Prove $\bigg\lfloor\frac{\lfloor x \rfloor }{m}\bigg\rfloor =\bigg\lfloor\frac{x }{m}\bigg\rfloor $ 
Prove $\bigg\lfloor\frac{\lfloor x \rfloor }{m}\bigg\rfloor =\bigg\lfloor\frac{x  }{m}\bigg\rfloor $ where $x\in \mathbb R , x\geqslant 0$ and $m\in \mathbb N$

What I did:
Two cases: 


*

*$x\in \mathbb Z_{\geqslant0}$

*$x\notin \mathbb Z_{\geqslant0}$

For case 1: there is noothing to prove $x=$ to some $k\in \mathbb Z_{\geqslant0}$
$$\bigg\lfloor\frac{k }{m}\bigg\rfloor $$

case 2:
I took example let $x=4.5$ and let $m=5$ so  $\bigg\lfloor\frac{\lfloor 4.5 \rfloor }{5}\bigg\rfloor \overset{?}=\bigg\lfloor\frac{4.5  }{5}\bigg\rfloor $
$$\bigg\lfloor\frac{4 }{5}\bigg\rfloor=0$$
$$\bigg\lfloor\frac{4.5  }{5}\bigg\rfloor =\bigg\lfloor\frac{9  }{10}\bigg\rfloor =0$$
How can I prove case 2 more formally?

Let $y=\bigg\lfloor\dfrac{x }{m}\bigg\rfloor$ 
then $y$ is an integer and $ym \le x \lt (y+1)m$
so, since $ym$ is an integer, we have $ym \le \lfloor x \rfloor \le x \lt (y+1)m$
and so $\bigg\lfloor\dfrac{\lfloor x \rfloor }{m}\bigg\rfloor = y = \bigg\lfloor\dfrac{x }{m}\bigg\rfloor$ 
