convert ceil to floor Mathematically, why is this true?
$$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a+b-1}{b}\right\rfloor$$
Assume $a$ and $b$ are positive integers.
Is this also true if $a$ and $b$ are real numbers?

$$\left\lceil \frac { a }{ b }  \right\rceil =\frac { a }{ b } +1-\left( \frac { a }{ b }  \mod\ 1 \right) ,\quad \quad   (1)$$
$$\left\lfloor \frac { a+b-1 }{ b }  \right\rfloor =\frac { a+b-1 }{ b } -\left( \frac { a+b-1 }{ b } \mod\ 1 \right) =\frac { a }{ b } +1-\frac { 1 }{ b } -\left( \frac { a }{ b } -\frac { 1 }{ b } \mod\ 1 \right) ,\quad \quad (2)$$
For the sake of simplicity write $$\left( \frac { a }{ b } \mod\ 1 \right) =k$$
The equations $(1)$ and $(2)$ are equal: $$\frac { a }{ b } +1-k=\frac { a }{ b } +1-\frac { 1 }{ b } -k+\left( \frac { 1 }{ b } \mod\ 1 \right) $$
Finally, $$\frac { 1 }{ b } =\left( \frac { 1 }{ b } \mod 1 \right) .$$ For $b>1$ your claim holds. 
I think that something is missing. Until the next edit...
