Chartrand Mathematical Proofs 3e Exercise 5.22 I am self-studying this book, and have got stuck on this question:
5.22 Let $S=\left\{ p+q\sqrt{2}:p,q\in\mathbb{Q}\right\}$ ,
  $T=\left\{ r+s\sqrt{3}:r,s\in\mathbb{Q}\right\}$ 
 . Prove that $S\cap T=\mathbb{Q}$.
My attempt so far:
Assume, to the contrary, that $S\cap T\neq\mathbb{Q}$.
Then $\exists x\in S\cap T$
  , such that $x\notin\mathbb{Q}$
 , or $\exists y\in Q$
 , such that $y\notin S\cap T$
Case 1.
Then $x\in S\bigwedge x\in T$
So $x=p+q\sqrt{2}\bigwedge x=r+s\sqrt{3}$
  , $p,q,r,s\in\mathbb{Q}$
$p+q\sqrt{2}=r+s\sqrt{3}$
Not sure where to go from here. I feel like the fact that there are no integers p and q such that $p\sqrt{2}=q\sqrt{3}$ is relevant somehow, but (if true), I can't figure out how to use that.

First of all, $S\supseteq\mathbb{Q}$ and $T\supseteq \mathbb{Q}$, so $S\cap T\supseteq\mathbb{Q}$. Quoting your post:

Then $\exists x\in S\cap T$
    , such that $x\notin\mathbb{Q}$
   , or $\exists y\in \mathbb{Q}$
   , such that $y\notin S\cap T$

The second case will not happen! 
As for your attempt for the first case, you are on the right track. So you have $p+q\sqrt{2} = r+s\sqrt{3}$. Write this as $p-r = s\sqrt{3}-q\sqrt{2}$. Now my hint to you is to square both sides and use the fact that $\sqrt{6}$ is irrational!
