression of the sum?
Adding the equations together gives:
Rearranging and remembering that an = a1 + (n − 1)d, we get:
8Introduction to Abstract Mathematics
Geometric Series
2 n-1 nnG 1+x +x + +x::= +x
What is the closed form expression of Gn?
2 n-1 nnG 1+x +x + +x::= +x
2 3 n n+1nxG x +x +x + +x +x=
GnxGn= 1 xn+1
n+1
n
1- xG =
1- x
9Introduction to Abstract Mathematics
Infinite Geometric Series
n+1
n
1- xG =
1- x
Consider infinite sum (series)
2 n-1 n i
i=0
1+x+x + +x + =x + x
n+1n
nn
1-lim x 1limG
1- x 1-=
x=
for |x| < 1 i
i=0
1x =
1- x
11Introduction to Abstract Mathematics
Principle of Mathematics induction
(chapter 4.2-4.4 of the book
Let P(n) be a property that is defined for integers n, and let a be a fixed integerSuppose the following two statements are true:
1. P(a) is true.2. for all integers k ≥a, if P(k) is true then P(k+1) is true.
Then the statement P(n) is true for all integers n ≥a
• Domino effect
Inductive sets of integer.A subset of positive integers S is called inductive set if k S then k +1 S.
Principle of mathematical induction states that if a S then all integers greater than or equal to a also are in S.
12Introduction to Abstract Mathematics
Method of Proof by Mathematical Induction
Consider a statement of the form
”For all integers n ≥ a, a property P(n) is true.”
To prove such a statement, perform the following two steps:
Step1 (Basic step): Show tah the property is true for n =a ( P(a) is true.)
Step2 ( Inductive step): show that for all integers k ≥ a,If property is true for n = k then it is true for n = k+1
( k ≥ a) (P(k)P(k + 1))
Let P(n) be the property “ n cent can be obtained using 3 cent and 5 cent coins”.Proposition(4.2.1): P(n) is true for all integers n ≥ 8.
13Introduction to Abstract Mathematics
The Induction Rule when a =1
1 and (from n to n +1),
proves 1, 2, 3,….
P(1), P(n)P(n +1)
(m N) P(m)
Like domino effect…
For any n>=1
Very easy to prove
Much easier to prove with P
(n) as an assumption.
14Introduction to Abstract Mathematics
Statements in green form a template for inductive proofs.
Proof: (by induction on n)
The induction hypothesis, P(n), is:
Proof by Induction
Let’s prove:
15Introduction to Abstract Mathematics
Proof by Induction
Base Case (n = 0):
11
1
r
r
0 12 0
? 11
11
rr r r
r
Wait: divide by zero bug!
This is only true for r 1
121.::
1(
11)P n r
rr r r
r
nn
12 1
11
nn r
r r rr
Theorem: 1. r
16Introduction to Abstract Mathematics
Induction Step: Assume P(n) for some n 0 and prove P(n + 1):
( ) 121.
11
1
rr r rr
r
+1+1
nn
Proof by Induction
Have P (n) by assumption:
So let r be any number 1, then from P (n) we have
12 1
11
rr r r
r
nn
How do we proceed?
17Introduction to Abstract Mathematics
Proof by Induction
111
11
nn nrr r r
r
+1 n
adding r n+1 to both sides,
1 1
( ) 1
1 ( 1)
1
1
1
n nr r r
r
r
r
+1n
But since r 1 was arbitrary, we conclude (by UG), that( ) 1
21.1
11
rr r rr
r
+1+1
nn
which is P (n+1). This completes the induction proof.
19Introduction to Abstract Mathematics
Proving a Property
Base Case (n = 1):
Induction Step: Assume P(i) for some i 1 and prove P(i + 1):
20Introduction to Abstract Mathematics
Proving an Inequality
Base Case (n = 3):
Induction Step: Assume P(i) for some i 3 and prove P(i + 1):
21Introduction to Abstract Mathematics
ParadoxProposition: All horses are the same color.
Proof: (by induction on n)
Induction hypothesis:
P(n) ::= any set of n horses have the same color
Base case (n =1): true!
