Answer to Question #215983 in Discrete Mathematics for GATO

Question #215983

Prove that if n is an integer and 3n + 2 is even, then n is even using

a) a proof by contraposition.

b) a proof by contradiction


1
Expert's answer
2021-07-13T08:53:05-0400

a) Assume the negation of conclusion:

Let nn is an integer. Assume that nn is odd. By definition of odd numbers, exists integer kk such that n=2k+1.n=2k+1.

Hence


3n+2=3(2k+1)+2=6k+5=2(3k+2)+13n+2=3(2k+1)+2=6k+5=2(3k+2)+1

So by definition of odd numbers, 3n+23n+2 is odd number. Thus we prove the negation of hypothesis.

Therefore we prove that if nn is an integer and 3n+23n+2 is even, then nn is even using a proof by contraposition.


b)

Let nn is an integer. Assume that 3n+23n+2 is even and nn is odd. Then by definition of odd numbers, there exists an integer kk such that n=2k+1.n=2k+1. Hence


3n+2=3(2k+1)+2=6k+5=2(3k+2)+13n+2=3(2k+1)+2=6k+5=2(3k+2)+1

So by definition of odd numbers, 3n+23n+2 is odd number, that contradicts our assumption that 3n+23n+2 is even.

Hence, it is not the case that 3n+23n+2 is even and nn is odd.

Therefore we prove that if nn is an integer and 3n+23n+2 is even, then nn is even using a proof by contradiction.



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