induction

# 7 quiz inductive step suppose that bn holds then 12 22

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Unformatted text preview: Quiz Inductive step: Suppose that B(n) holds. Then 12 + 22 + ... + n2 + (n+1)2= n(n+1)(2n+1)/6 + (n+1)2 Factoring out (n+1) on the right hand side yields (n+1)(n(2n+1)+6(n+1))/6 = (n+1)(2n2 +7n+6)/6 One easily verifies that this is equal to (n+1)(n+2)(2(n+1)+1)/6 Thus, B(n+1) holds. Therefore, the proof follows by induction on n. 8 Tip How can you verify whether your algebra is correct? Use http://www.wolframalpha.com [Not allowed in any exams, though. Sorry!] 9 What’s Wrong? 10 Billiard Balls “Theorem”: All billiard balls have the same color. Proof: By induction, on the number of billiard balls. Induction basis: Our theorem is certainly true for n=1. What’s wrong? Induction step: Assume the theorem holds for n billiard balls. We prove it for n+1. Look at the first n billiard balls among the n+1. By induction hypothesis, they have the same color. Now look at the last n billiard balls. They have the same color. Hence all n+1 billiard balls have the same color. 11 Weird Properties of Positive Integers “Theorem”: For all positive integers n, we have n=n+1. “Proof”: Suppose that the claim is true for n=k. Then k+1 = (k) + 1 = (k+1) + 1 by induction hypothesis. Thus, k+1=k+2. What’s wrong? Therefore, the theorem follows by induction on n. 12 Maximally Weird! “Theorem”: For all positive integers n, if a and b are positive integers such that max{a,b}=n, then a=b. Proof: By induction on n. The result holds for n = 1, i.e., if max {a, b} = 1, then a = b = 1. Suppose it holds for n, i.e., if max {a,b} = n, then a = b. Now consider max {a, b} = n + 1. Case 1: a − 1 ≥ b − 1. Then a≥b. Hence a=max{a,b}=n+1. Thus a − 1 = n and max {a − 1, b − 1} = n. By induction hypothesis,...
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