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?

Jon Ericson
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    Consider this: you can match (2,4) with 3, (6,8) with 5, (10,12) with 7, and so on. There are twice as many even numbers as odd numbers. So one third of all positive integers are odd. – David Mitra Jan 23 '12 at 20:54
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    My point being, you run into problems when talking about the relative size of infinite sets. Google "countable sets" for more information. – David Mitra Jan 23 '12 at 20:56
  • There is a countable number of even integers, and a countable number of odd integers, but there is also a countable number of integers which are a multiple of 10. I guess it is hard to talk about the half of infinity in general. – utdiscant Jan 23 '12 at 20:56
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    I am having a hard time getting this point across, so pardon me if it sounds rude or belittling: Plato and various other philosophers are humans just as we are. As a result, I believe we should be as critical of their ideas as we are our own. (Excluding, of course, some criticism due to their extreme lack of methods we have today.) More to my point, I don't fully understand the point of this: "The philosophical point is that there exists an Idea (or Form) called Odd and odd numbers are merely specific instances of Odd. They are not, themselves, identical to the concept of Oddness.(???)" – 000 Jan 23 '12 at 21:01
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    Why is there a distinction made between odd, in the context of numbers, and some other esoteric and undefined sense of odd? I'm not saying it's not valid, I'm just saying I don't see the use. "there is an uncountable number of integers which are a multiple of 10." @utdiscant, can you elaborate on this? – 000 Jan 23 '12 at 21:02
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    @user22144: That would be an excellent question for one of my usual haunts: [philosophy.stackexchange.com](http://philosophy.stackexchange.com/). Plato is not always easy to understand--especially in our aristotelian world. – Jon Ericson Jan 23 '12 at 21:08
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    You may want to read [this thread](http://math.stackexchange.com/questions/49034/is-infinity-an-odd-or-even-number). – Asaf Karagila Jan 23 '12 at 21:20
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    @Asaf: The question (and it's top answer) do indeed help. I might be able to provide my own answer shortly. – Jon Ericson Jan 23 '12 at 21:28
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    Note the **odd** thing that $1$ is not considered odd. Indeed it was not even considered to be a *number*, since number meant something built up from two or more $1$'s. – André Nicolas Jan 23 '12 at 21:29
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    @JonEricson, stackexchange just keeps getting better and better... **:)** http://philosophy.stackexchange.com/questions/2148/with-the-encouragement-of-a-user-i-pose-this-plato-question-for-you-all – 000 Jan 23 '12 at 23:22
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    Also, getting back to the original question: The implicit assumption is that you start from 0, but why would that be? If you pair 0 with 1, then obviously you'd pair -2 with -1. (It doesn't hurt that this happens to be how things end up settling into place in 2's complement, cf "the programmer in me.") – fluffy Jan 24 '12 at 01:10
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    See also this MathOverflow question, concerning two inequivalent concepts of "half" or at least "even" when it comes to infinite sets, namely, a set that can be cut-in-half versus a set that can be cut-into-pairs. http://mathoverflow.net/questions/69461/on-the-difference-between-two-concepts-of-even-cardinalities-is-there-a-model-of – JDH Jan 24 '12 at 01:25
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    [Positive and negative zero](http://en.wikipedia.org/wiki/Signed_zero) introduce problems of their own... – sarnold Jan 24 '12 at 02:33
  • [Here is a nice, short (1m18s) video of Hilberts Hotel](http://www.youtube.com/watch?feature=player_embedded&v=faQBrAQ87l4), which is related to the question. – user unknown Jan 24 '12 at 02:08

8 Answers8


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.

Brian M. Scott
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    +1. Re the first paragraph, it might be worthwhile to point out the set of perfect squares is also countably infinite, so in the sense of set theory, "there are as many squares as there are natural numbers". In stark contrast, their asymptotic density is $0$; in this sense, "most numbers are not perfect squares". – Srivatsan Jan 23 '12 at 21:34
  • What can be said about the asymptotic density of prime numbers? – Giorgio Jan 24 '12 at 17:59
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    @Giorgio: The [prime number theorem](http://en.wikipedia.org/wiki/Prime_number_theorem) says that $$\lim_{x\to\infty}\frac{\pi(x)}{x/\ln x}=1\;,$$ where $\pi(x)$ is the number of primes $\le x$. This implies that in the long run $\pi(x)/x$ behave like $1/\ln x$: it tends to $0$, which is the asymptotic density of the primes. – Brian M. Scott Jan 24 '12 at 18:33
  • Thanks, I did not know the prime number theorem. – Giorgio Jan 24 '12 at 18:53
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    @BrianM.Scott Asymptotic density is really interesting. I didn't know about it! Thanks! – Pedro Apr 30 '12 at 00:49
  • but the squares must take up at least 1/4 of the naturals, because every natural can be written as a sum of 4 squares. – sperners lemma Oct 24 '12 at 11:51
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    @spernerslemma: No the asymptotic density of the squares is $0$, not $\frac14$, as is easily proved from the definition of asymptotic density. It isn’t true that if $\underbrace{A+\ldots+A}_k=\Bbb N$, then the asymptotic density of $A$ is $\ge\frac1k$. – Brian M. Scott Oct 24 '12 at 11:55
  • @BrianM.Scott, is there a way you could measure sets A so that kA=N imply |A| >= 1/k? – sperners lemma Oct 24 '12 at 12:06
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    @spernerslemma: Off the top of my head I don’t know whether such a measure could be constructed in a consistent fashion. – Brian M. Scott Oct 24 '12 at 12:09
  • @BrianM.Scott, I guess consistency would hold if we just had forall A,B,C: if kA contains B, and hB contains C then khA should contain C? That seems to hold due to associativity. – sperners lemma Oct 24 '12 at 12:13

