- #1
QuietMind
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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?