High-school level proof that if $x_i$ are rationals and $\sqrt{x_1}+\dots\sqrt{x_n}$ is rational, then each $\sqrt{x_i}$ is rational.

Question

I'm having a hard time with the following problem:

Let $x_1,x_2...x_n$ be rational numbers. Prove that if the sum $\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}$ is rational, then all $\sqrt{x_i}$ are rational. Show, that the assumption for $x_i$ to be rational is necessary.

The only thing that I came up with is how to show this for n=2. Maybe there's some analogy for bigger n-s as well. If we assume that $\sqrt{x_1}+\sqrt{x_2}$ is rational then so must be $\sqrt{x_1}-\sqrt{x_2}$ (their product is rational). By adding those two together we get that both $\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_1}-\sqrt{x_2} = 2\sqrt{x_1}$ and $\sqrt{x_1}+\sqrt{x_2}-\sqrt{x_1}+\sqrt{x_2} = 2\sqrt{x_2}$ are rational which implies the rationality of both $\sqrt{x_1}$ and $\sqrt{x_2}$

I'm still in high school so I'd appreciate if you could keep the hints and answers on my level. Any help is appreciated. Thank you.

BTW: I've also tried to prove by contradiction and induction. Both attempts didn't work..

Answer 1

Here I show how to generalize your argument for $\,n=2\,$ to general n. It uses very simple field theory. Since you write that you are in high-school so wish to avoid field theory, below I explain what is needed, and work through a special case of the linked proof for motivation.

As with many inductive proofs, they key is to strengthen the inductive hypothesis, which here means proving the statement not only for rational numbers $\,\Bbb Q\,$ but also for larger "number systems" of real numbers that are obtained by adjoining square roots of positive numbers.

For example $\,\Bbb Q(\sqrt 5)\,$ denotes the reals obtainable by (field) arithmetic on rationals $\,\Bbb Q\,$ and $\,\sqrt 5\,$, where field arithmetic consists of the operations of addition, multiplication and division $\,a/b,\, b\neq 0.\,$ It is easy to show that the reals obtainable by iterating these operations are exactly those writable in the form $\,a+b\sqrt{5}\,$ for $\,a,b\in \Bbb Q\,$ (for division we can rationalize the denominator). We can iterate this construction, e.g. adjoining $\,\sqrt 3\,$ to $\,F = \Bbb Q(\sqrt 5)$ to get $\,F(\sqrt 3)\,$ with numbers $\,a+b\sqrt 3\,$ for $\,a,b\in \Bbb Q(\sqrt 5)$. This step-by-step construction of such towers of number systems proves very handy for inductive proofs (a special case of structural induction).

For motivation, we show how the induction step works to reduce the case $\,n\!=\!3\,$ to $\,n\!=\!2\,$ (your result). The induction step in the general proof works exactly the same way.

Suppose $\sqrt 2 + \sqrt 3 + \sqrt 5 = q\in \Bbb Q.\,$ It suffices to show one summand $\in \Bbb Q\,$ since then the sum of the other two is in $\,\Bbb Q\,$ so induction (your $n=2$ proof) shows they too are in $\,\Bbb Q$.

$\,\sqrt 2 + \sqrt 3 = q-\sqrt 5 \in \Bbb Q(\sqrt 5) = \{ a + b\sqrt 5\ : a,b\in\Bbb Q\}\ $ so by induction $\,\sqrt 2,\sqrt 3\in \Bbb Q(\sqrt 5)\,$ so

$$\begin{align} \sqrt{2}\ =\ a_2 + b_2 \sqrt{5},\ \ \ a_2,b_2\in \Bbb Q\\ \sqrt{3}\ =\ a_3 + b_3 \sqrt{5},\ \ \ a_3,b_3\in \Bbb Q \end{align}\qquad$$

If $\,b_3 < 0\,$ then $\, a_3 = \sqrt 3 - b_3\sqrt 5 = \sqrt 3 +\! \sqrt{5b_3^2}\in \Bbb Q\,$ so induction ($n\!=\!2\:\!$ case) $\Rightarrow \sqrt 3\in\Bbb Q,\,$ contradiction. Similarly if $\,b_2 < 0\,$ we obtain a contradiction $\,\sqrt 2\in\Bbb Q$.

Else all $\,b_i \ge 0\,$ so $\,q = \sqrt 2\! +\! \sqrt 3\! +\! \sqrt 5 = a_2\!+\!a_3+(b_2\!+\!b_3\!+\!1)\sqrt 5\,\Rightarrow\,\sqrt 5 \in \Bbb Q\ $ by solving for $\,\sqrt 5,\,$ using $\,b_2\!+\!b_3\!+\!1 \neq 0\,$ by all $\,b_i\ge 0.\ $

Thus in every case some summand $\in \Bbb Q,\,$ which completes the proof.

Answer 2

The case $n=3$ is actually very easy and requires no field theory, explicit or implicit.

If $\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}=a$ is rational, then moving $\sqrt{x_3}$ to the right-hand side and squaring we get $$ 2\sqrt{x_1x_2} = a_1-2a\sqrt{x_3}, $$ where $a_1=a^2+x_3-x_1-x_2$ is rational. Squaring again, $$ a_2 = -4aa_1\sqrt{x_3} $$ with $a_2=4x_1x_2-4a^2x_3-a_1^2$. Since $a>0$, it follows that either $\sqrt{x_3}$ is rational, or $a_1=a_2=0$. In the former case we are done, in the latter case $x_1x_2=a^2x_3$ and also $a^2+x_3-x_1-x_2=0$. Excluding $a^2$ we then get $x_3\in\{x_1,x_2\}$. Thus, either $\sqrt{4x_1}+\sqrt{x_2}$, or $\sqrt{x_1}+\sqrt{4x_2}$ is rational, and the claim follows by induction.

In the same way it should be possible to show that if none of the products $x_ix_j$ is a square, and $\alpha_1\sqrt{x_1}+\alpha_2\sqrt{x_2}+\alpha_3\sqrt{x_3}$ is rational, then in fact all summands are rational. (Here $\alpha_i$ and $x_i\ge 0$ are rational.)

In fact, I strongly suspect that one should be able to prove the general claim, with $n$ summands, using nothing but induction. The trick is, the induction should be by the rank of the group generated (multiplicatively) by $x_1,\dotsc,x_n$, not by $n$. This, however, would be a little longer to describe in details.