Re: less-than-uncool preschool
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Reply #74 –

Consider the set ℤ × (ℤ \ {0}) - that is, the set of all ordered pairs of integers, excluding those in which the second coordinate is zero. I'm going to define a relation on this set, and again I'll use the "∼" notation for relatedness according to this relation.

Let *a*, *b*, *c*, *d* ∈ ℤ, where *b* ≠ 0 and *d* ≠ 0. Let's say that

(*a*, *b*) ∼ (*c*, *d*)

whenever *ad* = *bc*.

This relation is reflexive: (*a*, *b*) ∼ (*a*, *b*) because *ab* = *ba*.

This relation is symmetric: If (*a*, *b*) ∼ (*c*, *d*), then *ad* = *bc*, so *cb* = *da*, so (*c*, *d*) ∼ (*a*, *b*).

This relation is transitive: If (*a*, *b*) ∼ (*c*, *d*) and (*c*, *d*) ∼ (*e*, *f*), then *ad* = *bc* and *cf* = *de*, and we have two cases. If *c* = 0, then *bc* = *cf* = 0, so *ad* = *de* = 0. But *d* ≠ 0, so the product law of arithmetic tells us that *a* = *e* = 0, and we have *af* = *be* = 0. If *c* ≠ 0, then we have *adcf* = *bcde*, so by the law of cancellation (for multiplication), taking into account that *cd* ≠ 0, we again have *af* = *be*. Either way, (*a*, *b*) ∼ (*e*, *f*).

So we have an equivalence relation.

Now consider the set of equivalence classes that we can derive from this relation.

Let's treat these equivalence classes as numbers in their own right. For a suitable choice of how to define the operations of addition and multiplication on these equivalence classes, they turn out to have all the familiar properties of the rational numbers. So let's go ahead and *define* a "rational number" as one of these equivalence classes.

I'll use the notation *a*/*b* as a shorthand for the equivalence class [(*a*, *b*)]. (Peeking ahead toward stuff I haven't yet defined, we might think of this equivalence class as the "quotient" resulting from dividing *a* by *b*.)

In this notation *a*/*b*, the number before the "/" is the "numerator", while the number after the "/" is the "denominator".