Adv Alegbra HW Solutions 11

# Adv Alegbra HW Solutions 11 - 11 Base Step. If n = 1, then...

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11 Base Step .If n = 1, then 4 n + 1 + 5 2 n 1 = 16 + 5 = 21 , which is obviously divisible by 21. Since our inductive step will involve two predecessors, we are obliged to check the case n = 2. But 4 3 + 5 3 = 64 + 125 = 189 = 21 × 9. Inductive Step . 4 n + 2 + 5 2 n + 1 = 4 · 4 n + 1 + 5 2 · 5 2 n 1 = 4 · 4 n + 1 + ( 4 · 5 2 n 1 4 · 5 2 n 1 ) + 5 2 · 5 2 n 1 = 4 ( 4 n + 1 + 5 2 n 1 ) + 5 2 n 1 ( 5 2 4 ). Now the last term is divisible by 21; the f rst term, by the inductive hypoth- esis, and the second because 5 2 4 = 21. 1.21 For any integer n 2, prove that there are n consecutive composite num- bers. Conclude that the gap between consecutive primes can be arbitrarily large. Solution. The proof has nothing to do with induction. If 2 a n + 1, then a is a divisor of ( n + 1 ) ! ; say, ( n + 1 ) != da for some integer d .I t follows that ( n + 1 ) !+ a = ( d + 1 ) a , and so ( n + 1 ) !+ a is composite for all a between 2 and n + 1. 1.22 Prove that the f rst and second forms of mathematical induction are equiv- alent; that is, prove that Theorem 1.4 is true if and only if Theorem 1.12 is true. Solution. Absent. 1.23 (
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## This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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