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 forx=1! What's up with that?