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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..

Bill Dubuque
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EngineerInProgress
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    Could you clarify your proof for the case of $n=2$? – ShapeOfMatter Jan 14 '19 at 19:32
  • Why are the indices in your title $j+1$ etc.? – J.G. Jan 14 '19 at 19:49
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    I tried using induction the following way: for n=1 this is trivial. Our induction hypothesis looks like this: If $\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}$ is rational then all $\sqrt{x_i}$ are rational. Now i would like to that that if $\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}+\sqrt{x_{n+1}}$ is rational then so are all $\sqrt{x_i}$. I cannot see how the induction hypothesis would help to be honest. If the sum $\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}+\sqrt{x_{n+1}}$ is rational we cannot conclude that $\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}$ is rational as well so as to use the hypothesis. – EngineerInProgress Jan 14 '19 at 19:49
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    Some more experienced user edited that for some reason... let me change it back to what it was. – EngineerInProgress Jan 14 '19 at 19:50
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    Related: https://math.stackexchange.com/q/30687/42969. – Martin R Jan 14 '19 at 19:51
  • Just a second ago someone in the comments proposed an approach using field extensions. I'm still in high school so I'd appreciate if you could keep the hints and answers on my level. Thanks :) – EngineerInProgress Jan 14 '19 at 19:57
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    I would be legitimately surprised (and delighted) to see a good high school-level proof of this fact. Where did you come across the problem? – Steven Stadnicki Jan 14 '19 at 20:17
  • I was just about to say the same thing as @StevenStadnicki did. It seems to me to be naturally a question about field extensions, and I've been trying to come up with a high-school level proof, with no success at all. – saulspatz Jan 14 '19 at 20:19
  • This was one of a tasks from a polish worksheet with tasks preparing for college entrance exams. Surprisingly, the other problems on the very same worksheet are very simple (for example find all integer pairs (x,y) that satisfy the equation $((x+1)(y-1)-1)^2=(x+2)^2+y^2$ ) – EngineerInProgress Jan 14 '19 at 20:25
  • In this answer I show how to generalize the argument you gave for $,n = 2,$ to general $n$. It uses only very basic field theory and could easily be rewritten at high-school level (you may be able to understand it already). – Bill Dubuque Jan 14 '19 at 22:09
  • @Steven See my prior comment (perhaps you have time to translate it to high-school language). – Bill Dubuque Jan 14 '19 at 22:28
  • Here come the Pythagoreans... – Math Enthusiast Jan 15 '19 at 02:39

2 Answers2

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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.

Bill Dubuque
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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.

W-t-P
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