Question

Find all \(x\) for which
\(\left| x - \left| x-1 \right| \right| = \lfloor x \rfloor\)

The answer is suppose to be in a range, but when I put in that it can be all real numbers, that is wrong.

Answer 1

x=1 and x= 1/3

Answer 2

I would try using the definition of the absolute value function to rewrite the LHS into something that you can easily match up with the RHS.
\[|x|=\begin{cases}x&\text{for }x>0\\0&\text{for }x=0\\-x&\text{for }x<0\end{cases}\]
This means
\[\begin{align*}|x-|x-1||&=\begin{cases}x-|x-1|&\text{for }x-|x-1|>0\\0&\text{for }x-|x-1|=0\\-x+|x-1|&\text{for }x-|x-1|<0\end{cases}\\[2ex]

&=\begin{cases}x-|x-1|&\text{for }x>\dfrac{1}{2}\\[1ex]0&\text{for }x=\dfrac{1}{2}\\[1ex]-x+|x-1|&\text{for }x<\dfrac{1}{2}\end{cases}
\end{align*}\]
You also have
\[\begin{align*}
|x-1|&=\begin{cases}x-1&\text{for }x>1\\0&\text{for }x=1\\-x+1&\text{for }x<1\end{cases}
\end{align*}\]

Answer 3

Putting all this together, you get
\[\begin{align*}
|x-|x-1||&=
\begin{cases}
1&\text{for }x>1\\x&\text{for }x=1\\[1ex]2x-1&\text{for }\dfrac{1}{2}<x<1\\[1ex]0&\text{for }x=\dfrac{1}{2}\\[1ex]-2x+1&\text{for }x<\dfrac{1}{2}
\end{cases}
\end{align*}\]

Answer 4

Now, given that
\[\lfloor x\rfloor=\begin{cases}&\vdots\\
0&\text{for }0\le x<1\\
1&\text{for }1\le x<2
\\&\vdots\end{cases}\]
you have the isolated solution \(x=\dfrac{1}{2}\) (since both LHS and RHS are zero) and the range \([1,2)\).