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?

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
