
x1 = 1  x
Hey forum, I'm dealing with this one
$\displaystyle x1 = 1  x$
I use the definition of the absolute value of an arbitrary argument which gives
$\displaystyle x1 = \{ x1 \ , \te{ \ if \ } x1 \geq 0 \ \Leftrightarrow \ x \geq 1 \ \ \ \ (1) \\ (x1) \ , \te{ \ if \ } x1 < 0 \ \Leftrightarrow \ x < 1 \ \ \ \ (2) \\$
We see that it is true for the second condition, because
$\displaystyle (x1) = 1x \\ x + 1 = 1 x \\ 1 x = 1 x$
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?

Re: x1 = 1  x
First case:
$\displaystyle x1=1x$
$\displaystyle 2x=2$
$\displaystyle x=1$
Second case:
$\displaystyle (x1)=1x$
$\displaystyle x+1=1x$
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 $\displaystyle x\leq{1}$

Re: x1 = 1  x

Re: x1 = 1  x
Another way of looking at it: z is defined as "z if [itex]z\ge 0[/itex], z if z< 0".
Obviously 1 x= (x 1) so we must have x 1< 0.