What is the Simplest Explanation for the Countability of the Integers? What is the simplest (or at least simple to understand) if one wanted to explain why the set of Integers has the same cardinality as the set of natural numbers to students who have a vague idea of why sets such as the whole numbers, even numbers, odd numbers have the same cardinality as that of the natural numbers (by establishing a 1-1 correspondence), but have had no courses in set theory or topology---so don't know of the arguments that underlie things like the countable union of countable sets is countable.
I do not want to iterate something like $Z$ is a subset of $Q$ and therefore countable. I want to try to explain it directly.   
It seems that a straightforward 1-1 argument doesn't apply here.
 A: There is a straightforward bijection, though - contra your last sentence - gotten by "interleaving:"
$$0, -1, 1, -2, 2, -3, 3, ...$$
I think this is readily understandable. It's when we look at the rationals that things get difficult.
A: Well, countability of a set means that one can tabulate the elements of the set in a list like this:
$$\begin{array}{cccc} 
0 & 1 & 2 & 3 & \ldots \\
a_0 & a_1 & a_2 & a_3 &\ldots
\end{array}$$
This holds for the natural numbers,
$$\begin{array}{cccc} 
0 & 1 & 2 & 3 & \ldots \\
0 & 1 & 2 & 3 &\ldots
\end{array}$$
the natural numbers without 0,
$$\begin{array}{cccc} 
0 & 1 & 2 & 3 & \ldots \\
1 & 2 & 3 & 4 &\ldots
\end{array}$$
the even natural numbers,
$$\begin{array}{cccc} 
0 & 1 & 2 & 3 & \ldots \\
0 & 2 & 4 & 6 &\ldots
\end{array}$$
the odd natural numbers,
$$\begin{array}{cccc} 
0 & 1 & 2 & 3 & \ldots \\
1 & 3 & 5 & 7 &\ldots
\end{array}$$
the integers,
$$\begin{array}{cccccccc} 
0 & 1 & 2  & 3 & 4 & 5 & 6 & 7& \ldots \\
0 & 1 & -1 & 2 &-2 & 3 &-3 & 4 & \ldots
\end{array}$$
and also the rational numbers by the 1st Cantor diagonalization argument.
A: Here's one:
$1\mapsto0$
$2\mapsto1$
$3\mapsto-1$
$4\mapsto2$
$5\mapsto-2$
$6\mapsto3$
$7\mapsto-3$
$\vdots$
A: If these students trust that the evens and odds both have the same cardinality as $\mathbb{N}$, then I don't see why a 1-1 correspondence doesn't work. Consider $f: \mathbb{N}\to\mathbb{Z}$ defined such that $f(0)=0$ and $f(2n)=n$, while $f(2n+1)=-(n+1)/2$.
In other words, the even numbers get mapped to $\mathbb{N}$, the odd numbers get mapped to $-\mathbb{N}$, and $0$ maps to $0$.
