This is best done universally, i.e. using the universal definition of floor
$\rm\qquad\qquad\qquad\qquad\ \ \ k\,\le\, \lfloor x\rfloor \color{#c00}{\iff} k\,\le\, x,\quad {\rm for\ all}\,\ k\in\Bbb Z$
Lemma $\rm\qquad\quad\color{#0a0}{\lfloor r/n\rfloor}\, =\, \lfloor{\lfloor r\rfloor}/n\rfloor\ \ $ for $\rm\ \ 0
Proof $\rm\quad\quad\quad\quad\quad \ \ \ k\ \le \lfloor{\lfloor r\rfloor}/n\rfloor$
$\rm\quad\quad\quad\quad\quad\color{#c00}\iff\quad k\ \le\ \:{\lfloor r\rfloor}/n$
$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor r\rfloor\qquad\!\! by\,\ n>0$
$\rm\quad\quad\quad\quad\quad\color{#c00}\iff\ \ nk\ \le\,\ \ \ r\qquad\, by\,\ n\in\Bbb Z$
$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ r/n\quad by\,\ n>0 $
$\rm\quad\quad\quad\quad\quad\color{#c00}\iff\ \ \ \ k\ \le\ \ \color{#0a0}{\lfloor r/n\rfloor}\quad {\bf QED} $
Yours is special case $\rm\ r = x/a,\,\ n = b.$