If m and n are off then is mn even?

No, mn is not even if m and n are odd.

If m and n are odd, then mn is odd as well.

Proof:

If m is odd, then it is in the form m = 2p+1, where p is some integer.

So if p = 0, then m = 1. If p = 1, then m = 3, and so on.

Similarly, if n is odd then n = 2q+1 for some integer q.

Multiply out m and n using the distribution rule

m*n = (2p+1)*(2q+1)

m*n = 2p(2q+1) + 1(2q+1)

m*n = 4pq+2p+2q+1

m*n = 2( 2pq+p+q) + 1

m*n = 2r + 1

note how I replaced the "2pq+p+q" portion with r. So I let r = 2pq+p+q, which is an integer.

The result 2r+1 is some other odd number as it fits the form 2*(integer)+1

Therefore, multiplying any two odd numbers will result in some other odd number.

Examples:

So there is no way to have m*n be even if both m and n are odd.

The general rules are as follows

The proof of the other two cases would follow a similar line of reasoning as shown above.

what graph

step-by-step explanation: