DCP 1244 Discrete Mathematics Lecture 5: Mathematical Induction

DCP 1244 Discrete Mathematics Lecture 5: Mathematical Induction

Tsung Tai Yeh

2021

Outline

I Mathematical Induction I Strong Induction I Common Mistakes

Introduction

I What is the formula of the sum of all positive integers ? (1 + 2 + ...+ n = ?)

- 1 = 1 - 1 + 2 = 3 - 1 + 2 + 3 = 6 - 1 + 2 + 3 + 4 = 10

I Is it reasonable to guess that the sum is n(n + 1)/2 ? I We can usemathematical induction to prove our guess.

Mathematical Induction

I Firstly, we define a propositional function P(n) - p(n): ”The sum of all positive integers”.

- Our target is to show ∀nP(n) is true.

I Next, we are going to show the following two statements to be true:

- P(1), calledbasic step

- P(n) → P(n + 1), calledinductive step, where domain of n is all positive integers.

I If both can be shown true, then we can conclude that ∀nP(n) is true.

I Why is the mathematical induction valid ?

Well-Ordering Property

I Well-Ordering Property:

- Every nonempty set of nonnegative integers has at least element.

- We can use Well-Ordering Property to prove the correctness of mathematical induction.

I Four axiom about Z⁺

- The number 1 is a positive integer.

- If n ∈ Z⁺, then n + 1, the successor of n is also a positive integer.

- Every positive integer other than 1 is the successor of a positive integer.

- Well-Ordering Property

Correctness of Mathematical Induction

I Theorem: Mathematical induction → the well-ordering property.

- We prove this by contradiction.

- Suppose that the well-ordering property were false.

- Let S be a counter-example: a nonempty set of nonnegative integers that contains no smallest element.

- Let P(n) be the statement ”i 6∈ S ” for all i ≤ n.

- Show that P(n) is true for all n that will contradict the assertion that S is nonempty.

- P(0) must be true, because if 0 ∈ S then clearly S would have a smallest element, namely 0.

Correctness of Mathematical Induction

I Theorem: Mathematical induction → the well-ordering property.

- Suppose that P(n) is true, so that i 6∈ S for all i = 0, 1, 2, .., n.

- If n + 1 6∈ S , then n + 1 would be the smallest element of S , which contradicts our assumption.

- Therefore, n + 1 6∈ S .

- Thus, we have shown by the principle of mathematical induction that P(n) is true for all n, which means that there can be no elements of S .

- This contradicts our assumption that S 6= ∅.

- Our proof by contradiction is complete.

Mathematical Induction Example

I Let P(n): The sum of first n positive odd integers is n². - We hope to use mathematical induction to show ∀nP(n) is true.

I The basic stepis P(1), which is :

- P(1): The sum of first 1 positive odd integers is 1². - This is obviously true.

I The inductive stepis ∀n(P(n) → P(n + 1))

- To show it is true, we focus on an arbitrary chosen k, and see if P(k) → P(k + 1) is true.

Mathematical Induction Example

I Suppose that P(k) is true.

- P(k): The sum of first k positive odd integers is k². - This implies 1 + 3 + ... + (2k − 1) = k².

I Then, we have:

- 1 + 3 + ... + (2k − 1) + (2k + 1) = k²+ (2k + 1) = (k + 1)². I so that P(k + 1) is true if P(k) is true.

Mathematical Induction Example

I We show the conditional statement P(k + 1) → P(k) is true.

- P(k) is a premise.

- An easier way to show P(k + 1).

I Once basic step and inductive step are proven, by mathematical induction, ∀nP(n) is true.

Mathematical Induction

I Mathematical Induction is - A powerful proof techniques.

- Leverage two statements, but it can imply infinite number of cases to be correct.

- Aim to find new theorems.

- Firstly, have a theorem (by guessing)

- Second, the induction is used to confirm the theorem is correct.

