prove that a necessary and sufficient condition for mn to be odd?

Question

is that both m and n are odd

Answer 1

Necessity "-->" and sufficiency "<---".

The statement we want to prove:
mn is odd if and only if m and n are odd.

Necessity "-->"

Assume mn is odd. {We want to prove m and n are odd}.
We can prove this by contradiction. Assume that it is NOT the case that m and n are odd. That means we have the following three cases:

Case I: m is even, n is even. Then
m = 2i, n = 2j for some integers i and j.

mn = (2i)(2j)
mn = 4ij
mn = 2(ij), and clearly mn is even. This is a contradiction for this case.

Case 2: m is odd, n is even. Then
m = 2i + 1, n = 2j, and
mn = (2i + 1)(2j)
mn = 4ij + 2j
mn = 2(2ij + j), showing that mn is even. We've obtained another partial contradiction.

Case 3: m is even, n is odd. (Since multiplication is commutative, the proof is similar to case 2 that mn is also even).

Therefore, if both m and n are NOT odd, mn is even; this is a contradiction. It therefore follows that m and n are odd.

Sufficiency. "<---"

Assume m and n are both odd. Then

m = 2i + 1
n = 2j + 1

mn = (2i + 1)(2j + 1)
mn = 4ij + 2i + 2j + 1
mn = 2(2ij + i + j) + 1, showing mn is an odd number.

Therefore, since necessity and sufficiency has been proven, it follows that

mn is odd if and only if m and n are odd.

Answer 2

What if one of them is even?
For instance m=2k for an integer k:

mn = 2k*n=(k * n) + (k * n)
That's even!

If OTOH both are odd m=2k+1 n=2p+1

mn = (2k + 1)(2p + 1) = 4kp + 2p + 2k + 1 = 2(2kp +p + k) + 1

A number greater by 1 than an even number is odd.

Answer 3

The only way to get an odd number for a multiplication answer is to have both factors be odd.

even times even yields even
odd times even yields even
even times odd yields even

ODD times ODD yields ODD