Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don't think is possible, but how to show it?


Squaring we get, $m=a^2+n-2a\sqrt n\implies \sqrt n=\frac{a^2+n-m}{2a}$ which is rational

  • $\begingroup$ @SujaanKunalan, thanks for your feedback $\endgroup$ – lab bhattacharjee Aug 1 '13 at 17:52

Assume $m$ is a non-square integer. Then $\sqrt{m}$ is irrational, and if $x=\sqrt{m}+\sqrt{n}$, then




If $x$ is rational, then the LHS is also rational. However the RHS is irrational, contradiction, so $x$ is irrational.

Same argument as here and here. It should be put in the FAQ :-)


If $\sqrt{n} + \sqrt{m}$ is rational, then since
($\sqrt{n} + \sqrt{m})(\sqrt{n} - \sqrt{m}) = n - m,$
$\sqrt{n} - \sqrt{m}$ is rational. Thus
$\sqrt{n}, \sqrt{m}$ are rational, n,m are squares.

  • $\begingroup$ Just wanted to say that this is a surprisingly slick solution, and deserves more recognition. $\endgroup$ – platty Nov 28 '18 at 6:44
  • $\begingroup$ @platty This is very well-known and probably has been mentioned here over a hundred times, for example see this 7-year-old answer for a more general perspective. $\endgroup$ – Bill Dubuque yesterday

Nice way to see thinks

Assume that, $$(\sqrt{n}+\sqrt{m})=\frac{p}{q}$$ Then we have $$(\sqrt{n}+\sqrt{m})=\frac{p}{q}\in\Bbb Q \implies n+m+2\sqrt{nm} =(\sqrt{n}+\sqrt{m})^2 =\frac{p^2}{q^2}\in\Bbb Q\\\implies \sqrt{nm} =\frac{n+m}{2}+\frac{p^2}{2q^2}\in\Bbb Q $$

But if $ nm $ is not a perfect square then $\sqrt{nm}\not \in\Bbb Q ,$ (This can be easily prove using the fundamental theorem of number theory: Decomposition into prime numbers) Hence in this case we have $$\sqrt{n}+\sqrt{m}\not \in\Bbb Q$$

Remark $~~~~~$1. $mn$ can be a perfect square even though neither $n$ nor $m$ is a perfect square. (see the example below)

  1. We can still have $\sqrt{n}+\sqrt{m}\not\in \Bbb Q$ even if $mn$ is perfect square.(see the example below)

Example: $n= 3$ and $ m = 12$ are not perfect square and $ nm = 36 =6^2.$ Moreover, $$\sqrt{n}+\sqrt{m} = \sqrt{3}+\sqrt{12} =3\sqrt 3 \not \in\Bbb Q$$


It's easy to show that if the result is rational, it has to be natural.

Hint: $$m=(\lfloor\sqrt{m}\rfloor+\epsilon)^{2}=\lfloor\sqrt m \rfloor ^2+2\epsilon \lfloor\sqrt m \rfloor + \epsilon^2$$ $$n=(\lceil \sqrt n \rceil-\epsilon)^2=\lceil \sqrt n \rceil^2-2\epsilon\lceil \sqrt n \rceil+\epsilon^2$$ Subtract these two from each other and show that $\epsilon$ has to be rational.

  • $\begingroup$ m+n+2*sqrt(m*n) has to be rational $\endgroup$ – Mahdi Sep 22 '14 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.