The rationals form a field

By [L1], each axiom may be verified on fixed representatives.

Associativity of $+$: both $((a,b)+(c,d))+(e,f)$ and $(a,b)+((c,d)+(e,f))$ have representative $(adf+cbf+ebd,bdf)$.

Commutativity of $+$: $(ad+cb,bd)$ and $(cb+ad,db)$ are equal pairs.

Additive identity: $(a,b)+(0,1)=(a\cdot 1+0\cdot b,b\cdot 1)=(a,b)$.

Additive inverses: since $b\mathrm{\ne }0$, [L3] gives $b^2\mathrm{\ne }0$. Thus $(a,b)+(-a,b)=(ab-ab,b^2)=(0,b^2)\sim (0,1)$.

Associativity and commutativity of $\cdot$: $((ac)e,(bd)f)=(a(ce),b(df))$ and $(ac,bd)=(ca,db)$.

Multiplicative identity: $(a,b)(1,1)=(a,b)$; and $1\mathrm{\ne }0$ since $1\cdot 1\mathrm{\ne }0\cdot 1$.

Distributivity: $(a,b)((c,d)+(e,f))$ has representative $(a(cf+ed),b(df))$, while $(a,b)(c,d)+(a,b)(e,f)$ has representative $(acbf+aebd,b^{2}df)=(b(acf+aed),b(bdf))$; cancelling the common nonzero factor $b$ these are equivalent.

Inverses: $[(a,b)]=0$ iff $a\cdot 1=0\cdot b$ iff $a=0$; for $a\mathrm{\ne }0$, $(a,b)(b,a)=(ab,ba)\sim (1,1)$.

All field axioms hold: $\mathbb{Q}$ is a field.