question on Floor function: $[na] = [a]n?$

Question

let $n\in \mathbb{N}, a \in \mathbb{R}$.

What can I then say about the gauß-function or floor-function:

$[an]$ ?

I have to show: $\left[\frac{[na]}{n}\right] = [a] := max\{ z \in \mathbb{Z}: z \le a\}$.

Well, if [na] = n[a] since n is a natural number, then: $\left[\frac{[na]}{n}\right] = \left[ \frac{n[a]}{n} \right] = [ [a] ] = [a]$ ?

Answer 1

Assuming $\;a\ge0\;$ :

Write $\;na=k+t\;,\;\;k\in\Bbb N\;,\;\;t\in [0,1)\;$ , then:

$$\lfloor na\rfloor=k\implies\frac kn=\frac k{\frac ka-\frac ta}=\frac{ak}{k-t}=a\frac{k}{k-t}$$

But since

$$\;1\le\frac k{k-t}<2\;$$

we're then done...

Answer 2

This is straightforward using the universal property of the floor function, viz. $$\rm k\le \lfloor r \rfloor \color{#c00}\iff k\le r,\ \ \ for\ \ \ k\in \mathbb Z,\ r\in \mathbb R$$ Therefore for $\rm\:0 < n\in \mathbb Z,\ a\in \mathbb R,\ $
$${\rm\begin{eqnarray} &\rm k &\le&\:\rm\ \lfloor \lfloor na \rfloor / n\rfloor \\ \color{#c00}\iff& \rm k &\le&\ \ \rm \lfloor na \rfloor / n \\ \iff& \rm nk &\le&\ \ \rm \lfloor na \rfloor \\ \color{#c00}\iff& \rm nk &\le&\ \ \rm na \\ \iff& \rm k &\le&\ \ \rm a \\ \color{#c00}\iff& \rm k &\le&\ \ \rm \lfloor a \rfloor \\ \\ \Rightarrow\ \ \rm \lfloor \lfloor na\!\!&\rm \rfloor / n\rfloor\ &=&\rm\ \ \lfloor a\rfloor \end{eqnarray}\quad\!}$$