# 2.4 Mathematical Induction

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1 2.4 Mathematical Induction

2 What Is (Weak) Induction? The Principle of Mathematical Induction works like this:

3 What Is (Weak) Induction? The Principle of Mathematical Induction works like this: We want to show some statement P(n) is true for all n n 0.

4 What Is (Weak) Induction? The Principle of Mathematical Induction works like this: We want to show some statement P(n) is true for all n n 0. We show some base case P(n 0 ) is true.

5 What Is (Weak) Induction? The Principle of Mathematical Induction works like this: We want to show some statement P(n) is true for all n n 0. We show some base case P(n 0 ) is true. We assume P(k) is true for some k n 0.

6 What Is (Weak) Induction? The Principle of Mathematical Induction works like this: We want to show some statement P(n) is true for all n n 0. We show some base case P(n 0 ) is true. We assume P(k) is true for some k n 0. We show P(k + 1) is true.

7 What Is (Weak) Induction? The Principle of Mathematical Induction works like this: We want to show some statement P(n) is true for all n n 0. We show some base case P(n 0 ) is true. We assume P(k) is true for some k n 0. We show P(k + 1) is true. So, basically, we are trying to show P(k) P(k + 1) is a tautology for any choice k n 0.

8 The Idea Behind Induction How to think of induction...

9 The Idea Behind Induction How to think of induction... If a baby can take the first step...

10 The Idea Behind Induction How to think of induction... If a baby can take the first step... If you can step on the first rung of a ladder...

11 When To Use (Weak) Induction We use mathematical induction when... we have a series and we are trying to prove the general formula.

12 When To Use (Weak) Induction We use mathematical induction when... we have a series and we are trying to prove the general formula. we are trying to prove a counting problem.

13 When To Use (Weak) Induction We use mathematical induction when... we have a series and we are trying to prove the general formula. we are trying to prove a counting problem. there seems to be a pattern that seems to change in a fixed manner.

14 First (Famous) Example Example Prove that if n is a positive integer, then n = n i = n(n + 1) 2

15 First (Famous) Example Example Prove that if n is a positive integer, then n = n i = n(n + 1) 2 Let P(n) be the proposition that the sum of the first n positive integers is n(n+1) 2. We will show this is true by mathematical induction.

16 First (Famous) Example Example Prove that if n is a positive integer, then n = n i = n(n + 1) 2 Let P(n) be the proposition that the sum of the first n positive integers is n(n+1) 2. We will show this is true by mathematical induction. Base Case:

17 First (Famous) Example Example Prove that if n is a positive integer, then n = n i = n(n + 1) 2 Let P(n) be the proposition that the sum of the first n positive integers is n(n+1) 2. We will show this is true by mathematical induction. Base Case: P(1)

18 First (Famous) Example Example Prove that if n is a positive integer, then n = n i = n(n + 1) 2 Let P(n) be the proposition that the sum of the first n positive integers is n(n+1) 2. We will show this is true by mathematical induction. Base Case: P(1) P(1) is true because 1 = 1(1+1) 2.

19 First (Famous) Example Example Prove that if n is a positive integer, then n = n i = n(n + 1) 2 Let P(n) be the proposition that the sum of the first n positive integers is n(n+1) 2. We will show this is true by mathematical induction. Base Case: P(1) P(1) is true because 1 = 1(1+1) 2.

20 First (Famous) Example Inductive Step: We assume that P(k) holds for any positive integer k. That is, we assume k = k i = k(k + 1) 2

21 First (Famous) Example Inductive Step: We assume that P(k) holds for any positive integer k. That is, we assume k = We want to show P(k + 1) is true. k i = k(k + 1) 2

22 First (Famous) Example Inductive Step: We assume that P(k) holds for any positive integer k. That is, we assume k = We want to show P(k + 1) is true. k i = k(k + 1) 2 Aside: P(k + 1) is the same thing as the formula we have but we replace k with k + 1. So our goal is to arrive at k+1 = (k + 1)(k + 2) 2

