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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 $\lvert x\rvert$

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    $\begingroup$ You have to be careful when "splitting" $\sqrt{x\cdot x}$ into $\sqrt{x}\sqrt{x}$ since the latter is only well defined for $x\geq 0$ (and in this case $x=|x|$) $\endgroup$
    – Christian
    Commented Jun 11, 2016 at 18:04
  • $\begingroup$ $\sqrt {w/v}=\sqrt {w}/\sqrt {v} $ **ONLY** if you know w and v are positive. You don't know this so you can't say this. But you do know. $\sqrt {x/(1/x)}=\sqrt {x^2}$. And $\sqrt {x^2}=|x|$. If x < 0 then |x| is not x. $\endgroup$
    – fleablood
    Commented Jun 11, 2016 at 18:13
  • $\begingroup$ Go through this with x = -4 and see what happens. $\endgroup$
    – fleablood
    Commented Jun 11, 2016 at 18:14
  • $\begingroup$ Oh! Now I get it @fleablood. Forbidden operation. Thanks. $\endgroup$
    – Smejki
    Commented Jun 11, 2016 at 20:45
  • $\begingroup$ Not so much forbidden as mistaken assumptions. If x is positive, root (x)^2 = root (x^2)=x. no harm no foul. If x is negative root (x^2) = -x = (if root (-1)=i is allowed) root (x)^2. But need to realize neg value of x is possible. $\endgroup$
    – fleablood
    Commented Jun 11, 2016 at 21:09

3 Answers 3

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

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  • $\begingroup$ I lived with the impression that there is great difference between $(\sqrt{x})^2$ and $\sqrt{x^2}$ while the latter turns any x to positive, the former is undefined for negative x. And $\sqrt{x}\cdot\sqrt{x} = (\sqrt{x})^2$ not $\sqrt{x^2}$ Am I mistaken here? $\endgroup$
    – Smejki
    Commented Jun 11, 2016 at 19:05
  • $\begingroup$ Yes, you are right - $\sqrt{x^2}$ and $(\sqrt{x})^2$ are not the same functions because first one is defined for all $x\in\mathbb{R}$ and the second one is defined only for $x\ge 0$. Moreover, if we consider $\sqrt{x}$ as $\mathbb{R}\to\mathbb{C}$ function, it will be also defined for negative $x$ as $\sqrt{x} = i\sqrt{-x}$ where $i$ is imaginary unit. And then $(\sqrt{x})^2 = x$ for all $x\in\mathbb{R}$, which is still not the same to $|x| = \sqrt{x^2}$. $\endgroup$
    – Anton Grudkin
    Commented Jun 11, 2016 at 19:18
  • $\begingroup$ with the help of you and fleablood I realized where I made the mistake. Thanks $\endgroup$
    – Smejki
    Commented Jun 11, 2016 at 20:55
  • $\begingroup$ For pos x, root (x squared) = root (x squared = x but if x is neg root x is undefined and x squared is positive but root (x squared) is also pos so =|x|. Just have to be careful. If complex numbers are allowed then root (-1) =i is acceptable but the "splittin" doesn't work. For x <0 root (x) = i root (|x|) But root (ab) needn't equal root(a)root (b). It does in absolute value terms but the signs might have different results. $\endgroup$
    – fleablood
    Commented Jun 11, 2016 at 21:17
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Notice that you can do these properties only if they $(a,b)$ are positive.

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

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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.

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