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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?

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  • $\begingroup$ I think the addition comes from the addition of real numbers. $\endgroup$ – Aniruddha Deshmukh Nov 7 '18 at 9:42
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    $\begingroup$ Is some of your post missing? You say "as given below", but there is no corresponding definition below. $\endgroup$ – Sam Streeter Nov 7 '18 at 9:45
  • $\begingroup$ Thank you to those that edited my original post. $\endgroup$ – user150203 Nov 7 '18 at 9:52
  • $\begingroup$ @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? $\endgroup$ – user150203 Nov 7 '18 at 9:56
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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.

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    $\begingroup$ 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 $\endgroup$ – Sam Streeter Nov 7 '18 at 9:50
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    $\begingroup$ @SamStreeter - Thanks for that. I will look into it :-) $\endgroup$ – user150203 Nov 7 '18 at 9:52
  • $\begingroup$ @Jose - Thank you for your post. So, am I correct to infer that the definition of addition is axiomatic? $\endgroup$ – user150203 Nov 7 '18 at 9:55
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    $\begingroup$ @DavidG No. It is defined. I provided a definition, right?! $\endgroup$ – José Carlos Santos Nov 7 '18 at 9:57
  • $\begingroup$ @JoséCarlosSantos - Yes you did. I was just clarifying to make sure I hadn't misinterpreted you. This topic has consumed me, so I hope I don't come across as rude if I ask the obvious! haha $\endgroup$ – user150203 Nov 7 '18 at 9:58
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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.

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  • $\begingroup$ I do not deny that, but forgetting that extension. If we just consider the set of Rationals as a composition of integers. Then is the definition of addition an axiom or is it derived? $\endgroup$ – user150203 Nov 7 '18 at 9:54