For $n,k \in \mathbb{N}$, prove that $\left \lceil \frac{n}{k} \right \rceil=\left \lfloor \frac{n-1}{k} \right \rfloor +1 $.

Question

For $n,k \in \mathbb{N}$, prove that $\left \lceil \frac{n}{k} \right \rceil=\left \lfloor \frac{n-1}{k} \right \rfloor+1 $.

Answer 1

The number $a = \lfloor \frac{n-1}{k} \rfloor$ is the unique nonnegative integer satisfying $n-1 = ak + r$ with $0 \leq r < k$. This condition on $r$ implies the fraction $\frac{r+1}{k}$ is in the interval $[1/k, 1]$. Therefore, $$ \left\lceil \frac{n}{k} \right\rceil = \left\lceil \frac{ak+r+1}{k} \right\rceil = a + \left\lceil \frac{r+1}{k} \right\rceil = a + 1 . $$

Answer 2

If $k\mid n$, then this result is immediate.

Suppose $k\nmid n$, then $\lceil \frac{n}{k}\rceil = 1 + \lfloor \frac{n}{k} \rfloor$.

Lastly, can you reason why $\lfloor \frac{n}{k} \rfloor = \lfloor \frac{n-1}{k} \rfloor$.