Is $\sqrt{2+\sqrt{5}}$ included in the extension field $\mathbb{Q}(\sqrt5)$?
The motivation for this question is the understanding that the square root of a Complex Number $(\sqrt{a_0+b_0i})$ is itself a Complex Number, $(a_1+b_1i)$.
(I know. I'm comparing $\mathbb{C}$, an extension of $\mathbb{R} $, to $\mathbb{Q}(\sqrt5)$, an extension of $\mathbb{Q}$.)
I've tried the following:
$$m+n\sqrt5 = \sqrt{2+\sqrt{5}}$$
and solving for $m$ and $n$.
In my journey, I've found the minimal polynomial $x^4-4x^2-1$. I don't think there exist Rational numbers, $m$ and $n$.
Am I correct in saying so?
Taking from where you left off: $(m+n\sqrt{5})^2 = 2+\sqrt{5}\implies m^2+2mn\sqrt{5}+5n^2=2+\sqrt{5}\implies m^2+5n^2=2, 2mn=1$. Thus $m^2n^2=\dfrac{1}{4}$, and $m^2+\dfrac{5}{4m^2} = 2\implies 4m^4-8m^2+5=0\implies 4(m^2-1)^2=-1$. This equation has no solutions in $\mathbb{R}$, and thus has no solution in $\mathbb{Q}$. So back to the main question,$\sqrt{2+\sqrt{5}}\notin \mathbb{Q}(\sqrt{5})$.