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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.

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    There is one little useful fact: if $x$ is algebraic and $y$ is transcendental, then $x+y$ is transcendental; so is $xy$, unless $x=0$. So, for example, $\pi+\sqrt2$ and $\pi\sqrt2$ are both transcendental.2011-12-15

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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.

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    If Ian knows how to prove that $\sqrt{35}$ is irrational, it is enough to note that $\sqrt{35}=\frac12\left((\sqrt5+\sqrt7)^2-12\right)$, which would be rational if $\sqrt 5+\sqrt 7$ were.2013-09-04
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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

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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 $2$.