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I came across this answer to this question: Closed form expression for $n_j$ defined by $n_j=\lceil n_{j-1}/b \rceil$

I was hoping someone could clarify the following step:

$q−1 \leqslant \dfrac{(P−1)}{b} < \dfrac{x}{ab} ⩽ \dfrac{P}{b} ⩽ q $,

which says

$\left\lceil\dfrac{x}{ab}\right\rceil = > \left\lceil\dfrac{1}{b}\left\lceil\dfrac{x}{a}\right\rceil\right\rceil $.

I realize that the inequality is bounded between q-1 and q, so intuitively the ceiling equality makes sense. Testing real examples also shows that the equality is true. However, I'm a bit lost on how this is rigorously shown from the inequality. Any help would be very much appreciated!

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That's exactly the same thing Daniel Fischer did in the first inequality of the answer you are asking about, just with different numbers.

Consider a real number $\alpha$. Then by definition $\lceil \alpha \rceil$ is the smallest integer greater than or equal to $\alpha$. In other words, $n = \lceil \alpha \rceil$ if and only if $n$ is an integer and $$ n - 1 < \alpha \leq n. $$ Now, the inequality you have tells you that $$ \color{red}{q-1} \color{gray}{\leq \frac{P-1}{b}} \color{red}{\pmb{<} \frac{x}{ab}} \color{gray}{\leq \frac{P}{b}} \color{red}{\leq q} $$ hence $$ q = \left\lceil\frac{x}{ab}\right\rceil. $$ On the other hand, the same inequality also tells you that $$ \color{green}{q-1} \color{gray}{\leq \frac{P-1}{b}} \color{green}{\pmb{<}} \color{gray}{\frac{x}{ab} \leq} \color{green}{\frac{P}{b} \leq q} $$ and since by definition $P = \left\lceil\frac{x}{a}\right\rceil$ we conclude that $$ q = \left\lceil\frac{P}{b}\right\rceil = \left\lceil \frac{1}{b} \left\lceil\frac{x}{a} \right\rceil \right\rceil. $$

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  • $\begingroup$ Thanks a lot for breaking this down. I really appreciate it. $\endgroup$ – lstbl Dec 29 '15 at 22:14

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