Infinity Question: What is the ratio of odd to even numbers?

I was reading a discussion on another message board concerning the series 1/1 - 1/2 + 1/3 - 1/4 + 1/5… Apparently, the sum of this series can be different values, depending on how the series is ordered. This lead to a couple different observations, one of which was this:

Suppose I order the natural numbers in a sequence like so:

1, 3, 2, 5, 7, 4, 9, 11, 6…(4i-3), (4i-1), 2i…

In this sequence, every natural number is acocunted for and has a particular place. Additionally, each even number is paired with two odd numbers, so obviously the ratio of even to odd numbers is 1:2. Equally obviously, I could construct a series in a different way to get any ratio of even:odd numbers that I desire. However, this conclusion bothers me mightily, since I’m used to even/odd numbers occuring alternately, with a 1:1 correspondence. So is the ratio of even:odd numbers undefined? Or am I overlooking something?

COROLLARY: Just previous to the even:odd observation, someone commented on the puzzle, How many threes? The puzzle is: What percentage of all integers contains at least one instance of the digit three? and the official answer is: 100%, the rationale being that the percentage of numbers with threes in them rises as the number of digits in a number rises. The percentage of numbers containing the digit three can be expressed as 1 - (.9)^n, where n is the number of digits. It reaches 99% at about the point where n has 42 digits. Since there are always more natural numbers greater than any given number than there are less than that number, it’s easy to see that, on average, all natural number contain the digit three.

However, this explanation relies on ordering numbers in the usual, counting way. If I construct a series like:

1, 3, 2, 13, 4, 23, 5, 30, 6, 31…[ith number not containing 3], [ith number containing 3]…

then, again, all natural numbers are accounted for, and it’s easy to see that only 50% of natural number contain the digit three.

Is the official answer of the “puzzle of the threes” still valid? Or not? For the same reason as the ratio of odd to even numbers is undefined?

According to Cantor, the number of odd numbers is Aleph null. So is the number of even numbers. So is the number of integers, the number of integers with the digit 3, and the number of rational numbers. (Not intuitive, but the proof is quite convincing).

There are more irrational numbers; that number is C (not to be confused with the C as the speed of light).

Here’s the deal. Any infinite subset of a countable set is also countable. So if you take every other odd number, you can still match this with the even numbers. That’s all that’s going on in that list–they’re displaying two 1-1 correspondences at the same time.

Or, if you prefer ordered pairs, you can think of it as a 1-1 correspondence between the even numbers and a certain subset of ordered pairs of odd numbers (which isn’t easy to characterize, especially first thing in the morning).

Either way, there’s no particular paradox.

I doubt that the original answer to the “puzzle of the threes” was ever valid. As you’ve demonstrated, the set of numbers containing the digit three can be placed in 1-1 correspondence with its complement, so it’s the same size. FWIW, it doesn’t always make sense to talk about the ratios of sizes of infinite sets, and I think that’s what gets you in trouble here.

The way to “compare” two infinite sets (sets with an infinite number of points) is to see if you can line them up in a way that there is a one-to-one correspondance between them. That is, is there some correspondance such that for every element of set A, there is exactly one element of set B that corresponds to it; and vice versa.

Thus, you can set up a one-to-one correspondance of the odd integers and the even integers, as follows:

1 3 5 7 …
2 4 6 8 …

For every odd integer n, we have the corresponding even integer n+1; and for every even integer m, we have the corresponding odd integer n-1.

Thus, we have set up a one-to-one correspondance between the two infinite sets, and thus we conclude that they are the same “size”.

We can also set up a correspondance between the set of ALL integers and the set of even integers, as follows:

0 1 2 3 …
0 2 4 6 …

This is somewhat counter-intuitive, since you would think that there are “more” integers than there are even integers. But it is not so, and this illustrates the difficulty of using intuition when dealing with infinite sets. The fact is that there are the same “number” of integers as there are even integer, and that number is called Alef-naught. It is somehow the smallest size for an infinite set.

When speaking of the ratio of two numbers, you really need to have finite numbers to deal with. The ratio of infinite sets is tricksy.

For instance, we’ve just shown that the sets of odd integers and the set of even integers are the same size, because we CAN set up a one-to-one corresondance. That should imply that the ratio of even:odd is 1.

However, we could equally easily set up a one-to-two correspondance, as follows:
For every odd number n, assign the even numbers 2n and 2n+2.
1 3 5 7 … (Set of odd integers

2 6 10 14 … (Set of even integers, aligned
4 8 12 16 … two even for each odd)

Now, for every odd integer, we have assigned two even integers, in a unique fashion. Thus, you could argue that the “ratio” of odd/even is 1:2.

