1. 1 Introduction to Abstract Mathematics Chapter 4: Sequences and Mathematical Induction Instructor: Hayk Melikya melikyan@nccu.edu 4.1- Sequences. 4.2, 4.3 Mathematical Induction 4.4 Strong Mathematical induction and WOP

2. 2 Introduction to Abstract Mathematics Sequences A sequence (informally) is a collection of elements (objects or numbers usually infinite number of) indexed by integers. a 1, a 2, a 3, …, a n, … 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 a k is called a term, where k is called an index Sequences can be computed using an explicit formula: a k = k / (k + 1) for k > 0 Finding an explicit formula given initial terms of the sequence

3. 3 Introduction to Abstract Mathematics Summation

4. 4 Introduction to Abstract Mathematics A Telescoping Sum

5. 5 Introduction to Abstract Mathematics Product

6. 6 Introduction 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) Factorial How to estimate n!?

7. 7 Introduction to Abstract Mathematics Arithmetic Series Given n numbers, a 1, a 2, …, a n with common difference d, i.e. a i+1 - a i =d. What is a simple closed form expression of the sum? Adding the equations together gives: Rearranging and remembering that a n = a 1 + (n − 1)d, we get:

8. 8 Introduction to Abstract Mathematics Geometric Series What is the closed form expression of G n ? G n  xG n =1  x n+1

9. 9 Introduction to Abstract Mathematics Infinite Geometric Series Consider infinite sum (series) for |x| < 1

10. 10 Introduction to Abstract Mathematics Some Examples

11. 11 Introduction 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 integer Suppose 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.

12. 12 Introduction 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.

13. 13 Introduction 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.

14. 14 Introduction 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:

15. 15 Introduction to Abstract Mathematics Proof by Induction Base Case (n = 0): Wait: divide by zero bug! This is only true for r  1 Theorem:

16. 16 Introduction to Abstract Mathematics Induction Step: Assume P(n) for some n  0 and prove P(n + 1): Proof by Induction Have P (n) by assumption: So let r be any number  1, then from P (n) we have How do we proceed?

17. 17 Introduction to Abstract Mathematics Proof by Induction adding r n+1 to both sides, But since r  1 was arbitrary, we conclude (by UG), that which is P (n+1). This completes the induction proof.

18. 18 Introduction to Abstract Mathematics Summation Try to prove:

19. 19 Introduction to Abstract Mathematics Proving a Property Base Case (n = 1): Induction Step: Assume P(i) for some i  1 and prove P(i + 1):

20. 20 Introduction to Abstract Mathematics Proving an Inequality Base Case (n = 3): Induction Step: Assume P(i) for some i  3 and prove P(i + 1):

21. 21 Introduction to Abstract Mathematics Paradox Proposition: 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

22. 22 Introduction 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)

23. 23 Introduction 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

24. 24 Introduction 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 ≤ b Suppose 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

25. 25 Introduction 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 = p 1  p 2    p s m = q 1  q 2    q t

26. 26 Introduction to Abstract Mathematics

27. 27 Introduction to Abstract Mathematics Prime Products …So n = k  m = p 1  p 2    p s  q 1  q 2    q t is a prime product.  This completes the proof of the induction step. Every integer > 1 is a product of primes.

28. 28 Introduction to Abstract Mathematics Available stamps: 5¢5¢3¢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?

29. 29 Introduction 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¢:

30. 30 Introduction 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!

31. 31 Introduction 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.

32. 32 Introduction 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!

33. 33 Introduction to Abstract Mathematics Proof: suppose Theorem: is irrational …can always find such m, n without common factors… why always? Well Ordering Principle By WOP,  minimum |m| s.t. so where |m 0 | is minimum.

34. 34 Introduction to Abstract Mathematics but if m 0, n 0 had common factor c > 1, then and contradicting minimality of |m 0 | 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.

35. 35 Introduction 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 m 0  C. 4.Reach a contradiction (somehow) – usually by finding a member of C that is < m 0. 5.Conclude no counterexamples exist. QED Well Ordering Principle in Proofs

36. 36 Introduction to Abstract Mathematics Induction (Proving equation) v For any integer n>=2, v Proof: v We prove by induction on n. v Let P(n) be the proposition that

37. 37 Introduction to Abstract Mathematics v Base case, n=2: So P(2) is true. v Inductive step: Suppose that P(n) is true for some n>=2. So,

38. 38 Introduction to Abstract Mathematics v Then, for n>=2, v By induction, P(n) is true for all integers n>=2. By the inductive hypothesis

39. 39 Introduction to Abstract Mathematics Induction (Divisibility) v For any integer n>=1, is divisible by 6 v Proof: v We prove by induction on n. v Base case, n=1: is divisible by 6. is divisible by 6. So it is true for n=1.

40. 40 Introduction to Abstract Mathematics v Inductive step: Suppose that for some n>=1, is divisible by 6 v Then, v Either n+1 or n+2 is even, so the last term is divisible by 6. Therefore is divisible by 6. v By induction, is divisible by 6 for all integers n>=1 By the inductive assumption

41. 41 Introduction to Abstract Mathematics Induction (Proving inequality ) v For any integer n>=4, v Proof: We prove by induction on n. v Base case, n=4: So the claim is true for n=4.

42. 42 Introduction to Abstract Mathematics v Inductive step: Assume that for some n>=4 Assume that for some n>=4 v Then, v By induction, for all integer n>=4. By the inductive hypothesis By assumption, n>=2

43. 43 Introduction 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 induction 5. (4.2) Prove formulas for the sums of sequences using induction 6. (4.3) Use induction to prove inequalities 7. (4.3) Use induction to prove results with divisibility 8. (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)