…
Inductive case:Assume any n horses have the same color.Prove that any n+1 horses have the same color.
…
First set of n horses have the same color
Second set of n horses have the same color
22Introduction to Abstract Mathematics
What is wrong?
Proof that P(n) → P(n+1)
is false if n = 1, because the two
horse groups do not overlap.
First set of n=1 horses
n =1
Second set of n=1 horses
Paradox
(But proof works for all n ≠ 1)
23Introduction to Abstract Mathematics
Prove P(1).
Then prove P(n+1) assuming all of
P(1), …, P(n) (instead of just P(n)).
Conclude (n.)P(n)
Strong Induction
1 2, 2 3, …, n-1 n.
So by the time we got to n+1, already know all
of
P(1), …, P(n)
Strong induction
Ordinary induction
equivalent
24Introduction to Abstract Mathematics
Principle of Strong Mathematical Induction
Let P(n) be a property that is defined for integers n, and let a and b be a fixed integers with a ≤ bSuppose the following two statements are true:
1. P(a), P(a+1), . . . And P(b) are true.
2. for all integers k ≥b, if P(i) is true for all integers i from a through k then P(k+1) is true.Then the statement P(n) is true for all integers n ≥a
25Introduction to Abstract Mathematics
Every integer > 1 is a product of primes.
Prime Products (revisited)
Proof: (by strong induction)
Base case is easy.
Suppose the claim is true for all 2 <= i < n.
Consider an integer n.
In particular, n is not prime.
So n = k·m for integers k, m where n > k,m >1.
Since k,m smaller than n,
By the induction hypothesis, both k and m are product of primes
k = p1 p2 ps
m = q1 q2 qt
27Introduction to Abstract Mathematics
Prime Products
…So
n = k m = p1 p2 ps q1 q2 qt
is a prime product.
This completes the proof of the induction step.
Every integer > 1 is a product of primes.
28Introduction to Abstract Mathematics
Available stamps:
5¢ 3¢
Theorem: Can form any amount 8¢
Prove by strong induction on n > 0.
P(n) ::= can form (n +7)¢.
Postage by Strong Induction
What amount can you form?
29Introduction to Abstract Mathematics
Postage by Strong Induction
Base case (n = 1):
(1 +7)¢:
Inductive Step: assume (m +7)¢ for 1 m n,
then prove ((n +1) + 7)¢
cases:
n +1= 1, 9¢:
n +1= 2, 10¢:
30Introduction to Abstract Mathematics
case n +1 3: let m =n 2.
now n m 0, so by induction hypothesis have:
(n 2)+8
= (n +1)+8+
3
Postage by Strong Induction
We’re done!
In fact, use at most two 5-cent stamps!
31Introduction to Abstract Mathematics
Postage by Strong Induction
Given an unlimited supply of 5 cent and 7 cent stamps,
what postages are possible?
Theorem: For all n >= 24,
it is possible to produce n cents of postage from 5¢ and 7¢ stamps.
32Introduction to Abstract Mathematics
Every nonempty set ofnonnegative integers
has a least element.
Familiar? Now you mention it, Yes.
Obvious? Yes.
Trivial? Yes. But watch out:
Well Ordering Principle
Every nonempty set of nonnegative rationals
has a least element.
NO!
Every nonempty set of nonnegative integers
has a least element.NO!
33Introduction to Abstract Mathematics
Proof: suppose
2mn
Theorem: is irrational2
…can always find such m, n without common factors…
why always?
Well Ordering Principle
By WOP, minimum |m| s.t. 2 .mn
so
0
0
2m
n where |m0| is
minimum.
34Introduction to Abstract Mathematics
0
0
/2
/m cn c
but if m0, n0 had common factor c > 1, then
and contradicting minimality of |m0|0 0/m c m
Well Ordering Principle
The well ordering principle is usually used in “proof by contradiction”.
• Assume the statement is not true, so there is a counterexample.
• Choose the “smallest” counterexample, and find a even smaller counterexample.
• Conclude that a counterexample does not exist.
