TITLE: What is $\mathbb{Q}$?
QUESTION [6 upvotes]: When we say set of rationals $\mathbb{Q}$, which of the following does it refer to?
$$\left\{\frac{p}{q}~|~p,q\in\mathbb{Z},q\neq 0\right\}$$
or
$$\left\{\left[\frac{p}{q}\right]~|~p,q\in\mathbb{Z},q\neq 0\right\}$$
where $[a]$ denote the equivalence class under the equivalence relation: 
$$\frac{p}{q}\sim \frac{r}{s}\qquad \text{if}~ps=qr.$$
In other words, the second set is just the fractions with $\gcd(p,q)=1$.
The equivalent question can be: are $\frac{1}{2}$ and $\frac{2}{4}$ distinct rational numbers?
While showing $\mathbb{Q}$ is countable, using snake diagram arguments, they use first set, i.e. all expressions of the form $\frac{p}{q}$. And then an enumeration of these expressions is done. This is acceptable proof of countability if our definition is the first set. However if we take the second definition, then I feel these proofs (of countability) are incomplete: one also needs to show that 

Each of the equivalence class $\left[\frac{p}{q}\right]$ is countable. -- follows as one can give a one-to-one correspondence with $\mathbb{Z}$.
Countable union of countable sets us countable -- a proof for this is similar to the snake diagram argument.

REPLY [6 votes]: Forget about naive mathematics. What exactly $\frac{p}{q}$ is? If it is the same as the ordered pair $(p,q)$ then why bother with a different symbol? No, it's something else.
And indeed, you start with $X=(\mathbb{Z}\times\mathbb{Z})\backslash(\mathbb{Z}\times\{0\})$ and on that set you define relation
$$(p,q)\sim (r,s)\text{ iff }ps=qr$$
and then what you call a fraction is 
$$\frac{p}{q}:=[(p,q)]$$
$$\mathbb{Q}:=X/\sim$$
So as you can see $[\frac{p}{q}]$ doesn't even make sense. $\frac{p}{q}$ already is an equivalence class. At least formally.
And under this construction we have $(1,2)\sim (2,4)$ and so $\frac{1}{2}=\frac{2}{4}$. Here the equality sign "$=$" means "equal as sets". And there is no ambiguity here.
So formally rationals are sets. But that should not be surprising. I mean integers are sets as well. And naturals. And relations. And functions. And ordered pairs. At the end of the day (almost) everything in math is a set.

While showing $\mathbb{Q}$ is countable, using snake diagram arguments, they use first set, i.e. all expressions of the form $\frac{p}{q}$.

Or more precisely what I defined earlier as $X$. Yes, typical proofs show that $X$ is countable. But then $\mathbb{Q}$ is at most countable, since it is a surjective image of $X$ (via the quotient function $X\to X/\sim$). It can be easily shown that $\mathbb{Q}$ cannot be finite since $(x,1)\sim (y,1)$ if and only if $x=y$. And so it has to be infinite countable.

However if we take the second definition, then I feel these proofs (of countability) are incomplete: one also needs to show that 

Each of the equivalence class $\left[\frac{p}{q}\right]$ is countable. -- follows as one can give a one-to-one correspondence with $\mathbb{Z}$.


No, whether equivalence classes are countable or not is irrelevant. For example consider $Y/\sim$ where $x\sim y$ for all $x,y$. Then $Y/\sim$ has exactly one element even though the single equivalence class (being $Y$) can be arbitrarly large.
Of course every equivalence class of our original $X$ has to be at most countable since it is a subset of $X$ which (as we've established) is countable. It is precisely infinite countable because $(p,q)\sim(np,nq)$ for any $n\in\mathbb{Z}\backslash\{0\}$ and $(np,nq)=(mp,mq)$ if and only if $n=m$.


Countable union of countable sets us countable -- a proof for this is similar to the snake diagram argument.


As before: irrelevant to the problem.
In fact your points 1. and 2. are useful to show that if $\mathbb{Q}$ is countable then so is $X$ (being a countable union of countable sets). But that's not what you're trying to prove.