In short, the term “ratio” is just not easily definable for infinite sets in a way that is consistent with the term for finite sets.

((Damn, it’s hard to make things line up!))

[Edited by C K Dexter Haven on 09-25-2001 at 08:28 AM]

OK, let me readjust my thinking on this. So, for any infinite, countable set of numbers (which is what Aleph null implies, er, right?), it doesn’t make any sense to say that this set has exactly the same number of elements, or twice as many, or whatever, as compared to another set.

So there are the same number of odd numbers as even numbers in the sense that both sets number Aleph null; however, one can’t say that there are the same number of odds versus evens in the sense that odds and evens are inherently “paired,” if that makes sense.

And, moreover, the phrasing of the “threes” puzzle (“What percentage of all integers contains at least one instance of the digit three?”) is meaningless, since both sets, “numbers containing the digit three” and “numbers not containing the digit three” have Aleph null elements. Is that a fair summary?

And on preview, I see responses by ultrafilter and Dex that, I think, reinforce my interpretation above. OK, I think this makes sense. Thanks, guys. I guess this thread is something of an echo of other “infinity-question” threads that pop up in GQ every month or so (that I’ve even participated in before). And I thought I had a new spin on it…

<< And, moreover, the phrasing of the “threes” puzzle (“What percentage of all integers contains at least one instance of the digit three?”) is meaningless, since both sets, “numbers containing the digit three” and “numbers not containing the digit three” have Aleph null elements. Is that a fair summary? >>

Correct, Zut, pretty much. Unless one defines what is meant by “percentage” in some peculiar way, as a limit of an infinite sequence.

For example, you could ask what percentage of integers from 0 to 1000 contain the number 3, and you could answer that. You could ask what percentage of integers from 0 to 10,000 contain the number 3, and you could answer that. You could ask what percentage of integers from 0 to N (where N is any finite number), and presumably you could answer that.

You thus could generate some sort of sequence of the percentage of integers less than N containing 3, and you could see whether that sequence has a finite limit as N goes to infinity (gets larger and larger.)

I don’t know the answer to that offhand. I suspect there is no limit to that sequence. The percent of numbers with a 3 in them seems to be pretty fuzzy. For instnce, if you look at numbers from 1 to 2999 you get a much lower percent than if you look at numbers from 1 to 3999, since you’ve just added 1000 numbers that all contain 3.

So it is possible to define ratios and percentages for infinite sets as the limit of a sequence of ratios or percentages for finite sets as the finite sets approach infinity. (This would give you a ratio of 1:1 for odd:even, for instance.)

This isn’t quite correct. It does make sense to say that two infinite sets have the same number of elements, or that set A has fewer elements than set B, or that set A has more elements than set B.

Just in case you’re not familiar with it, “iff” is shorthand for “if and only if”.

Of course, what we mean by this isn’t quite as simple as it is for finite sets (although the definitions used here will work for finite sets). Set A and set B have the same number of elements iff there is a 1-1 correspondence (or bijection, the term I prefer) between their elements.

The definitions of the other two statements rely on the notion of a proper subset. Set X is a subset of set Y iff every element of X is an element of Y, and X is a proper subset of Y iff Y has elements that X does not. Anyhow, set A has fewer elements than set B iff there is a bijection between A and a proper subset of B, but there is not a bijection between A and B. In this case, B has more elements than A. If there is a bijection between A and a proper subset of B, then A has at most as many elements as B, and B has at least as many elements as A.

I don’t have my set theory book with me, so I’m only 99.44% sure about this.

I couldn’t believe the amazingly bad posts on this thread until I went to the other MB that zut pointed to. It is even worse.

Folks, the series zut refers to is extremely well known and is called the “Alternating Harmonic Series”. Do a google and find out more.

  1. This series does in fact converge. (Conditionally but not absolutely, just two techinical types of convergence.) This is a consequence of “Leibniz’s Theorem on Alternating series.” Basically, if the terms consistently go towards zero and alternate in sign, the series converges.

All this is given in my old Calc book: “Advanced Calculus” by D.V. Widder, 1961, p. 293.

On the net, try:

The arbitrary re-arranging that people are doing is horribly, horribly, bad math. Do not ever do this on alternating series without applying some proofs first that your re-arrangement is provably correct. Naive commutativity fails completely in these examples. You have to be at least 2nd year calculus level to do this right. All your intuitions are wrong.

  1. It converges to the natural log of 2 (as given on the other MB). When done in the normal order. Re-ordering means that you are specifying a different series with a different limit.

I see nothing wrong with the posts by RealityChuck, Dex, and ultrafilter, they seem perfectly fine to me.

