<s>[INST] 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.
 [/INST] Alternative approach:
In a problem like this, my first step is to clear the denominator.
$\displaystyle \frac{\sqrt{3 + \sqrt{5}}}{\sqrt{10} + \sqrt{2}} \times \frac{\sqrt{10} - \sqrt{2}}{\sqrt{10} - \sqrt{2}} = \frac{1}{8} \times \left[\sqrt{3 + \sqrt{5}}\right] \times \left[\sqrt{10} - \sqrt{2}~\right].$
Therefore, the problem reduces to an examination of
$$ \left[\sqrt{3 + \sqrt{5}}\right] \times \left[\sqrt{10} - \sqrt{2}~\right].\tag1$$
Since both of the factors in (1) above are positive, the problem reduces to determining whether there exists positive rational numbers $a,b$ such that $a^2 \times b^2$ equals the square of the expression in (1) above.
This resolves to determining whether there exists positive integers $c,d,e,f$ such that
$$\frac{c^2 e^2}{d^2 f^2} = \left[3 + \sqrt{5}\right] \times \left[12 - 2 \sqrt{20}\right].$$
$$\left[3 + \sqrt{5}\right] \times \left[12 - 2 \sqrt{20}\right] = \left[3 + \sqrt{5}\right] \times \left[12 - 4 \sqrt{5}\right] = 16.$$
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