Rational numbers definition recursive?

Question

I have little to no experience with number theory, so it is somewhat difficult for me to communicate this question, but here goes:

My understanding of rational numbers is that they are generated by integers (I may already be wrong here). The definition I was given years ago in school was as follows: "a rational number is a fraction of two integers." This was fine and good, but after giving it some thought, I ran into a logical hurdle. I take "a fraction of two integers" to be equivalent to "the division of two corresponding integers." The problem for me is in the definition of division. I take division to be the backwards mapping of multiplication (maybe wrong). Provided that I only have integers at my disposal, multiplication would be defined as the product of two integers, which itself would be an integer. Therefore, you would have a mapping of two integer inputs into a single integer output. Conversely, division would be the opposite of this process, and would seem to only be defined for a subset of pairs of integers, say [4 and 2 -> 2], but not [1 and 2 -> .5]. So it seems recursive to me to define rational numbers as the division of two integers, because division, (as I defined it, which could be incorrect) only maps back from the product of two integers. Rational numbers (it seems) would have to exist already for m/n to be valid for any m which isn't composite of n.

Again, this could all just be garbage reasoning, but I don't understand how you can define a numbers system by an operation which only applies if the number system already exists. There is probably a gaping hole in my mathematical reasoning here, so please be patient with me.

Answer 1

I take "a fraction of two integers" to be equivalent to "the division of two corresponding integers."

Not really. "Division" is an operation, a fraction is an entity (an element of some set).

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There are several mathematically rigorous (and equivalent, of course) methods of defining fractions. Most of them assume that you already defined the set $\mathbb Z$.

The best way I know of to define rational numbers is by defining them as equivalence classes on $\mathbb Z\times\mathbb Z\setminus\{0\}$. The process is as follows:

1. You define a relation $\sim $ on $\mathbb Z\times\mathbb Z\setminus\{0\}$ as $$\forall a,b,c,d\in\mathbb Z:(a,b)\sim(c,d)\iff ad=cb$$
2. You take the set of equivalence classes and define that to be $\mathbb Q$, i.e. $\mathbb Q = \mathbb Z/\sim$
3. You define adition on $\mathbb Q$ as $$[(a,b)] + [(c,d)] = [ad + bc,bd]$$ and multiplication as $$[(a,b)]\cdot [(c,d)] = [(ac, bd)].$$ It is relatively simple to show that the two operations are well defined.
4. You define the inequality relation on $\mathbb Q$ as $$[(a,b)]\leq [(c,d)]\iff a\cdot |d|\cdot \mathrm{sign}(b)<c\cdot|b|\cdot\mathrm{sign}(d) $$ and again, you can show that this is a well defined partial ordering on $\mathbb Q$.

Once you have that, you can show that the mapping $$\mathbb Z\to\mathbb Q\\ x\mapsto [(x,1)]$$

is an injection that respects all the structures you have so far (it's a injective ring homomorphism that maintains the ordering). Once that is done, you basically equate $\mathbb Z$ with $\{[(x,1)]| x\in\mathbb Z\}$ and you refer to the elemens of $\mathbb Q$ as $\frac{a}{b}$ instead of $[(a,b)]$.

Answer 2

More logically correct is this: The set of rational number is the collection of "equivalence classes of pairs or integer such that the second member of the pair is not 0" with equivalence defined by "(a, b) is equivalent to (c, d) if and only if ad= bc". We further define addition of two rational numbers, x and y, by "take (a, b) as a representative of x (that is, a pair (a, b) in the equivalence class x) and (c, d) as a representative of y (again (c, d) is in the equivalence class y). "x+ y" is the equivalence class containing (ad+ bc)/bd." Similarly we define multiplication by "xy is the equivalence class containing (ac, bd). Note that, since neither b nor d can be 0, bd is not 0.

We then think or the "fraction" $\frac{a}{b}$ as meaning the equivalence class containing the pair (a, b). It is easy to see that, for example, the equivalence class containing (1, 2) also contains (2, 4), (4, 8), etc., all the fractions that are "equal" to 1/2.