Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 1 - Section 1.7 - Introduction to Proofs - Exercises - Page 91: 18

Answer

See the solution.

Work Step by Step

(a) To prove a conditional sentence by contraposition, we assume the negation of the consequent, and prove the negation of the antecedent. $Proof.$ Assume $n$ is an odd integer. (Note that since we assumed $n$ is an integer, in order to prove the antecedent of our statement is false, we must show $3n+2$ is odd.) Since $n$ is odd, $n=2k+1$ for some integer $k$. Then $3n+2=3(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+1$, where $3k+1$ is an integer. Hence, $3n+1$ is odd. Since we have proven the contrapositive of the statement, the statement is true. (b) To prove a conditional statement by contradiction, we will assume the antecedent is true and the consequent is false, and show that this assumption leads to a contradiction. $Proof.$ Assume the statement is false. That is, assume $n$ is an integer, $3n+1$ is even, and $n$ is odd. From the proof above, we know that since $n$ is odd, $3n+2$ must be odd as well. Thus we have $3n+2$ is even and $3n+2$ is odd, a contradiction. Thus the statement cannot be false, which means the statement must be true.
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