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: $\left[\frac{[na]}{n}\right] = \left[ \frac{n[a]}{n} \right] = [ [a] ] = [a]$ ?

  • $\begingroup$ Try to write $a=b+c$ where $b \in \mathbb{Z}$ and $c\in [0,1)$ then you see you cannot say $[na]=n[a]$ $\endgroup$ – Kaladin Mar 8 '14 at 14:54
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    $\begingroup$ Watch it: $$\left\lfloor 4\frac13\right\rfloor=\left\lfloor\frac43\right\rfloor=1\neq 0=4\cdot 0=4\left\lfloor\frac13\right\rfloor$$ $\endgroup$ – DonAntonio Mar 8 '14 at 14:56
  • $\begingroup$ Ok! Thank you. But how can I then proof the equation above.. ? $\endgroup$ – Vazrael Mar 8 '14 at 15:03
  • $\begingroup$ For inspiration you might want to look at math.stackexchange.com/questions/376720/… $\endgroup$ – fgp Mar 8 '14 at 15:15

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,\ $
$${\rm\begin{eqnarray} &\rm k &\le&\:\rm\ \lfloor \lfloor na \rfloor / n\rfloor \\ \color{#c00}\iff& \rm k &\le&\ \ \rm \lfloor na \rfloor / n \\ \iff& \rm nk &\le&\ \ \rm \lfloor na \rfloor \\ \color{#c00}\iff& \rm nk &\le&\ \ \rm na \\ \iff& \rm k &\le&\ \ \rm a \\ \color{#c00}\iff& \rm k &\le&\ \ \rm \lfloor a \rfloor \\ \\ \Rightarrow\ \ \rm \lfloor \lfloor na\!\!&\rm \rfloor / n\rfloor\ &=&\rm\ \ \lfloor a\rfloor \end{eqnarray}\quad\!}$$

  • $\begingroup$ Hi Bill Dubuque. Thank you for this straightforward proof. How can I conclude the solution from $k \le ⌊⌊na⌋/n⌋ ⟺ k \le ⌊a⌋$ ? $\endgroup$ – Vazrael Mar 9 '14 at 15:08
  • $\begingroup$ @K.L. Specializing $\,k = \lfloor a\rfloor\,$ yields $\,\lfloor a\rfloor \le \lfloor \lfloor na\rfloor/n\rfloor .\,$ The reverse inequality follows similarly. See also here. $\endgroup$ – Bill Dubuque Mar 9 '14 at 15:23

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...

  • $\begingroup$ This means $⌊\frac{⌊na⌋}{n}⌋ = ⌊\frac{k}{n}⌋ = ... = ⌊a\frac{k}{k-t}⌋$ and this has to be ⌊a⌋. But this isn't true? $\endgroup$ – Vazrael Mar 8 '14 at 15:26
  • $\begingroup$ @K.L. But read the last line with inequalities in my answer. What do you deduce when you multiply it by $\;0\le a\;$ ?! $\endgroup$ – DonAntonio Mar 8 '14 at 15:27
  • $\begingroup$ Take a step back, and look at this geometrically. $[a]n$ means you round $a$ down, and multiply with $n$. Now look at $[an]$. $an$ is obviously larger than $[a]n$ - so the question is, can it exceed $[a]n$ far enough to be rounded down to a different integer? It can - if the fractional part of $a$ is larger than $\frac{1}{n}$, $an - [a]n \geq 1$, and so $[an] > [a]n$. $\endgroup$ – fgp Mar 8 '14 at 15:30
  • $\begingroup$ @DonAntonio. If I do this, I get $ a ≤ a\frac{k}{k−t} < 2a $. I'm really not sure what to do next. Additionally, why can I assume a as a not-negative number? $\endgroup$ – Vazrael Mar 8 '14 at 15:42
  • $\begingroup$ You can't assume $\;a\ge \;$, @K.L.: I do assume that. You'd have to figure out a similar way for negative one. And $$a\le a\frac k{k-t}<2a\implies \left\lfloor a\frac k{k-t}\right\rfloor =\lfloor a\rfloor$$ $\endgroup$ – DonAntonio Mar 8 '14 at 15:57

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