How to show that $\sqrt{3 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{3 + \sqrt{2}})$ Is there a simple way to show that $\sqrt{3 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{3 + \sqrt{2}})$ (assuming this is true), using Galois theory? I have tried to show this by computational means but the calculations are horrible. 

Let $\,K\,$ denote $\,\mathbb Q(\sqrt{3+\sqrt2})\,$ for convenience, then observe that
$$x=\sqrt{3+\sqrt2}$$
$$\Rightarrow\quad x^2-3=\sqrt2\qquad\ \ \ $$
$$\Rightarrow\quad x^4-6x^2+7=0\quad$$
$$\Rightarrow\quad[\,K:\mathbb Q\,]=4\qquad\ \ \ $$
First, $\,(\sqrt{3+\sqrt2})^2-3=\sqrt2\in K$
Now if $\,\sqrt{3-\sqrt2}\in K,$ then $\,\sqrt{3+\sqrt2}\cdot\sqrt{3-\sqrt2}=\sqrt7\in K$
Thus $\,K=\mathbb Q(\sqrt2,\sqrt7)$,$\,$ and this means that
$$\qquad\qquad\exists\ a,b,c,d\in\mathbb Q\ :\ a+b\sqrt2+c\sqrt7+d\sqrt{14}=\sqrt{3+\sqrt2}\qquad(\#)$$
We must have $\,c=d=0\quad(\%)$
Hence $\,\sqrt{3+\sqrt2}\in\mathbb Q(\sqrt2)\,$, which is impossible because $$[\mathbb Q(\sqrt2):\mathbb Q]=2<\mathbb [\mathbb Q(\sqrt{3+\sqrt2}):\mathbb Q]$$
As a result, $\,\sqrt{3-\sqrt2}\notin K$

$\left.\right.$
$\textbf{Proof}$ of $(\%):$
Square both sides of $(\#)$, we get
$$(a^2+2b^2+7c^2+14d^2-3)+2(ab+7cd-1)\sqrt2+2(ac+2bd)\sqrt7+2(ad+bc)\sqrt{14}=0$$
Now we must have
\begin{align*}
W&=a^2+2b^2+7c^2+14d^2-3=0\\
X&=ab+7cd-1=0\\
Y&=ac+2bd=0\\
Z&=ad+bc=0
\end{align*}
First, if $\,c=0,\,d\neq0$, we have $\,a=b=0\,$ by $Y$ and $Z$. Then $X$ will become $\,-1=0$, which is obviously a contradiction.
Next, if $\,c\neq0,\,d=0$,  the same argument as above will lead to a contradiction.
Hence suppose $\,c,d\neq0$, then we have the following two cases:
(1) $b\neq0:$
$$c=-\frac{ad}b\ \ (\text{by }Z)\ \ \Rightarrow\ \ \frac db(a^2-2b^2)=0\ \ (\text{by }Y)\ \ \Rightarrow\ \ a=\pm\sqrt2b\notin\mathbb Q\ \ (\text{contradiction})$$
(2) $b=0:$
$$a=0\ \ (\text{by }Z)\ \ \Rightarrow\ \ c=\frac1{7d}\ \ (\text{by }X)\ \ \Rightarrow\ \ 98d^4-21d^2=-1\ \ (\text{by }W)\ \ \Rightarrow\ \ d\notin\mathbb Q\ \ (\text{contradiction})$$
As a result, we must have $\,c=d=0$
$\left.\right.$
(This answer has been improved by the suggestions from Kenny Wong)
