Absolute Values and Floors Problem:

Find all $x$ for which
  $$\left| x - \left| x-1 \right| \right| = \lfloor x \rfloor.$$Express your answer in interval notation.

How would I do this? There are lots of absolute value signs and a floor sign that I can't handle...
 A: If $x\ge 1$ then $|x-|x-1||=|x-x+1|=|1|=1=\lfloor x\rfloor\iff x\in[1,2[$
If $x<1$ then $|x-|x-1||=|x+x-1|=|2x-1|=\lfloor x\rfloor$


*

*If $0\le x<1$ then $|2x-1|=\lfloor x\rfloor=0\iff x=\frac 12$ 

*If $x<0$ then $|2x-1|>0$ and $\lfloor x\rfloor<0$ so no solutions



Finally the solutions are $\{\frac 12\}\cup[1,2[$

A: $x=1$ is a solution.
if $x>1$ it becomes
$$1=\lfloor x \rfloor $$
whose solution is $1 <x <2$.
if $x <1$, it is
$$|2x-1|=\lfloor x \rfloor $$
if $x\geq 0,5$ we get
$$2x-1=0$$
or $$x=0,5 $$
now if $x \leq 0,5$, it becomes
$$1-2x =\lfloor x \rfloor$$ and because
$$A-1 <\lfloor A \rfloor  \leq A ,$$
$$x-1 <1-2x\leq x $$ or
$$\frac {1}{3}\leq x <\frac {2}{3}$$
and $\lfloor x \rfloor =0\implies x=0,5$.
finally, we should have
$$x\in \{0.5\}\cup [1,2 [$$
A: Zero is not a solution.
There can be no negative solutions because if $x<0$ then
$$ \vert x - |x-1|\vert=\vert x-(1-x) \vert= \vert 2x-1\vert=1-2x>0$$0$
and for $x<0$, $1-2x>\lfloor x\rfloor$.
Thus the only solutions would have to be positive.
$x=1$ is a solution.
If $x>1$ then $\vert x - |x-1|\vert=1$ so the only solutions for $x>1$ lie on the interval $(1,2)$.
So the only remaining question is whether there are any solutions on the interval $(0,1)$
$x=\frac{1}{2}$ is clearly a solution, so you should look for solutions on the intervals $\left(0,\frac{1}{2}\right)$ and $\left(\frac{1}{2},1\right)$.
Can you take it from here?
