## § Operations with modular fractions

- Quick note on why it's legal to perform regular arithmetic operations on fractions $a/b$ as operations on $ab^{-1}$ where $ab^{-1} \in \mathbb Z/pZ$.
- The idea is that we wish to show that the map $a/b \mapsto ab^{-1}$ is a ring homomorphism $\phi: \mathbb Q \to \mathbb Z/p \mathbb Z$.
- The proof: (i) the map $Z \rightarrow Z/pZ$ is a ring homormophism, (ii) map from an integral domain to a field always factors through the field of fractions of the domain, we get a map $\phi: \mathbb Q \rightarrow \mathbb Z/ p \mathbb Z$. So from abstract nonsense, we see that $\phi$ will be a well defined ring.hom.
- More down to earth: let's check addition multiplication, and multiplicative inverse. All else should work automagically.
- For addition, we wish to show that $\phi(a/b + c/d) = \phi(a/b) + \phi(c/d)$. Perform the calculation:

\begin{aligned}
&\phi(a/b + c/d) \\
&=\phi((ad + bc)/bd) \\
&= (ad + bc)(bd)^{-1}\\
&= abb^{-1}d^{-1} + bcb^{-1}d^{-1} \\
&= ad^{-1} + cd^{-1} \\
&= \phi{a/d} + \phi{c/d} \\
\end{aligned}
- For multiplication, we wish to show that $\phi(a/b \cdot c/d) = \phi(a/b) \cdot \phi(c/d)$:

\begin{aligned}
&\phi(a/b \cdot c/d) \\
&=\phi{ac/bd}
&= ac(bd)^{-1} \\
&= acd^{-1}b^{-1} \\
&= ab^{-1} \cdot cd^{-1} \\
&= \phi{a/b} \cdot \phi{c/d} \\
\end{aligned}
- For inverse, we wish to show that $\phi(1/(a/b)) = \phi(a/b)^{-1}$:

\begin{aligned}
&\phi(1/(a/b))
&=\phi{b/a}
&= ba^{-1}
&= (ab^{-1})^{-1}
&= \phi(a/b)^{-1}
\end{aligned}
Thus, we can simply represent terms $a/b$ in terms of $ab^{-1}$ and perform arithmetic as usual.