Is Modular Arithmetic Notation Good?

Question

When we write $$a = b\, (\mathrm{mod}\, n)$$ we mean $a$ and $b$ belong to the same equivalence class. This is a symmetric property, so writing $$a \equiv_n b$$ is intuitive. On the other hand, we also treat $(\mathrm{mod}\, n)$ like an operator: that is, given some $b$, we write $b\, (\mathrm{mod}\, n)$ to denote the remainder of $b$ upon division by $n$. In this setting, it is not necessarily true that $a = b\, (\mathrm{mod}\, n)$, even if $a$ and $b$ belong to the same equivalence class. Disagreements in notation are never good, so my friend writes $\lfloor b\rfloor_n$ to mean the remainder after division by $n$.

There is now a very clear distinction between congruence mod $n$ and the modulo operator. However, this makes me think, what if there is a deep reason why the common notation is good? To me the current conventions seem bad, but there must be reason it is popular. Why do people use the current notation?

Answer 1

Nobody uses $a\mod b$ as an operator. That's an exaggeration, but basically that notation is not as popular as you seem to think. It is non-standard and discouraged. Programmers sometimes use it because they're used to the % operator in many programming languages, but I don't think I've ever seen someone write $a\mod b$ on math.SE without a mathematician telling them that's not really how the notation works.

Sometimes it's useful to have a notation for "the unique $x\in \{0, ... b-1\}$ such that $a\equiv x\mod b$", and in that case writing $a\mod b$ might not be an unreasonable way of doing it as long as explain the notation beforehand. $\lfloor a \rfloor_b$ seems reasonable as well. There is no standard notation for this operation.

Answer 2

I've also wondered what the simplest notation would be. Some alternative notation uses $mod(a,n)$. So then we can write the same identity multiple ways: \begin{align} ab \: (mod \: n) &= (a \: (mod \: n))(b \: (mod \: n)) \: (mod \: n) \\ mod(ab,n) &= mod(mod(a,n)mod(b,n),n)\\ ab &\equiv_n mod(a,n)mod(b,n) \end{align} Using $mod(a,n)$ requires one to remember which comes first (the $a$ or the $n$). Alternative notation might use $mod_n(a)$ as a replacement for $mod(a,n)$. So we would have a fourth way to express the same identity: $$ ab \equiv_n mod_n(a)mod_n(b)$$ This consistently keeps $n$ as a subscript.

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Anyone who wants to use notation $\equiv_n$ or $mod_n(a)$ can simply define them in one line at the start of their work.