23 Consider k+1 i = k i + (k + 1)

24 Consider k+1 i = = k i + (k + 1) k(k + 1) 2 + k + 1

25 Consider k+1 i = = k i + (k + 1) k(k + 1) = + k k(k + 1) 2(k + 1) + 2 2

26 Consider k+1 i = = k i + (k + 1) k(k + 1) = + k k(k + 1) 2(k + 1) = k2 + 3k + 2 2

27 Consider k+1 i = = k i + (k + 1) k(k + 1) = + k k(k + 1) 2(k + 1) = k2 + 3k (k + 1)(k + 2) = 2

28 Consider k+1 i = = k i + (k + 1) k(k + 1) = + k k(k + 1) 2(k + 1) = k2 + 3k (k + 1)(k + 2) = 2 This last equation shows that P(k + 1) is true under the assumption that P(k) is true. So, by mathematical induction, we have shown that that n n(n + 1) i = n Z + 2

29 Union/Intersection Example Example Let A 1, A 2,..., A n be any n sets. Show that ( n ) A i = n A i

30 Union/Intersection Example Example Let A 1, A 2,..., A n be any n sets. Show that Note: Does this look familiar? ( n ) A i = n A i

31 Union/Intersection Example Example Let A 1, A 2,..., A n be any n sets. Show that ( n ) A i = n Note: Does this look familiar? This is just an extension of DeMorgan s law. A i

32 Union/Intersection Example Example Let A 1, A 2,..., A n be any n sets. Show that ( n ) A i = n Note: Does this look familiar? This is just an extension of DeMorgan s law. A i Let P(n) be the predicate that the equality holds for any n sets. We will prove by mathematical induction that for all n 1, P(n) is true.

33 Union/Intersection Example Base Case:

34 Union/Intersection Example Base Case: P(1):

35 Union/Intersection Example Base Case: P(1): Certainly A 1 = A 1 is true.

36 Union/Intersection Example Base Case: P(1): Certainly A 1 = A 1 is true. Inductive Step: We assume P(k) ( k ) A i = k is true. We want to show P(k + 1) is true. A i

37 Union/Intersection Example Consider ( k+1 ) A i = A 1 A 2... A k A k+1

38 Union/Intersection Example Consider ( k+1 A i ) = A 1 A 2... A k A k+1 = (A 1 A 2... A k ) A k+1

39 Union/Intersection Example Consider ( k+1 A i ) = A 1 A 2... A k A k+1 = (A 1 A 2... A k ) A k+1 = (A 1 A 2... A k ) A k+1

40 Union/Intersection Example Consider ( k+1 A i ) = A 1 A 2... A k A k+1 = (A 1 A 2... A k ) A k+1 = (A 1 A 2... A k A k+1 ( k ) = A i A k+1

41 Union/Intersection Example Consider ( k+1 A i ) = A 1 A 2... A k A k+1 = (A 1 A 2... A k ) A k+1 = (A 1 A 2... A k A k+1 ( k ) = A i A k+1 = ( k+1 ) A i

42 Union/Intersection Example Consider ( k+1 A i ) = A 1 A 2... A k A k+1 = (A 1 A 2... A k ) A k+1 = (A 1 A 2... A k A k+1 ( k ) = A i A k+1 = ( k+1 A i ) So, we have shown that P(k + 1) is true when we assume P(k) is true, and so by mathematical induction we have proven our statement.

43 Inequalities Example Use mathematical induction to prove the inequality n < 2 n for all positive integers n.

44 Inequalities Example Use mathematical induction to prove the inequality n < 2 n for all positive integers n. Let P(n) be the proposition that n < 2 n.

45 Inequalities Example Use mathematical induction to prove the inequality n < 2 n for all positive integers n. Let P(n) be the proposition that n < 2 n. Base Case:

46 Inequalities Example Use mathematical induction to prove the inequality n < 2 n for all positive integers n. Let P(n) be the proposition that n < 2 n. Base Case: P(1)

47 Inequalities Example Use mathematical induction to prove the inequality for all positive integers n. n < 2 n Let P(n) be the proposition that n < 2 n. Base Case: P(1) Since 1 < 2 1 = 2, P(1) is true.

48 Inequalities Example Use mathematical induction to prove the inequality for all positive integers n. n < 2 n Let P(n) be the proposition that n < 2 n. Base Case: P(1) Since 1 < 2 1 = 2, P(1) is true. Inductive Step: Assume P(k) is true. That is, assume k < 2 k for k Z +. We want to show that P(k + 1) is true.

49 Inequalities Consider k + 1 < 2 k + 1

50 Inequalities Consider k + 1 < 2 k k + 2 k

51 Inequalities Consider k + 1 < 2 k k + 2 k = 2 2 k

52 Inequalities Consider k + 1 < 2 k k + 2 k = 2 2 k = 2 k+1

53 Inequalities Consider k + 1 < 2 k k + 2 k = 2 2 k = 2 k+1 This shows that P(k + 1) is true, based on the assumption that P(k) is true. Therefore, by mathematical induction, we have shown that for all n Z +, n < 2 n.

