# Independence of radicals: First-principles proof of special case

Reposting from MathStackexchange, original post is here, but got no answer.

I've known this problem for a long time:

Problem. Show that the number $\alpha=\sqrt{1} + \sqrt{2} + \ldots + \sqrt{n}$ is irrational for $n\geq 2$.

but I haven't been able to find a solution from first principles (in the sense of a high-school math olympiad kind of proof, not using advanced theory; so for example you can observe using the theory of algebraic integers that if $\alpha$ is rational, it must be integer, but I would consider that too heavy of an apparatus). I was wondering if anybody knows one/can come up with one?

Solutions that use some heavier theory, but not too much, are also welcome.

Update: I'm aware of solutions proceeding by Galois theory, etc. but my reason to believe this has an elementary solution is that it was in a rather interesting and high-quality list of high-school olympiad preparation problems that I found on a math forum.

• I would be surprised if there exists a proof that avoid the language of field extensions. What is the list of problems you are talking about? – Francesco Polizzi Oct 19 '16 at 5:35
• I know a proof that only uses basic field theory, but no Galois theory and no fact about the ring of algebraic integers. If you are interested, I can share it here. BTW I heard this problem in a high school math camp. – GH from MO Oct 19 '16 at 20:09
• Yes, that'd be great! – amakelov Oct 19 '16 at 20:11
• @FrancescoPolizzi: the only copy of this list that I have is unfortunately in Bulgarian, and it can be found here. The problem appears as #1 on page 5. Curiously, the list was compiled by Vesselin Dimitrov, who you might have heard of in relation to the ABC conjecture. Another curious thing is that while the list was compiled sometime around 2005, one of the problems in it appeared on the Romanian Master in Mathematics contest in 2010 (however, I only realized that after the competition :) ) – amakelov Oct 19 '16 at 21:30

Theorem. Let $K$ be a field of characteristic different from $2$. Let $x_1,\dots,x_n$ be arbitrary elements in an arbitrary field extension of $K$ such that the square of each $x_i$ lies in $K$, the sum $x_1+\dots+x_n$ lies in $K$, and no subsum of $x_1+\dots+x_n$ vanishes (including the full sum). Then each $x_i$ lies in $K$.
Proof. We proceed by induction on $n$. For $n=1$ the statement is trivial. Now we assume that $n>1$ and the statement holds with $n-1$ in place of $n$. Let $x_1,\dots,x_n$ satisfy the conditions of the theorem. If any $x_i$ lies in $K$, then by the induction hypothesis all the other $x_j$'s lie in $K$, and we are done. So we can assume that no $x_i$ lies in $K$. The sum $x_2+\dots+x_n$ lies in the field extension $K(x_1)$, hence by the induction hypothesis each $x_i$ lies in this field extension. That is, $x_i=a_i+b_ix_1$ for some $a_i,b_i\in K$. By our assumption, each $b_i$ is nonzero. Furthermore, $x_i^2=(a_i^2+b_i^2x_1^2)+2a_ib_ix_1$ lies in $K$, whence $2a_ib_ix_1$ also lies in $K$. This forces that $a_i=0$, because otherwise $2a_ib_i\neq 0$ and $x_1=(2a_ib_ix_1)/(2a_ib_i)\in K$, a contradiction. We conclude that $x_i=b_ix_1$. As a result, $x_1+\dots+x_n=(b_1+\dots+b_n)x_1$ is a nonzero element of $K$, hence $b_1+\dots+b_n$ is also a nonzero element of $K$. But then $x_1\in K$, a contradiction again.
Corollary. Let $a_1,\dots,a_n$ be positive rational numbers such that the sum $\sqrt{a_1}+\dots+\sqrt{a_n}$ is rational. Then each term $\sqrt{a_i}$ is rational.
Proof. Apply the theorem with $K:=\mathbb{Q}$ and $x_i:=\sqrt{a_i}$. Note that no subsum of $x_1+\dots+x_n$ vanishes (including the full sum), because the $x_i$'s are positive.