Confusing $\sqrt{2}$ is irrational proof
Let $a, b$ be arbitrary positive integers.
Then $2b^{2}$ is divisible by 2 an odd number of times, while $a^{2}$ is divisible by 2 an even number of
times, so $2b^{2} \neq a^{2}$ and therefore $|2b^{2}a^{2}| \geqq 1.$
If $\frac{a}{b} \gt 3- \sqrt{2}$ then $\frac{a}{b} \gt 3-2\sqrt{2}\gt\frac{1}{6}$
If $\frac{a}{b} \leqq 3-\sqrt{2}$ then
$|\sqrt{2} - \frac{a}{b}| = \frac{|2b^{2}-a^{2}|}{b^{2}(\sqrt{2}+\frac{a}{b})} \geqq \frac{1}{b^2(\sqrt{2}+\frac{a}{b})} \geqq \frac{1}{3b^{2}}$
I get that $2b^{2} \neq a^{2}$ because $\sqrt{2} \neq \frac{a}{b}$ I also get that $|2b^{2}a^{2}|\geq1$ because were letting a, b be arbitrary positive integers.
I don't understand where the divisible by 2 an odd and even number of times comes from.
I notice that 3 is the smallest integer such that $3-\sqrt{2} $ and $3-2\sqrt{2} $return positive values.
The "divisible by $2$ an odd and even number of times" comes from the fact that if $a$ is divisible by $2$ $n$ times (i.e. $a=2^n \cdot m,m\in\mathbb{Z}),$ then $a^2$ is divisible by $2^{2n},$ and so it is divisible an even number of times. Likewise, $b^2$ is divisible by $2$ and even number of times, so $2b^2$ is divisible by $2$ an odd number of times.
Also, the inequality for $|2b^2a^2|$ could be simplified to $>$ since $a$ and $b$ are positive integers and the smallest possible value of $|2b^2a^2|$ is $2.$
You didn't say that you don't understand the last step so I assume you do (it's just using previous inequalities and the difference of squares formula).