Question on a constructive proof of irrationality of $\sqrt 2$

Question

Here is the constructive proof of $\sqrt 2 \not \in \mathbb Q$ found on this page :

Given positive integers $a$ and $b$, because the valuation (i.e., highest power of 2 dividing a number) of $2b^2$ is odd, while the valuation of $a^2$ is even, they must be distinct integers; thus $|2 b^2 - a^2| \geq 1$. Then

$$\left|\sqrt2 - \frac{a}{b}\right| = \frac{|2b^2-a^2|}{b^2(\sqrt{2}+a/b)} \ge \frac{1}{b^2(\sqrt2 + a / b)} \ge \frac{1}{3b^2},$$

the latter inequality being true because we assume $\frac{a}{b} \leq 3- \sqrt{2}$ (otherwise the quantitative apartness can be trivially established).

I don't understand why the first equality holds: why is it possible to divide by $\sqrt 2 + a/b$, since it is not yet known whether this number is zero... ?

Thank you in advance for your comments !

Answer 1

$\sqrt{2}$ and $a/b$ are both (strictly) positive, so $\sqrt{2} + a/b$ cannot be zero.

Answer 2

This equality cannot be zero because it is said that,

Given positive integers $a$ and $b$.

Thus a/b cannot be zero.