sum of irrational numbers - are there nontrivial examples?

Question

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?

Answer 1

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, $q=0$).

Answer 2

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...

Answer 3

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 $x^3 - 3x - 18 = (x-3)(x^2 + 3x + 6).$