Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$? 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:


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*$ \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?

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 $n\in\mathbb{N}$.
