How to solve the equation $3x-4\lfloor x\rfloor=0$ for $x\in\mathbb{R}$? As a homework, I was asked to solve this equation, $$(3x-4\lfloor x\rfloor=0),x\in \Bbb R$$ For $x\in \Bbb Z:\lfloor x\rfloor=x \implies x=0$ But for $x\not\in\Bbb Z : \lfloor x\rfloor=\frac 34x$ So now we know that $\frac 34x\in\Bbb Z$ and $x\in\Bbb R-\Bbb Z$, so maybe ? define a function such that : $$f:\begin{Bmatrix}(\Bbb R-\Bbb Z)\to \Bbb Z \\ x\mapsto \frac34x\end{Bmatrix}$$ Even if trying this did walk me into $x=\frac43$ I'm still left with no rigorous proof (An explanation is would be nice if possible). Any help would be appreciated. Thank you for your time.
 A: Let $x=n+\epsilon$, where $n\in\mathbb{Z}$ and $0\leq\epsilon<1$
Then the equation becomes $$3n+3\epsilon=4n\implies 3\epsilon=n$$
Therefore, $0\leq\frac n3<1\implies n=0,1,2$ and correspondingly, $\epsilon=0, \frac 13,\frac 23$
Therefore the solutions are $$x=0,\frac 43,\frac 83$$
A: rewrite this as $4\lfloor x \rfloor =3x$, so that $x=\frac{n}{3}$ for some integer $n$.
We now have $4\lfloor\frac{n}{3}\rfloor=n$
A: Hint:
You can rewrite $3x=4\lfloor x \rfloor$ as $3\left(\lfloor x \rfloor+\{x\}\right)=4\lfloor x \rfloor$, so $\lfloor x \rfloor=3\{x\}$.
Since $0\le \{x\}<1,\; 0\le \lfloor x \rfloor<3$  and therefore $\lfloor x \rfloor\in \{0, 1, 2\}$
A: $$3x=4\lfloor x\rfloor$$
Suppose that, for some integer, $n$, we know that $n \le x \lt n+1$. Then 
$\lfloor x\rfloor = n$ and 
$$x = \dfrac 43n$$ 
But we still need to make sure that 
\begin{array}{cccC}
   n &\le &x &\lt &n+1 \\
   n &\le &\dfrac 43n &\lt &n+1 \\
   3n &\le &4n &\lt &3n+3 \\
   0 &\le &n &\lt &3 \\
\end{array}
So $n \in \{0,1,2\}$; that is, $x \in \left\{ 0,\dfrac 43,\dfrac 83 \right\}$
