sum of irrational numbers - are there nontrivial examples? 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?

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