I would appreciate if somebody could help me with the following problem:

Q: How to proof? $(n,k\in\mathbb{N})$



closed as off-topic by user223391, user284331, Lord Shark the Unknown, Wouter, Claude Leibovici Feb 21 '18 at 6:55

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    $\begingroup$ Are you missing something important in the problem statement? e.g. "... if and only if $k$ divides $n$"? $\endgroup$ – Excluded and Offended Mar 29 '14 at 8:39
  • $\begingroup$ This is not correct. Take for example $n=3$ and $k=4$. $\endgroup$ – Marc Mar 29 '14 at 8:42

We have $$\frac nk-1<\left\lfloor \frac{n}{k} \right\rfloor\le \frac nk$$ and $$\frac{1-n}k\le-\left\lfloor \frac{n-1}{k} \right\rfloor<\frac{1-n}k+1$$ so adding these inequalities gives $$\frac1k-1<\underbrace{\left\lfloor \frac{n}{k} \right\rfloor-\left\lfloor \frac{n-1}{k} \right\rfloor}_{\in\Bbb Z}<\frac1k+1$$ so $$\left\lfloor \frac{n}{k} \right\rfloor-\left\lfloor \frac{n-1}{k} \right\rfloor\in\{0,1\}$$ Notice that the two values are possible. For example: $k=n=1$ gives the value $1$ and $n=2$ and $k=3$ gives the value $0$.

  • $\begingroup$ Dear Smai, do you know any article, book or... in which we can have almost all about the floor function and its properties? (+1) $\endgroup$ – mrs Aug 19 '14 at 15:13

You cannot prove what is not true. Plugging in $n<k$ will result in $$\left\lfloor \frac{n}{k} \right\rfloor-\left\lfloor \frac{n-1}{k} \right\rfloor=0$$


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