Prove that the rational numbers are countable. 
 

I see that the hint give me the answer, so if I were to prove the statement I will write the hint but I do not think that the proof is so trivial like this. could anyone show me how he/she will write the proof in terms of this hint please.
Also I have another question from where did $Q_{0}$ come? if $k=0$ we can not divide by 0 in the definition for $Q_{k}$, so may be $Q_{0}$ comes from $j \in \mathbb{N}$, but our union is over k which belongs to $\mathbb{N}$, so k may be equal 0. I got confused, could anyone explain this for me please?  
 A: If you are allowed to use that a countable union of countable sets is countable then things are not so difficult:
Put $A(i)=\{\frac{i}{j}| j \in \mathbb{Z}$ \ $\{0\}\}$ which is a countable set because $\mathbb{Z}$ \ $\{0\}$ is countable.
Thus we have $\mathbb{Q}=\bigcup_{i \in \mathbb{Z}}A(i)$ which is a countable union of countable sets.
This is not exactly the way your book proposes but its very similar.
I believe you get the idea.
A: The other answer works but it's not intuitive and the formula given falls from the sky.
The initial idea is correct. That every positive rational number can be put in lowest terms, and that these representations inject into $\mathbb{N} \times \mathbb{N}$ means that all we have to do is show this is countable, and apply the fact that the union of two countable sets is countable (this can be done by weaving together the sets, the same way the odds and evens can be woven together), and then add in $0$.
Here is a more natural map. We will inject $\mathbb{N} \times \mathbb{N} \to \mathbb{N}$. Since there is an obvious injection in the reverse direction, the two are bijective by Cantor-Bernstein.
The injection is given by mapping $(a,b) \mapsto 2^a3^b$. Since every such number will be unique (because $2$ and $3$ are prime), this is an injection. 
So we have a sequence of injections $\mathbb{Q} \to \mathbb{N} \times \mathbb{N} \to \mathbb{N}$, and an obvious injection $\mathbb{N} \to \mathbb{Q}$ given by the inclusion, and so again by Cantor-Bernstein, we have a bijection, and so the positive rationals are countable.
To include the negative rationals, use the argument we outlined above. We have shown that the positive rationals are countable. The same argument shows the negative rationals are countable. Now fix an ordering of each of them, so we have what amount to sequences $(a_1, a_2, a_3, ...)$ and $(b_1, b_2, b_3)$ where $b_i = -a_i$. Then the sequence $(a_1, b_1, a_2, b_2, ...)$ will do the job.
Lastly, we need to add $0$. But we know that a countable set union a singleton is still countable. So this completes the proof.
This argument has the nice property that it tells us how to prove that any finite product of countable sets is countable, all we have to do is just use more prime numbers.
A: As Omnomnonom said, we just need to prove that $A=\Bbb N\times \Bbb N $ is countable.
for this, consider the map
$$f:(i,j)\mapsto n=\frac {(i+j)(i+j+1)}{2}+j$$
$f $ is a bijection from $A $ to $\Bbb N $.
in fact for $n\in \Bbb N $ there exist a unique $(i,j) $ given by
$$j=n-\frac {d (d+1)}{2} $$
with
$$d=\lfloor \frac {-1+\sqrt {8n+1}}{2} \rfloor $$
and
$$i=d-j $$
such that $f (i,j)=n $
done.

$f $ is the triangle numbering.

according to this, the members of $A $ are classified as
$$(0,0),(1,0),(0,1),(2,0),(1,1),(0,2),(3,0).... $$
