Question

Prove that if n or m is an odd integer, then n*m is an even integer.

Proposed proof: Suppose that n or m are even. Then n = 2k and m = 2j for some integers k and j. This shows that n*m = (2k)*(2j) = 4k*j. Therefore, n*m is even.

Answer 1

n=2k+1

m=2l

where k,l are integers.

Then:

n∙m=(2k+1)∙2l=4kl+2l=2∙(2kl+l)

Therefore, n∙m is even.