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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.

$alpha:=sqrt{2}+sqrt{3}notin U:= mathbb{Q}(sqrt{2})cupmathbb{Q}(sqrt{3})iffalpha notin mathbb{Q}(sqrt{2})$ and $alpha notin mathbb{Q}(sqrt{3})$

Proof: $alpha notin mathbb{Q}(sqrt{2})$

Assume $exists_{r,sin mathbb{Q}}r+ssqrt{2}=sqrt{2}+sqrt{3}$

$iff sqrt{3}=r+(s-1)sqrt2$

$iff 3=r^2+2r(s-1)sqrt2+2(s-1)^2$

Which implicates $3=0$ or $sqrt3in mathbb{Q}$ 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

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 $mathbb{Q}[sqrt{2}]$.

Well, $alphanotin {Bbb Q}(sqrt 2)$ is equivalent to $sqrt 3notin {Bbb Q}(sqrt 2)$.
On the contrary, $sqrt 3 = a+bsqrt 2$ for some rational numbers $a,b$. Squaring gives $3 = a^2+2absqrt 2 + 2b^2$. Then $sqrt 2$ could be written as a rational number, which is impossible as it is irrational.

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.

Instead of that just try to check is this $U:=mathbb{Q}(sqrt 2)cup mathbb{Q}(sqrt 3)$ a Group?

Let us take $sqrt2,sqrt3in U$ [Both exists] but $sqrt2+sqrt3notin U$.

So definitely it cannot be a Subfield of $mathbb{R}$.