Rewriting repeated integer division with multiplication

Question

In many programming languages, such as C and C++, integer division of positive numbers is defined by simply ignoring the remainder. $5 / 2 == 2$.

In general, is it true of positive integers $a$, $b$, and $c$ that

$(a / b) / c$

will always give the same result as

$a / (b * c)$

We can assume that $b$ and $c$ can be multiplied with no overflow.

Answer 1

This is very easy using the universal property of the floor function, viz. $$\qquad \rm n\le \lfloor r \rfloor \iff n\le r,\ \ \ for\ \ \ n\in \mathbb Z,\ r\in \mathbb R$$ Thus for $\rm\:0 < c\in \mathbb Z,\ r\in \mathbb R,\ $ (e.g. $\rm\:r = a/b\in\mathbb Q\:$ in your case) $$\rm\begin{eqnarray} &\rm n &\le&\:\rm\ \color{#c00}{\lfloor \lfloor r \rfloor / c\rfloor} \\ \iff\ & \rm n &\le&\ \ \rm \lfloor r \rfloor / c \\ \iff\ & \rm cn &\le&\ \ \rm \lfloor r \rfloor \\ \iff\ & \rm cn &\le&\ \ \rm r \\ \iff\ & \rm n &\le&\ \ \rm r/c \\ \iff\ & \rm n &\le&\ \ \rm \color{#0a0}{\lfloor r/c \rfloor} \\[0.5em] \Longrightarrow\ \ \rm \color{#c00}{\lfloor \lfloor r\rm \rfloor}&\color{#c00}{/ c\rfloor\ }&=&\rm\ \ \color{#0a0}{\lfloor r/c\rfloor} \end{eqnarray}$$

since both are $\:\!\le\:\!$ each other (by choosing $\,\rm n\,$ equal to each above).

For $\rm\:r = a/b\:$ we get your special case $\rm\ \lfloor \lfloor a/b \rfloor / c\rfloor = \lfloor a/(bc)\rfloor. $

Remark $ $ If you know a little category theory you can view this universal property of floor as a right adjoint to inclusion, e.g. see Arturo's answer here, or see any textbook on category theory. Or, equivalently, it can be viewed as a Galois connection. But, of course, one need not know any category theory to understand the above proof. Indeed, I've had success explaining this (and similar universal-inspired proofs) to bright high-school students.

Answer 2

Note that $a$ has a unique representation as $k_1b+r_1$ where $0\leq r<b$. Here, the number $k_1$ is the result of of the integer division $a/b$. Also, $a$ has a unique representation as $l(bc)+s$, where $0\leq s<bc$, so that $a/(b\cdot c)=l$.

Now $k_1$ can be represented as $k_2c+r_2$, with $0\leq r_2<c$, giving that $(a/b)/c=k_2$. Note that $a=k_2(bc)+r_2b+r_1$ and that $$0\leq r_2b+r_1\leq(c-1)b+r_1<(c-1)b+b=cb.$$ By unicity of $l$ and $s$, it follows that $k_2=l$. In other words, $(a/b)/c$ is indeed the same number as $a/(b\cdot c)$.