Final answer:
To prove that 'mn is even if and only if m is even or n is even,' one needs to realize that multiplying an even number by any integer results in an even product. If mn is even, then according to the fundamental theorem of arithmetic, at least one of the factors must include the prime factor 2, meaning that either m or n must be even.
Step-by-step explanation:
The question asks us to show that mn is even if and only if m is even or n is even. To prove this, we need to consider two separate cases. First, assume that m is even; this means that m can be expressed as 2k, where k is some integer. When we multiply 2k by any integer n, the product is 2(kn), which is clearly an even number. Secondly, if n is even, it can be written as 2l for some integer l, and multiplying any integer m by 2l will give us 2(ml), again an even number. Therefore, if either m or n is even, mn will certainly be even.
If mn is even, it means that mn can be expressed as 2p for some integer p. Since 2 is a prime number, it must be a factor of either m or n for their product to be even, according to the fundamental theorem of arithmetic. Thus, either m or n must be even if mn is even.
For example, let m=4 (an even number) and n=3 (an odd number). The product mn=4*3=12, which is even, illustrating that when one of the numbers is even, the product is even.