If n^3 is odd, where n is positive integer, then n is odd. Help by contradiction

Answer 1

Suppose that n³ is odd and assume, by contradiction, that n is even.

An even number is one of the form 2k, where k is an integer.

If n is even, then n = 2k, for some integer k, and so n³ = (2k)³= 8k³.

If we set k' = 4k³, then n³ = 8k³ = 2 × 4k³ = 2k', and so n³ is also even!

But n³ is odd, which is a contradiction since no number can be both even and odd.

Since n cannot be even because it implies a contradiction, it follows that n must be odd.

QED