sum of irrational numbers - are there nontrivial examples?

Solution 1:

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

Solution 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...