I know that the sum of two (or more) irrational numbers can be rational. For example, both $\sqrt{2}$ and $1-\sqrt{2}$ are irrational numbers, but their sum is rational.

Also I know that $\sqrt{m}$ is either integer or an irrational number for any natural number $m$. In other words, it can not be a rational number unless it is an integer.

I have no problem to prove the rationality/irrationality of the square root of any natural number. Using "direct proof" for rational numbers, and using "proof by contradiction" for irrational numbers.

I could not combine what I know together to answer the following questions:

Can $$\sum_{k=1}^{n}\sqrt{k}=\sqrt{1}+\sqrt{2}+\sqrt{3}+\dots+\sqrt{n}$$ be rational for any integer $n>1$? If yes, then what is the least such $n$?

Can it be an integer for any integer $n>1$? If yes, then what is the least such $n$?

Note: If it can not be an integer, does not mean it can not be rational.

Any help would be really appreciated.

  • 2
    $\begingroup$ You can show that $\sqrt1+\cdots+\sqrt n$ is an algebraic integer (i.e., is the root of a monic polynomial with integer coefficients). Consequently it cannot be rational without being an integer. $\endgroup$ – Barry Cipra Sep 14 '19 at 21:42

We first prove that if $x,y\in\mathbb{Q}$ are such that $\sqrt{x},\sqrt{y}\notin\mathbb{Q}$, then$\sqrt{x}+\sqrt{y}\notin\mathbb{Q}$. Let us suppose the contrary, we have $$ \sqrt{x}=\frac{(\sqrt{x}+\sqrt{y})^2+x-y}{2(\sqrt{x}+\sqrt{y})}\in\mathbb{Q} $$ which is not. Let $n\geqslant 2$ and $S_n=\sum_{k=1}^n{\sqrt{k}}$, if $S_n-\sqrt{2}\in\mathbb{Q}$ then $S_n=(S_n-\sqrt{2})+\sqrt{2}\notin\mathbb{Q}$. If $S_n-\sqrt{2}\notin\mathbb{Q}$, by the lemma $S_n=(S_n-\sqrt{2})+\sqrt{2}\notin\mathbb{Q}$. In particular $S_n$ is not an integer for all $n\geqslant 2$.

EDIT : this doesn't work.

  • $\begingroup$ How are you applying the lemma? You need $S_n - \sqrt 2$ to be of the form $\sqrt x$. $\endgroup$ – Eric M. Schmidt Sep 14 '19 at 21:09
  • $\begingroup$ You are right it doesn't work. $\endgroup$ – Tuvasbien Sep 14 '19 at 21:32
  • $\begingroup$ the actual proof is not that trivial and hints toward it are linked in the comments above (it requires some Galois theory though); not sure if there is an elementary solution for $n \ge 5$ (not an ad hoc one where you just take the rational conjugates, build the minimal polynomial and show it has no integral solutions) $\endgroup$ – Conrad Sep 14 '19 at 21:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.