TAOCP Exercise 1.2.4-35

 TAOCP

Excise 1.2.4-35

$\lfloor \frac{ (x + m) }{n} \rfloor = \lfloor \frac{  \lfloor x \rfloor +m }{n} \rfloor$, where $x$ is real, $m$ and $n$ are integers, $n >0$

Proof:

Obviously  $\lfloor(x+m)/n \rfloor \geq \lfloor(\lfloor x \rfloor + m)/n \rfloor$, the only case that it is not equal is $(\lfloor x \rfloor +m)/n$and $(x+m)/n$ lies on different sides of an integer $k$, which means 

$$1 > k -(\lfloor x \rfloor + m ) / n > 0$$  

and 

$$(x+m)/n - k > 0$$ ......(1)

furthermore, $\lfloor x \rfloor + m$is integer, thus 

$$k - (\lfloor x \rfloor + m)/n \geq 1/n$$ ......(2)

from (1) and (2)

$(x+m)/n -k + k - (\lfloor x \rfloor + m)/n = (x- \lfloor x \rfloor)/n > 1/n$  which is impossible. so only equal relation is valid.