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. \begin{equation} \mathbb{Q} = \bigg\{ \frac{p}{q}:p,q \in \mathbb{Z} , q \neq 0 \bigg\} \end{equation} We may equivalently define the following as a definition of the rationals. \begin{equation} \mathbb{Q} = \bigg\{ \frac{a}{b}:(a,b) \in \mathbb{Z} \times \mathbb{Z} \setminus \{ 0 \} \bigg\} \end{equation}
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.
Says who?
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.