It is easy to see that $\sqrt{2}$ and $\sqrt{2}+\sqrt{3}$ are irrational. So $\sqrt{2}+\sqrt{3} + \sqrt{4}$ is irrational. But what about $\sqrt{2}+\sqrt{3} + \sqrt{4} + \sqrt{5}$? I suspect that $$\sum_{n=2}^{k} \sqrt{n}$$ is always irrational, is it true and is there a simple way to proof that?
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$\begingroup$ @PeterForeman I don't disagree, but most likely the OP doesn't quite understand the terminology used in the question and answers there. They likely don't know what a $\mathbb Q$-vector space means, for example. $\endgroup$– YiFanOct 12, 2019 at 11:29
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$\begingroup$ Not quite a duplicate, as a sequence of roots of adjacent integers are added, not (typiccally) primes,. $\endgroup$– coffeemathOct 12, 2019 at 11:31
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$\begingroup$ See this post which provides an answer to your question. $\endgroup$– Peter ForemanOct 12, 2019 at 11:38
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1 Answer
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$\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$ cannot be rational. Let $p$ be the greatest prime in $[2,n]$. By quadratic reciprocity and Dirichlet's theorem there is some huge prime $P\equiv 1\pmod{4}$ such that $p$ is a quadratic non-residue $\!\!\pmod{P}$, while the primes less than $p$ are quadratic residues $\!\!\pmod{P}$. It follows that $\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$ does not belong to $\mathbb{F}_{P}$ but it belongs to a quadratic extension of $\mathbb{F}_P$. In particular it cannot be rational.
answered Oct 12, 2019 at 16:33
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$\begingroup$ I don't have the skills to understand the answer so I'm not sure that Exchange stack etiquette requires me to vote it. On the other hand, it seems a pity to me because there is a risk that good answers will go into oblivion. $\endgroup$ Oct 13, 2019 at 10:36