Prove that mn^k is even if k > 0

Question

Let n, m, k be integers where n is even and m is odd. Prove that mn^k is even if k > 0.

I know intuitively that n^k can just be expressed as an even number, because n is even. But is that a valid proof?

Answer 1

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 × n^k = m * (2a)^k = m × 2^k × a^k = 2(m × 2^k-1 × a^k) [this shows that m × n^k = 2(some integer)]

Here we have shown the definition of even: since m × 2^k-1 × a^k is an integer, m × n^k is even.

Did you catch where we used the fact that k > 0? It was when we claimed that m × 2^k-1 × a^k is an integer; this is not the case if k ≤ 0. For example, if k = 0, then m × 2^k-1 × a^k = m × 2^-1 × a⁰ = m × (1/2) × 1 = m/2, which is not an integer since m is odd.