Is it possible to do modulo of a fraction

Solution 1:

One can perform arithmetic of fractions mod $\,m\,$ as long as all fractions $\,a/b\,$ have denominator $\,b\,$ coprime to $\,m,\,$ since then, by Bezout, $\,b\,$ is invertible mod $\,m\,$ so the fraction has the unique denotation $\,x = a/b = ab^{-1}$ (the unique solution of $\,bx = a).\,$ The usual rules of fraction arithmetic remain true (as long as one restricts to such fractions), e.g. from a prior answer:

Hint $\,\ {\rm mod}\ 13\!:\ \dfrac{41}7 \equiv \dfrac{28}7 \equiv 4\ \ $ by $\ \ 41\equiv 41\!-\!13 \equiv 28$

Alternatively $\,\ \dfrac{41}{7}\equiv\dfrac{(-2)(-1)}{-6}\equiv \dfrac{-2}{-2}\dfrac{12}3\equiv 4\ \ $ by $\ \ \begin{eqnarray}41&&\!\!\equiv\ \ 2\\ 7 &&\!\!\equiv -6\end{eqnarray}$

Alternatively $\,\ \dfrac{41}{7}\equiv \dfrac{2}7\equiv \dfrac{4}{14}\equiv \dfrac{4}1\ $ by Gauss's Algorithm.

Such twiddling (adding/subtracting the modulus from numerator or denominator till things divide or factor nicely) works quite well for small numbers. For larger numbers one can invert the denominator by the Extended Euclidean Algorithm, or Gauss's algorithm if the modulus is prime.

Beware $\ $ The use of fractions in modular arithmetic is valid only when the denominator is invertible, i.e. coprime to the modulus. Otherwise the quotient need not be unique, for example mod $\rm\:10,\:$ $\rm\:4\,x\equiv 2\:$ has solutions $\rm\:x\equiv 3,8,\:$ so the "fraction" $\rm\:x \equiv 2/4\pmod{10}\,$ cannot designate a unique solution of $\,4x\equiv 2.\,$ Indeed, the solution is $\rm\:x\equiv 1/2\equiv 3\pmod 5,\,$ which requires canceling $\,2\,$ from the modulus too, since $\rm\:10\:|\:4x-2\iff5\:|\:2x-1.\:$

Generally the grade-school rules of fraction arithmetic apply universally (i.e. in all rings) where the denominators are invertible. This fundamental property will be clarified conceptually when one learns in university algebra about the universal properties of fractions rings and localizations.

Solution 2:

One natural way to define the modular function is $$a \pmod b = a-b\left\lfloor \frac{a}{b}\right\rfloor$$ where $\lfloor \cdot \rfloor$ denotes the floor function. This is the approach used in the influential book Concrete Mathematics by Graham, Knuth, Patashnik.

This will give you $1/2 \pmod 3 = 1/2$.