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 × 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} × a^{0 } = m × (1/2) × 1 = m/2, which is *not* an integer since m is odd.