Prove that for all $a,b \in\mathbb Z$, if $a+b$ is even then $a-b$ is even.
I started off by assuming that a and $b$ are both odd integers, $2k+1$ and $2m+1$ such that their sum is $2(k+m+1)$ which is an even integer. I then concluded that $a-b$ is $2(k-m)$ which is also even, hence the statement is true.
My question is, is this enough to prove that the statement is true or do I also have to proof it when both $a$ and $b$ are even, and when one of $a$ and $b$ is even and the other is odd. Thanks.
Best Answer
I think this is definitely a good start, but it is always a good idea to cover all possible cases, which means mentioning even and even, but also even and odd. I hope this helps!
Another way to proof it, is without using cases: If $a+b$ is even, than there exists a whole number $n$ such that $a+b=2n$. If you substract $2b$ from both sides, you'll see that the result is $a-b=2n-2b=2(n-b)$. Which is even and that is exactly what you were trying to prove.