Proving the exponential laws for rational exponents I have managed to prove the usual exponent laws for integer exponents, using induction on the natural numbers.  I am now trying to prove that the laws also hold for rational exponents.  I'm starting by trying to prove the law $x^nx^m = x^{n+m}$, where $n = \frac a b$ and $m = \frac c d$.  
Using the following definition:
$x = y^{\frac a b} \iff x^b = y^a$
I can prove it as follows:
$x^{\frac a b}x^{\frac c d} = x^{\frac {ad} {bd}}x^{\frac {bc} {bd}} = (x^{ad})^{\frac 1 {bd}}(x^{bc})^{\frac 1 {bd}} = (x^{ad}x^{bc})^{\frac 1 {bd}} = (x^{ad+bc})^{\frac 1 {bd}} = x^{\frac {ad + bc} {bd}} = x^{{\frac a b}+{\frac c d}}$
Aside from the definition, I have proved no other things about rational exponents.  So, the problem I have is that I use the following assumptions in this proof:


*

*$x^{\frac a b} = (x^a)^{\frac 1 b}$

*$(xy)^{\frac a b} = x^{\frac a b}y^{\frac a b}$


Both assumptions I need to prove but I cannot figure out how.  Can anyone help?
Edit:  I now realize the proof of (1.) is trivial: $y = (x^a)^{\frac 1 b} \iff y^b = (x^a)^1 = x^a \iff y = x^{\frac a b}$.  For (2.) as far as I get is proving that $(xy)^{\frac a b} = (x^ay^a)^{\frac 1 b}$
 A: Let $x, y$ be not negative, $a=1$ and $b$ odd.
$$y=x^{\frac{1}{b}} \overset{def.}\Leftrightarrow y^b=x \Leftrightarrow y= \sqrt[b]{x}.$$
Now let $x, y$ not negative, $a$ be arbitrary and $b$ odd.
$$y=x^{\frac{a}{b}} \overset{def.}\Leftrightarrow y^b = x^a \Leftrightarrow \sqrt[b]{y^b} = \sqrt[b]{x^a} \Leftrightarrow y = \sqrt[b]{x^a} $$
Since $(\sqrt[b]{x^a})^b = x^a$ and by your proof for integer exponents $((\sqrt[b]{x})^a)^b = (\sqrt[b]{x})^{ab} = x^a$, we have $(\sqrt[b]{x^a})=(\sqrt[b]{x})^a$.
Now let $x, y$ be not negative and $b, d$ odd.
$$x^{\frac{a}{b}}x^{\frac{c}{d}}=x^{\frac{ad}{bd}}x^{\frac{bc}{bd}} = (\sqrt[bd]{x})^{ad} (\sqrt[bd]{x})^{bc} \overset{\text{form the integer result}}=(\sqrt[bd]{x})^{ad+bc}= x^{\frac{ad+bc}{bd}}.$$
For even denominators you basically do the same, but pay attention because even roots are not unique ($x^2 = 2 \Rightarrow x= \sqrt{2}$ or $- \sqrt{2}$).
Also note that you have to place proper restrictions on the domain of $x$ and $y$, as for example you might not want to have $\sqrt{-1}$.
