# Thread: |x-1| = 1 - x

1. ## |x-1| = 1 - x

Hey forum, I'm dealing with this one

$\displaystyle |x-1| = 1 - x$

I use the definition of the absolute value of an arbitrary argument which gives

$\displaystyle |x-1| = \{ x-1 \ , \te{ \ if \ } x-1 \geq 0 \ \Leftrightarrow \ x \geq 1 \ \ \ \ (1) \\ -(x-1) \ , \te{ \ if \ } x-1 < 0 \ \Leftrightarrow \ x < 1 \ \ \ \ (2) \\$

We see that it is true for the second condition, because

$\displaystyle -(x-1) = 1-x \\ -x + 1 = 1 -x \\ 1 -x = 1- x$

Therefore, the answer must be that x must be smaller than 1. But the equation is also true for x=1 ! What's up with that?

2. ## Re: |x-1| = 1 - x

First case:

$\displaystyle x-1=1-x$

$\displaystyle 2x=2$

$\displaystyle x=1$

Second case:

$\displaystyle -(x-1)=1-x$

$\displaystyle -x+1=1-x$

So we get $\displaystyle x<{1}$ from case 2, and $\displaystyle x=1$ from case 1. Put them together to give the whole range of solutions and you get that $\displaystyle x\leq{1}$

Thanks!!!

4. ## Re: |x-1| = 1 - x

Another way of looking at it: |z| is defined as "z if $z\ge 0$, -z if z< 0".
Obviously 1- x= -(x- 1) so we must have x- 1< 0.