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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2$\pi + x = 1{}{}$ – Jul 15 '14 at 03:33
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1The positive irrationals? How about $\sqrt2+(17-\sqrt2)$? – Gerry Myerson Jul 15 '14 at 03:35
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3All 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 03:36
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Hmm, maybe my question as asked doesn't capture the essence of what I really want to know about the existence of. See, $\sqrt{2} + (17 - \sqrt{2})$ still requires the additive inverse $-\sqrt{2}$ of $\sqrt{2}$ to form the (second) addend. – Drake P Jul 15 '14 at 03:40
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2Thomas Andrews' comment makes it clear: EVERY example is like that. – Lee Mosher Jul 15 '14 at 03:43
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3Inescapable, $x+y=r$ iff $y=r+(-x)$. – André Nicolas Jul 15 '14 at 03:44
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To put it another way, every number is $\sqrt{2}$ plus some number. – Lee Mosher Jul 15 '14 at 03:44
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1Okay, well I suppose that does answer it, thank you! Apparently what I though might exist indeed cannot. – Drake P Jul 15 '14 at 03:45
2 Answers
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$$ \underbrace{{}\quad\pi\quad{}}_{\text{an irrational number}} + \underbrace{\Big( 10-\pi\Big)}_{\text{an irrational number}} \text{is rational.} $$
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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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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1Nice! 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 06:15