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If I were to construct the field of rationals from integers, I might start with the Cartesian product of the integers, something like this: $S = \{(a, b) : a, b \in \mathbb{Z}, b \neq 0\}$.

$S$ is now a set of ordered pairs that look like rational numbers (eg, $(1, 2)$, $(4, 4)$, $(5, 5)$), except I now need at least an additive identity (which I lost by saying $b \neq 0$ to avoid division by 0), so I can define the addition operation on $S$ to be $(ad + bc, bd)$. I can now use $(0,1)$ as the additive identity (eg, $(0,1) + (4,5) = (0(5) + 4(1), 1(5)) = (4, 5)$).

Next, I at least need additive inverses, so I can just invert the first element of the ordered pair: $(-4, 5) + (4, 5) = (-20 + 20, 25) = (0, 25)$. This seems like my additive identity if I think of $\frac{0}{1} = \frac{0}{25}$, but I actually need to set up an equivalence relation to make those two equal, such as $(a, b) \sim (c, d) \leftrightarrow ad = bc$.

That is as far as I've gotten. Does that sound right so far? I do not want to proceed if I am misunderstanding something at this point.

tau
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    All good so far. All you need now is to define multiplication. – 5xum Jan 20 '20 at 07:35
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    You've more or less have it. But I'd put the equivalence classes early on. – fleablood Jan 20 '20 at 07:35
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    Yes, the quotient $S/\sim$ "is" already $\Bbb Q$. You have to show that the addition and multiplication defined are well-defined, i.e., sum and product donot depend on tha member of equivalence class picked ... – Hagen von Eitzen Jan 20 '20 at 07:35
  • oh right, i need the equivalence classes to make sure that my operation is even an operation (well defined). thank you! – tau Jan 20 '20 at 07:42
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    What about writing your pairs in a more visual way, say vertically aligned with a bar in between ? You could write identities such as

    $$-\frac ab\sim\frac{-a}b,$$ $$\frac ab+\frac cd\sim\frac{ad+bc}{bd},$$ $$\frac ab\frac cd\sim\frac{ac}{bd}$$

    and so on.

     –  Jan 20 '20 at 07:49
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    See e.g. here (and its links) for the general fraction field construction. – Bill Dubuque Jan 20 '20 at 08:08
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    You are not misunderstanding anything. Everything is going as it should. Although, in playing and working it, you are addressing issues as you come across and discover them rather than the proper order (you have to define addition before you can declare what the additive inverse is; and you have to define what the elements are [with equivalence classes] before you can do anything) but basically you are doing everything correctly. – fleablood Jan 20 '20 at 18:09

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