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Get Full Access to A Transition To Advanced Mathematics - 7 Edition - Chapter 1.5 - Problem 12
Get Full Access to A Transition To Advanced Mathematics - 7 Edition - Chapter 1.5 - Problem 12

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# Assign a grade of A (correct), C (partially correct), or F (failure) to each ISBN: 9780495562023 335

## Solution for problem 12 Chapter 1.5

A Transition to Advanced Mathematics | 7th Edition

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Problem 12

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justifyassignments of grades other than A.(a) Suppose m is an integer.Claim. If is odd, then m is odd.Proof. Assume that is not odd. Then is even and forsome integer k. Thus is a perfect square; that is, is an integer.If is odd, then for some integer n, which meansm2 = Thus is m2 2k = (2n + 1)2 = 4n2 + 4n + 1 = 2(2n2 + 2n) + 1.2k = 2n + 1 2k 2k 2km m2 = 2k m2 2m212 z < 1.0 < x < y <5odd, contrary to our assumption. Therefore must be even.Thus if is not odd, then m is not odd. Hence if is odd, then mis odd.(b) Suppose t is a real number.Claim. If t is irrational, then is irrational.Proof. Suppose is rational. Then where p and q areintegers and Therefore, where p and are integersand so t is rational. Therefore, if t is irrational, then isirrational.(c) Suppose x and y are integers.Claim. If x and y are even then is even.Proof. Suppose x and y are even but is odd. Then, for someinteger Therefore, The left sideof the equation is even because it is the sum of even numbers. However,the right side, 1, is odd. Since an even cannot equal an odd, we have acontradiction. Therefore, is even.(d) Suppose a, b, and c are integers.Claim. If a divides both b and c, then a dividesProof. Assume that a does not divide Then there is no integerk such that However, a divides b, so for someinteger m; and a divides c, so for some integer n. ThusTherefore is an integer satisfyingThus the assumption that a does not divide isfalse, and a does divide1

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##### ISBN: 9780495562023

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