Addition in $R_S$ is well defined Let $R$ be a commutative ring with $1 \neq 0$ and suppose S is a multiplicatively closed subset of $R \backslash {\{\, 0 \,\} }$
containing no zero divisors. We have the relation ∼ defined on $R × S$ via $(a, b) ∼ (c, d) ⇔ ad = bc$.
Let $R_S$ denote the set of equivalence classes $\frac{a}{b}$ 
of $(a,b)$.
I want to show that addition in $R_S$, i.e., $\frac{a}{b} + \frac{c}{d}= \frac{ad+bc}{bd}$, is well-defined.
Not sure how to approach this. Any help would be appreciated. Thanks!

There is a standard method: Pick $(a^\prime, b^\prime)$ with $(a,b) \sim (a^\prime, b^\prime)$ (hence $ab^\prime = a^\prime b$) and show $\frac{ad+bc}{bd} = \frac{a^\prime d+b^\prime c}{b^\prime d}$. This is really straight forward.
Since the addition is symmetric, there is no need to also allow an other representative for the other summand. It would just make the calculation more unconvenient.
