Finding multiplicative factors of $p\in[\sqrt{2},2)\cap\mathbb{Q}$

Question

Given $p\in[\sqrt{2},2)\cap\mathbb{Q}$, how to find $q,r\in(0,\sqrt{2})\cap\mathbb{Q}$ such that $p=qr$ ?

Context: When constructing $\mathbb{R}$ with Dedekind cuts, this question arises when trying to prove that $\sqrt{2}\cdot\sqrt{2}=2$.

Answer 1

Pick any rational $q\in(\sqrt{p},\sqrt{2})$.

Then $q^2>p$. Let $r=p/q<q$. So $rq=p$, we've chosen $q$ so that $q\in(0,\sqrt{2})$ and we've chosen $r>0$ with $r<q<\sqrt{2}$.

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You can find a specific $q$ as follows.

If $p=\frac{p_1}{p_2}$ with $p_i$ integers, then $2-p\geq\frac{1}{p_2}$.

Find a positive solution to the Pell-like equation $a^2-2b^2=-1$ with $b^2>p_2$.

Then $$2-\frac{a^2}{b^2}=\frac{1}{b^2}<\frac{1}{p_2}=2-p.$$

So $p<\frac{a^2}{b^2}<2$.

Set $q=\frac{a}{b}$.