Why is the square root of $x$ divided by $x$ equal to $1$ divided by the square root of $x$ I do not understand why this is the case
$$
\frac{\sqrt{3}}{3} = 0.57735\dots
$$
and
$$
\frac{1}{\sqrt{3}} = 0.57735\dots
$$
MIND = BLOWN 
 A: Well, $$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3}\sqrt{3}} =\frac{\sqrt{3}}{\sqrt{3}}\frac{1}{\sqrt{3}} =\frac{1}{\sqrt{3}}.$$
It's really just the fact that $$\frac{a}{a^2} =\frac{1}{a}.$$
A: Write $3 = (3^{1/2}\cdot 3^{1/2})$ and then simplify your fraction. :)
A: Let's consider a general case, for some integer $x$. (In fact, it's true for any real number $x$.) We have
$$ x^{1/2} / x = x^{1/2} \cdot x^{-1} = x^{-1/2}. $$
A: All the algebra above is correct, though if you haven't learned exponent rules yet that might not be intuitive. Here's how I think about it; Consider the fraction $\dfrac{x^a}{x^b}$. You have $a$ $x$'s on the top and you have $b$ $x$'s on the bottom. Division means "taking away", so we take away $b$ $x$'s from the top, which used to have $a$ of them. So now the top has $a-b$ $x$'s and the bottom is now 1, so
$$\dfrac{x^a}{x^b} = x^{a-b}$$
Notice how I never actually said $a$ and $b$ were whole numbers. This logic still applies if they are fractions or any real number. So $\dfrac{\sqrt{x}}{x} = x^{1/2 - 1} = x^{-1/2} = \dfrac{1}{\sqrt{x}}$.
A: $$\frac {\sqrt {x}}x=\frac 1{\sqrt x}$$
since $$\frac ab=\frac cd \iff ad=bc$$
with $bd \neq0$

In general 
$$\frac {\sqrt[a] {x^b}}x=\frac 1{\sqrt[a] {x^{a-b}}}$$
$($for $x>0)$
