Let $S = \{p + q\sqrt{2} \mid p, q ∈ \Bbb Q\}$ and $T = \{r + s\sqrt{3} \mid r, s ∈ \Bbb Q\}.$ Prove: $S ∩ T = \Bbb Q$.

Question

Given the following question:

Let $S = \{p + q\sqrt{2} \mid p, q ∈ \Bbb Q\}$ and $T = \{r + s\sqrt{3} \mid r, s ∈ \Bbb Q\}$. Prove: $S ∩ T = Q$.

How am I suppose to show that every $p + q\sqrt{2}$ pair and $r + s\sqrt{3}$ pair appears in the set of Rational Numbers? What proof technique would be optimal here, I was thinking a direct proof would work the best.

If $S ∩ T = \Bbb Q$ then both $S$ and $T$ must contain every rational number, correct?

Answer 1

You must show that if $ p + q\sqrt{2}=r + s\sqrt{3} $ then $q=s=0$. Try starting from $$ q\sqrt{2}=r -p+ s\sqrt{3} $$