Sorry if this is a daft question, but if we consider the set of rationals under the 'addition' operator we can form an Abelian Group.
I'm curious: in that situation, is the definition of addition (as given below) derived or axiomatic?
I ask as without knowing the multiplicative inverse of rationals, how can we derive the definition of addition? Moreover, shouldn't they be independent of one another?
On
Its a general construction: Each integral domain $R$ can be extended to a field $K$ by using the equivalence relation on $R\times (R\setminus \{0\})$, $$(a,b)\sim (c,d) :\Longleftrightarrow ad = bc.$$ The equivalence class of $(a,b)$, $b\ne 0$, is $$\frac{a}{b} = \{(c,d)\mid (a,b)\sim (c,d)\},$$ the set of all fractions with equal value. The quotient set $$K = \{\frac{a}{b}\mid a,b\in R,b\ne 0\}$$ becomes a field with the operations $$\frac{a}{b} + \frac{c}{d} := \frac{ad+bc}{bd}$$ and $$\frac{a}{b} \cdot \frac{c}{d} := \frac{ac}{bd}.$$ The mapping $\phi:R\rightarrow K:a\mapsto \frac{a}{1}$ is a ring monomorphism so that $R$ can be viewed as a subring of $K$. Hope it helps.
The addition is defined. More precisely, if you define $\mathbb Q$ as the set of equivalence classes of $\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})$ with respect to the equivalence relation$$(a,b)\sim(c,d)\text{ if and only if }ad=bc,$$then you define$$\bigl[(a,b)\bigr]\times\bigl[(c,d)\bigr]=\bigl[(ad+bc,bd)\bigr].$$This only requires that you know how to multiply (and add) integers.