Proof that $\sqrt{2}$ is irrational

If $m$ and $n$ are both even you can still divide both by 2. Absurd we already divided them by all integers possible greater than 1. Thus, at least 1 of them is odd.

If both m and n are divisible by the same integer greater than 1, divide both by the largest such integer.

If the they are both even then you didn't divide by the bigger integer. Because you can still divide both by $2$.

I'd have worded it differently. I'd have said "If they have any factors in common divide both by that factor and keep repeating until they have no factors in common".

It means the same thing but it just makes it more clear that in the end, they will have no factors in common. And if the have no factors in common, they can't have $2$ in common. And the can't both be even.

If the square root of $2$ is rational, then by definition it can be written as a ratio of two whole numbers. But then, as long as both numbers are even, you can divide both sides by $2$, and get two new numbers that have the same ratio.

Now, these two new numbers may still be even of course, but that means that you just divide them both by two again. And you cannot indefinitely keep dividing both sides, because the numbers are finite, and they keep getting smaller and smaller, so there must be some point where it is no longer true that both sides are even.

Therefore, if the square of of two can be expressed as a ratio at all, then it is also possible to express it as a ratio of two whole numbers where not both sides are even.