Prove that for any integers m and n, if m is odd and n is odd, then mn is odd.?

m is odd ... so m = 2a + 1
n is odd ... so n = 2b + 1 ... where a and b are integers

mn = (2a + 1)(2b + 1) <<< expand
mn = 4ab + 2a + 2b + 1
mn = 2(2ab + a + b) + 1

2(2ab + a + b) is even (because 2 is a factor) thus
2(2ab + a + b) + 1 is odd

When m is an odd integer and n is also an odd integer,
let m = 2M + 1 and n = 2N + 1,
where M and N are integers.

mn = (2M + 1)(2 N + 1)
= 2 MN + 2M + 2N + 1
= 2 × (MN + M + N) + 1

For all integers M and N, 2 × (MN + M + N) must be an even integers.
Then, 2 × (MN + M + N) + 1 must be an odd integer.

Since mn = 2 × (MN + M + N) + 1, then mn must be an odd integer.

m=2h+1, n=2k+1, some integers h,k, so
mn = (2h+1)(2k+1) = 4hk + 2h + 2k + 1 = 2(2hk + h + k) + 1 = even + 1 = odd

By the fundamental theorem of arithmetic, every integer greater than 1 is either a prime number, or can be written as the product of prime numbers. The smallest prime number, and the *only* even prime number, is 2. Any number that has 2 as one of its prime factors is even; any number that *does not* have 2 as one of its prime factors is odd.

Ergo, if m and n are odd, then neither m nor n has 2 as one of its prime factors. Multiplying the two numbers together will produce another number, mn, which *also* does not have 2 as one of its prime factors, and is therefore odd.

I hope that helps. Good luck!