# Proving the Set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators.

I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!

Assume $$\mathbb{Q}$$ is the union of disjoint equivalence classes over ratios of integers. Let the set of rational numbers be defined as such. $$$$\mathbb{Q} = \bigg\{ \frac{p}{q}:p,q \in \mathbb{Z} , q \neq 0 \bigg\}$$$$ We may equivalently define the following as a definition of the rationals. $$$$\mathbb{Q} = \bigg\{ \frac{a}{b}:(a,b) \in \mathbb{Z} \times \mathbb{Z} \setminus \{ 0 \} \bigg\}$$$$

Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $$\mathbb{Q}$$ cannot be the union of disjoint equivalence classes over ratios of integers.

• What? Can you come up with a pair $(a,b)\in\mathbb Z\times\mathbb Z\setminus\{0\}$ where equivalence classes cannot be defined in the usual sense? – YiFan Jan 4 '19 at 6:15

You seem to be confusing the fact that an equivalence relation on $$A$$ is a certain subset of $$A\times A$$ with the fact that in this case $$A$$ itself is a cartesian product.
Here the equivalence relation is a certain subset of $$(\mathbb Z\times(\mathbb Z\setminus\{0\}))\times(\mathbb Z\times(\mathbb Z\setminus\{0\}))$$ and there's nothing that prevents such a relation from being reflexive.
More precisely, the relation is $$\{ ((a,b),(p,q)) \mid aq=pb \}$$ which is reflexive because $$((a,b),(a,b))$$ is in the relation for every $$a\in\mathbb Z$$, $$b\in\mathbb Z\setminus\{0\}$$, because $$ab=ab$$ is always true.