Please forgive my cursory knowledge of modular arithmetic. My question is straightforward: is there a property of modular arithmetic that simplifies $\frac{a}{b}\mod n$?
I attempted using the fact that $a\cdot b\mod n \equiv ((a\mod n)\cdot(b\mod n))\mod n$ but found that $\frac{a\cdot n}{n} \mod n \equiv a \mod n$ does not hold.
Thank you. :)
It depends on what $\frac ab \mod n$ means.
Consider $\frac 84 \mod 6$. If you mean that to mean $q \mod 6$ where $q = \frac 84 = 2\in \mathbb Z$ then this is nothing more or less then $2 \mod 6$. No issue.
But if $q = \frac 13 \not \in \mathbb Z$ then $q\mod 6$... what do you mean?
You could, but probably don't, mean $[q]= \{q + k6|k \in \mathbb Z\}$ in many analysis/calculus/geometry context this make sense. In particular $\theta \mod 2\pi =\{\theta + 2k\pi\}$ has obvious practical applications.
But we almost never mean this in number theory. In number theory it is assume the we are working in the number system $\mathbb Z/n\mathbb Z = \{$ the set of complete residue systems $\mod n$. In which case $[x]$ when $x$ is not an integer simply makes no sense.
But In the case where $a*x \equiv 1 \mod n$ then $x$ acts practically as a multiplicative inverse to $a$ and we wish to say $x = a^{-1}\mod n$ (which means, by definition, that $ax \equiv 1 \mod n$. And we do right that as $x\equiv \frac 1a \mod n$.
Exampe $3*5 \equiv 1 \mod 14$ so $3 \equiv \frac 15 \mod 14$. It's important to realize that $\frac 15$ has NOTHING to do with the rational number $0.2$ but simply means ... the $x$ so that $x*5 \equiv 1 \mod 14$.
So in this case $\frac ab \mod n$ means the number $x$ so that $bx \equiv a \mod n$.....
.... if such a residue class exists
.... and if it is distinct.
There are two catches: 1) it might not exists; $\frac 32 \mod 6$ is is the integer $x$ so that $2x \equiv 3 \mod 6$. There is no solution to that. 2) there might be multiple solutions; $2*6\equiv 8*6\equiv 3 \mod 9$ so $\frac 36 \mod 9$ would be ill-defined.
But $\frac ab \mod n$ will have a distinct solution if and only if $b$ and $n$ are relatively prime. Then there will be a solution to $bx = 1 + kn$ and it is unique $\mod n$. And $\frac ab \equiv ax\mod n$ would be the unique value.
If $\gcd(b,n)=d\ne 1$ and $a$ is not a multiple of $d$ then there is no solution to $bx = a + kn$ and $d|b$ and $d|n$ but $d\not \mid a$. And if $a = md$ then $bx = a +kn$ will have a solution. But $b(x + \frac nd) =bx + \frac bd*n \equiv bx \equiv a \mod n$ is also a solution.
tl;dr
In number theory $\frac ab$ means the unique residue solution to $bx \equiv a \mod n$ which only exists if $\gcd(b, n) =1$ and which is equal to $ay\mod n$ where $y\equiv \frac 1b \mod n$ is the unique solution to $by \equiv 1 \mod n$.