Prove $a^2$ is even $\implies$ $a$ is even. It was proven via contradiction by my friend.
Here is the proof in question, which uses proof by contradiction
Assume towards a contradiction that $a$ is odd. Let $a = 2k+1$, then $a^2 = 4k^2 + 4k + 1 = 2(2k^2+2k)+1$ Therefore $a^2$ is odd. Contradiction.
To me, this proof in question seems incorrect because it shows that if $a$ is odd then $a^2$ is odd. We have shown that the converse of what we want to contradict is a contradiction. I feel like this proof by contradiction is not mathematically correct or vigorous enough, and for it to be correct we would have to start from $a^2$ being even, working our way down to $a$ being odd, and then finding a contradiction.
Am I correct? Or am I just confused about something?
Best Answer
Your friend has done the contrapositive. He has not proved it by contradiction exactly. The contrapositive of a statement is: if the conclusion is false, then the premise is false. You friend assumed the conclusion did not apply, namely that $a^2$ is not even (which means it's odd), and went on to prove that $a$ is not even (which means it's odd).
Contradiction proceeds the following way: Assume that the conclusion is false but the premise is true. Then, come up with a false statement logically (a false statement could be like $0 = 1$ or $3$ divides $2$). Hence, we must assume both that $a^2$ is even and $a$ is odd, and come up with a false statement.
A proof by contradiction would say : Let $a^2$ be even and $a$ be odd. We will derive a contradiction.
Note that $a^2-a$ is an odd number, since it is an odd number subtracted from an even number. However, $a^2-a =a(a-1)$. Since $a$ and $a-1$ are consecutive numbers, one of them is an even number. Hence, since $a^2-a$ has an even number as a factor, it is also an even number. But then $a^2-a$ is both even and odd, which is not possible. Hence, the original statement "Let $a^2$ be even and $a$ be odd", must be false. Hence, if $a^2$ is even, $a$ too must be even.