Let $\displaystyle \bold{Z} \times (\bold{Z}- \{0 \}) = \{(a,b): a,b \in \bold{Z}, b \neq 0 \} $. This is a countable set. Now let us suppose that we have a function $\displaystyle f: \bold{Z} \times (\bold{Z}- \{0 \}) \to \bold{Q} $ defined by $\displaystyle f(a,b) := a/b $. Then $\displaystyle f(\bold{Z} \times (\bold{Z}- \{0 \})) = \bold{Q} $ which is at most countable. Since $\displaystyle \bold{Q} $ in infinite, then it must be countable. Then we can arrange the rational numbers in a sequence: $\displaystyle \bold{Q} = \{a_{0}, a_{1}, a_{2}, a_{3}, \ldots \} $.

How would you come up with a sequence, such that every element of the sequence is different from every other element, and the elements of the sequence exhaust $\displaystyle \bold{Q} $?