I've read this on Rudin's Principles of Mathematical Analysis:
1.1 Example We now show that the equation
$$p^2=2$$
is not satisfied by any rational $p$. If there were such a $p$, we could write $p=m/n$ where $m$ and $n$ are integers that are not both even. Let us assume this is done, then it implies
$$m^2=2n^2$$
My doubt is: Why not both even? I was thinking that it has some relation with $m^2=2n^2$ but this premise is given before the enunciation of $m^2=2n^2$. I guess there is some property for rational numbers that has some relation with this, but I'm unaware of. The question may be trivial but I can't figure it out. I guess I understand the rest of the proof quite well, but I'm stuck at this statement.
Best Answer
If they are both even, we can divide the equation by $4$, getting $m'^2=2n'^2$ with $m=2m'$ and $n=2n'$. We can keep doing this until at least one is not even. As each has a finite number of factors of $2$, we can only do it a finite number of times.