Solving $4\lfloor x \rfloor=x+\{x\}$. What is wrong in my solution? Can anyone tell? $\lfloor x \rfloor$ = Floor Function, and $\{x\}$ denotes fractional part function

Solve for $x$
$$4\lfloor x \rfloor= x + \{x\}$$

$ x = \lfloor x \rfloor + \{x\}$
$\implies x - \{x\} = \lfloor x \rfloor$
$\implies 4x - 4\{x\} = x + \{x\}$
$\implies 3x = 5\{x\}$
Note that ,
$\implies 0 \leq  \{x\}<1$
$\implies 0 \leq 5\{x\} < 5 \iff 0 \leq 3x <5 \iff\boxed{ 0 \leq x < \dfrac{5}{3}}$
But the answer is $ x = 0$
 A: You have correctly deduced that $\color{blue}{0\leq x<\frac 53}$. To get the solution, the given equation $4\lfloor x\rfloor=x+\{x\}$ must also be satisfied.
Note that $x=1$ does not satisfy the given equation. If $x>1$, then $\lfloor x\rfloor=1$ and hence $4=x+\{x\}<\frac 53+1=8/3$ (by the blue colored inequality) , which is not possible.
Hence it follows that $0\leq x<1$.
Since, $\lfloor x\rfloor=0$, it follows that $\color{red}{x+\{x\}=0}$ (due to the given equation).
The identity $x=\lfloor x\rfloor+\{x\}$ gives $x=\{x\}$, which along-with the red-colored equation gives: $x=0$.
A: How did you get from $$4\lfloor x \rfloor = x + \{x\}$$
to
$$x=\lfloor x\rfloor + \{x\}?$$
The two equations are not equivalent!

But OK, ignoring the very awkward way the solution is written, there is nothing really wrong with your solution, in the sense that you correctly prove that if $4\lfloor x \rfloor = x + \{x\}$, then $0\leq x<\frac53$. However, your mistake is then saying that every $x$ satisfying the equation $0\leq x < \frac53$  is also a solution to the original equation, and that is simply not true. For example, $x=\frac12$ does not satisfy the equation.
A: As mentioned elsewhere, your conclusion is correct, but inconclusive. It might be easier to see what is going on if we write everything in terms of $\lfloor x\rfloor$ and $\{x\}$:
$$
\begin{align}
4\lfloor x\rfloor
&= x+\{x\}\tag{1a}\\
&=\lfloor x\rfloor+2\{x\}\tag{1b}\\
\lfloor x\rfloor
&=\tfrac23\{x\}\tag{1c}\\
0&\le\lfloor x\rfloor\lt\tfrac23\tag{1d}\\
\lfloor x\rfloor&=0\tag{1e}\\
\{x\}&=0\tag{1f}\\
x&=0\tag{1g}
\end{align}
$$
Explanation:
$\text{(1a)}$: copied equation
$\text{(1b)}$: $x=\lfloor x\rfloor+\{x\}$
$\text{(1c)}$: solve $\text{(1a)}$ and $\text{(1b)}$ for $\lfloor x\rfloor$
$\text{(1d)}$: $\text{(1c)}$ and $0\le\{x\}\lt1$
$\text{(1e)}$: $\lfloor x\rfloor$ is an integer
$\text{(1f)}$: combine $\text{(1c)}$ and $\text{(1e)}$
$\text{(1g)}$: $x=\lfloor x\rfloor+\{x\}$
A: $0 \leq x <\frac 53 $ is correct. This implies that $[x]=0$ or $[x]+1$. In the first case  we get $x+\{x\}=0$ which implies $x=0$. Suppose $[x]=1$. Then we have $1 \leq x <\frac 53 $ from which  we get $4<\frac 5 3 +1$ which is wrong. Hence we must have $x=0$.
A: We have :
$$4 \lfloor x \rfloor = x + \{x\} = \lfloor x \rfloor + 2 \{x\}$$
then :
$$2 \{x\} = 3 \lfloor x \rfloor \in \mathbb{Z}$$
We deduce that :
$$\{x\} = 0 \text{ or } \{x\} = \dfrac{1}{2}$$

*

* If $\{x\} = 0$ then $3 \lfloor x \rfloor = 2 \{x\} = 0$. It means that $\lfloor x \rfloor = 0$ then $x = \lfloor x \rfloor + \{x\} = 0$.

* If $\{x\} = \dfrac{1}{2}$ then $3 \lfloor x \rfloor = 2 \{x\} = 1$.  It means that $\lfloor x \rfloor = \dfrac{1}{3}$ wich is impossible because $\lfloor x \rfloor \in \mathbb{Z}$.

The unique solution is $x = 0$.
