(1)

\(m\in\mathbb{Z}\)とする。

\(n=2m+1\)のとき

\begin{align*} \left\lfloor \frac{n}{2}\right\rfloor +\left\lceil \frac{n}{2}\right\rceil & =\left\lfloor \frac{2m+1}{2}\right\rfloor +\left\lceil \frac{2m+1}{2}\right\rceil \\ & =\left\lfloor m+\frac{1}{2}\right\rfloor +\left\lceil m+\frac{1}{2}\right\rceil \\ & =m+m+1\\ & =2m+1\\ & =n \end{align*}

\(n=2m\)のとき

\begin{align*} \left\lfloor \frac{n}{2}\right\rfloor +\left\lceil \frac{n}{2}\right\rceil & =\left\lfloor \frac{2m}{2}\right\rfloor +\left\lceil \frac{2m}{2}\right\rceil \\ & =m+m\\ & =2m\\ & =n \end{align*}

-

これより、与式は成り立つ。

(1)-2

\(m\in\mathbb{Z}\)とする。

\begin{align*} \left\lceil \frac{n}{2}\right\rceil +\left\lfloor \frac{n}{2}\right\rfloor & =2\frac{n}{2}-\mod\left(2\frac{n}{2},1\right)-\mzp_{0,1}\left(0,1;2\mod\left(\frac{n}{2},1\right)\right)+\left|\sgn\mod\left(\frac{n}{2},1\right)\right|\\ & =n-\mzp_{0,1}\left(0,1;2\mod\left(\frac{n}{2},1\right)\right)+\left|\sgn\mod\left(\frac{n}{2},1\right)\right|\\ & =\begin{cases} n-\mzp_{0,1}\left(0,1;2\mod\left(\frac{2m+1}{2},1\right)\right)+\left|\sgn\mod\left(\frac{2m+1}{2},1\right)\right| & n=2m+1\\ n-\mzp_{0,1}\left(0,1;2\mod\left(\frac{2m}{2},1\right)\right)+\left|\sgn\mod\left(\frac{2m}{2},1\right)\right| & n=2m \end{cases}\\ & =\begin{cases} n-\mzp_{0,1}\left(0,1;1\right)+1 & n=2m+1\\ n-\mzp_{0,1}\left(0,1;0\right)+0 & n=2m \end{cases}\\ & =n \end{align*}

(2)

\begin{align*} \left\lfloor n+x\right\rfloor & =n+x-\mod\left(n+x,1\right)\\ & =n+x-\mod\left(x,1\right)\\ & =n+\left\lfloor x\right\rfloor \end{align*}

(3)

\begin{align*} \left\lceil n+x\right\rceil & =n+x-\mod\left(n+x,-1\right)\\ & =n+x-\mod\left(x,-1\right)\\ & =n+\left\lceil x\right\rceil \end{align*}