Are half of all numbers odd?

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?

Answer

The basic problem with your argument is that the concept of half of an infinite set is not well-defined. It’s possible to pair up the even integers with the odd integers with none left over in either set, and if we were talking about finite sets, that would be a demonstration that each was half of the whole set of integers. However, it’s also possible to pair up the multiples of $100$, say, with all the rest of the integers with none left over in either set, and the multiples of $100$ are obviously only part of the set of even numbers. Clearly, then, this kind of pairing argument cannot lead to any very useful notion of half of the set of integers.

There is a notion of asymptotic density of a set of positive integers that does a pretty good job of capturing many people’s intuitive sense of what half (or any other fraction) of the set of positive integers should mean. Let $A\subseteq\mathbb{Z}^+$; for each $n\in\mathbb{Z}^+$, $$\frac{|A\cap\{1,2,\dots,n\}|}n$$ is the fraction of the first $n$ positive integers that are in the set $A$. If this fraction approaches a limit $\alpha$ as $n$ increases, i.e., if $$\lim_{n\to\infty}\frac{|A\cap\{1,2,\dots,n\}|}n=\alpha\;,$$ the set $A$ is said to have asymptotic density $\alpha$. (Note that the limit need not exist: not all sets of positive integers have asymptotic densities. It’s not hard to show that if $A$ is the set of even integers or the set of odd integers, its asymptotic density is $1/2$, so it’s meaningful to say that each of these sets comprises half of the positive integers in terms of asymptotic density. Similarly, for each positive integer $n$, the set of multiples of $n$ can easily be shown to have density $1/n$, exactly as one would wish.

Added: In the comments Srivatsan points out an excellent example: the perfect squares can be paired up one-for-one with the integers, but their asymptotic density is $0$: they get sparser and sparser as you get further away from $0$. Thus, in one sense there are just as many perfect squares as there are integers, and yet in another sense almost every positive integers is a non-square.

Attribution
Source : Link , Question Author : Jon Ericson , Answer Author : Brian M. Scott

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