Absolute Values and Floors

Question

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...

Answer 1

$x=1$ 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 $$x\in \{0.5\}\cup [1,2 [$$

Answer 2

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 $\{\frac 12\}\cup[1,2[$

Answer 3

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?