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It is trivial to show that the set of irrational numbers is not closed under addition. Just choose an irrational number $p$ and add it to its additive inverse $-p$ to get $0\in\mathbb{Q}$. However, I have yet to see a (non-trivial) example of a rational sum $(p + q) \in \mathbb{Q}$ of two irrational numbers $p$ and $q$ where $q \ne -p$. Can anyone provide such an example, or is this not possible? A similar question might be whether the set of positive irrational numbers is closed under addition.

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$\pi + x = 1{}{}$ –  user61527 Jul 15 '14 at 3:33
The positive irrationals? How about $\sqrt2+(17-\sqrt2)$? –  Gerry Myerson Jul 15 '14 at 3:35
All of them will be of the form $a+b=q$ where $a,b$ are irrational and $q$ is rational. So it will always be of the form $a,q-a$ where $a$ is irrational and $q$ is rational. –  Thomas Andrews Jul 15 '14 at 3:36
Thomas Andrews' comment makes it clear: EVERY example is like that. –  Lee Mosher Jul 15 '14 at 3:43
Inescapable, $x+y=r$ iff $y=r+(-x)$. –  André Nicolas Jul 15 '14 at 3:44

2 Answers 2

up vote 0 down vote accepted

In base 2:

$$ x=0.1\,0\,1\,00\,1\,000\,1\,00001\underbrace{00000}_5\,1\,\underbrace{000000}_6\, 1\underbrace{0000000}_7\, 1\ldots\ldots\ldots $$ $$ y=0.0\,1\,0\,11\,0\,111\,0\,1111\,0\,\underbrace{11111}_5\,0\,\underbrace{111111}_6\,0\,\underbrace{1111111}_7\,0\,\ldots\ldots\ldots $$

$x+y$ is rational.

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Nice! And this works in any base $b \ge 2$, such that (rational) sum $x + y = \frac{1}{b-1}$ by geometric series. –  Drake P Jul 15 '14 at 6:15

$$ \underbrace{{}\quad\pi\quad{}}_{\text{an irrational number}} + \underbrace{\Big( 10-\pi\Big)}_{\text{an irrational number}} \text{is rational.} $$

Any time you have $x$ and $y$ both positive irrational numbers and $x+y=r$ rational, then clearly you have $y = r-x$, so necessarily your pair is $x$ and $r-x$.

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