While they are both countable, there is a more applicable mechanism for measuring how many natural numbers have a certain property. If $A \subset \mathbb{N}$, let $$A_n = \text{number of elements in A less or equal to n}$$ then we can define density of $A$ as $$\operatorname{density}(A) = \lim_{n\to\infty} \frac{A_n}{n}$$Then the odd numbers have density $\frac{1}{2}$.

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    I don't have the mathematical chops to arrive at that answer, but it makes sense now that I see it. It's a sort of probabilistic idea, right? – Jon Ericson Jan 23 '12 at 21:39
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    @Jon: I guess you can view it so; it's just extrapolating from our intuition that how "dense" certain numbers are in $\mathbb{N}$ corresponds to how many of them are in first $n$ numbers, where $n$ ranges over large values of $n$. –  Jan 24 '12 at 18:52
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    Eh, I worded my last comment a bit clumsily. The point I was trying to make is that this density is much less a statement about natural numbers itself, and much more a statement about this particular ordering of natural numbers. –  Jan 25 '12 at 15:44
  • @user5501 Then again, there is a unique well-behaved ordering on the natural numbers, intrinsic to the additive structure. – Bart Michels Aug 26 '18 at 16:37

It's already mentioned in the comments that you will encounter some problems when talking about half of the elements of an infinite set. There are ways to compare the size of infinite sets.

See: http://en.wikipedia.org/wiki/Cardinality

Two sets have the same cardinality (=size), if there exists a bijection between them. The natural numbers and all other sets with the same cardinality are called countably infinite. That means roughly that we can find a way of enumerating them (and additionally they are infinite). It is easy to give a bijection from the natural numbers to the even numbers. Just associate to a number $n$ the even number $2n$, so we conclude that the cardinality of the even numbers is the same as the cardinality of all natural numbers! The same is true for the odd numbers, so in set theory the two sets you want to compare are considered equally large!

That being said, there are other notions of size for sets of natural numbers. One of them is natural density: http://en.wikipedia.org/wiki/Natural_density

Instead of "counting" all even numbers at once you restrict the set of numbers you look at to natural numbers $\leq n$ and calculate the proportion $x_n$ of even numbers in this set. $$x_n =\frac{\text{even numbers below }n}{n}$$ The density is then defined as the limit of the $x_n$ and you will find that it is indeed $\frac{1}{2}$, just as the natural density of the odd numbers.

There are many interesting theorems about the densities of certain subsets of the natural numbers like Dirichlet's theorem on primes in arithmetic progressions. One instance of the theorem tells us for example that "half of the primes are congruent to 1 modulo 4 and the other half is congruent to 3".

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There is an infinite amount of even numbers.

There is an infinite amount of odd numbers.

How do we compare infinities? Is one infinity larger than the other?

We look at the set of even numbers, say A, and the set of odd numbers B. Two sets are said to have the same cardinality if there is a bijection between them. We can find such a bijection, so these "infinities" are the same.

It might also be worth mentioning that there are "the same number of" fractions as there are whole numbers.

The Chaz 2.0
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  • I just had to look up "bijection" so I'm obviously still in an early learning stage. But what do we map zero with in the set of odd numbers? (I think this is my non-platonic "proof", by the way.) – Jon Ericson Jan 23 '12 at 21:17
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    @Jon: Map each even integer $2n$ to $2n+1$, the next larger odd integer, and each odd integer $2n+1$ to $2n$, the next smaller even integer. – Brian M. Scott Jan 23 '12 at 21:31
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    Map $0$ to $1$, $2$ to $3$, $4$ to $5$, $6$ to $7$, and so on. – André Nicolas Jan 23 '12 at 21:34
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    you can in fact compare infinities. $\lim_{x\to\infty}\frac{x+1}{2x} = \frac 1 2$ so clearly the bottom infinity is larger than the top infinity. As result of this, your last statement about fractions is also erroneous. There are many more fractions than there are whole numbers (in fact, infinitely more) – wnafee Jan 24 '12 at 00:43
  • @wnafee I didn't say that we can't compare infinities. Also, my last statement was in the sense that $|\mathbb{Q}| = |\mathbb{N}|$. – The Chaz 2.0 Jan 24 '12 at 01:20
  • By bijection, are we referring to the Peano successor function? (Or is that a particular *way* of representing this bijection?) – 000 Jan 25 '12 at 02:19
  • That function is usually assumed to be *injective*, half of bijective, if you will. Definitely research cardinality. André has given one such bijection. – The Chaz 2.0 Jan 25 '12 at 02:32

[The other answers have more than answered my question, so my goal in self-answering is to make sure I understand what's been said so far. I would appreciate corrective comments if I make any mathematical mistakes. I'm heavily indebted to JDH's excellent answer to the question "Is infinity an odd or even number?"]

