I was doing dimensional analysis for a problem, and I noticed something curious about the remainder operator (sometimes referred to as the $mod$ operator).
Letʼs recap how arithmetic operators interact with dimensional analysis. Say the dimensions of $a$ are $A$, and the dimensions of $b$ are $B$. For instance, $a$ could be an areal measure and $A$ would be $L^2$, while $b$ might be a momentum and $B$ would then be $MLT^(-1)$.
Given such quantities $a$ and $b$, itʼs well known that the dimensions of $a*b$ are $A*B$, and those of $a/b$ are $A/B$. Now, how about the dimensions of $a+b$ and of $a-b$? That was a trick question: it doesnʼt make sense or add or subtract $a$ and $b$ unless they have the same dimensions and are expressed in the same unit. You cannot add a length and a time, and furthermore, you cannot even add two length quantities if one were expressed in metres and the other in kilometres. Likewise with subtraction.
So whatʼs the story with the remainder operator? What are the dimensions of $a mod b$? Because of the association with $a div b$, my intuition told me it might be $A/B$. Letʼs test that out.
Imagine Iʼm walking towards a bus stop where thereʼs a bus every $10$ minutes (in the direction I want to go). When Iʼm far away, I see a bus go past, and it would take me, say, $27$ more minutes to get to the stop. How long should I expect to wait for my next bus? The answer, of course, is $10 - (27 mod 10)$ minutes or $3$ minutes. In general, except for the improbable case of my arriving at the stop simultaneous with a bus, if the time-period (the reciprocal of frequency) of buses were $b$ (with dimension $T$) and the time it took me to get to the stop were $a$ (with dimensions $T$), the time I should wait for would be $b - a mod b$, with dimensions $T$. The fact that this expression makes sense must mean that both $b$ and $a mod b$ have dimensions of $T$, and are in the same units.
So that answers what the dimensions of $a mod b$ must be: the same as the dimensions of $b$. But how can the dimensions be independent of the dimensions of $a$? Actually, theyʼre not. $a mod b$ is $a - kb$, with $k$ chosen such that the result is in $[0, b[$. So the dimensions of $a mod b$ are the same as those of $a$, which then implies the following:
Like addition and subtraction, the remainder operator is only defined on quantities with the same dimensions and expressed in the same unit, with the result also with the same dimensions and in the same unit.