Solving $\lfloor {2x} \rfloor + \lceil {x} \rceil - 4x = 0$ 
I need to find:
$A=\{x \in \mathbb{R}\vert\,\lfloor {2x} \rfloor + \lceil {x} \rceil - 4x = 0\}$

Let $x \in \mathbb{R}$. Then
$$\begin{align*}
\lceil {x} \rceil =4x- \lfloor {2x} \rfloor
&\iff
4x- \lfloor {2x} \rfloor -1       <     x         \leq        4x -\lfloor {2x} \rfloor\\
&\iff
\bigg[ 4x- \lfloor {2x} \rfloor -1      <      x   \bigg]       \land      \bigg[ x       \leq      4x -\lfloor {2x} \rfloor  \bigg]\\
&\iff 
(3x-1      <       \lfloor {2x} \rfloor ) \land (\lfloor {2x} \rfloor       \leq      3x)
\end{align*}
$$
Now define $B=\{x\in \mathbb R\mid(3x-1<\lfloor {2x} \rfloor \}$ and $C=\{x\in\mathbb R\mid\lfloor {2x} \rfloor       \leq      3x\}$. Then $A=B\cap C$.

My confusion:
$-\dfrac{2}{3} \in (B \cap C)$ but $-\dfrac{2}{3} \notin A$
In which step have I gone wrong?


The first step missed the condition that $\lceil x\rceil$ is an integer
The first step in your equivalence transformation as quoted below is wrong.

$$
\lceil x\rceil =4x- \lfloor 2x\rfloor \iff
4x- \lfloor 2x\rfloor -1<x\le 4x- \lfloor 2x\rfloor
$$

The definition of $\lceil x \rceil$ is "the least integer that is not smaller than $x$". Or, what is equivalent, "the integer that is $\ge x$ and $<x+1$". Hence, the first transformation should have been
$$
\lceil x\rceil =4x- \lfloor 2x\rfloor \\
\iff 4x- \lfloor 2x\rfloor \text{ is an integer and }
x\le 4x- \lfloor 2x\rfloor < x +1
$$
A full solution
$\lceil x\rceil =4x- \lfloor 2x\rfloor$ implies $4x$ is an integer. There are three cases.

*

*$x$ is an integer.
Then $x = 4x-2x$.
$x=0$

*$x$ is not an integer but $2x$ is an integer.
Then $\lceil x\rceil = x+\frac12$.
$x +\frac12 = 4x-2x$.
$x=\frac12$.

*$2x$ is not an integer but $4x$ is an integer, i.e., $\lfloor 2x\rfloor = 2x-\frac12$. There are two subcases.

*

*$x$ is $\frac14$ plus an integer, i.e., $\lceil x\rceil = x+\frac34$.
$x+\frac34=4x-(2x-\frac12)$
$x=\frac14$

*$x$ is $\frac34$ plus an integer, i.e., $\lceil x\rceil = x+\frac14$.
$x+\frac14=4x-(2x-\frac12)$
$x=-\frac14$
Hence $x\in\{-\frac14, 0, \frac14, \frac12\}$.
