# Adv Alegbra HW Solutions 9 - n = 3 k m = 3 t m where both k...

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9 we record the inductive hypothesis: ( 1 + x ) n 1 + nx . Multiplying both sides of this inequality by the positive number 1 + x preserves the inequality: ( 1 + x ) n + 1 = ( 1 + x )( 1 + x ) n ( 1 + x )( 1 + nx ) = 1 + ( n + 1 ) x + nx 2 1 + ( n + 1 ) x , because nx 2 0. 1.17 Prove that every positive integer a has a unique factorization a = 3 k m , where k 0 and m is not a multiple of 3. Solution. Model your solution on the proof of Proposition 1.14. Replace even by multiple of 3 and odd by not a multiple of 3. We prove this by the second form of induction on a 1. The base step n = 1 holds, for 1 = 3 0 · 1 is a factorization of the desired kind. For the inductive step, let a 1. If a is not a multiple of 3, then a = 3 0 a is a good factorization. If a = 3 b , then b < a , and so the inductive hypothesis gives k 0 and an integer c not divisible by 3 such that b = 3 k c . It follows that a = 3 b = 3 k + 1 c , which is a factorization of the desired kind. We have proved the existence of a factorization. To prove uniqueness, suppose that
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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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