Irrational number to an irrational power may be rational

This is an interesting problem. To solve it I should find two irrational numbers r and s such that r^s is rational.

I am not sure I am able to do that. However, I am confident that the following argument does solve the problem.

As we know, √2 is irrational. In particular, √2 is real and also positive. Then √2^√2 is also real. Which means that it is either rational or irrational.

If it's rational, the problem is solved with r = √2 and s = √2.

Assume √2^√2 is irrational. Let r = √2^√2 and s = √2. Then r^s = (√2^√2)^√2 = √2^√22 = √2² = 2. Which is clearly rational.

Either way, we have a pair of irrational numbers r and s such that r^s is rational. Or do we? If we do, which is that?

(There's an interesting related problem.)

References

1. T. Gowers (ed.), The Princeton Companion to Mathematics, Princeton University Press, 2008, p. 157

titled "Irrational number to an irrational power may be rational" you seek an irrational number which when raised to an irrational power is rational. In your proof of existence, you raise the square root of 2 to the square root of 2, and ask whether it is rational or irrational. This is answered on page 216 of Herstein's "Topics in Algebra", 2nd edition where we learn that in 1934, Gelfond and Schneider both proved (independently) that if numbers a and b are both algebraic and b is irrational, then a to the b is transcendental. Since you no doubt knew this, there remains to ask why you didn't mention this in your statement.

To make it clear, what Mario says implies that √2^√2 is transcendental because √2 is algebraic: it's one of the roots of the polynomial equation x² - 1 = 0. Thus it is not rational and the first choice for r and s should be discarded.