Sum of irrational numbers Well, in this question it is said that $\sqrt[100]{\sqrt3 + \sqrt2} + \sqrt[100]{\sqrt3 - \sqrt2}$, and the owner asks for "alternative proofs" which do not use rational root theorem. I wrote an answer, but I just proved $\sqrt[100]{\sqrt3 + \sqrt2} \notin \mathbb{Q}$ and $\sqrt[100]{\sqrt3 - \sqrt2} \notin \mathbb{Q}$, not the sum of them. I got (fairly) downvoted, because I didn't notice that the sum of two irrational can be either rational or irrational, and I deleted my (incorrect) answer. So, I want help in proving things like $\sqrt5 + \sqrt7 \notin \mathbb{Q}$, and $(1 + \pi) - \pi \in \mathbb{Q}$, if there is any "trick" or rule to these cases of summing two (or more) known irrational numbers (without rational root theorem).
Thanks.
 A: To prove that $\sqrt5+\sqrt7$ is irrational: 
$\sqrt 5+\sqrt 7=\frac{a}{b}$
$\frac{a^2}{b^2}=12+\sqrt{35}$
$\frac{a^2-12b^2}{b^2}=\sqrt{35}$
$35=\frac{(a^2-12b^2)^2}{(b^2)^2}$
$35|a^2-12b^2$
$35^2|(a^2-12b^2)^2$
$35^2|(b^2)^2$
Both the numerator and denominator are multiples of an even power of 2. Contradiction.
The method can be extended to many other sums of nth roots.
A: Assume $\sqrt a + \sqrt b = \dfrac pq$ where $p,q\in \mathbb{Z}$ and $a,b $ rational; $\sqrt a$ irrational
$$q[\sqrt a + \sqrt b] = p$$
$$q[a-b] =p[\sqrt a - \sqrt b]$$
$$\frac{q}{p}[a-b] = \sqrt a - \sqrt b $$
Therefore, $$\sqrt a + \sqrt b  + \sqrt a - \sqrt b  = \frac pq + \frac qp(a-b)$$
$$2\sqrt a = \frac{p^2 + q^2(a-b)}{qp}$$
$2\sqrt a = \dfrac{p^2 + q^2(a-b)}{qp}$   which is a contradiction because $\sqrt a$ is an irrational number and so cannot be written as a ratio of integers
A: Here is a useful trick, though it requires a tiny bit of field theory to understand: If $\alpha + \beta$ is a rational number, then $\mathbb{Q}(\alpha) = \mathbb{Q}(\beta)$ as fields.  In particular, if $\alpha$ and $\beta$ are algebraic, then the degrees of their minimal polynomials are equal.
So, for example, we can see at a glance that $\sqrt{5} + \sqrt[3]{7}$ is irrational, because $\sqrt{5}$ and $\sqrt[3]{7}$ have algebraic degree $2$ and $3$ respectively.
Note that this trick doesn't work on your original example, because $\alpha=\sqrt[100]{\sqrt{3} + \sqrt{2}}$ and $\beta=\sqrt[100]{\sqrt{3} - \sqrt{2}}$ do have the same degree. But we can also use field theory: since $\alpha \beta = 1$, if $\alpha+\beta$ is rational then $\alpha$ and $\beta$ satisfy a rational quadratic.  However, $\alpha^{100}= \sqrt{3}+\sqrt{2}$ already has degree $4$ over $\mathbb{Q}$, so $\alpha$ certainly has degree bigger than $2$.
