Proof by Contradiction Involving Irrational Numbers and Sets

Question

Please help me with the following proof. I'm meant to use proof by contradiction to solve it.

Let $S=\{p+q\sqrt{2}:p,q\in\Bbb Q\}$ and $T=\{r+s\sqrt{3}:r,s\in\Bbb Q\}$ Prove that $S\cap T=\Bbb Q$.

Assume, to the contrary, that $S\cap T \not = \Bbb Q$. Then, $S\cap T \not\subseteq\Bbb Q$ or $\Bbb Q\not\subseteq S\cap T$.

Let $S\cap T \not\subseteq\Bbb Q$. Then, there exists some element $x$ such that $x\in S\cap T \land x\not\in\Bbb Q$.

$\implies x\in S\land x\in T\land x\not\in\Bbb Q \implies p+q\sqrt{2}=x=s+r\sqrt{3}, p,q,s,t\in\Bbb Q \land x\not\in\Bbb Q$

I'm not sure where to go from here. I know that if I let $q,r=0$, then $p+q\sqrt{2}=x=r+s\sqrt{3}\in\Bbb Q$, which contradicts the assumption that $x\not\in\Bbb Q$. However, I'm not sure if this is sufficient.

For the converse, let $\Bbb Q\not\subseteq S\cap T$. Then, there exists some element $x$ such that $x\in\Bbb Q \land x\not\in S\cap T$.

$\implies x\in\Bbb Q \land (x\not\in S \lor x\not\in T) \implies (x\in\Bbb Q \land x\not\in S) \lor (x\in\Bbb Q \land x\not\in T)$

Here, again, I'm not sure how to continue. I think I need to show that a contradiction arises in both the case that $(x\in\Bbb Q \land x\not\in S)$ and the case that $(x\in\Bbb Q \land x\not\in T)$, but I'm not certain. Similar to what I wrote above, in both of these cases, I could let $q,r=0$, which would imply $x\in\Bbb Q$, and produce a contradiction, but I don't know if this is correct, or enough.

Answer 1

Nomenclature note: it’s not a “converse”; you have an “or”, and are proving what happens to each of the two disjuncts. “Converses” are implications going the other way, which is not what you have here.

For your first one: note that hyou have $p+q\sqrt{2}=r+s\sqrt{3}$. Note also that at least one of $q$ and $s$ must be nonzero (otherwise, you have a rational number). Say $q\neq 0$. Then you can rewrite this as $$\sqrt{2} = \frac{1}{q}\left( r-p + s\sqrt{3}\right).$$ Try squaring both sides and see what happens, remembering that neither $\sqrt{2}$ nor $\sqrt{3}$ are rational.

As for the other... note that if $q\in\mathbb{Q}$, then $q=q+0\sqrt{2}\in S$ and $q=q+0\sqrt{3}\in T$, so you certainly have $\mathbb{Q}\subseteq S\cap T$. That inclusion holds.

Answer 2

Assume, to the contrary, that $S\cap T \not = \Bbb Q$. Then, there exists $p+q\sqrt{2}=r+s\sqrt{3}$

$p-r=s\sqrt{3}-q\sqrt{2}$

$(p-r)^2=3s^2-2sq\sqrt{6}+2q^2$

$\sqrt{6}= ({3s^2+2q^2-(p-r)^2})/2sq$

${RHS\in\Bbb Q\,LHS \not = \Bbb Q }$

Contradiction, if otherwise $s=q=0$

Answer 3

If $x\in S\cap T$ then $x\in \Bbb Q.$

Let $x=p+q\sqrt 2=r+s\sqrt 3$ with $p,q,r,s\in \Bbb Q.$ Then $$(2x-p-r)(-r+p)=(x-p)^2-(x-r)^2=2q^2-3s^2\in \Bbb Q.$$ (i). If $p\ne r$ then $x=\frac {1}{2}\left(p+r+\frac {2q^2-3s^2}{(-r+p)}\right)\in \Bbb Q.$

(ii). If $p=r$ then $q\sqrt 2=x-p=x-r=s\sqrt 3,$ so $q\sqrt 2=s\sqrt 3.$ Now $q=0$ implies $x=p\in \Bbb Q.$ But $q\ne 0$ implies $\sqrt {2/3}=s/q\in \Bbb Q,$ which is impossible.