Prove that $ \left\lfloor{\frac xn}\right\rfloor= \left\lfloor{\lfloor{x}\rfloor\over n}\right\rfloor$ where $n \ge 1, n \in \mathbb{N}$ 
Prove that $ \left\lfloor{\frac xn}\right\rfloor= \left\lfloor{\lfloor{x}\rfloor\over n}\right\rfloor$ where $n \ge 1, n \in \mathbb{N}$ and $\lfloor{.}\rfloor$ represents Greatest Integer $\mathbf{\le x}$ or floor function 

I tried to prove it by writing $x = \lfloor{x}\rfloor + \{x\} $ where $ \{.\}$ represents Fractional Part function and $ 0 \le \{x\} < 1$
So we get,
$ \lfloor{\frac xn}\rfloor= \lfloor{{\lfloor x\rfloor\over n}+ {\{x\}\over n}}\rfloor \tag{1}$
Then I tried to use the property,
$\lfloor{x+y}\rfloor =\begin{cases}
\lfloor x\rfloor + \lfloor y\rfloor& \text{if $0\le \{x\} + \{y\}$} < 1 \tag{2}\\
1+ \lfloor x\rfloor + \lfloor y\rfloor & \text{if $1\le \{x\} + \{y\}$} < 2 \\
\end{cases} $
So if I can prove $(1)$ = first case of $(2) $
I’ll have ,
$ \lfloor{\frac xn}\rfloor= \lfloor{{\lfloor x\rfloor\over n}}\rfloor+ \lfloor{\{x\}\over n}\rfloor = \lfloor{\lfloor{x}\rfloor\over n}\rfloor$ as the second term will come out to be zero. However, I am unable to prove this.

Can someone help me out with this proof by showing me how $\mathbf(1)$= first case of $\mathbf (2)$ and proving the question using this method and also giving a clear proof using a simpler method

 A: Simpler method, using only that :


if $m$ is an integer and $y$ is a real then $y \ge m \Longleftrightarrow \lfloor y \rfloor \ge m$.


\begin{align*}
\big\lfloor \frac{x}{n}\big\rfloor &=\max\big\{k\in \mathbb{Z}\ |\ \frac{x}{n} \ge k\big\}=\max \big\{k\in\mathbb{Z}\ |\ x \ge nk\big\}\\
&=\max \big\{k\in\mathbb{Z}\ |\ \lfloor x\rfloor \ge nk\big\}=\max\big\{k\in\mathbb{Z}\ |\ \frac{\lfloor x\rfloor}{n}\ge k\big\} = \big\lfloor\frac{\lfloor x\rfloor}{n}\big\rfloor.
\end{align*}
A: The OP's proof method is a good one.  It remains to calculate $$\left\{\frac{\lfloor x \rfloor}{n}\right\}$$
and
$$\left\{\frac{\{x \}}{n}\right\}$$
Note that the first one is at most $\frac{n-1}{n}$, and the second is strictly less than $\frac{1}{n}$, hence their sum is strictly less than $1$.

More details, as requested:
Since $\lfloor x\rfloor$ is an integer, write it as $qn+r$, where $q,r$ are integers and $0\le r\le n-1$.  This is the division algorithm.  Now $\frac{\lfloor x \rfloor}{n}=\frac{qn+r}{n}=q+\frac{r}{n}$.  This has fractional part $\frac{r}{n}$, which is at most $\frac{n-1}{n}$.
Now, $0\le \{x\}<1$, so $0\le \frac{\{x\}}{n}<\frac{1}{n}$.  It is its own fractional part (since it's between $0$ and $1$), which is strictly less than $\frac{1}{n}$.
A: Consider by the archimedian principal there are a unique integers $k,m$ so that $kn \le kn + m\le   x< kn + m + 1 \le (k+1)n$.
So $\frac {[x]}n = k + \frac mn$
So $\{\frac {[x]}n\} = \frac mn\le  \frac {n-1}n$ and $\frac {\{x\}}n < \frac 1n$ so $\{\frac {\{x\}}n\} = \frac {\{x\}}n < \frac 1n$.
So $\{\frac {[x]}n\} + \{\frac {\{x\}}n\} = \frac mn + \frac {\{x\}}n < \frac {n-1}n + \frac 1n = 1$.