35Introduction to Abstract Mathematics
To prove ``n. P(n)’’ using WOP:
1. Define the set of counterexamples
C ::= {n | ~ P(n)}
2. Assume C is not empty.
3. By WOP, have minimum element m0 C.
4. Reach a contradiction (somehow) –
usually by finding a member of C that is < m0 .
5. Conclude no counterexamples exist. QED
Well Ordering Principle in Proofs
36Introduction to Abstract Mathematics
Induction (Proving equation)Induction (Proving equation)
For any integer n>=2,For any integer n>=2,
Proof:Proof: We prove by induction on n .We prove by induction on n .
Let P(n) be the proposition thatLet P(n) be the proposition that
n
n
n 2
111
3
11
2
11
222
n
n
n 2
111
3
11
2
11
222
37Introduction to Abstract Mathematics
Base case, n=2:Base case, n=2:
So P(2) is true.So P(2) is true.
Inductive step:Inductive step:
Suppose that P(n) is true for some n>=2. So,Suppose that P(n) is true for some n>=2. So,
22
12
4
3
2
11
2
n
n
n 2
111
3
11
2
11
222
38Introduction to Abstract Mathematics
Then, for n>=2,Then, for n>=2,
By induction, P(n) is true for all integers n>=2. By induction, P(n) is true for all integers n>=2.
2222 1
11
11
3
11
2
11
nn
21
11
2
1
nn
n
2
2
1
2
2
1
n
nn
n
n
1
2
2
1
n
n
By the inductive hypothesis
39Introduction to Abstract Mathematics
Induction (Divisibility)Induction (Divisibility)
For any integer n>=1, is divisible by 6For any integer n>=1, is divisible by 6
Proof:Proof:
We prove by induction on n .We prove by induction on n .
Base case, n=1:Base case, n=1:
is divisible by 6.is divisible by 6.
So it is true for n=1.So it is true for n=1.
52 nn
6511 2
40Introduction to Abstract Mathematics
Inductive step:Inductive step: Suppose that for some n>=1, is Suppose that for some n>=1, is divisible by 6divisible by 6
Then,Then,
Either n+1 or n+2 is even, so the last term is divisible by 6.Either n+1 or n+2 is even, so the last term is divisible by 6.
Therefore is divisible by 6.Therefore is divisible by 6.
By induction, is divisible by 6 for all integers n>=1 By induction, is divisible by 6 for all integers n>=1
52 nn
511 2 nn
621 2 nnn 6262 223 nnnnn
683 23 nnn )633(5 23 nnnn
knnk integer somefor )633(6 2 2136 nnk
By the inductive assumption
511 2 nn
52 nn
41Introduction to Abstract Mathematics
Induction (Proving inequalityInduction (Proving inequality))
For any integer n>=4,For any integer n>=4,
Proof: We prove by induction on n .Proof: We prove by induction on n .
Base case, n=4:Base case, n=4:
So the claim is true for n=4.So the claim is true for n=4.
2! nn
2416244321!4
42Introduction to Abstract Mathematics
Inductive step:Inductive step:
Assume that for some n>=4Assume that for some n>=4
Then,Then,
By induction, for all integer n>=4. By induction, for all integer n>=4.
2! nn
21!1! 1 nnnnn
nn 21
21 n
By the inductive hypothesis
By assumption, n>=2
2 ! nn
43Introduction to Abstract Mathematics
1. (4.1) Expand a sequence/sum/product from sequence/sum/product notation, and vice versa
2. (4.1) Rewrite a sum by separating off and adding on the last term
3. (4.1) Know the definition of a factorial.4. (4.2) Know the method of proof by induction5. (4.2) Prove formulas for the sums of sequences using induction6. (4.3) Use induction to prove inequalities7. (4.3) Use induction to prove results with divisibility8. (4.4) Use strong induction on recursively defined sequences.
Practice Problems: (Section 4.1) 1, 2, 10, 11, 19-22, 34, 35, 43, 46(Section 4.2) 4, 6, 10, (11-15)(Section 4.3) 8-10, 16-20(Section 4.4) 1-8, 14, 17
Chapter 4 (Sections 4.1, 4.2, 4.3, 4.4)