1. ## 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?

2. 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 $\displaystyle x^{\sqrt{2}} = 2$ for x:

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

Now you just have to show that $\displaystyle \sqrt{2^{\sqrt{2}}}$ is irrational.

3. 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 $\displaystyle 2=e^{\ln(2)}$

4. 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.