If $ f(x) = \left\{\frac{3x}{2}\right\}\;,$ Where $\{x\}=x-\lfloor x \rfloor$. Then real solution of $f(f(x))=0$ 
If $\displaystyle f(x) = \left\{\frac{3x}{2}\right\}\;,$ Where $\{x\}=x-\lfloor x \rfloor$. Then no. of solution of the equation
$f(f(x))=0$  and $f(f(f(x)))=0$ and $f(f(f(f(x))))=0$.

$\bf{My\; Try::}$ We can write  $\displaystyle f(x)= \left\{\frac{3x}{2}\right\}=\frac{3x}{2}-\lfloor \frac{3x}{2} \rfloor  = \left\{\begin{matrix}
\displaystyle \frac{3x}{2} \;,& 0\leq x<\displaystyle \frac{2}{3} \\\\ 
\displaystyle \frac{3x}{2}-1 \;,& \displaystyle \frac{2}{3}\leq x<\displaystyle 1 
 & 
\end{matrix}\right.$
Now for Solution of $f(f(x)) = 0.$ We have used Two cases.
$\bullet$ If $0\leq x<\displaystyle \frac{2}{3} .$ Then we use $\displaystyle f(x) = \frac{3x}{2}.$ So $\displaystyle f(f(x)) = 0\Rightarrow \frac{9x}{4}=0\Rightarrow x=0$..
$\bullet$ If $\displaystyle \frac{2}{3}\leq x<\displaystyle 1  .$ Then we use $\displaystyle f(x) = \frac{3x}{2}-1.$ So $\displaystyle f(f(x)) = 0\Rightarrow \frac{9x}{4}-\frac{5}{2}=0\Rightarrow x=\frac{10}{9}$.
So we get only one solution $x=0$ for $f(f(x))=0$
Is my solution is Right, If not , Then plz explain me how can i solve it.
 A: Note that $f(f((\frac23)^2))=0$, so $0$ is not the only solution.

$\bullet$ If $0\leq x<\displaystyle \frac{2}{3} .$ Then we use $\displaystyle f(x) = \frac{3x}{2}.$ So $\displaystyle f(f(x)) = 0\Rightarrow \frac{9x}{4}=0\Rightarrow x=0$..

The mistake you are making is that even when $0\le x\le \frac23$, $f(x)$ may not be in the same interval. Such is the case with $(\frac23)^2$, it is strictly between $0$ and $\frac23$, but $f((\frac23)^2)$ is $\frac23$, which is not.
Can you see why this does not yield every solution?

Here is how I found all the solutions to $f(f(x))=0$.
$f(f(x))=0$ if and only if $\frac32f(x)$ is an integer. This is the case if and only if $f(x)$ is some integer multiple of $\frac23$. $f$ can only return values in $[0,1)$, however, so $f(x)$ must be either $0$ or $\frac23$.
$f(x)=0$ if and only if $\frac32x$ is an integer:
$$\frac32x=n\implies x=\frac23n$$
And so, $x=\frac23n$ is a solution to $f(f(x))=0$ for every integer $n$.
$f(x)=\frac23$ if and only if $\frac32x$ is $n+\frac23$ for some integer $n$:
$$\frac32x=n+\frac23\implies x=\frac23n+\left(\frac23\right)^2$$
And so, $x=\frac23n+\left(\frac23\right)^2$ is a solution to $f(f(x))=0$ for every integer $n.$
In conclusion, the solutions to $f(f(x))=0$ are all numbers of the forms  $\frac23n+\left(\frac23\right)^2$ and $\frac23n$ for integer $n$.
