Reposting from MathStackexchange, original post is [here](https://math.stackexchange.com/questions/1849410/independence-of-radicals-first-principles-proof-of-special-case), 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. 



Independence of radicals: First-principles proof of special case
Upon the OP's request, I share the proof (using only basic field theory) of the following more general result.


**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.


