In this proof extracted from the Wikipedia
A classic proof by contradiction from mathematics is the proof that the square root of 2 is irrational. If it were rational, it could be expressed as a fraction $a/b$ in lowest terms, where a and b are integers, at least one of which is odd. But if $a/b = \sqrt 2$, then $a^2=$ $2b^2$. Therefore $a^2$ must be even. Because the square of an odd number is odd, that in turn implies that $a$ is even. This means that $b$ must be odd because $a/b$ is in lowest terms. On the other hand, if $a$ is even, then $a^2$ is a multiple of $4$. If $a^2$ is a multiple of $4$ and $a^2=2b^2$, then $2b^2$ is a multiple of $4$, and therefore $b^2$ is even, and so is $b$. So $b$ is odd and even, a contradiction. Therefore the initial assumption—that $\sqrt 2$ can be expressed as a fraction—must be false.
Knowing that a proof by contradiction you assume P and Not(Q) what's P and what's not Q in this proof?