Hey fellas. I'm stuck on the following question. I'm quite ashamed to say I've spent about an hour and half on it, and my paper is virtually blank.

Question:

If $\displaystyle z$ is a real number, then $\displaystyle z^n$ is defined recursively for non-negative integers n: $\displaystyle z^0 = 1$, and for $\displaystyle n \ge 0, z^{n+1}=z . z^n $. Also $\displaystyle z^{-n} = \frac{1}{z^n} $, for any natural number n.

If $\displaystyle z$ is a positive real number, and $\displaystyle b$ a non-zero integer, then we define the positive $\displaystyle b^{th}$ root $\displaystyle y=z^{\frac{1}{b}} $ to be the positive real number $\displaystyle y $ such that $\displaystyle y^b = z$.

1)Is it possible for a positive real number to have two different $\displaystyle b^{th} $ roots?

For integers a,b, where b doesn't equal 0, and a real number $\displaystyle z>0$, we define $\displaystyle z^{\frac{a}{b}} = (z^a)^{\frac{1}{b}}$

2)If a,b,c,d are integers and b,d do not equal 0, $\displaystyle ad=bc$, and $\displaystyle z>0$, use these definitions to prove that $\displaystyle z^{\frac{a}{b}} = z^{\frac{c}{d}} $

Please help!!