Why is $\sqrt{x/x^{-1}}$ OR $\sqrt{x/{1/x}}$ = $\lvert x\rvert$ and not just x

Question

I have this task: Find equal expression to square root of fraction of x and its inverted value (this is translated from my mother tongue so I'm sorry if I've used incorrect terms). Anyway the starting point is clear reduce this:

$\sqrt{\frac{x}{x^{-1}}}$

The correct answer is $\lvert x\rvert$ and I don't know how to get there. It seems I am missing some correct thinking at the very last step.

So this is what I do (even the slightest mindsteps included):

$\sqrt{\frac{x}{x^{-1}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{\frac{\frac{x}{1}}{\frac{1}{x}}} = \frac{\sqrt{\frac{x}{1}}}{\sqrt{\frac{1}{x}}} = \frac{\frac{\sqrt{x}}{\sqrt{1}}}{\frac{\sqrt{1}}{\sqrt{x}}} = \frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sqrt{x}}} = {\frac{\sqrt{x}}{1}}\cdot{\frac{\sqrt{x}}{1}} = \sqrt{x}\cdot\sqrt{x} = (\sqrt{x})^2 $ which I believe $=x$

OR

$\sqrt{\frac{x}{x^{-1}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{\frac{\frac{x}{1}}{\frac{1}{x}}} = \frac{\sqrt{\frac{x}{1}}}{\sqrt{\frac{1}{x}}} = \frac{\frac{x^\frac{1}{2}}{1}}{\frac{1}{x^\frac{1}{2}}} = {\frac{x^\frac{1}{2}}{1}}\cdot{\frac{x^\frac{1}{2}}{1}} = x^\frac{1}{2}\cdot x^\frac{1}{2} = x^{\frac{1}{2}+\frac{1}{2}} = x^1 = x $

If this is correct then how $(\sqrt{x})^2 \neq x$ and instead$(\sqrt{x})^2 =\lvert x\rvert$ ?

EDIT: so the answer is that this split $\sqrt{\frac{\frac{x}{1}}{\frac{1}{x}}} = \frac{\sqrt{\frac{x}{1}}}{\sqrt{\frac{1}{x}}}$ is forbidden since it is valid for positive x only. And x is unknown. What I should have done is this less fancy but correct $\sqrt{\frac{\frac{x}{1}}{\frac{1}{x}}} = \sqrt{{\frac{x}{1}}\cdot{\frac{x}{1}}} = \sqrt{{x}\cdot{x}} = \sqrt{x^2}$ which indeed is $\lvert x\rvert$

Answer 1

The point is $\sqrt{x^2} = |x|$ because the square root is always positive. So if $x < 0$ then $\sqrt{x^2} = -x$ (for example $\sqrt{(-2)^2} = \sqrt{4} = 2 = -(-2)$), i.e. $$ \sqrt{x^2} = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} $$ which is exactly definition of $|x|$.

In your case $\sqrt{x/x^{-1}} = \sqrt{x/(1/x)} = \sqrt{x^2} = |x|$.

Answer 2

Notice that you can do these properties only if they $(a,b)$ are positive.

$$\sqrt { \frac { a }{ b } } =\frac { \sqrt { a } }{ \sqrt { b } } ,\quad \sqrt { ab } =\sqrt { a } \sqrt { b } $$

Answer 3

Functions are usually defined to have only one "out-value". Sometimes when defining a function (for example for the purpose of expressing the solutions of an equation) one has to pick a branch. This basically means to pick one of the solutions to the equation as out-value for the function.

For example the equation

$$x^2 = k$$

When solving this equation, we write $$x = \pm \sqrt{k}$$

Where $\sqrt{\cdot}$ is usually defined to be a branch so that is the positive value which squared equals $k$.

If we choose that branch, then $\sqrt{.}$ will always be positive.