Proving $\left\lfloor \frac{\left\lfloor a/b \right\rfloor}{c} \right\rfloor=\left\lfloor\frac{a}{bc}\right\rfloor$ for positive integer $a$, $b$, $c$

Question

How can we prove the following? $$\left\lfloor \frac{\left\lfloor \dfrac{a}{b} \right\rfloor}{c} \right\rfloor = \left\lfloor \frac{a}{bc} \right\rfloor$$ for $a,b,c \in \mathbb{Z}^+$

I don’t know if I’m doing something wrong, but I can’t prove it even though I’m pretty sure it’s true.

Obviously, because the concept of algebra isn’t aware of the fact that we are restricting the variables to positive integers, and given my assumption that the equality doesn’t necessarily hold for non-integers, an element of non-algebraic problem solving is needed, i.e. making a change to the expression given our knowledge of that condition, which then allows for algebraic maneuvers that show that the equality holds. I think that’s what I’m missing.

Thanks.

Answer 1

Hint:

Write remainder theorem: $$a = bc\cdot k+r$$ where $0\leq r<bc$.

Answer 2

$$a/b = \lfloor a/b \rfloor + \{a/b\} \implies$$ $$\frac{a/b}c = \frac{\lfloor a/b \rfloor}c+ \frac{\{a/b\}}c = \left \lfloor \frac{\lfloor a/b \rfloor}c\right \rfloor + \left \{\frac{\lfloor a/b \rfloor}c\right\}+ \frac{\{a/b\}}c $$

So to finish we can prove that for any $q \in \mathbf{Q}^+$ and $p \in \mathbf{Z}^+$ we have

$$ \left \{\frac{\lfloor q \rfloor}p\right \}+ \frac{\{q\}}p < 1 \iff \left\{\frac{np+r}p\right\}+ \frac{\{q\}}p =\frac{r}p + \frac{\{q\}}p < 1$$

where $0 \leq r < p$, but $\frac{\{q\}}p < \frac{1}{p}$ and we are done.