54 Harmonic Series Example Definition The harmonic series, H j, is the series j 1 i =

55 Harmonic Series Example Definition The harmonic series, H j, is the series j 1 i = Example Prove for any nonnegative integer n, H 2 n 1 + n 2

56 Harmonic Series Example Let P(n) be the proposition that H 2 n 1 + n 2.

57 Harmonic Series Example Let P(n) be the proposition that H 2 n 1 + n 2. Base Case:

58 Harmonic Series Example Let P(n) be the proposition that H 2 n 1 + n 2. Base Case: P(0)

59 Harmonic Series Example Let P(n) be the proposition that H 2 n 1 + n 2. Base Case: P(0) P(0) is true because H 2 0 = H 1 =

60 Harmonic Series Example Let P(n) be the proposition that H 2 n 1 + n 2. Base Case: P(0) P(0) is true because H 2 0 = H 1 = Inductive Step: Assume P(k) is true. That is, assume H 2 k 1 + k 2, whenever k is a nonnegative integer.

61 Harmonic Series Example Let P(n) be the proposition that H 2 n 1 + n 2. Base Case: P(0) P(0) is true because H 2 0 = H 1 = Inductive Step: Assume P(k) is true. That is, assume H 2 k 1 + k 2, whenever k is a nonnegative integer. We want to show that if P(k) is true, then P(k + 1), which states that H 2 k k+1 2, is also true.

62 Harmonic Series Example Consider H 2 k+1 = k k k+1

63 Harmonic Series Example Consider H 2 k+1 = k k k+1 = H 2 k k k+1

64 Harmonic Series Example Consider H 2 k+1 = k k k+1 = H 2 k k k+1 ( 1 + k 2 ) k k+1

65 Harmonic Series Example Consider H 2 k+1 = k k k+1 = H 2 k k k+1 ( 1 + k 2 ) k ( 1 + k ) + 2 k 2 2 k k+1

66 Harmonic Series Example ( 1 + k )

67 Harmonic Series Example ( 1 + k ) = k 2

68 Harmonic Series Example ( 1 + k ) = k 2 = 2 + k + 1 2

69 Harmonic Series Example ( 1 + k ) = k 2 = 2 + k = 1 + k + 1 2

70 Harmonic Series Example ( 1 + k ) This establishes the proof. = k 2 = 2 + k = 1 + k + 1 2

71 Divisibility Example Example Prove that n 3 n is divisible by 3 whenever n Z +.

72 Divisibility Example Example Prove that n 3 n is divisible by 3 whenever n Z +. Let P(n) denote the proposition n 3 n is divisible by 3.

73 Divisibility Example Example Prove that n 3 n is divisible by 3 whenever n Z +. Let P(n) denote the proposition n 3 n is divisible by 3. Base Case:

74 Divisibility Example Example Prove that n 3 n is divisible by 3 whenever n Z +. Let P(n) denote the proposition n 3 n is divisible by 3. Base Case: P(1) P(1) is true because = 0 and 0 is divisible by 3.

75 Divisibility Example Example Prove that n 3 n is divisible by 3 whenever n Z +. Let P(n) denote the proposition n 3 n is divisible by 3. Base Case: P(1) P(1) is true because = 0 and 0 is divisible by 3. Inductive Step: Assume P(k) is true, which is to say that if k Z +. k 3 k is divisible by 3. We want to show that P(k + 1) is true, where P(k + 1) is (k + 1) 3 (k + 1).

76 Divisibility Example Now, (k + 1) 3 (k + 1) =

77 Divisibility Example Now, (k + 1) 3 (k + 1) = k 3 + 3k 2 + 3k + 1 k 1

78 Divisibility Example Now, (k + 1) 3 (k + 1) = k 3 + 3k 2 + 3k + 1 k 1 (k 3 k) + 3(k 2 + k)

79 Divisibility Example Now, (k + 1) 3 (k + 1) = k 3 + 3k 2 + 3k + 1 k 1 (k 3 k) + 3(k 2 + k) k 3 k is divisible by 3 by our inductive hypothesis. And, since k 2 + k is an integer for any integer k, we have that 3(k 2 + k) is also divisible by 3. It follows that (k 3 k) + 3(k 2 + k) is divisible by 3, completing the proof.

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