In proof by contradiction, we want to change the statement “if P then Q” to “if NOT Q then NOT P.” These two statements are equivalent, so if we can prove the latter, then we have proved the former. Consider the following theorem.
Theorem: For any integer $latex x$, if $latex x^2$ is odd, then $latex x$ is odd.
We can assign the statements above to P (the hypothesis) and Q (the conclusion) as follows.
P: $latex x^2$ is odd
Q: $latex x$ is odd.
If we are going to change this statement to “if NOT Q then NOT P,” then, we have to find the opposite of P and opposite of Q.
For Q, what is the opposite of “x is odd”? This can mean “$latex x$ is NOT odd” or equivalently, “x is even.” This is also the same with P. The statement NOT P is the same as “$latex x^2$ is NOT odd” or equivalently “$latex x^2$ is even.” Therefore, the statement in the form of “if NOT Q, then NOT P” can be stated as follows.
Theorem: If $latex x$ is even, then $latex x^2$ is even.
If $latex x$ is even, then it is divisible by 2, which means that x is a product of 2 and some integer $latex k$. So, $latex x = 2k$.
Squaring, we have
$latex x^2 = (2k)^2 = 4k^2$.
Now, $latex 4k^2$ is a multiple of 2 since $latex k$ is an integer. Therefore, $latex 4k^2$ is divisible by 2 which means that it is even.
Since $latex 4k^2$ is even, $latex x^2$ is also even and this proves the theorem.
Notice that the restated theorem (or the contrapositive) is a lot easier to prove that its equivalent statement above.