1 Introduction to Abstract Mathematics Chapter 4: Sequences and Mathematical Induction Instructor: Hayk Melikya [email protected] 4.1- Sequences. 4.2,

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


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


Summation


A Telescoping Sum


Product


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!?


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:


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


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


Some Examples


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.


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.


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.


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:


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


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?


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.


Summation

Try to prove:


Proving a Property

Base Case (n = 1):

Induction Step: Assume P(i) for some i 1 and prove P(i + 1):


Proving an Inequality

Base Case (n = 3):

Induction Step: Assume P(i) for some i 3 and prove P(i + 1):


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


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)


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


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


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



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.


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?



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¢:


case n +1 3: let m =n 2.

now n m 0, so by induction hypothesis have:

(n 2)+8

= (n +1)+8+

3


We’re done!

In fact, use at most two 5-cent stamps!



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.


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!


Proof: suppose

2mn

Theorem: is irrational2

…can always find such m, n without common factors…

why always?


By WOP, minimum |m| s.t. 2 .mn

so

0

0

2m

n where |m0| is

minimum.


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


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.


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


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


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


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


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 .


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


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


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 .


So the claim is true for n=4.So the claim is true for n=4.

2! nn

2416244321!4


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


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)

1 Introduction to Abstract Mathematics Chapter 4: Sequences and Mathematical Induction Instructor: Hayk Melikya [email protected] 4.1- Sequences. 4.2,

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