How do I show that the following map establishes a bijection between $\mathbb Q$ and $\mathbb Z \times \mathbb{Z}_{>0}$ Define $$f:\mathbb Z \times \mathbb{Z}_{>0} \to \mathbb{Q}$$ by $$f(p, q) = \frac{p}{q} $$
Edit: Is there an explicit map then that is a bijection?
 A: It doesn't; $f$ is not injective:
$f(2, 4) = \frac{2}{4} = \frac{1}{2} = f(1, 2)$
A: You asked for an explicit map that is a bijection.  
The map $f$ that you've defined is a surjection, but it's not injective.  How can we make it injective?  Well, we first construct a bijection $g$ from $\mathbb Z_{>0}$ to $\mathbb Z\times\mathbb Z_{>0}$
\begin{align}
\color{#c00}{g(1)=}(0,&1)\\
\color{#0a0}{g(2)=}(-1,&1)\\
\color{#0a0}{g(3)=}(0,&2)\\
\color{#0a0}{g(4)=}(1,&1)\\
\color{#00a}{g(5)=}(-2,&1)\\
\color{#00a}{g(6)=}(-1,&2)\\
\color{#00a}{g(7)=}(0,&3)\\
\color{#00a}{g(8)=}(1,&2)\\
\color{#00a}{g(9)=}(2,&1)\\
g(10)=(-3,&1)
\end{align}
Can you see how the pattern continues?
Now $f\circ g$ is a surjection from $\mathbb Z_{>0}$ to $\mathbb Q$.  How do we convert it into a bijection?  Define a function $h\colon \mathbb Z_{>0}\to\mathbb Z_{>0}$ by
$$
h(n)=\min\{m\colon\textrm{ There exists a sequence }m_1<\dots<m_n=m\textrm{ such that }\\f(g(m_1)),\dots,f(g(m_n))\textrm{ are distinct.}\}
$$
Exercise: Show that $f\circ g\circ h$ is a bijection from $\mathbb Z_{>0}$ to $\mathbb Q$.
Then our required bijection from $\mathbb Z\times\mathbb Z_{>0}$ to $\mathbb Q$ is given by
$$
f\circ g\circ h\circ g^{-1}
$$
This should illustrate the general point: bijections from $\mathbb Z$ (and similar sets) to $\mathbb Q$ exist, but in general they are difficult to construct, they don't have nice formulae and they aren't intuitive.  
