# Irrationality

• Feb 16th 2010, 04:40 PM
Gusbob
Irrationality
Question:
For irrational numbers x,y show that x^y is not necessarily irrational.

How would I go about this question? I thought about proof by contradiction but don't think it will work because x^y could also be irrational. Do I just find a counter-example?
• Feb 16th 2010, 04:48 PM
icemanfan
Quote:

Originally Posted by Gusbob
Do I just find a counter-example?

This seems to be a convenient way to go about answering the question.

Solving $x^{\sqrt{2}} = 2$ for x:

$x = 2^{\frac{1}{\sqrt{2}}} = 2^{\frac{\sqrt{2}}{2}} = \sqrt{2^{\sqrt{2}}}$

Now you just have to show that $\sqrt{2^{\sqrt{2}}}$ is irrational.
• Feb 16th 2010, 05:27 PM
Drexel28
Quote:

Originally Posted by Gusbob
Question:
For irrational numbers x,y show that x^y is not necessarily irrational.

How would I go about this question? I thought about proof by contradiction but don't think it will work because x^y could also be irrational. Do I just find a counter-example?

I'm not entirely sure what you're entitle to assume. Easies example $2=e^{\ln(2)}$
• Feb 16th 2010, 07:06 PM
Bacterius
Statement : For irrational numbers x,y show that x^y is not necessarily irrational.

This statement is equivalent to disproving the statement : For irrational numbers x,y show that x^y is necessarily irrational.

Therefore, one counter-example (such as the one given by Drexel) is enough to disprove the latter, and thus to prove the original statement.