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induction

# induction - CSCE 222 Discrete Structures for Computing...

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CSCE 222 Discrete Structures for Computing Proof by Induction Dr. Hyunyoung Lee Based on slides by Andreas Klappenecker 1

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Motivation Induction is an axiom which allows us to prove that certain properties are true for all positive integers (or for all nonnegative integers, or all integers >= some fixed number) 2
Induction Principle Let A(n) be an assertion concerning the integer n. If we want to show that A(n) holds for all positive integer n, we can proceed as follows: Induction basis: Show that the assertion A(1) holds. Induction step: For all positive integers n, show that A(n) implies A(n+1). 3

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Standard Example For all positive integers n, we have A(n): 1+2+...+n = n(n+1)/2 Induction basis: Since 1 = 1(1+1)/2, the assertion A(1) is true. Induction step: Suppose that A(n) holds. Then 1+2+...+n+(n+1) = n(n+1)/2 + n+1 = (n 2 + n+2n+2)/2 = (n+1)(n+2)/2, hence A(n+1) holds. Therefore, the claim follows by induction on n. 4
The Main Points We established in the induction basis that the assertion A(1) is true. We showed in the induction step that A(n+1) holds, assuming that A(n) holds. In other words, we showed in the induction step that A(n)->A(n+1) holds for all n >= 1. 5

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Example 2 Theorem : For all positive integers n, we have 1+3+5+...+(2n-1) = n 2 Proof. We prove this by induction on n. Let A(n) be the assertion of the theorem. Induction basis: Since 1 = 1 2 , it follows that A(1) holds. Induction step: Suppose that A(n) holds. Then 1+3+5+...+(2n-1) + (2n+1) = n 2 + 2n+1 = (n+1) 2 holds. In other words, A(n) implies A(n+1). 6
Quiz Theorem : We have 1 2 + 2 2 + ... + n 2 = n(n+1)(2n+1)/6 for all n >= 1.

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