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I know that the sum of irrational numbers does not have to be irrational. For example $\sqrt2+\left(-\sqrt2\right)$ is equal to $0$. But what I am wondering is there any example where the sum of two irrational numbers isn't obviously rational like an integer and yet after, say 50 digits after the decimal point it turns out to be rational. Are there any such examples with something non-trivial going on behind the scenes?

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Take any complicated rational $r$ and some irrational $i$. Then $j=r-i$ is irrational and $j+i=r$. –  Alexander Sep 20 '13 at 17:36
Relevant: Sum of irrational radicals is irrational? asks if there are nontrivial examples where $\sqrt[n]a + \sqrt[m]b$ is rational; the answer is no. –  MJD Sep 20 '13 at 17:39
My thougts exactly, after I posted this. But I am interested more in the cases like pi+e. –  Adam Sep 20 '13 at 17:41
That is a much harder question! Nobody knows if $\pi+e$ is rational. –  MJD Sep 20 '13 at 17:42
It's a much harder question to ask such things about transcendental numbers. (And we don't even know whether $ \ \gamma \ $ or $ \ \zeta(5) \ $ are irrational...) –  RecklessReckoner Sep 20 '13 at 17:43
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3 Answers

The examples below are not quite what you're asking for (because the results, while not obviously rational, wind up being nice integers), but they may nonetheless be amusing:

$$\sqrt{3 \; + \; 2\sqrt{2}} \; - \; \sqrt{3 \; - \; 2\sqrt{2}} \; = \; 2$$

$$\sqrt[3]{3\sqrt{21} \; + \; 8} \; - \; \sqrt[3]{3\sqrt{21} \; -\; 8} \; = \; 1$$

$$\sqrt[3]{2 \; + \; \sqrt{5}} \; - \; \sqrt[3]{-2 \; + \; \sqrt{5}} \; = \; 1$$

$$\sqrt[3]{10 \; + \; \sqrt{108}} \; - \; \sqrt[3]{-10 \; + \; \sqrt{108}} \; = \; 2$$

$$\sqrt[3]{9 \; + \; 4\sqrt{5}} \; + \; \sqrt[3]{9 \; - \; 4\sqrt{5}} \; = \; 3$$

How can these be verified? In the first example, squaring, then rearranging, then squaring works. In the second example, show that the numerical expression is a solution to $x^3 + 15x - 16 = 0$ and then show that $x=1$ is the only real solution to this cubic equation by observing that $x^3 + 15x - 16 = (x-1)(x^2+x+16).$ The third example winds up being the only real solution to $x^3 + 3x - 4 = (x-1)(x^2 + x + 4),$ the fourth example winds up being the only real solution to $x^3 + 6x - 20 = (x-2)(x^2 + 2x + 10),$ and the fifth example winds up being the only real solution to $x^3 - 3x - 18 = (x-3)(x^2 + 3x + 6).$

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Is it just me, or a single pattern is evident in all those examples? –  Adam Sep 24 '13 at 22:25
@Adam: I suspect there is some type of "preservation of field extension over the rationals under a certain quadratic conjugation operation" that holds for each of these equations, but I don't know enough algebra (or have time now to research the issue) to be more explicit. If you pose this as a question, then I'm sure you'll get several good answers, which I would also be interested in seeing. –  Dave L. Renfro Sep 25 '13 at 11:34
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Choose two irrational numbers $x$ and $y$ randomly, say independently and each according to the Lebesgue measure on $(0,1)$. Then $x+y$ is irrational with full probability. (Proof: For every real number $z$, $P[x+y=z]=0$, hence $P[x+y\in\mathbb Q]=\sum\limits_{z\in\mathbb Q}P[x+y=z]=0$, QED.)

In this sense, the sum of "almost every" pair of irrational numbers is irrational. Note that, to begin with, "almost every" real number is irrational...

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I didn´t understand the proof, but I upvoted the answer as I think it brings something new to the discussion. –  Adam Sep 20 '13 at 18:01
Now, I understand the proof, nice. –  Adam Sep 20 '13 at 18:03
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If $a+b=q$, where $a,b\notin\mathbb{Q}$ and $q\in \mathbb{Q}$, then $a=q-b$, so just choose a rational number whith sufficiently long period of decimals and you will get what you want. On the other hand, this is still quite trivial, since here we just sum up $b$ and $q-b$ (in your question, $q=0$).

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