Proving $\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$ for $a,b>1$ I'm trying to prove rigorously the following:

$$\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$$   for integers $a,b \geq 1$ and real $x$.

So far I haven't gotten far. It's enough to prove this instead:

$$\left\lfloor \frac{z}{c} \right\rfloor = \left\lfloor \frac{\lfloor z \rfloor}{c} \right\rfloor$$   for integers $c \geq 1$ and real $z$

since we can just put $z=\lfloor x/a \rfloor$ and $c=b$.

It is trivial when done universally, i.e. using the universal definition of floor
$\begin{align}  k\,&\le\, \lfloor x\rfloor \color{#c00}{\iff} k\,\le\, x,\quad {\rm for\ all}\,\ k\in\Bbb Z\\[1em]
{\bf Lemma}\qquad\quad\color{#0a0}{\lfloor r/n\rfloor}\,& =\, \lfloor{\lfloor r\rfloor}/n\rfloor\ \ {\rm 
for}\ \ 0<n\in\Bbb Z,\,\ r\in\Bbb R\\[.6em]
{\bf Proof}\qquad\qquad\quad\ \ \  k \,&\le \lfloor{\lfloor r\rfloor}/n\rfloor\\[.4em]
\color{#c00}\iff\quad k\ & \le\ \:{\lfloor r\rfloor}/n\\[.2em]
\overset{\times\, n}\iff\ \ nk\ &\le\ \ \lfloor r\rfloor\qquad {\rm by}\,\  n>0\\[.4em]
\color{#c00}\iff\ \ nk\ &\le\,\ \ \ r\qquad\ \  {\rm  by}\,\  n\in\Bbb Z\\[.2em]
\overset{\div\,n}\iff\ \ \ \ k\ &\le\:\ \ \ r/n\quad\  {\rm by}\,\  n>0\\[.4em]
\color{#c00}\iff\ \ \ \  k\ &\le\ \ \color{#0a0}{\lfloor r/n\rfloor}\quad {\small\bf QED}
\end{align}$
Yours is special case $\ r = x/a,\,\ n = b.$
Re: universality: see the links in comments below for more general (category-theoretic) viewpoints.
