TITLE: Are half of all numbers odd?
QUESTION [127 upvotes]: Plato puts the following words in Socrates' mouth in the Phaedo dialogue:

I mean, for instance, the number three, and there are many other examples. Take the case of three; do you not think it may always be called by its own name and also be called odd, which is not the same as three? Yet the number three and the number five and half of numbers in general are so constituted, that each of them is odd though not identified with the idea of odd. And in the same way two and four and all the other series of numbers are even, each of them, though not identical with evenness. (104a-b)

The philosophical point is that there exists an Idea (or Form) called Odd and odd numbers are merely specific instances of Odd.  The are not, themselves, identical to the concept of Oddness.
But I bolded a throwaway phrase that prompts my question: Are half of all integers odd?

Plato likely did not consider one to be odd nor did he likely consider either zero or negative integers among his set of "numbers in general".  I don't see his proof (if he had or was aware of one), but I would imagine it be something along the lines that for every even number N there is an odd number N+1.  Therefore half of all numbers greater than 1 are odd.
I'm aware that zero is even, which makes me think there is one extra even number than odd numbers.  My thinking is that if half of all positive numbers are odd and half of all negative numbers are odd, than leaving out zero, half of all integers are odd.  But when you add in the only non-positive, non-negative number, which is even, you have an extra even number.  (The programmer in me wants to add the concepts of -0 and +0 for symmetry.  The rest of me thinks that's nuts!)
However, I think my imagined platonic proof still works as long as zero is definitely even and won't work if it's either odd or neither parity.
Are either of these proofs valid?

REPLY [7 votes]: I actually think that there are good reasons that show that half of all integers are odd.  If you ask a lot of mathematicians this question, they'd give you some silly example showing you that, for example, only $\frac{1}{1000}$ of all numbers are odd.  They'd do this by constructing a bijection like the following mapping every odd number to a distinct sequence of $999$ even numbers: 
$$
\phi(1)=(2,4,6,8,\dots,1998) \\
\phi(3)=(2000,2002,\dots,3998) \\
\dots
$$
However, this approach only uses the combinatorial structure of the integers, and ignores the algebraic structure.  If we require $\phi$ to be not just a bijection from the odd numbers to some sequence of even numbers, but instead to be a group homomorphism whose kernel is the even numbers, we have only one choice: 
$$
\phi(n)=n\mod 2
$$
For finite groups, we know that $|G|/|\ker\phi| = |\phi(G)|$.  If we pretend that this applies to infinite groups as well, then that means that there are twice as many integers as there are even numbers, since the image of our homomorphism is the two-element group $\mathbb{Z}_2$.  Therefore, half of all integers are odd.  
Another approach we could take is to use the orbit-stabilizer theorem, which states that for a finite group $G$, a set $X$ and an action $\psi:G\times X\to X$, we have $|G|=|\mathrm{orb}_\psi(x)||\mathrm{stab}_\psi(x)|$ for any $x\in X$, where $\mathrm{orb}_\psi(x)$ is the orbit of $x$ under $\psi$ and $\mathrm{stab}_\psi(x)$ is the stabilizer of $x$ under $\psi$.  If we pretend this applies to infinite groups, we can take $G=\mathbb{Z}$, $X=\{\{\textrm{odd numbers}\}, \{\textrm{even numbers}\}\}$, and let $G$ act on $X$ by addition.  Then the orbit of the set of odd numbers is the whole of $X$, which has size $2$, and the stabilizer of that set is precisely the set of even numbers.  So we get 
$$
|\mathbb{Z}| = 2\left|\{\textrm{even numbers}\}\right|
$$
and therefore, half of all integers are odd.  
Another, more combinatorial approach, is as follows.  We might not want to take full advantage of the algebraic structure of $\mathbb{Z}$, but the example at the top showing that the odd numbers occupy only $\frac{1}{1000}$ of the integers seems silly for a nother reason: the images of the odd numbers $\phi(n)$ are too far away from the odd numbers themselves!  One way of using this is to define the notion of density, which has been done above.  Consider instead the following approach.  Let the integers be the vertices of a graph, and join two integers if they are adjacent.  So $2$ and $3$ are joined by an edge, while $2$ and $4$ are not.  (This makes sense, because adjacent numbers are in some sense close together.  It would be silly to connect the number $3$ to the number $2000$.)  
Then this graph is bipartite, and its vertex sets are precisely the set of odd numbers and the set of even numbers.  Moreover, the graph is regular, as each vertex has precisely two neighbours.  A simple counting argument shows that all finite regular bipartite graphs have vertex sets that are the same size.  If we pretend that this applies to infinite regular bipartite graphs (notice a theme here?) then we see that the set of odd numbers and the set of even numbers are the same size, so half of all integers are odd.