The localisation relation is an equivalence relation and fraction arithmetic is well defined

Reflexivity has witness $1$, and symmetry uses the same witness with the negative equation. If $u(r{s}^{\mathrm{\prime }}-r^{\mathrm{\prime }}s)=0$ and $v(r^{\mathrm{\prime }}{s}^{\mathrm{\prime }\mathrm{\prime }}-r^{\mathrm{\prime }\mathrm{\prime }}{s}^{\mathrm{\prime }})=0$, then $uv{s}^{\mathrm{\prime }}\in S$ and the identity $s^{\mathrm{\prime }}(r{s}^{\mathrm{\prime }\mathrm{\prime }}-r^{\mathrm{\prime }\mathrm{\prime }}s)=s^{\mathrm{\prime }\mathrm{\prime }}(r{s}^{\mathrm{\prime }}-r^{\mathrm{\prime }}s)+s(r^{\mathrm{\prime }}{s}^{\mathrm{\prime }\mathrm{\prime }}-r^{\mathrm{\prime }\mathrm{\prime }}{s}^{\mathrm{\prime }})$ shows $uv{s}^{\mathrm{\prime }}(r{s}^{\mathrm{\prime }\mathrm{\prime }}-r^{\mathrm{\prime }\mathrm{\prime }}s)=0$; hence the relation is transitive.

Suppose $(r,s)\sim (\rho ,\sigma )$ with witness $u$. For fixed $(a,t)$, multiplying the relation by $t^2$ shows $(rt+as,st)\sim (\rho t+a\sigma ,\sigma t)$, and multiplying it by $at$ shows $(ra,st)\sim (\rho a,\sigma t)$. Changing the second representative in the same way proves both operations are well defined.

Associativity, commutativity, and distributivity follow by expanding the displayed fraction formulas over common denominators. The classes $0\mathrm{/}1$ and $1\mathrm{/}1$ satisfy the zero and identity laws, and additive inverses are $(-r)\mathrm{/}s$.

The formulas give ${\lambda }_S(r+r^{\mathrm{\prime }})={\lambda }_S(r)+{\lambda }_S(r^{\mathrm{\prime }})$, ${\lambda }_S(r{r}^{\mathrm{\prime }})={\lambda }_S(r){\lambda }_S(r^{\mathrm{\prime }})$, and ${\lambda }_S(1)=1\mathrm{/}1$. Finally $(s\mathrm{/}1)(1\mathrm{/}s)=s\mathrm{/}s=1\mathrm{/}1$, since $(s,s)\sim (1,1)$ with witness $1$.