Proving or Disproving rational raised to rational is rational number

1. Oct 20, 2005

jhson114

Im trying to either prove or disprove that if a and b are rational numbers, then a^b is also rational. I tried doing it with a contradiction, but i cant seem to correctly arrive at a solution. this is how i started the problem

defn of rational number: a,b = {m/n: m,n are all nonzero integers}
1. a^b is irrational (hypothesis/assumption)
2. b^b is irrational (from 1)
3. (m/n)^(m/n) (from defn. of rational number)
4. [m^(m/n)]/[(n^(m/n)] (algebra)

i'm stuck right here. i need to prove that an integer raised to a rational number is either rational or irrational. any inputs will be really helpful. thank you

2. Oct 20, 2005

devious_

Use a counter-example. Have you considered $$\sqrt{2}$$?

3. Oct 20, 2005

jhson114

i dont understand what you mean by "consider square-root of two". can you be more specific?

4. Oct 20, 2005

Tide

And then there is $1^1$.

5. Oct 20, 2005

Tide

He means $2^{1/2}$.

6. Oct 20, 2005

jhson114

oh i see. 2^(1/2) is irrational number, which disproves the above statement. however like Tide said, 1^1 is a rational number. but since there's a one statement that made it false, it makes the entire statement false, right?

7. Oct 20, 2005

Tide

All it takes is a single counterexample to disprove a theorem!

8. Oct 20, 2005

jhson114

thank you very much for all your help Tide and devious. :)