Prove that $|-x| = |x|$ Using only the definition of Absolute Value:
$\left|x\right| =  \begin{cases} x & x> 0 \\
-x & x < 0 \\
0 & x = 0,\end{cases}$
Prove that $|-x| = |x|.$
This seems so simple, but I keep getting hung up. I use the definition insert $-x$ into the definition, but I end up with:
$\left|-x\right| =  \begin{cases} -x & x> 0 \\
-(-x) & x < 0 \\
0 & x = 0\end{cases}$
which doesn't make sense to me. It certainly doesn't equal $|x|,$ does it?
I would use $|x| = \sqrt{x^2}$ but I am supposed to prove that identity later in the problem set. What am I doing wrong?
 A: Rather, you should have $$\lvert-x\rvert=\begin{cases}-x & -x>0\\-(-x) & -x<0\\0 & -x=0.\end{cases}$$ Can you take it from there?
A: Here's an even quicker way.  Use the property $|ab|=|a||b|$ on $|-x|$.
A: Good question! You need to flip the $>$ and $<$ signs in the definition.
It's for the same reason that you need to flip the sign when you multiply both sides of $a > b$ by $-1$, getting $-a < -b$.
A: $\left|-x\right| =  \begin{cases} -x & \bbox[5px,border:2px solid #F0A]{-x> 0} \\
-(-x) & \bbox[5px,border:2px solid #F0A]{-x < 0} \\
0 & \bbox[5px,border:2px solid #F0A]{-x = 0}\end{cases}$
A: Proof by cases:
$1)$ $x \gt 0$, and so $-x \lt 0$
$|x|=x, |-x|=-(-x)=x$
$2)$$x=0$ (clear)
$3)$ $x \lt 0$, and so $-x \gt 0$
then $|x|=-(x)$
$|-x|=-(x)$
QED
A: Your substitutions show that $|-x|=-|x|$. which is false.
The error comes from the fact that when introducing 
the minus sign you did not swap the sense of inequalities
once swapped
$|-x|= -x\quad  (x<0)$
$|-x| = x\quad  (x>0)$
$|-x| = 0\quad  (x=0)$
you see that $|-x|$ definition matches that of $|x|$ again.
A: You can just do case analysis. Check the cases of when $x>0$, $x=0$ and $x<0$
