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Unformatted text preview: Homework 4 Solutions Kyle Chapman October 25, 2011 69.7 Here, we first proove the identity for n = 0 and n = 1. ( α β ) / √ 5 = (1 1) / √ 5 = 0 ( α 1 β 1 ) / √ 5 = ( α β ) / √ 5 = √ 5 / √ 5 = 1 Now, for any n ≥ 2, then knowing it is true for all previous cases proves that in particular it is true of the two previous cases. For the induction, we first note that α 2 = α + 1 and β 2 = β + 1. These are both verified by direct computation. Now we perform the following computation to prove the inductive step. ( α n +1 β n +1 ) / √ 5 = ( α n 1 ( α 2 ) β n 1 ( β 2 )) / √ 5 = ( α n 1 ( α + 1) β n 1 ( β + 1)) / √ 5 = ( α n + α n 1 β n β n 1 ) / √ 5 = ( α n β n ) / √ 5 + ( α n 1 β n 1 ) / √ 5 Thus, the recursion holds and so the formula holds for every natural number. 89.2 First note that by Proposition 10.3, d = sa + tb for some s,t ∈ Z . Further note that ( nb ) a = ( na ) b . Here, choose n large enough so that s + nb >...
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This note was uploaded on 12/26/2011 for the course MATH 8 taught by Professor Bigelow during the Fall '08 term at UCSB.
 Fall '08
 Bigelow

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