For numbers greater than one, which is the set Socrates refers to, even numbers are those that can be divided on the two sides of a scale so that the scale balances. If I have 10 identical coins (an even number), I can put 5 on each side. But if I add (or subtract) one coin from the total, the scale will not balance. (Something is "odd" about the number.)

Moving up a level of abstraction, we notice that an even number of any identical object will balance: 10 coins, 10 jars of olive oil, 10 grains of sand, etc. So depending on your point of view, evenness is a property of 10 or 10 is an instance of Even. Even when we consider things that we can't reasonably weigh on a scale, such as imaginary objects, can still be labeled as even if there are an even number of them.

So what happens if you take numbers and metaphorically divide them between the two pans of a scale: even numbers in one and odd numbers in the other? It depends on the range of numbers you consider. [2...3] will balance, but [2...4] won't. But you can always add one more number to the range and make the scale balance again, so [2...5] does balance.

Now Socrates didn't have a concept of infinity, so he probably was thinking of arbitrarily large numbers. If you happen to pick an odd number of numbers, than one or the other pan on the hypothetical scale will be tip. But since the range is arbitrary, you can always pick just one more number and restore the balance. Nothing about zero or infinity or negative numbers changes this: if you want half of all numbers to be odd, just make sure your range includes an even number of numbers.

When you start talking about infinity or an infinite number of numbers and so on, things get weird and you can't really apply everyday intuition. There are any number of paradoxical-seeming results that you can arrive at if you start messing around with infinite series. For instance, you can have a hotel with an infinite number of rooms, an infinite number of guests, and plenty of room for an infinite number of new guests. More to the point, you can match an infinite number of odd numbers to an infinite number of even numbers at any ratio you like without running out of either type of number. So saying that half of all numbers is odd is just as true as saying that 1 in a thousand numbers are odd.

However, we can still look back at our scale and for large ranges of numbers, the scale either balances or very-nearly balances for any particular range you pick. Sure there is one extra even number in the range [2...1,000], but it would take a pretty accurate scale to determine which pan is weighed down more. And the larger the range, the smaller the difference. Mathematically, this way of thinking corresponds to the concept of natural density. Thankfully, the natural density of the odd numbers turns out to be half when measured this way.

As an aside, Wikipedia mentions that:

This notion can be understood as a kind of probability of choosing a number from the set A. Indeed, the asymptotic density (as well as some other types of densities) is studied in probabilistic number theory.


As long as we steer away from the (very intriguing) notion of infinity, it seems that Socrates was right to say that half of the numbers are odd.

Jon Ericson
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I think the OP has the idea of two infinite sets A and B having the "same number of elements" if $card(n, A)/card(n, B) \to 1$ as $n \to \infty$, where $card(n, C)$ is the number of elements of $C$ that are less than $n$. In this sense, the evens and odds have the same number of elements.

In this sense, there are more evens than squares, and this is an equivalence relation.

Of course, this leads to the idea of the density of sequences (Schnirliman (sp?), natural, logarithmic, ...) which has many interesting results.

(Added a bit later: I see that Lovre beat me by 3 minutes.)

marty cohen
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Your question brings up your confusion around the problem of "0" and suggests that this might lead one to think that there is one extra even number.

While this is erroneous and can be proven mathematically as others have above, there is also a simpler explanation which stems from a philosophical question: Why do you assume that the "center" of the natural numberline is "0"?

The concept of "center" is something that works with finite groups. But the numberline is infinite. (What is the midpoint of an infinite ordered set?)

If we had randomly placed the number "-1" as the "center" (ie. there are infinite numbers less than it and larger than it on either side), would that have led you to think that there is one extra odd number?

The point is to make you realize that, in reality, mathematicians' choice of "0" being at the center of every numberline we draw is just a tradition and is just an agreed upon matter. It has no "scientific" basis and should not affect the way you look at things

Now, from a mathematical perspective: $\lim_{x\to\infty}\frac{x}{x+1} =\lim_{x\to\infty}\frac{x}{x} =1$

So even if you have one extra value, it wont affect the overall result because we're dealing with non-finite sets!

So I guess the summary of what I'm trying to tell you is that your initial problem of how to deal with the extra "0" is not actually a problem after all :). You can safely (mathematically) say that half of all numbers are odd.

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    I think there is a plausible argument that could be made that 0 is the centre of the number line by symmetry and/or the fact that it is the unique fixed point under negation. This all depends of course on how you define "centre" :-) – mikera Jan 24 '12 at 01:52
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    the point is to realize that 0 being the "center" is not scientifically based, and hence if you randomly decided that -1 is now your new "center" it cannot possibly mean that you now have one extra odd number (based on the line of thinking that the question suggested) – wnafee Jan 24 '12 at 21:14

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.

John Gowers
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