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.

A vivid conceptual proof: $\:$ If $\rm\:[a,b]\:$ contains a unique integer $\rm\:k\:$ then clearly $\rm\ \lceil a\rceil\: =\: k\: =\: \lfloor b\rfloor\:.$  
This applies to $\rm\:\ a = n/m,\:\ b = (n+m-1)/m\:.\:$ Notice that $\rm\:[a,b]\:$ contains a unique integer since $\rm\:m\:$ divides exactly one of the consecutive $\rm\:m\:$ integers $\rm\:n,\:n+1,\:\cdots,\:n+m-1\:.$
Note $\ $ This problem is exercise 3.12 in Graham; Knuth; Patashnik: Concrete Mathematics. Curiously they overlook this simple solution, instead giving essentially the solution in my other answer here.
