Prove that $\sqrt{\frac{1+\sqrt5}2}$ is not in $\mathbb Q(\sqrt5)$

Question

Prove that $$\sqrt{\frac{1+\sqrt5}2} \not \in \mathbb Q(\sqrt5)$$

I think I should start taking the next equation $\sqrt{\frac{1+\sqrt5}2}=p+q\sqrt5$ but I don't see how to continue from here

Answer 1

Let $K = \mathbb Q(\frac{1+\sqrt{5}}{2})$. Note that $N_{K/\mathbb Q}(\frac{1+\sqrt{5}}{2}) = -1$. If $\alpha = \sqrt{\frac{1+\sqrt{5}}{2}} \in K$, then

$$-1 = N_{K/\mathbb Q}(\alpha^2) = N_{K/\mathbb Q}(\alpha)^2.$$

Since $N_{K/\mathbb{Q}}(\alpha) \in \mathbb Q$, this is a contradiction.

Answer 2

$\sqrt{\frac{1+\sqrt5}{2}}=p+q\sqrt5$
$\frac{1+\sqrt5}{2}=(p+q\sqrt5)^2=(p^2+5q^2)+2pq\sqrt5$
$p^2+5q^2=\frac12=2pq$
$p,q\neq 0$
$(p/q)^2-2(p/q)+5=0$
let $r=p/q,r\in\Bbb Q$
$r^2-2r+5=0$
$0\leq(r-1)^2=-4$
contradiction