Equation with ceiling

Question

I have to show that: $$ \lceil{ \frac{\lceil{ \frac{n}{a} \rceil }}{b}} \rceil=\lceil {\frac{n}{ab} } \rceil$$ That`s what I tried: If $n=ka, k \in \mathbb{Z}$: $ \lceil{ \frac{n}{a}} \rceil=\frac{n}{a}$,therefore the relation is satisfied. But how can I show it,when $n \neq ka$ ?

Answer 1

Write $\lceil \frac{n}{a}\rceil=\frac{n}{a}+\frac{\nu}{a}$, where $0\le \nu< a$. Now $$\left\lceil \frac{\lceil \frac{n}{a}\rceil}{b} \right\rceil=\left\lceil \frac{ \frac{n}{a} + \frac{\nu}{a}}{b} \right\rceil=\left\lceil \frac{n}{ab}+\frac{\nu}{ab}\right\rceil ~~~(\star)$$

Now, there is some integer $t$ such that $(t-1)ab<n\le (t)ab$; specifically $t=\lceil \frac{n}{ab}\rceil$. If we can show that $(\star)=t$ then we are done.