This is expanding on this question: Show that if $n$ and $k$ are positive integers, then $\lceil \frac{n}{k} \rceil = \lfloor \frac{n - 1}{k} \rfloor + 1$ as I'm unclear on how to solve this statement...
I'm not clear on the answer provided in the textbook, and I'm not able to solve it, this is how far I get: There are unique integer $b$ and real number $r$, with $0 \leq r \lt 1$ such that $\frac{n}{k} = b + r$, therefore $n = k(b+r)$, and $n-1 = k(b+r) -1$, so $\frac{n-1}{k} = b + r - \frac{1}{k}$
Case i) $\lceil \frac{n}{k} \rceil = b + r$ when $r = 0$
Then $b = \lfloor b + 0 - \frac{1}{k}\rfloor + 1$, since $b$ is an integer
$b = b + \lfloor -\frac{1}{k} \rfloor + 1$, since $k$ is an integer, definition of $\lfloor - \frac{1}{k} \rfloor = -1 $, therefore $b = b$
Case ii)$\lceil \frac{n}{k} \rceil =b +1$ when $0 \lt r \lt 1$
Then $b+1 = \lfloor b + r - \frac{1}{k}\rfloor + 1$, and this is where I get stuck. Since $b$ is an integer $b + 1 = b + \lfloor r - \frac{1}{k} \rfloor + 1$
Ok, I think I can apply the division algorithm in this fashion: If $n$ and $k$ are positive integers then there are unique integers $q$ and $r$, with $0 \leq r \lt k$, such that $n = kq + r$. Let $n - 1 = kq + r -1$ divide both equations by $k$ such that
$\frac{n}{k} = q + \frac{r}{k}$ and $\frac{n-1}{k} = q +\frac{r-1}{k}$
Case i) $r =0$ as per above. $q = q$
Case ii) $r \gt 0$
LHS: $\lceil \frac{n}{k}\rceil = \lceil q + \frac{r}{k}\rceil = q + 1$
RHS: $\lfloor \frac{n-1}{k}\rfloor + 1 = \lfloor q + \frac{r-1}{k}\rfloor + 1 = q + \lfloor \frac{r-1}{k}\rfloor + 1 = q + 0 + 1 = $LHS
Since r is greater than 0 and an integer it must be an integer between 1 and $k-1$, therefore $\lfloor \frac{r-1}{k} \rfloor$ will always evaluate to 0