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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