Huh? zut mentioned in the OP the fact that the alternating harmonic series can be made to converge to any real number by rearrangement. The whole point of this thread was whether or not there is any meaningful way to define ratios of (the size of) infinite subsets of natural numbers, so that those ratios are invariant under reorderings of the natural numbers.

What can I say? I was attempting to explain, not just show off how much I know.

It seems obvious that the integer numbers can be arranged in ordered increasing sequence [odd, even, odd, even, odd] etc (1, 2, 3, 4, 5…), so that there seems to be equal numbers of odd and even integers as we add more and more sequential integers to the list. There is, however, another way of ordering integers.

It is clear that every even number consists of an odd number times some power of two. Thus 2 = 12, 4=122, 6=32, 10=52, 12=32*2 and so on. Every even number can thus be associated with one, and only one, odd number (anyone want to prove this?). So, for every odd number we can collect all the even numbers which are that odd number times some power of two, giving collections:
(1, 2, 4, 8, 16, 32…)
(3, 6, 12, 24, 48…)
(5, 10, 20, 40, 80…)
(7, 14, 28, 56, 112…)

etc, where each collection contains integers not present in any of the other collections.

Thus with every odd number we can associate an infinity of even numbers!

BobGomersall, what you said looks correct. The “anyone want to prove this?” statement seems to me to follow directly from the Fundamental Theorem of Arithmetic (unique prime factorization).

This is a zombie thread. The original conversation took place almost exactly ten years ago. However, since at least some of the participants are still around, I’ll take the liberty of replying to something that stuck out to me:

What you can do is take two infinite sets of whole numbers, compare the sizes of those sets less than N, and look at what happens when N gets very large. This is the idea behind the Prime Number Theorem, which, in a rough, informal sense, says that 1/ln(N) of the first N whole numbers are prime, as N gets large.

I think that what BobGomersall is saying is that every number can be broken down into a unique set of prime components.

3, 23, 223, 2223, etc
5, 25, 225, 2225 etc.

That works even is the the original odd term is not prime because it can be broken down into primes. 9 = 3*3

9, 233, 2233, 2223*3

This is the fundamental theorum of arithmetic

As expected with something that fundamental it has myriads of applications.

Just to clarify, I was not breaking down odd numbers into prime factors. My example showed the sets for odd numbers 1, 3, 5 and 7. The next set is
(9, 18, 36, 72, 144…).

None of these numbers can be members of any other odd number set. I was only showing there are alternative ways to count even and odd numbers, which do not indicate the set of odd integers seems to contain as many members as the set of even integers.

I actually came upon this as part of investigations into the Collatz function which of course has no explicit links to prime numbers but does depend upon the ‘difference’ between odd and even numbers.

You said:

Both Thudlow Boink (whose post I didn’t see) and I agree that this is a direct consequence of the fundamental theorem of arithmetic. That means an explicit link to prime numbers, i.e. every even number is composed of a unique set of prime factors. And so is even odd number, of course.

And furthermore, we can simply add one to each odd number to get a corresponding even number. This gives us a map that associates with each even number an infinite set of even numbers, and no two of these sets overlap. If I follow your phrasing, I might be tempted to say that this doesn’t suggest that the set of even numbers is the same size as itself, but of course that’s silly. The essential takeaway here is that trying to reason intuitively about infinite sets simply doesn’t work well.

To truly understand the paradoxes of the infinite, how does the following scenario end?

A smart monkey and a dumb ape are on a desert island that has an infinite pile of coconuts, labeled (this is important) 1, 2, 3, 4,… Each animal has a cave. During the first half hour, the ape collects all the nuts labeled 1, 2, …, 1000 and piles them in his cave. The monkey does nothing. During the next quarter hour (he’s learning to do it faster), the ape collects all the nuts labeled 1001, 1002, …, 2000, while the monkey goes into the ape’s and steals the nut labeled 1. During the next 7.5 minutes, the ape collects 2001, …, 3000 while the monkey steals number 2. Each interval takes half the time of the preceding one and in each interval, the ape collects the next 1000 nuts and the monkey steals the next one (in order) from the ape’s cache. What is the situation at the end of one hour?

The monkey has all the coconuts (he has numbers 1, 2, 3, … obviously) and therefore the ape has none.

I disagree. I’ll just leave it at that.

The situation at the end of the hour is unspecified in the terms of the problem, and can be anything.

I wonder whether, after having now read 10 more years worth of Ultrafilter’s (and others’) posts, ftg would be willing to revise his sentiment here…

I know it was ten years ago but my curiosity remains insatiable.