Prove that $\sqrt{a_1}\pm \sqrt{a_2}\pm\ldots\pm \sqrt{a_n}$ is irrational for $a_1,\ldots,a_n\in\mathbb{N}$ and $a_i\neq m^2$

Asked Dec 02 '16 at 21:14

Active Dec 02 '16 at 21:28

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Consider a sequence $a_1,a_2,\ldots,a_n$ such that $a_1,a_2,\ldots a_n\in\mathbb{N_{\ge 1}}$, $a_1\neq a_2\neq \ldots\neq a_n$ and for every $1\le i\le n$, $a_i$ is square-free.

$$ \sqrt{a_1}\pm \sqrt{a_2}\pm\ldots\pm \sqrt{a_n}$$

is an irrational number.

This question originates from solving a AoPS Olympiad problem (prove that $\sqrt{10}-\sqrt{6}-\sqrt{5}+\sqrt{3}$ is irrational) - I was wondering if this could be generalized to sequences of arbitraly length.

edited Dec 02 '16 at 21:28

asked Dec 02 '16 at 21:14

Are $a_i$ all pairwise distinct? – Levent Dec 02 '16 at 21:16
Yes, I'll modify the question. – Dec 02 '16 at 21:18
4

This is false as stated, consider $\sqrt{18}-\sqrt{8}-\sqrt{2}$. It is true once you assume $a_i$ are squarefree, however. – Wojowu Dec 02 '16 at 21:25
Modified the question appropriately. – Dec 02 '16 at 21:26
(you still have to assume $a_i\neq 1$; earlier it was contained in the assumption that they aren't squares) – Wojowu Dec 02 '16 at 21:27
Modified again, thanks for helpful comments! – Dec 02 '16 at 21:29
1

$a_1\neq a_2\neq \ldots\neq a_n$, I don't think it means what you want. You should write $i \neq j \Rightarrow a_i \neq a_j$. – Tacet Dec 02 '16 at 21:31
Well, $\sqrt{x} + \sqrt{y} = q \implies \sqrt{xy} = (q-x-y)/2$ is rational. but x and y being squarefree and inequal makes this impossible. Can probably generalize for induction. – fleablood Dec 02 '16 at 21:31
See this question and this answer. – dxiv Dec 02 '16 at 23:18

Prove that $\sqrt{a_1}\pm \sqrt{a_2}\pm\ldots\pm \sqrt{a_n}$ is irrational for $a_1,\ldots,a_n\in\mathbb{N}$ and $a_i\neq m^2$

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