Is the notion of multiplicative order defined for rational numbers as well as integers?

Question

The multiplicative order of $a$ modulo $N$ refers to the smallest exponent $m$ such that $a^m\equiv 1\mod N$, for $a$, $m$, and $N$ integers. Can we define something similar for all rational $a$? Do the properties remain the same?

Answer 1

Aight, wow this got long, and it's kinda rough, but I'm not going to edit it rn, since I might've spent a bit too long writing it and I've got to get back to my proper work, so sorry :/

The first question you should maybe ask is the following: Can we take rational numbers modulo $N$? Well, it depends on what you mean by mod $N$. If you think about modular arithmetic as working with remainders when you divide by $N$, then in the rationals it becomes necessary to ask: What do we mean by dividing by $N$?

I'm not quite sure about your background, but I'm assuming based on your question that you aren't too familiar with abelian groups/rings and their quotients, so I'll write this at an elementary level.

If $p$ is a rational number, is division expressed by $$p = (p/N)\cdot N + 0,$$ or perhaps is it $$p = q\cdot N + r,$$ where we require $q\in\Bbb{Z}$ and $0\le r < N$. Both of these are possible ways to define division with remainder in $\Bbb{Q}$. The first however doesn't produce an interesting system of modular arithmetic, since all the remainders are zero. Thus we have to use the second definition in order to get a meaningful notion of modular arithmetic. (Luckily, this comes out to the definition you gave in the comment above, since $r$ as defined here is equal to $p-N\lfloor\frac{p}{N}\rfloor$.)

Thus let's define $\Bbb{Q}/N\Bbb{Z}$ (the set of rationals mod $N$), where $N$ doesn't have to be an integer, it can be any rational, I'm just leaving it as $N$ for consistency of notation, to be the set $\newcommand\set[1]{\left\{{#1}\right\}}\set{q\in\Bbb{Q}:0\le q < N}$. This is analagous to how we can define $\Bbb{Z}/N\Bbb{Z}$ (the set of integers mod $N$) to be the set $\set{n\in\Bbb{Z}:0\le n< N}$. Now the nice thing about $\Bbb{Z}/N\Bbb{Z}$ is that it has nice notions of addition and multiplication. In particular, these definitions satisfy the property that if we define $r:\Bbb{Z}\to\Bbb{Z}/N\Bbb{Z}$ to be the map that takes an integer $n$ to its remainder on division by $N$, then $r(n+m)=r(n)+r(m)$, and $r(nm)=r(n)r(m)$.

What about our $\Bbb{Q}/N\Bbb{Z}$? Well we would want that the following identity holds: $$\newcommand\of[1]{\left({#1}\right)}r\of{N\cdot\frac{1}{N}} = r(N)r\of{\frac{1}{N}}$$ however if we assume this holds, then (when $N>1$) $$1=r(1)=r\of{N\cdot \frac{1}{N}}=r(N) r\of{\frac{1}{N}} = 0\cdot \frac{1}{N} = 0,$$ which is not true.

Thus we can't have a nice notion of multiplication in our $\Bbb{Q}/N\Bbb{Z}$. (Abstractly the reason for this is that $N\Bbb{Z}$ is not an ideal of $\Bbb{Q}$, and there in fact can't be a nontrivial quotient of $\Bbb{Q}$ with a nice multiplication since $\Bbb{Q}$ is a field and has no nontrivial ideals.)

The core problem is that $1/N$ will be nonzero in $\Bbb{Q}/N\Bbb{Z}$, but $N$ is 0 in $\Bbb{Q}/N\Bbb{Z}$, and you can't divide by zero.

Thus briefly, there isn't a notion of order for rational numbers, at least not generally, or in quite the way you meant.

However, we can salvage things, sort of!

The barrier for defining multiplication on $\Bbb{Q}/N\Bbb{Z}$ is essentially that we would have to have a multiplicative inverse for 0, which is impossible. So what if we just disallow dividing by things that have a common divisor with $N$? Let $$S=\set{m\in\Bbb{Z}: \operatorname{gcd}(m,N)=1},$$ and let's define $S^{-1}\Bbb{Z}$ to be the following subset of $\Bbb{Q}$: $$\set{r\in\Bbb{Q}: \exists_{m\in S}: mr\in \Bbb{Z}}.$$ Intuitively, we can think of $S^{-1}\Bbb{Z}$ as rationals that can be expressed by fractions with denominators in $S$. You can check that this set is closed under addition and multiplication and hence forms a ring. Now we can take this ring mod $N$, (but this time, the notion of congruence is more like $x\equiv y$ if $x-y \in NS^{-1}\Bbb{Z}$) and I claim that everything works well, and we get a proper multiplication in the sense that it doesn't matter if we multiply things before or after we reduce mod $N$. And now, as long as $a$ can be expressed as a fraction with a denominator relatively prime to $N$, we can compute it's order mod $N$.

However, it will turn out that $S^{-1}\Bbb{Z}/NS^{-1}\Bbb{Z}$ is actually the same as $\Bbb{Z}/N\Bbb{Z}$, so every element of $S^{-1}\Bbb{Z}$ with no common factor with $N$ can be mapped to an invertible element of $\Bbb{Z}/N\Bbb{Z}$ with the same order, which behaves essentially the same way.