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Unformatted text preview: n 0. 1.3 Show, for all n 1, that 10 n leaves remainder 1 after dividing by 9. Solution. This may be rephrased to say that there is an integer q n with 10 n = 9 q n + 1. If we dene q 1 = 1, then 10 = q 1 + 1, and so the base step is true. For the inductive step, there is an integer q n with 10 n + 1 = 10 10 n = 10 ( 9 q n + 1 ) = 90 q n + 10 = 9 ( 10 q n + 1 ) + 1 . Dene q n + 1 = 10 q n + 1, which is an integer. 1.4 Prove that if 0 a b , then a n b n for all n 0. Solution. Base step . a = 1 = b , and so a b . Inductive step . The inductive hypothesis is a n b n ....
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 Fall '11
 KeithCornell

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