-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – YiFan Jan 4 '19 at 6:15
4
$\begingroup$

Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).

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.

$\endgroup$
  • $\begingroup$ Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated. $\endgroup$ – UmamiBoy Jan 4 '19 at 5:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.