I am just trying to understand proofs by the contrapositive method. I do know that the contrastive method is taking the negation of the second argument implies the negation of the first argument.
For example: Show that the square of an even number is an even number using contrapositive proofs.
I don't know how to even begin! I know that we have to prove that if a number is odd, then the square of that number is odd.
But how can i put that in a proof form?
If the original statement is "if $P$, then $Q$," then the contrapositive statement is "If not $Q$, then not $P$." So, in your case, the original claim reads, "If $n$ is even, then $n^2$ is also even." The contrapositive would then read, "If $n^2$ is not even, then $n$ is not even," or equivalently (since if a number is not even, it is odd): "If $n^2$ is odd, then $n$ is odd." Let's prove this. If $n^2$ is odd, then $n^2 = 2m + 1$ for some (nonnegative) integer $m$. Then $2m = n^2 - 1 = (n-1)(n+1)$. But since the LHS is even, so must be the right. But $n-1$ and $n+1$ share the same parity--i.e., they are either both odd or both even. Since an odd number times an odd number is odd, it follows that $n-1$ is even, hence $n$ itself is odd.
It's an unnecessarily complicated proof, of course--for the direct proof is that if $n$ is even, then $n = 2m$ for some integer $m$, hence $n^2 = (2m)^2 = 4m^2$ which is obviously even.
There is an inaccuracy in your formulation of the problem. The contrapositive of $A\implies B$ is $\text{not}(B)\implies\text{not}(A)$ rather than the other way around. Thus, proving that "if $n$ is odd then $n^2$ is odd" is contrapositive of the statement that "if the square of a number is even then the number itself is even" rather than the statement you cited.
To show the contrapositive, assume $n$ is odd so that $n=2k+1$. Then $n^2=4k^2+2k+1$ and therefore also odd, q.e.d.
