If you raise an irrational number to a rational power, it is possible to get something rational. For instance, raise Sqrt[2] to the power 2 and you'll get 2.

But what happens if you raise an irrational number to an irrational power? Can this ever be rational?

The answer is yes, and we'll prove it *without having to find specific numbers that do the trick*!

Theorem. There exist irrational numbers A and B so that A^{B} is rational.

Proof. We know that Sqrt[2] is irrational. So, if A=Sqrt[2] and B=Sqrt[2] satisfy the conclusion of the theorem, then we are done. If they do not, then Sqrt[2]^{Sqrt[2]} is irrational, so let A be this number. Then, letting B=Sqrt[2], it is easy to verify that A^{B}=2 which is rational and hence would satisfy the conclusion of the theorem. QED.

This proof is *non-constructive* because it (amazingly) doesn't actually tell us whether Sqrt[2]^{Sqrt[2]} is rational or irrational!

**The Math Behind the Fact:**

Actually, Sqrt[2]^{Sqrt[2]} can be shown to be irrational, using something called the Gelfond-Schneider Theorem (1934), which says that if A and B are roots of polynomials, and A is not 0 or 1 and B is irrational, then A^{B} must be irrational (in fact, transcendental).

But you don't need Gelfond-Schneider to construct an explicit example, assuming you know transcendental numbers exist (numbers that are not roots of non-zero polynomials with integer coefficients). Let x be any transcendental and q be any positive rational. Then x^{log_x(q)}=q so all we have to show is that log_x(q) is irrational. If log_x(q)=a/b then q=x^{a/b}, implying that x^{a}-q^{b}=0, contradicting the transcendentality of x.

**How to Cite this Page:**

Su, Francis E., et al. “Rational Irrational Power.” *Math Fun Facts*. <https://www.math.hmc.edu/funfacts>.

**References**:

R. Vakil, A Mathematical Mosaic, 1996.

**Fun Fact suggested by**:

Ravi Vakil