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

a) Assume the negation of conclusion:

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

Hence

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

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

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

b)

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

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

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

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

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