Let p, q be two irrational numbers. Can p^{q} be rational? Can p^{q} be irrational?

Denote log_{2}3 by x. As 2 to an integer power can't be equal to 3 to another integer power, x is irrational. At the same time, the square root of 2 is an irrational number, which, to the power x, gives the irrational number square root of 3, and to the power 2x gives the rational number 3.

Fix some (positive, for simplicity) irrational number *x*. All its powers are distinct. Further, there are uncountably many irrational numbers, so *x* has uncountably many irrational powers. However, there are only countably many rational numbers, so at least one irrational power of *x* is an irrational number. Analagously for the other way, consider 2. There are uncountably many irrational powers of 2, so at least one of them (say 2^{y}) is an irrational number (say *z*). Then, *z ^{1/y}* = 2 is a rational number that is an irrational power of an irrational number.

Let's proof that p^{q} can be rational. If √2^{√2} is rational, we may directly get our conclusion since we can use irrational numbers p = q = √2. Otherwise, we can use p = √2^{√2} and q = √2. In this case p^{q} = (√2^{√2})^{√2} = √2^{2} = 2 is rational.

Let's proof that p^{q} can be irrational. If √2^{√2} is irrational, we may directly get our conclusion since we can use irrational numbers p = q = √2. Otherwise, we can use p = √2 and q = 1 + √2. In this case p^{q} = √2^{√2} ⋅ √2 is a product of a rational and and irrational number and must be irrational.

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Last revised August 2003