Unformatted text preview: n = 3 k m = 3 t m , where both k and t are nonnegative and both m and m are not multiples of 3; it must be shown that k = t and m = m . We may assume that k ≥ t . If k > t , then canceling 3 t from both sides gives 3 k − t m = m . Since k − t > 0, the left side is a multiple of 3 while the right side is not; this contradiction shows that k = t . We may thus cancel 3 k from both sides, leaving m = m . 1.18 Prove that F n < 2 n for all n ≥ 0, where F , F 1 , F 2 , . . . is the Fibonacci sequence. Solution. The proof is by the second form of induction. Base step : F = < 1 = 2 and F 1 = 1 < 2 = 2 1 . (There are two base steps because we will have to use two predecessors for the inductive step.) Inductive step :...
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 Fall '11
 KeithCornell
 Logic, Mathematical Induction, Negative and nonnegative numbers, Natural number

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