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.
Answer
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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