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 de±ne 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 . De±ne 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
 Mathematical Induction, Inductive Reasoning, Natural number, inductive hypothesis, inductive step

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