Division of Compound Units Why are compound units of speed,density etc. expressed in terms of a numerator and a denominator? For example, m/s, kg/m3 etc.
I've always understood division as either the inverse of multiplication, splitting a collection of objects into n equal parts or into parts of size n.
So, are units such as m/s actually metres divided by seconds? If so, does it conform to my understanding of division expressed above? In other words, is the word per the same thing as division? e/g Metres per second? Why? How?
 A: I think this could be understood as splitting "a collection of objects into equal parts".
For example, if you split the distance you travel as many times as is the units of time you completed the journey in, you get parts of the length you travel in one unit of time. Your speed.
Similarly, if you have a certain mass, and split it into as many parts as is the units of volume the mass takes up, you get parts of the mass in one unit of volume. Your mass density.
And so on.
A: $
\newcommand{\km}{{\mathrm {km}}}
\newcommand{\h}{{\mathrm h}}
\newcommand{\kmh}{{ \km / \h }}
\newcommand{\hkm}{{ \h / \km }}
\newcommand{\mh}{{\mathrm {mh}}}
$
The reason why a compound unit is used (or exists) (for example $\kmh$ as kilometers per hour) is because it is often intuitive to think in those terms when trying to answer a problem that deals with distances and time.
Compound units are arbitrary and there is no reason why we are not using $\hkm$ other than linguistics or intuition.
For some problem sets, it can be easier to deal in $\hkm$. As an example, given that:


*

*You know the distances between the two cities

*You know the speed at which the train goes

*You want to know how long it will take for the train to go from one city to the other


The usual formula that comes up is:
$$ t = {d \over v} $$
To add some variation to the original problem, say we have three train models, with different top speed each.
$$\begin{align*}
  v_A = 100 \kmh \\
  v_B = 200 \kmh \\
  v_C = 300 \kmh 
\end{align*} $$
How long does it take for each train to travel $500 \mathrm{km}$"?
Using division:
$$ t_x = {d \over v_x} $$
$$
\begin{align*}
 t_A &= {{500 \mathrm {km}} \over 100 \kmh} = 5 \mathrm h \\
 t_B &= {{500 \mathrm {km}} \over 200 \kmh} = 2.5 \mathrm h \\
 t_C &= {{500 \mathrm {km}} \over 300 \kmh} = 1.\overline6 \mathrm h
\end{align*} $$
If instead we introduce the concept of slowness ($l$, from latin lentus) as the inverse of speed:
$$ \newcommand{\mhkm}{{\mathrm {mh} / \mathrm {km}}} 
\begin{align*} 
 v^-1_A =& l_A = {10 \mhkm} \\
 v^-1_B =& l_B = {5 \mhkm} \\
 v^-1_C =& l_C = {3.\overline3 \mhkm}
\end{align*} $$
Then the equation becomes:
$$ t_x = l_x \times d $$
$$
\begin{align*}
 t_A &
    = {10 \mhkm \times 500 \km} & 
    =& 5000 \mh &
    =& 5 \h & \\
 t_B &
    = {5 \mhkm \times 500 \km} &
    =& 2500 \mh &
    =& 2.5 \h & \\
 t_C & 
     = {3.\overline3 \mhkm \times 500 \km} &
     =& 1666.\overline6 \mh &
     =& 1.\overline6 \h &
\end{align*} $$
Now our problem is represented in terms of multiplications only. A human with a calculator might not see much difference, but a human with only a pen and paper will have a much easier time with multiplications, especially when the numbers have more than a few decimals, plus, in many other contexts, multiplication involves 
Also, on a computer, multiplications have significant performance advantages over division, especially when dealing with integers or fixed-point numbers, therefore, given a small set of train speeds and a large set of distances, it is more efficient to compute the inverse of the speeds in advance and multiply those with the distances than directly dividing the distance by the speed. Most compilers perform such an optimization when dividing by a constant that is known at compile time, provided that the precision of the calculation is not affected.