Mathematical Induction Example 1

I Show that 1 + 2 + 2²+ ... + 2ⁿ= 2ⁿ⁺¹− 1, n ∈ Z⁺. - Let P(n) be 1 + 2 + 2²+ ... + 2ⁿ= 2ⁿ⁺¹− 1, n ∈ Z⁺. - Basis step: P(0) is true because 2⁰= 1 = 2¹− 1.

- Inductive step:

- Firstly, assume P(k) is true, where k ∈ Z⁺. - P(k) = 1 + 2 + 2²+ ... + 2^k = 2^k+1− 1.

- Second, if P(k) is true, then P(k + 1) is also true.

- 1 + 2 + 2²+ ... + 2^k + 2^k+1 = 2^(k+1)+1− 1 = 2^k+2− 1.

- Hence, we know that P(n) is true, n ∈ Z⁺ through the completion of basis and inductive steps.

Mathematical Induction Example 2

I Prove that n < 2ⁿ, n ∈ Z⁺. - Let P(n) be n < 2ⁿ, n ∈ Z⁺.

- Basis Step: assume P(1) is true, because 1 < 2¹ = 2.

- Inductive step:

- Firstly, P(k) is true, where k < 2^k and k ∈ Z⁺.

- Second, to show that if P(k) is true, then P(k + 1) is also true.

- k + 1 < 2^k + 1 ≤ 2^k + 2^k = 2 × 2^k = 2^k+1. - Thus, P(k + 1) is true, that is k + 1 < 2^k+1. - Hence, n < 2ⁿ is true, where n ∈ Z⁺, through the completion of basis and inductive steps.

Mathematical Induction Example 3

I- Prove that 2ⁿ< n!, where n ≥ 4 and n ∈ Z⁺. - Let P(n) be the proposition 2ⁿ< n!.

- Basis step: p(4) is true, because 2⁴ < 4!.

- Inductive step:

- Firstly, assume that P(k) is true, where k ≥ 4 and k ∈ Z⁺.

- Second, if P(k) is true, P(k + 1) is also true.

- 2^k+1 = 2 × 2^k =< 2 × k! < (k + 1)k! = (k + 1)!.

- Thus, P(k + 1) is true when P(k) is true.

- Hence, 2ⁿ< n! is true, where k ≥ 4 and k ∈ Z⁺, through the completion of basis and inductive steps.

Mathematical Induction Example 4

I Prove that n³− n is divisible by 3, where n ∈ Z⁺. - Let P(n) be ”n³− n is divisible by 3”.

- Basis step: P(1) is true because 1³− 1 = 0 is divisible by 3.

- Inductive step:

- Firstly, assume P(k) is true, where k ∈ Z⁺. - Second, if P(k) is true, P(k + 1) is also true.

- (k + 1)³− (k + 1) = (k³+ 3k²+ 3k + 1) − (k + 1) = (k³− k) + 3(k²+ k).

- Thus, P(k + 1) is true, because the first and second term is divisible by 3.

- Hence, n³− n is divisible by 3, where n ∈ Z⁺, through the completion of basis and inductive steps.

Mathematical Induction Example 5

I Prove that 7ⁿ⁺²+ 8²ⁿ⁺¹ is divisible by 57, where n ∈ Z⁺. - Let P(n) be ”7ⁿ⁺²+ 8²ⁿ⁺¹” is divisible by 57.

- Basis step: P(0) is true, because 7⁰⁺²+ 8^2·0+1= 57 is divisible by 57.

- Inductive step:

- Firstly, assume P(k) is true, where k ∈ Z⁺. - Second, if P(k) is true, P(k + 1), which is 7^(k+1)+2+ 8^2(k+1)+1, is also true.

- 7^(k+1)+2+ 8^2(k+1)+1= 7^k+3+ 8^2k+3=

7 · 7^k+2+ 8²· 8^2k+1= 7(7^k+2+ 8^2k+1) + 57 · 8^2k+1. - Both 7(7^k+2+ 8^2k+1) and 57 · 8^2k+1 are divisible by 57.

