- #1

QuietMind

- 42

- 2

## Homework Statement

True or false and why: If a and b are irrational, then ##a^b## is irrational.

## Homework Equations

None, but the relevant example provided in the text is the proof of irrationality of ##\sqrt{2}##

## The Attempt at a Solution

Attempt proof by contradiction. Say ##a^b## is rational, then we can express

##a^b = \frac{c}{d}##

for some integer c and d. Without loss of generality, we can choose c and d to express the fractions in lowest terms.

Then, ##a^b d = c## which implies that ##a^b## must be an integer. This seems to me like a suspicious statement. The problem reduces to whether or not we can have ##a^b = c## where c is an integer. How would I show this possible/ impossible, if indeed this is the correct path to a solution?