Proof that $\mathbb{Q}(\sqrt{2})\cup\mathbb{Q}(\sqrt{3})$ is not a subfield of $\mathbb{R}$ Í already have found a proof About the statment which I did not understand, I will write it down, so maybe someone can explain me the part that I did not understand. But if there are easier ways to prove the statement I would be delighted to know. 
$\alpha:=\sqrt{2}+\sqrt{3}\notin U:= \mathbb{Q}(\sqrt{2})\cup\mathbb{Q}(\sqrt{3})\iff\alpha \notin \mathbb{Q}(\sqrt{2})$ and $\alpha \notin \mathbb{Q}(\sqrt{3})$
Proof: $\alpha \notin \mathbb{Q}(\sqrt{2})$
Assume $\exists_{r,s\in \mathbb{Q}}r+s\sqrt{2}=\sqrt{2}+\sqrt{3}$
$\iff \sqrt{3}=r+(s-1)\sqrt2$
$\iff 3=r^2+2r(s-1)\sqrt2+2(s-1)^2$
Which implicates $3=0$ or $\sqrt3\in \mathbb{Q}$ or $\sqrt{3/2}\in \mathbb{Q}$ or $\sqrt{2}\in \mathbb{Q}$
I don't understand the last implication, I Need some help here.
Many Thanks 
 A: Well, $\alpha\not\in {\Bbb Q}(\sqrt 2)$ is equivalent to $\sqrt 3\not\in {\Bbb Q}(\sqrt 2)$.
On the contrary, $\sqrt 3 = a+b\sqrt 2$ for some rational numbers $a,b$. Squaring gives $3 = a^2+2ab\sqrt 2 + 2b^2$. Then $\sqrt 2$ could be written as a rational number, which is impossible as it is irrational.
A: Me neither - but if $r(s-1) \neq 0$ you can rewrite the last equation to obtain $$\mathbb{Q} \ni \frac{3-r^2-2(s-1)^2}{2r(s-1)} = \sqrt{2} \notin \mathbb{Q},$$ a contradiction. If $s=1$ we have $3=r^2$, if $r=0$ the equation reads $3=2(s-1)^2$ - both is impossible because of the uniqueness of prime factorization.
A: Instead of that just try to check is this $U:=\mathbb{Q}(\sqrt 2)\cup \mathbb{Q}(\sqrt 3)$ a Group?
Let us take $\sqrt2,\sqrt3\in U$ [Both exists] but $\sqrt2+\sqrt3\notin U$.
So definitely it cannot be a Subfield of $\mathbb{R}$.
A: We know that any element of $\mathbb{Q}[\sqrt{2}]$ can be uniquely written as $a + b \sqrt{2}$ with $a, b \in \mathbb{Q}$ (the unicity is important). So, if $\alpha \in \mathbb{Q}[\sqrt{2}]$, then so does $\alpha - \sqrt{2} = \sqrt{3} \in \mathbb{Q}[\sqrt{2}]$. So you can write $\sqrt{3} = r + s \sqrt{2}$. Squaring that equality, we get
$$3 = (r^2 + 2 s^2) + 2rs \sqrt{2} $$
By unicity, the number $3$ can be uniquely written as $3 + 0 \sqrt{2}$, so in the right hand side we get $r^2 + 2 s^2 = 3$ and $2rs=0$. From $rs = 0$ we get that $r=0$ or $s = 0$. If $s=0$, then $r^2 = 3$, which is impossible since $r$ is rational. And if $r = 0$, then $2s^2=3$ so $s^2 = \frac{3}{2}$, which is also impossible since $s$ is rational. So in summary, $\sqrt{3}$ is not in $\mathbb{Q}[\sqrt{2}]$.
