Does the definition of rational numbers already assumes an understanding of them? I mean, how do we know that the correct equivalence relation is the equality of the crossed products? Does it already assume an intuitive informal understanding of rationals? If yes, what would that intuition be?
 A: Addendum to @fleablood 's correct answer.
Throughout mathematics, formal definitions always come after intuitive understanding. The intent of the formal definition is to capture that understanding so we can reason carefully even in places we don't quite yet understand.
You can see this as early as Greek geometry. The elementary theorems there were known long before Euclid's codification.
Newton and Leibniz and their followers developed calculus long before the $\epsilon - \delta$ definition of a limit provided rigor.
In your question about rational numbers you do have the right intuition based on years of elementary school arithmetic, so you can see that the definition captures that intuition.
Students often find new mathematics difficult because they see formal definitions before they have developed much intuition. It's hard to understand the definition of a group before you've actually looked at  different kinds of groups in different contexts. I think courses in abstract algebra usually move to abstraction too soon - but if they didn't then they couldn't "cover as much". That tension is hard to resolve.
A: What do you mean "correct" equivalence relation?
In a way the answer to your question is "yes" because or goal is to take our intuitive understanding of rations and formalize it in a way so that our formulation fits our intuition.  But in a more important way the answer to your question is "no", because our goal is that are formalization does not rely on any intuition and can apply to other mathematical structures that may prove very different than our intuition.
Our formulation is the $\mathbb Q$ is the set of equivalence classs on the relation on $\mathbb Z\times \mathbb N$ that $(a,b)\sim (c,d) \iff ad=bc$.  And for that formulation all we need to prove is that $\sim$ is an equivalence relation which is fairly straightforward to show.
There is no question that it is the "correct" equivalence relation.  Correct for what?
We can show that if we do have an intuitive idea of what that rationals are that this equivalence fits our intuition.  If we intuitively have some idea that $\frac ab$ is $a\div b$ and $\div$ is the inverse of $\times$ then if $a\div b = c\div d$ then $(a\div b)\times b = (c\div d)\times b$ so $a = (c\div d)\times b$ so $a\times d =(c\div d)\times b \times d = [(c\div d)\times d]\times b= c \times d$.
And alternatively if $ad = bc$ and $d\ne 0;c\ne 0$ then $(ad)\div d = (bc)\div d$ so $a =(bc)\div d$ so $a\div c = [(bc)\div d]\div c= [(bc\div c)]\div d = b\div d$.
So yes, it fits our intuition.
