i need somebody review my proof of this exercise and correct me, pls.
Prove exist $x\in\mathbb{Q}, r_{1},r_{2}\in\mathbb{I}$ such that $r_{1}<x<r_{2}$
We know $-\sqrt{2}<1<\sqrt{2}$, Let $p,q\in\mathbb{Z}$ then:
Case 1: $\text{p,q>0}$
-$\sqrt{2}p<p<\sqrt{2}p$, $\Rightarrow -\sqrt{2}\frac{p}{q}<\frac{p}{q}<\sqrt{2}\frac{p}{q}$
Case 2: $p>0$,$q<0$
$-\sqrt{2}p<p<\sqrt{2}p, \Rightarrow -\sqrt{2}\frac{p}{q}>\frac{p}{q}>\sqrt{2}\frac{p}{q}$
Then exist $x=\frac{p}{q}$ such that $r_{1}=-\sqrt{2}\frac{p}{q}>x=\frac{p}{q}>\sqrt{2}\frac{p}{q}=r_{2}$
Your question is badly phrased. I think you wanted to say, for all $r_1,r_2\in\mathbb{R}\setminus\mathbb{Q}$ with $r_1<r_2$, there exists $x\in\mathbb{Q}$ such that $r_1<x<r_2$.
Pick $q\in\mathbb{Z}_{>0}$ such that $q\left(r_2-r_1\right)\geq 1$ (which exists by the Archimedean Property). By the Well Ordering Principle, let $p$ be the smallest integer with $p\geq qr_1$. Then, show that $qr_1<p<qr_2$.