Look at the Contrapositive: If n is even, then 3n + 2 is even...
Suppose that the conclusion is false, i.e., that n is even.
Then n = 2k for some integer k.
Then we have:
3n + 2 = 3(2k) + 2
3n + 2 = 6k + 2
3n + 2 = 2(3k + 1).
Thus 3n + 2 is even, because it equals 2j for an integer j = 3k + 1.
So 3n + 2 is not odd.
We have shown that ¬(n is odd) → ¬(3n + 2 is odd),
therefore, the contrapositive (3n + 2 is odd) → (n is odd) is also true.
Suppose that the conclusion is false, i.e., that n is even.
Then n = 2k for some integer k.
Then we have:
3n + 2 = 3(2k) + 2
3n + 2 = 6k + 2
3n + 2 = 2(3k + 1).
Thus 3n + 2 is even, because it equals 2j for an integer j = 3k + 1.
So 3n + 2 is not odd.
We have shown that ¬(n is odd) → ¬(3n + 2 is odd),
therefore, the contrapositive (3n + 2 is odd) → (n is odd) is also true.