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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??

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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.

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  • $\begingroup$ Oh it is my mistake. So it would be better to think 10000 instead of 9999. $\endgroup$ – J.U.math Dec 6 '16 at 2:13
  • $\begingroup$ @Edgar.W: yes, since this sort of method would work for $1/n$ for any $n$ odd. I'm afraid I don't have any great clarifying remarks for this sort of problem, since again, it can be quite hard in general! $\endgroup$ – Alex Wertheim Dec 6 '16 at 2:15
  • $\begingroup$ You said you could solve this type of problem using Galois theory, could you give me a link to it,please? $\endgroup$ – J.U.math Dec 6 '16 at 2:18
  • $\begingroup$ @Edgar.W: sure. Here is an example where you can prove that $\sqrt{5} \notin \mathbb{Q}(\sqrt[3]{2})$ using Galois-theoretic methods: math.stackexchange.com/questions/937071/… Of course, there are easier approaches in this case, e.g. degree arguments. $\endgroup$ – Alex Wertheim Dec 6 '16 at 2:22
  • $\begingroup$ I should point out that this is not a great example, since the argument that $[L:\mathbb{Q}] = 6$ is equivalent to the fact that $\sqrt{5} \notin \mathbb{Q}(\sqrt[3]{2})$, but you can see how the line of reasoning works nonetheless. $\endgroup$ – Alex Wertheim Dec 6 '16 at 2:28

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