Proof of simplification of radicals in fraction I'm not Even sure I'm using the correct terminology here, but I'm helping out my high school daughter with her algebra and was presented with the following rule:
$\sqrt3/\sqrt2 = \sqrt{3/2}$
Accepting that this is true (which a calculator did demonstrate), I should be able to step through a proof using the process for simplifying radicals in the denominator.  So, given:
$\sqrt3/\sqrt2$
I then multiply by the value of the denominator divided by itself:
$\sqrt3/\sqrt2 * \sqrt2/\sqrt2$
to get:
$(\sqrt3 * \sqrt2)/2$
which equals:
$\sqrt6/2$
And that's where I get stuck.  What can I do to get to:
$\sqrt{3/2}$
Other than using a calculator of course.  :)
 A: From my perspective (this is probably not the only way to arrive at the property you're asking about):
$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$
(with $a,b\ge 0$) is a special case of
$$\left(\frac{a}{b}\right)^x=\frac{a^x}{b^x}$$
(or, equivalently, $(ab)^x=a^xb^x$), the Distributive Property of Exponentiation over Division (or Multiplication).
The Distributive Property of Exponentiation over Multiplication, for integer exponents, comes from associativity and commutativity of multiplication:
$$(ab)^n=\underset{n\text{ factors of }ab}{\underbrace{(ab)\cdots(ab)}}=\left(\underset{n\text{ factors of }a}{\underbrace{a\cdots a}}\right)\left(\underset{n\text{ factors of }b}{\underbrace{b\cdots b}}\right)=a^nb^n$$
The property extends naturally to non-integer exponents.

original answer:
$\frac{\sqrt{6}}{2}=\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}}=\sqrt{\frac{3}{2}}$
Is that what you were looking for?
A: While it's a simple consequence of the distributive law for exponents, one can view it from a more general perspective. Namely, both $\rm\ \sqrt{3/2}\ $ and $\rm\ \sqrt{3}/\sqrt{2}\ $ are roots of $\rm\ x^2 - 3/2  $. But this polynomial has a unique positive root since it is increasing from $\rm\: -3/2\ $ to $\: +\infty\ $ on $\ [0,\infty)\:.\ $ Therefore this uniqueness theorem implies that the two roots are equal. One can frequently apply analogous techniques to more much complicated expressions involving messy nested radicals - inferences far removed from the laws of exponents. Thus I point out once again what I have emphasized in many varied posts here: uniqueness theorems provide powerful tools for proving equalities.
