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. 🙂
Best Answer
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?