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 $c=b$.

Let $\lfloor x/a \rfloor$ = c(say)=>x=ca+d 0≤d

Let $\lfloor c/b \rfloor$=e(say)=>c=be+f  0≤f

=>x=ca+d=(be+f)a+d=abe+af+d.
$\frac{x}{a}$=be+f+ $\frac{d}{a}$
$\frac{\frac{x}{a}}{b}$=e+$\frac{f}{b}$+$\frac{d}{ab}$
=>$\lfloor x/a/b \rfloor$ = e = $\lfloor c/b \rfloor$  = $\lfloor \lfloor x/a \rfloor /b \rfloor$
