What is $\mathbb{Q}$?

Question

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

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

Answer 1

$\Bbb Q$ is indeed the equivalence classes of fractions, where two fractions are equivalent iff they can both be expanded / simplified to the same fraction. So yes, $\frac12$ and $\frac24$, while distinct fractions, are considered the same rational number.

Answer 2

Formally, it is the latter. However, you will (very) often see people just write $\mathbb Q=\{p/q\mid p,q\in\mathbb Z,q\neq0\}$ instead, and this is because we already are assuming equivalence relations, in the sense that we immediately consider $2/4$ and $1/2$ to be the same number; i.e. $2/4=1/2$. But in order to formally construct this, we have to use the construction by equivalence classes, which puts equality onto different ordered pairs of numbers to make them fractions in the usual sense. So, $2/4$ and $1/2$ are distinct before you impose the equivalence relation, equivalent (equal) after you do so, and the reason why writing $\mathbb Q=\{p/q\mid p,q\in\mathbb Z,q\neq0\}$ is legitimate is because the equivalence is implicitly assumed.

Answer 3

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

1. 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 $n=m$.

2. 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.