Homework 4 Solutions - , . . . , k , where k 1. We want to...

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HOMEWORK 4: COMMENTS 1. Non-book problems (1) We proceed by strong induction on n . We need two base cases: a 1 = 2 1 +1, a 2 = 2 2 +1. Assume, as inductive hypothesis that a i = 2 i + 1 for integers i in the range 1 , . . . , k , where k 2. We want to show that a k +1 = 2 k +1 + 1. Note that a k +1 = 3 a k - 2 a k - 1 because k 2 , = 3(2 k + 1) - 2(2 k - 1 + 1) by inductive hypothesis = 3 · 2 k + 3 - 2 k - 2 = 2 · 2 k + 1 = 2 k +1 + 1 . Hence a n = 2 n + 1 for all n N . Book Problems. Problem 21 Solution. We proceed by strong induction on n . The base case is true by by hypothesis , i.e., x + 1 /x is an integer because we are assuming so. Assume, as inductive hypothesis that x i + 1 /x i is an integer for integers i in the range 1
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Unformatted text preview: , . . . , k , where k 1. We want to show that x k +1 + 1 /x k +1 is an integer. We have ( x k +1 + 1 /x k +1 ) = ( x k + 1 /x k )( x + 1 /x )-( x k-1 + 1 /x k-1 ) Now, the right hand side is an integer since ( x k + 1 /x k ) , ( x k-1 + 1 /x k-1 ), and ( x + 1 /x ) are all integers, the rst two by inductive hypothesis, the later by the original problems hypothesis. Hence the left hand side is an integer. By induction, x n + 1 /x n is an integer for all n N . Date : February 21th, 2008. 1...
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This note was uploaded on 06/15/2009 for the course MATH 74 taught by Professor Courtney during the Spring '07 term at University of California, Berkeley.

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