# Abelian Group of Rationals over addition - axiomatic or derived? [closed]

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?

• I think the addition comes from the addition of real numbers. – Aniruddha Deshmukh Nov 7 '18 at 9:42
• Is some of your post missing? You say "as given below", but there is no corresponding definition below. – Sam Streeter Nov 7 '18 at 9:45
• Thank you to those that edited my original post. – user150203 Nov 7 '18 at 9:52
• @AniruddhaDeshmukh - Yes, I agree it's defined elsewhere, but in just looking at the Rationals on their own. Is the definition of addition a derived operator or axiomatic? – user150203 Nov 7 '18 at 9:56

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.
• Just to add a little something for those curious about this viewpoint: this is an example of the localization of a ring. Here we are thinking of $\mathbb{Q}$ as the localization of the ring $\mathbb{Z}$ with respect to the multiplicative system $\mathbb{Z} \setminus \{0\}$. For more, see en.wikipedia.org/wiki/Localization_of_a_ring – Sam Streeter Nov 7 '18 at 9:50
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.