# Can $\sum_{n=2}^{k} \sqrt{n}$ be rational? [duplicate]

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?

• @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. Oct 12, 2019 at 11:29
• Not quite a duplicate, as a sequence of roots of adjacent integers are added, not (typiccally) primes,. Oct 12, 2019 at 11:31
• See this post which provides an answer to your question. Oct 12, 2019 at 11:38

$$\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$$ 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.