Existence of a function from $f : \mathbb Z^2 \rightarrow \mathbb Z$ I have a problem with the following question:

Does there exist a function $f : \mathbb Z^2 \rightarrow \mathbb Z$ that is one-to-one and onto, and hence invertible?
  (If yes, then $\mathbb Z^2$
  can be called "countably inﬁnite.") If such a function exists then give a
  clear description of it. If such a function does not exist then explain why not.

Now, I cannot think of any function that satisfies this. How can there be an inverse function from $\mathbb Z \rightarrow \mathbb Z^2 $? If a function of this type doesn't exist, how I can I explain so?
 A: It may be hard to find an explicit function with these properties, so instead try "proving that it exists".  For example, think of elements of $Z\times Z$ as points in the Euclidean plane, and show that there are only finitely many of then inside any circle, so that "there exists" a finite list giving all of them, etc.
A: Simple explicit example of $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ bijection is pictured below.
This isn't a very formal definition, but I would say that it should pass as a "clear description" (in particular there are no ambiguities or doubts about whether the function is really 1-to-1 and onto).
$\hspace{70pt}$
If you are looking for formulas, then the simplest way would be to go through $\mathbb{N}\times\mathbb{N} \to \mathbb{N}$ bijections. My favorite can be found here along with some other (even simpler) examples.
I hope this helps $\ddot\smile$
A: Regarding to @Andre's comment, why don't you employ the function $(n,m)\to 2^i3^j$ for showing that $$\mathbb N\times\mathbb N$$ is  infinitely countable.
A: There is a way to do this that seems to me more appropriate than relying on "explicit" functions.
First, you use the function:
$\phi: \mathbb{N}\times \mathbb{N} \rightarrow \mathbb{N}$
given by $\phi(n,m)=2^n.3^m$
This function is injective.(#1 why?) Here, we could invoke the Cantor-Bernstein-Schroeder theorem (with the way back being obvious), but we can use the fact that any infinite subset of a countable set is countable (#2 why?) (meaning we can make a bijection $\nu$ from it to $\mathbb{N}$)
So, call "$A$" the image of $\phi$. Then we have:
$\displaystyle \mathbb{N}\times \mathbb{N} \xrightarrow{\phi} A \xrightarrow{\nu} \mathbb{N}$
And the composition is a bijection from $\mathbb{N}\times \mathbb{N}$ to $\mathbb{N}$.
Now, a bijection from $\mathbb{Z}$ to $\mathbb{N}$ is easy (#3 why?) (Hint: just "count" the integers going forwards and backwards - formalize this as an exercise, if you don't know.)
It rests to show that there is a bijection from $\mathbb{Z}\times \mathbb{Z}$ to $\mathbb{N}\times \mathbb{N}$. But this follows from the bijection from $\mathbb{Z}$ to $\mathbb{N}$ (#4 why?)
A: You can first map $\mathbb Z^2 \rightarrow \mathbb N$ by using a $g$ that maps:
$$ (0,0) \rightarrow 0$$
$$ (1,0) \rightarrow 1$$
$$ (1,1) \rightarrow 2$$
$$ (0,1) \rightarrow 3$$
etcetera, in an outward spiral shape. Then you can map $\mathbb N \rightarrow \mathbb Z$ by using the function $h(x)$ that maps $h(2n) = n+1$ and $h(2n+1) = -n$.
The mapping $f=g\circ h$ has the properties you are looking for.
