Ryan B. answered 02/09/22
7+ years tutoring Logic and Discrete Mathematics
It's close to a valid proof, but there a few details to fill in.
Recall the definition: an integer x is even if there exists an integer 'a' such that x = 2a.
Assume n is even, m is odd, and k > 0. The fact that m is odd is actually irrelevant in this proof.
Here we apply the definition of even to deduce: there exists an integer 'a' such that n = 2a.
Therefore, m × nk = m * (2a)k = m × 2k × ak = 2(m × 2k-1 × ak) [this shows that m × nk = 2(some integer)]
Here we have shown the definition of even: since m × 2k-1 × ak is an integer, m × nk is even.
Did you catch where we used the fact that k > 0? It was when we claimed that m × 2k-1 × ak is an integer; this is not the case if k ≤ 0. For example, if k = 0, then m × 2k-1 × ak = m × 2-1 × a0 = m × (1/2) × 1 = m/2, which is not an integer since m is odd.