<s>[INST] $M=\{a+b\sqrt{2}: a,b \in \mathbb{Q} \}$ and $N=\{c+d\sqrt{3}: c,d \in \mathbb{Q}\}$. $M \cap N \subseteq \mathbb{Q}$. Let $M=\{a+b\sqrt{2}: a,b \in \mathbb{Q} \}$ and $N=\{c+d\sqrt{3}: c,d \in \mathbb{Q}\}$. 
Prove $M \cap N \subseteq \mathbb{Q}$.
 [/INST] What we have to prove is this:
If $a+b\sqrt 2 = c+d\sqrt 3$ with $a,b,c,d \in \mathbb Q$, then $b = d = 0$.
$$\begin{align}
a+b\sqrt 2 & = c+d\sqrt 3 \\
(a-c)^2 & = (d\sqrt 3 - b\sqrt 2)^2 \\
& = 3d^2-2bd\sqrt 6 + 2b^2
\end{align}$$
So
$$bd \sqrt 6 = \frac12(3d^2+2b^2-(a-c)^2)$$
is rational. We know $\sqrt 6$ is irrational, so we must have $bd = 0$. Suppose $b = 0$. Then $a = c + d\sqrt 3$, so $d\sqrt 3$ is rational, hence $d = 0$. Similarly, $d = 0 \implies b = 0$.
Hence $b = d = 0$.
</s>