# Solution to problem 2

## Problem 2

Let p, q be two irrational numbers. Can pq be rational? Can pq be irrational?

### My Solution

Denote log23 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.

### A Solution Submitted by Alexey Radul

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 2y) is an irrational number (say z). Then, z1/y = 2 is a rational number that is an irrational power of an irrational number.

### A Solution Submitted by Frederick Lewis

Let's proof that pq 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 pq = (√2√2)√2 = √22 = 2 is rational.

Let's proof that pq 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 pq = √2√2 ⋅ √2 is a product of a rational and and irrational number and must be irrational.

Last revised August 2003