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.

$\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}+\sqrt{10}}\!=\!\dfrac{\sqrt{\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)^2}}{2\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)}\!=\!\dfrac{\sqrt{\dfrac12}+\sqrt{\dfrac52}}{2\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)}\!=\!\dfrac12.$
