Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows: 
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
   \frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{10}} = \frac{a}{b} \\
   \frac{3+\sqrt{5}}{12 + 4\sqrt{5}} = \frac{a^2}{b^2} \\ 
   \frac{3+\sqrt{5}}{4(3+\sqrt{5})} = \frac{a^2}{b^2} \\
   \frac{1}{4} = \frac{a^2}{b^2}
\end{align*}
And because we get $\frac{1}{4}$ which is rational, we can conclude that the proof is right and there aren't any contradictions. Hence $\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{5}}$ is a rational number.
I guess my proof lacks something and I don't feel it's complete yet. I would appreciate any recommendations on how to improve my answer. Thanks in advance.

Consider
$$
x=\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}+\sqrt{10}}
$$
Then you know that $x>0$. Then
$$
x^2(\sqrt{2}+\sqrt{10})^2=3+\sqrt{5}
$$
that becomes
$$
x^2(2+10+2\sqrt{20})=3+\sqrt{5}
$$
and therefore, owing to $2\sqrt{20}=4\sqrt{5}$,
$$
4x^2=1
$$
so $x=1/2$.
