$M=\{a+b\sqrt{2}: a,b \in \mathbb{Q} \}$ 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 $M \cap N \subseteq \mathbb{Q}$.
 A: 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 $b = d = 0$.
A: Let $x \in M \cap N$.
Then $x \in M$ and $x \in N$.
$x \in M$ means that $x = a + b\sqrt{2}$ for some $a, b \in \mathbb{Q}$.
Also, $x \in N$ means that $x = c + d\sqrt{3}$ for some $c, d \in \mathbb{Q}$.
Therefore, for $x \in M \cap N$, the following must be true:
$$a + b\sqrt{2} = c + d\sqrt{3}$$
for some $a, b, c, d \in \mathbb{Q}$.
Equality is satisfied in $$a + b\sqrt{2} = c + d\sqrt{3}$$
if and only if
$$b = d = 0$$
and
$$a = c.$$
A: Given $M=\mathbb{Q(\sqrt 2)}$ and $N=\mathbb{Q(\sqrt 3)}$, where $\mathbb{Q(\sqrt 2)}$ and $\mathbb{Q(\sqrt 3)}$ are the smallest field extensions of $\mathbb{Q}$ containing both ($\mathbb{Q}$ and $\sqrt 2$ )and ($\mathbb{Q}$ and $\sqrt 3$) respectively.  
Let us consider $M\cap N$. If $x\in M\cap N,$ then $x\in M$ and $x\in N$, So  
$a+b\sqrt 2=r+s\sqrt 3,$ $ a,b,r,s\in \mathbb{Q}$  
$\implies a=r$ and $b=s=0$ because b,s $\in \mathbb{Q}$. The only other way possible was $b=\sqrt 3$ and $s=\sqrt 2$, which would violate the fact that b,s $\in \mathbb{Q}$. Thus $x=a=r$ and hence $x\in \mathbb{Q}$, proving that $M\cap N \subseteq\mathbb{Q}$. $\square$
