Prove that $|-x| = |x|$

Question

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?

Answer 1

$\left|-x\right| = \begin{cases} -x & \bbox[5px,border:2px solid #F0A]{-x> 0} \\ -(-x) & \bbox[5px,border:2px solid #F0A]{-x < 0} \\ 0 & \bbox[5px,border:2px solid #F0A]{-x = 0}\end{cases}$

Answer 2

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 $-a < -b$.

Answer 3

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?

Answer 4

Here's an even quicker way. Use the property $|ab|=|a||b|$ on $|-x|$.

Answer 5

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.

Answer 6

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