0
$\begingroup$

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

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

$$1=\left\lfloor\frac{n}{k}\right\rfloor-\left\lfloor\frac{n-1}{k}\right\rfloor$$

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, user284331, Lord Shark the Unknown, Wouter, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.