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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:

  1. $x\in \mathbb Z_{\geqslant0}$

  2. $x\notin \mathbb Z_{\geqslant0}$

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?

Error 404
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    Even if you prove this particular statement, you don't prove the general one. You need a proof for all $x$ and $m$. Use the fact that $a= \lfloor x \rfloor$ if and only if $a$ is an integer and $a \le x < a+1$. – Crostul Mar 29 '16 at 10:21

1 Answers1

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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 $\bigg\lfloor\dfrac{\lfloor x \rfloor }{m}\bigg\rfloor = y = \bigg\lfloor\dfrac{x }{m}\bigg\rfloor$

Henry
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