How is $\mathbb{Q}$ countably infinite?
The definition says all elements of the set must have a one-to-one relation to the natural numbers. I do not understand this.
How do the elements in $\mathbb{Q}$ have a one-to-one relation with natural numbers?
Because you can construct a mapping of the rationals into the integers.
One easy way is to map the positive rational $\dfrac{a}{b}$ to the integer $2^a 3^b$.
This shows that there are at least as many integers as there are positive rationals.
For the sake of completeness I will write one down here.
The rational numbers are arranged thus $$\frac {0}{1}, \frac {1}{1}, \frac {-1}{1},\frac {1}{2},\frac {-1}{2},\frac {2}{1 },\frac {-2}{1},\frac {1}{3},\frac {2}{3},\cdots $$
It is clear that every rational number will appear somewhere in this list. Thus it is possible to set up a bijection between each rational number and its position in the list, which is an element of $\mathbb N $.
Also see here. Hope it helps.
A roundabout way of showing this is to express the elements of $\mathbb{Q} - \{0\}$ as products of prime numbers. Labeling the prime numbers as $p_1, p_2, ...$ every nonzero rational number can be uniquely expressed as a product
$$p_1^{n_1} p_2^{n_2} \cdots$$
where $n_i$ are integers and for all but finitely many $i$, $n_i = 0$.
Your question becomes equivalent to asking why the set $A$ of sequences of integers $(n_1,n_2, ...)$ such that $n_i = 0$ for all but finitely many $i$, is countable.
To show this, let $A_j$ be the set of integer sequences $(n_1, ... , n_j, 0, 0, ...)$. Then $A_j$ is clearly in bijection with $\prod\limits_{i=1}^j \mathbb{Z}$, which is countable, and $A$ is equal to the union of the $A_j$. Now, you just have to show that a countable union of countable sets is countable.
