Final answer:
To prove that m is even given that n = m^2 is even, we use the properties of even numbers. An even number can be expressed as 2k, so n = 2k. By assuming m is odd and showing a contradiction, we conclude m must be even.
Step-by-step explanation:
If n = m2 and n is an even number, then to prove that m is also an even number, we can use the properties of even numbers. An even number is any integer that is divisible by 2, which can be represented as 2k, where k is an integer. Since n is given to be even, we have n = 2k. Now, let's assume that m is not even; then m would have to be odd. An odd integer can be expressed in the form 2j + 1, where j is any integer. Squaring m gives us m2 = (2j + 1)2 = 4j2 + 4j + 1, which is an odd number (sum of even and odd number). However, this contradicts our original statement that n is even. Therefore, our assumption that m is odd must be false, implying m must be even.