Proof: If aﾲ is Even, then a is Even

To prove that if aﾲ is even, then a is even, we can use a proof by contradiction.

Assume that a is an integer and aﾲ is even, but a is odd.

Since a is odd, we can express it as a = 2k + 1, where k is an integer.

Now, let's square a:

aﾲ = (2k + 1)ﾲ = 4kﾲ + 4k + 1 = 2(2kﾲ + 2k) + 1

We can see that aﾲ can be expressed as 2 times an integer plus 1, which implies that aﾲ is odd.

However, we have assumed that aﾲ is even. This creates a contradiction, as a number cannot be both even and odd simultaneously.

Therefore, our assumption that a is odd must be false. Thus, a must be even.

Hence, we have proven that if aﾲ is even, then a is even.