Prove if number is rational or irrational

Question

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.

Answer 1

$$\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}} =\frac{\sqrt{6+2\sqrt5}}{\sqrt{2}(\sqrt{2} + \sqrt {10})} = \frac{\sqrt{5}+1}{2+2\sqrt{5}}=\frac{1}{2}$$

Answer 2

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 $x=1/2$.

Answer 3

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

$$\frac{c^2 e^2}{d^2 f^2} = \left[3 + \sqrt{5}\right] \times \left[12 - 2 \sqrt{20}\right].$$

$$\left[3 + \sqrt{5}\right] \times \left[12 - 2 \sqrt{20}\right] = \left[3 + \sqrt{5}\right] \times \left[12 - 4 \sqrt{5}\right] = 16.$$

Answer 4

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