14 - LECTURE 14: INDUCTION GENERALIZED (6.2) There are two...

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LECTURE 14: INDUCTION GENERALIZED ( § 6.2) There are two kinds of people, those who finish what they start and so on. Robert Byrne 1. Generalization Induction works if we start our first step anywhere in Z . Result. For every nonnegative integer n we have 2 n > n . Proof. We work by induction on n 0. Base case: If n = 0 then the inequality is 1 > 0 which is true. Inductive step: Assume 2 k > k for some k 0. We want 2 k +1 > k + 1. We compute LHS = 2 k +1 = 2 · 2 k > 2 · k = k + k. If k 1 then k + k k + 1 = RHS and we are done. But we could have k = 0, in which case 2 k +1 = 2 > 1 = k + 1. So the inductive step works in either case. ± Result. For every integer n 5 , we have 2 n > n 2 . Let’s sketch how the inductive step might go. We assume 2 k > k 2 (for some k 5) and we want 2 k +1 > ( k + 1) 2 = k 2 + 2 k + 1. We compute LHS = 2 k +1 = 2 · 2 k > 2 k 2 . So we want 2 k 2 > k 2 + 2 k + 1. Is this true? We might be able to use the fact that k 5 to get this. In fact
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This note was uploaded on 03/08/2011 for the course MATH 334 taught by Professor Smith during the Spring '11 term at Vanderbilt.

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14 - LECTURE 14: INDUCTION GENERALIZED (6.2) There are two...

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