# Why does $\frac{|x|}{x^2}$ reduce to $\frac{1}{|x|}$?

This simplification confused me: $$.....=\frac{|x|}{x^2} = \frac{1}{|x|}$$

I get cancelling a degree of x, but why must you introduce the abs. val sign in the denominator? Is it because the left side is guarateed to be positive, so you must retain that in the final expression ?

• Yes that's why. Oct 14, 2016 at 12:34
• Another way of looking at it is that $x^2=|x|^2$. Then you just cancel one of the $|x|$ out.
– 5xum
Oct 14, 2016 at 12:40

If you want, you can look at is like this: $$\frac{|x|}{x^2} = \frac{|x|}{|x|^2} = \frac{1}{|x|}$$

• Of course $x\neq 0$, however; the op didn't say anyhing. Oct 14, 2016 at 12:38
• Ok, so I can't just "cancel" |x| against a bare x. I should really make them the same thing with your intermediate step, and then cancel identical |x|'s. Thanks! Oct 14, 2016 at 16:33
• No, you cancel $|x|$ with a "bare" $x$, not without knowing whether or not $x$ is negative. You're welcome! Oct 14, 2016 at 16:39
• Then why would I bother introducing the || on the bottom? I have cancelled a bare x, and there is one left. Your response makes no sense. Oct 15, 2016 at 20:53
• @JackOfAll My comment immediately before your most recent comment should have been (in response to your second comment) "No, you canceled |x| with a "bare" x, without knowing whether or not $x$ is negative. If $x \lt 0$, $\dfrac{|x|}{x \cdot x}$, and you simply canceled an x from top and bottom, you get $\frac{1}{x} \lt 0$. Oct 15, 2016 at 21:11

$$\frac{|x|}{x^2} = \frac{\sqrt{x^2}}{x^2}= \frac{1}{\sqrt{x^2}} = \frac{1}{|x|}$$

$\left|x\right|=x$ if $x>0$ and $\left|x\right|=-x$ if $x<0$

if $x>0$ we will have $\frac{x}{x^2}=\frac{1}{x}$

if $x<0$ we will have $\frac{-x}{x^2}=-\frac{1}{x}$

So, $\frac{\left|x\right|}{x^2}=\frac{1}{\left|x\right|}$.

It is because $$|x^2|=|x|^2.$$

Because $\forall x\in\Bbb{R},\, x^2=|x|^2$.

Indeed if $x\geq 0$ one has $|x|=x$ and the result follows immediately.

Now if $x\lt 0$ one has $|x|=-x$ and $|x|^2=(-x)(-x)=x^2$