<s>[INST] 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.
 [/INST] what about Mathematical induction prove :
n and m is natural number :
\begin{aligned} m=1 \rightarrow \left \lfloor \frac{1+n-1}{1} \right \rfloor = \left \lceil \frac{n}{1} \right \rceil \rightarrow true \end{aligned}
\begin{aligned}m=2 \rightarrow \left \lfloor \frac{2+n-1}{2} \right \rfloor = \left \lceil \frac{n}{2} \right \rceil \rightarrow true \end{aligned}
\begin{aligned} \left \lceil \frac{n}{2} \right \rceil = \begin{Bmatrix}
\frac{n}{2} \ n \ is \ even 
\ \
\frac{n+1}{2} \ n \ is \ odd
\end{Bmatrix}
\end{aligned}
\begin{aligned}
if \ m=k \rightarrow \left \lfloor \frac{k+n-1}{k} \right \rfloor = \left \lceil \frac{n}{k} \right \rceil 
\end{aligned}
\begin{aligned}
prove  \ m=k+1 \rightarrow \left \lfloor \frac{k+1+n-1}{k+1} \right \rfloor = \left \lceil \frac{n}{k+1} \right \rceil
\end{aligned}
\begin{aligned} 
\left \lfloor \frac{n-1}{k+1}+1 \right \rfloor = \left \lceil \frac{n}{k+1} \right \rceil 
\end{aligned}
\begin{aligned} 
\left \lfloor \frac{n-1}{k+1} \right \rfloor+1 = \left \lceil \frac{n}{k+1} \right \rceil 
\end{aligned}
... ? 
</s>