Are the positive irrational numbers closed under addition?

Clearly this is not the case for all irrational numbers, e.g. $\sqrt 2 + (- \sqrt 2)$ but what about all irrational numbers $x > 0$?

Intuitively I would think it is closed, but maybe there is some weird counterexample.

An alternative formulation of the question is, "Does the sum of any collection of positive irrational numbers ever yield a rational number?"

I was thinking a proof by contradiction, by taking $a_1, a_2, \dots a_n$ to be a collection of positive irrationals, and assume that $$\sum_{i=1}^n a_i = \frac pq, (p,q \in \mathbb Z, q\neq 0).$$ We can move $q$ to the left hand side, but that just kicks the can down the road since each product $q\cdot a_i$ is irrational but whether the sum of these terms is still irrational has yet to be established.

Edit: Well duh. It would've helped to realize that (eg.) $2-\sqrt 2$ is irrational! I shall however leave the question up since I don't think this has been asked on the site before, and I was looking for it so maybe someone else will too.

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    $\begingroup$ $\sqrt 2 +(2-\sqrt 2)=2$. $\endgroup$ Aug 11, 2022 at 6:05
  • $\begingroup$ Ah, I did not know that $2-\sqrt 2$ is also irrational. $\endgroup$ Aug 11, 2022 at 6:06
  • $\begingroup$ Is $n - x$ irrational for all $n \in \mathbb Z$ and all irrational numbers $x$? $\endgroup$ Aug 11, 2022 at 6:07
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    $\begingroup$ Yes. This is easy to justify intuitively if you think about their decimal expansions. Such an $x$ has a nonterminating and nonrepeating decimal expansion; multiplying it by $-1$ and adding $n$ will only change a few, finitely-many digits. $\endgroup$ Aug 11, 2022 at 6:09
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    $\begingroup$ If $n-x$ is rational, say $n-x=p/q$, then $x=n-p/q$ is rational. $\endgroup$
    – David
    Aug 11, 2022 at 6:10

1 Answer 1


No it is not. Take for example: $a = 3 -\sqrt{2}, b = \sqrt{2}+1$. Is $a+b$ still irrational ?


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