How to prove $\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor$? everybody, how can I prove that, for natural $m$ and $n$,
$$
\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor \; ?
$$
Thanks a lot.

The claim is false for real $n$ and $m$. Take $n = \pi$ and $m = -3$ for a counter-example. It isn't even true for integral $n$ and $m$. Take $n = 3$ and $m = -3$ for a counter-example.
Suppose $n$ and $m$ are positive integers. Let $\chi_{\pm,X}(x)$ denote the characteristic function of $X_{\pm}$ defined to be $1$ if $x \in X_{\pm}$ and $0$ otherwise. Write 
\begin{align}
\lceil \tfrac{n}{m} \rceil = \tfrac{n}{m} - \lbrace \tfrac{n}{m} \rbrace + \chi_{+, \mathbb{R} \setminus \mathbb{Z}}(\tfrac{n}{m}) 
\end{align}
and, similarly,
\begin{align}
\lfloor \tfrac{n + m - 1}{m} \rfloor = \tfrac{n}{m} + 1 - \tfrac{1}{m} - \lbrace \tfrac{n-1}{m} \rbrace
\end{align}
by the $1$-periodicity of the fractional part function. If the claim is to be true, then it must be that 
\begin{align}
\lbrace \tfrac{n}{m} \rbrace - \lbrace \tfrac{n - 1}{m} \rbrace =   \tfrac{1}{m} + \chi_{+, \mathbb{R} \setminus \mathbb{Z}}(\tfrac{n}{m}) - 1.
\end{align}
which is indeed an identity. The extra factors handle the case when $m$ divides $n$, in which case
\begin{align}
\lbrace \tfrac{n}{m} \rbrace - \lbrace \tfrac{n - 1}{m} \rbrace =  \tfrac{1}{m} + 0 - 1.
\end{align}
