# absolute value

• Feb 23rd 2009, 08:18 PM
dori1123
absolute value
How can I show that $x+x|y|-y-|x|y =0 \implies x=y$?

$|x| = \{ x$ if $x \geq 0$ and $-x$ if $x < 0\}$, right?

So I break this into 4 cases,
if $x,y \geq 0$, then $x+xy-y-xy = 0 \implies x=y$
if $x,y < 0$, then $(-x)+(-x)(-y)-(-y)-(-x)(-y) = -x+xy+y-xy = 0 \implies x=y$
I have trouble with cases $x \geq 0$ and $y <0$, and $x<0$ and $y \geq 0$, some help please.
• Feb 24th 2009, 08:12 AM
red_dog
If $x\geq 0, \ y\leq 0$, then we have $x-2xy-y=0$

But: $x\geq 0, \ -2xy\geq 0, \ -y\geq 0$

A sum of positive numbers is 0 if all the numbers are 0. Then $x=y=0$

In the same way you can proove if $x\leq 0, \ y\geq 0$