- Hence, 7ⁿ⁺²+ 8²ⁿ⁺¹ is divisible by 57.

Strong Induction

I Used when we cannot easily prove a result using mathematical induction.

I The inductive stepin the mathematical and strong induction:

- Mathematical induction: if P(k) is true, then P(k + 1) is also true.

- Strong induction: show that if P(j ) is true, then P(k + 1) is true, where k ∈ Z⁺, and j = 1, 2, ..., k.

Strong Induction

I To prove that P(n) is true, where P(n) is a propositional function and n ∈ Z⁺.

I Basis step:

- To verify that P(1) is true.

I Inductive step:

- Show that the conditional statement

(P(1) ∧ P(2) ∧ ... ∧ P(k)) → P(k + 1) is true, where k ∈ Z⁺.

Strong Induction Example 1

I Show that if n is an integer greater than 1, then n can be written as the product of primes.

- Let P(n) be the proposition that n can be written as the product of primes.

- Basis step:

- P(2) is true, because 2 can be written as the product of one prime, itself.

- Inductive step:

- Assume that P(j ) is true, where j ∈ Z⁺, and 2 ≤ j ≤ k.

- We need to show that P(k + 1) is true, that is, k + 1 is the product of primes.

- Case 1: k + 1 is prime, and P(k + 1) is true.

- Case 2: k + 1 is composite.

- Two integers a and b, where 2 ≤ a ≤ b < k + 1.

Strong Induction Example 2

I Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps.

- Let P(n) be the statement that postage of n cents can be formed using 4-cent and 5-cent stamps.

- Basis step: P(12), P(13), P(14), P(15) are true by using 4-cent or 5-cent stamps.

- Inductive step:

- P(j ) is true for 12 ≤ j ≤ k, where k ∈ Z⁺, k ≥ 15.

- To show P(k + 1) is true, we can form postage of k + 1 cents for k ≥ 15 from postage of k − 3 cents.

- P(k − 3) is true because k − 3 ≥ 12, that is, we can form postage of k − 3 cents just using 4-cent and 5-cent stamps.

Well-ordering Property

I Every nonempty set of nonnegative integers has a least element.

Well-ordering Property Example 1

I Prove that if a is a positive integer, then there are unique integers q and r with 0 ≤ r < d and a = dq + r .

- Let S be the set of nonnegative integers of the form a − dq, where q ∈ Z.

- The set S is nonempty because −dq can be made as large as desired. S has a least element r = a − dq₀.

- In the case, r < d , where r ∈ Z⁺.

- There would be a smaller non-negative element in S , a − d (q₀+ 1).

- Supposed that r ≥ d , and a = dq₀+ r , a = d (q₀+ 1) = (a − dq₀) − d = r − d ≥ 0.

- Hence, there are integers r and q, where 0 ≤ r < d .

Mistaken Proofs by Mathematical Induction

I In a group of n people, everyone has the same name.

- Let P(n) be in a group of n people, everyone has the same name.

- Basis step:

- In a group of 1 person, he/she have the same name.

- Inductive step:

- Assume P(k) denotes in a group of any k people, all of them have the same name, where k ∈ Z⁺.

- k + 1 people must have the same name, because the first k has the same name, and last k people also have the same name.

I What’s wrong in this proof ?

- There is no overlap between the first k and last k group of people, we cannot conclude that both of them have the same

Mistaken Proofs by Mathematical Induction

I 2 × n = 0, n ∈ Z⁺. - Basis step:

- 2 × n = 0 is true, when n = 0.

- Inductive step:

- It holds for k, we then consider k + 1.

- k + 1 = i + j , where i , j ∈ Z⁺.

- 2(k + 1) = 2(i + j ) = 2i + 2j = 0 + 0 = 0.

I What’s wrong in this proof ?

- It is not true for k = 0. The number 0 + 1 cannot be written as the sum of 2 strictly smaller non-negative integers, i , j .