I want to show that:

$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$

That's what I have tried:

  • $ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \in \mathbb{Z}: m \leq \frac{n}{2}\}$
  • $\left\lceil \frac{n}{2} \right\rceil= \min \{ m \in \mathbb{Z}: m \geq \frac{n}{2}\}$

If $n=2k,k \in \mathbb{Z}$,then: $\frac{n}{2} \mathbb{Z}$,so $$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n}{2}, \frac{n-2}{2}, \dots \right\}=\frac{n}{2} \\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n}{2}, \frac{n+2}{2}, \dots \right\}=\frac{n}{2}$$

Therefore, $ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$.

If $n=2k+1, k \in \mathbb{Z}$,then $\frac{n}{2} \notin \mathbb{Z}$.So:

$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n-1}{2}, \frac{n-3}{2}, \dots \right\}=\frac{n-1}{2}\\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n+1}{2}, \frac{n+3}{2}, \dots \right\}=\frac{n+1}{2}$$

Therefore, $ \lfloor \frac{n}{2}\rfloor + \lceil \frac{n}{2} \rceil=\frac{n-1}{2}+\frac{n+1}{2}=n$

Is there also an other way to show the equality or is it the only one?

  • 2
    $\begingroup$ Presumably, for $n$ an integer... $\endgroup$ – Thomas Andrews Jul 26 '14 at 17:08
  • $\begingroup$ Yes,I have to show that the inequality stands for each integer $n$. $\endgroup$ – evinda Jul 26 '14 at 17:11
  • $\begingroup$ @ThomasAndrews Is the way I showed the equality the only possible or is there also an other one? $\endgroup$ – evinda Jul 26 '14 at 23:07
  • $\begingroup$ What do you expect as "an other way" ? $\endgroup$ – Yves Daoust Jul 27 '14 at 8:36

Since the function $f(x)=\{x\}=x-\lfloor x\rfloor$ is periodic with period one, the function: $$ g(n) = \left\lfloor\frac{n}{2}\right\rfloor+\left\lceil\frac{n}{2}\right\rceil -n$$ is periodic with period $2$. Since $g(0)=g(1)=0$, $g(n)=0$ for every $n\in\mathbb{N}$.

  • $\begingroup$ Did you conclude this,because of the fact that $f$ is periodic? $\endgroup$ – evinda Jul 26 '14 at 23:27
  • 1
    $\begingroup$ Yes, it is exactly so. $\endgroup$ – Jack D'Aurizio Jul 26 '14 at 23:28
  • $\begingroup$ I understand..thank you!!! $\endgroup$ – evinda Jul 27 '14 at 12:51

For even $n$, $\lfloor\frac n2\rfloor=\lceil\frac n2\rceil=\frac n2$.

For odd $n$, $\lfloor\frac n2\rfloor=\frac{n-1}2$, and $\lceil\frac n2\rceil=\frac{n+1}2$.

  • $\begingroup$ So,is this the only way to show that the relation is true? $\endgroup$ – evinda Jul 26 '14 at 17:15
  • $\begingroup$ Er, your way is different, its uses bounds on number sets. What else did you expect, this is rather elementary ? $\endgroup$ – Yves Daoust Jul 26 '14 at 17:17
  • $\begingroup$ Could you explain me further which the difference is? :/ $\endgroup$ – evinda Jul 26 '14 at 17:18
  • $\begingroup$ You use bounds on number sets, this is unnecessary. $\endgroup$ – Yves Daoust Jul 26 '14 at 17:20
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    $\begingroup$ @YvesDaoust how did you prove the equality in your post? $\endgroup$ – Kaster Jul 26 '14 at 17:39

Well, here's another one. In fact, it is possible to prove that for any integer $n$ and any real number $\alpha$, we have $$n=\lfloor (1-\alpha) n\rfloor +\lceil \alpha n\rceil.$$

First, we have $$\alpha n\leq\lceil\alpha n\rceil,$$ therefore, $$n-\lceil\alpha n\rceil\leq (1-\alpha)n,$$ but then, since $n-\lceil\alpha n\rceil$ is an integer, this means $$n-\lceil\alpha n\rceil\leq \lfloor (1-\alpha)n\rfloor.$$

Similarly, we have $$\lfloor (1-\alpha)n\rfloor\leq (1-\alpha)n,$$ therefore, $$\alpha n\leq n-\lfloor (1-\alpha)n\rfloor,$$ but then again, $$\lceil\alpha n\rceil\leq n-\lfloor (1-\alpha)n\rfloor,$$ from the above, we get $$\lfloor (1-\alpha) n\rfloor +\lceil \alpha n\rceil\leq n\leq \lfloor (1-\alpha) n\rfloor +\lceil \alpha n\rceil.$$


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