As you know, it holds that $\sqrt3\notin\mathbb{Q}(\sqrt2)$ The reason for this is about degree of field. That is, on the contrary, if $\sqrt3\in\mathbb{Q}(\sqrt2)$ then $\sqrt3+\sqrt2\in\mathbb{Q}(\sqrt2)$
So, $2$ = [$\mathbb{Q}(\sqrt2)$ : $\mathbb{Q}$] = [$\mathbb{Q}(\sqrt2)$ : $\mathbb{Q}(\sqrt2 +\sqrt3)$] [$\mathbb{Q}(\sqrt2 +\sqrt3)$ : $\mathbb{Q}$] This leads to the following. $deg(\sqrt2 +\sqrt3 , \mathbb{Q}$) = 2 However, this is a contradiction because we know that the degree of the irreducible polynomial which is $x^4-10x^2+1 \in \mathbb{Q}[x]$ is 4.
Now, lets consider following case,
$\sqrt3\notin\mathbb{Q}$$(2^{\frac{1}{9999}})$
If we can not present specific irreducible polynomials as in the previous situation(In fact, can not you calculate 9999 squared by hand?), what argument can we prove this fact??
In general, this type of problem can be quite difficult, though methods from Galois theory are often effective. However, in your particular case, there is a neat observation we can make. Indeed, the minimal polynomial of $2^{1/9999}$ over $\mathbb{Q}$ is $X^{9999}-2$, which is irreducible by Eisenstein at $2$. Hence, the degree of the extension $\mathbb{Q}(2^{1/9999})$ over $\mathbb{Q}$ is $9999$. If we had $\sqrt{3} \in \mathbb{Q}(2^{1/9999})$, then we would have $\mathbb{Q}(\sqrt{3}) \subset \mathbb{Q}(2^{1/9999})$, whence multiplicativity of degree gives us $[\mathbb{Q}(2^{1/9999}):\mathbb{Q}] = [\mathbb{Q}(2^{1/9999}):\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}):\mathbb{Q})]$. But $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}] = 2$ does not divide $9999$, a contradiction. This type of reasoning is often quite useful, although again, these problems can be quite tricky in general: see this question, for example.