I would like to know how to prove that $\sqrt{2} + \sqrt{5} + \sqrt{7}$ is an irrational number. I know how to do the proof for a sum of two roots. Can I just define $\sqrt{2} + \sqrt{5} :=c$ and then prove that $c$ and $c + \sqrt{7}$ are both irrational, when I treat $c$ as a single number? If this way is kind of wrong or won't work, is there another way to do the proof?

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    $\begingroup$ But watch out: $\sqrt 2+\sqrt{8}-\sqrt{18}$ is rational! $\endgroup$ Apr 23, 2014 at 20:40
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    $\begingroup$ Prove. Prove. Prove. Prove. Prove. Prove. $\endgroup$
    – Pedro
    Apr 23, 2014 at 20:40
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    $\begingroup$ Treat, rather than tread. $\endgroup$ Apr 23, 2014 at 20:50
  • $\begingroup$ And $c$ is a single number. What else would it be? $\endgroup$ Apr 23, 2014 at 20:51

3 Answers 3


One way to prove that the sum of two square roots (e.g. $\sqrt 2 + \sqrt 5$) is irrational is to suppose not, and square once. Upon rearrangement, you'll find a contradiction. Here we can do the same thing, squaring twice with some algebra: Suppose $$r = \sqrt 2 + \sqrt 5 + \sqrt7$$ were rational; then

$$r - \sqrt7 = \sqrt2 + \sqrt5$$

Squaring both sides, you'll find that

$$r^2 + 7 - 2 r \sqrt7 = 2 + 5 + 2\sqrt{10}$$

Simplify, place both radicals one one side of the equation, and square again; try to get a contradiction from here.

Of course, this doesn't always work: There are triplets of numbers such that the corresponding $r$ is rational, so carefully identify where the contradiction is.


(A stronger result for reference)

If $a_i$ are postive integers, then $\sum \sqrt[n]{a_i}$ is rational iff $\sqrt[n]{a_i}$ is.

  • $\begingroup$ Edited. Thanks. $\endgroup$ Apr 23, 2014 at 21:28

You might want to take a look at www.thehcmr.org/issue2_1/mfp.pdf.


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