Prove, for N as a whole number, n^2 is even if n is even.

Question

For the proof, "prove for n as a whole number, n^2 is even if and only if n is even", if we were to prove it via contradiction, would the negative of the statement which we need to contradict be:

"n^2 is odd if and only if n is even" or

"n^2 is even if and only if n is odd".

Answer 1

The negation would be: there is at least one $n $ where either $n^2$ is odd while $n $ is even or where $n $ is odd and $n^2$ is even. All other statements, yours, the other answer, and the comments are way too strong. The negation requires only a single counter example.

Answer 2

($n^2$ is even and $n$ is odd) or ($n^2$ is odd and $n$ is even)

Answer 3

The proof of your original statement.... If $n^2$ is even then 2|$n^2$. As 2 is prime, 2 |n. So n is even. If n is even then $n^2$ is obviously even. Hence your statement.