.999.... (Repeating) is equal to one?

You can talk about the subject of math in a new thread instead of derailing the topic with comments like those. I sound hostile because you kept coming here to say that you don't understand what's going on when you weren't obligated to be here. It's offensive to people that are actually discussing the topic.

edited 11th Sep '11 12:50:49 AM by abstractematics

Eh... the lines longer than infinity are pretty weird, I'll admit. I did mention how confusing it is. But after working with a fair amount of stuff involving uncountably infinite sets, it stops being quite so counterintuitive; I guess I've just gotten used to it. (Except I'm still admittedly a little freaked out by the continuum hypothesis. Set theory is strange.)

I'm not offended. That might not mean much, though, since I'm pretty hard to offend.

edited 11th Sep '11 12:53:52 AM by Enthryn

The topic appears to be done, since no one seems to want to debate that .999 is, in fact, just one but more complex.

Or, I suppose, 1 expressed as a decimal... or rather, 1 as a fraction expressed as a decimal...

Ok, you know what, fine, nevermind.

A different way of representing the number 1. That's all it is.

I was under the impression that there weren't different levels, degrees, magnitudes, etc. of infinity. To say that something is twice as infinite as something else relies on the rational number two, but infinity is not a rational number and hence concepts that hold true with rational numbers aren't really applicable. It's one reason that the "proof" [1*infinity = infinity, 2*infinity = infinity, therefore 1*infinity = 2*infinity, therefore 1 = 2] isn't meaningful (others are things like "dividing by infinity is nonsensical").

I may have mistaken something, of course. I often do.

Infinity is not a rational number, but 2 is an extended rela number, so you're good.

...except for 2*infinity = infinity. For the next step up, you need infinity².

You're looking for the concept or cardinality. That is, some way of comparing infinities.

Actually, no. That's not correct. See a proof of the set of rationals being the same size as the set of integers, or of the set (N x N) being the same size as N (N being the set of integers). To go to the next level, you need 2^Infinity. This takes you to the cardinality of the real numbers C.

edited 11th Sep '11 6:35:53 AM by DarkConfidant

...R, you mean? It's true that it's isomorphic to C, but I normally think of it as R.

(And yes, 2^infinity. Derp.)

Cardinal arithmetic is one of the ways to extend the natural numbers with some sort of infinity, sure. However, it's not the only one.

It all depends on what you are trying to represent using a number. If you are reasoning in terms of classes of equipollency (that is, your idea of a number n is more or less "the class of all groups of n objects") then cardinal arithmetic is the way to go, and, then, if n is infinite then 2n = n.

If your idea of a number is, instead, more related with the informal concept of "counting" - that is, if you are willing to identify a number with the set of all numbers less than it, and 0 with the empty set - then you want to use ordinals numbers, or surreal numbers if you are fancy and also want negative and fractional infinities. Then it is not true that 2 * infinity = infinity; for example, if omega is the smallest infinite number (that is the ordinal corresponding to the set of all integers with their usual order), then 2 * omega is different from omega, and corresponds to the set

{ (n, b) : n is a natural number, b is 0 or 1}

with the ordering given by

(n, b) < (n', b') if and only if b < b' or (b = b' and n < n').

Or you could want to reason in terms of sequences of natural numbers, and say that two sequences are the same if and only if they coincide over all but finitely many elements; then you'd be working with non-standard arithmetic (I am skipping a lot_ of details here) and, again, the double of an infinite "number" in this sense is not the same number.

The lesson to take here, I think, is that intuitions which lead to the same exact structure in the finite case lead to rather different ones when you consider the infinite one. Hope this helps!

edited 11th Sep '11 6:51:43 AM by Carciofus

C refers to the cardinality of the real number set R. The c stands for continuum. I do not mean to invoke the set of complex numbers.

Except that 2*infinity is the same as infinity, because I can create a 1 to 1 correspondence of the sets {1, 3/2, 2, 5/2, ...} and {1, 2, 3, 4, 5}, despite the first set having twice as many objects in it in some naive sense. Furthermore, I can remove the fractional terms {3/2, 5/2, ...} and still have as many objects as I started with.

The allegory of the infinite hotel: In a hotel with an infinite number of rooms, the statements: "Every room is filled" and "We're booked to capacity" are not synonymous.

edited 11th Sep '11 7:03:02 AM by DarkConfidant

If you say that two sets have the same number of elements if and only if there exists a 1 to 1 correspondence between them (that is, if you are using cardinal numbers) then sure.

But if you say that two ordered sets (A, <) and (A', <') have the same number of elements if and only if there exists a 1 to 1 correspondence f: A -> A' which preserves the ordering, in the sense that

a < b if and only if f(a) <' f(b)

for all a, b in A, and if you define 2 * (A, <) as the set obtained by taking two copies of (A, <) and putting one after the other (that is, if you are working with ordinal arithmetic) then the 2 * (A, <) != (A, <) even if A is infinite.

For example, the smallest infinite ordinal number is

\omega = 1, 2, 3, 4, 5 ... ,

in this order. Then, by definition

2 * \omega = 1, 2, 3, 4, 5, ..., 1', 2', 3', ...

in this order, and it is not difficult to see that there exists no bijection between \omega and \omega' which preserves the order.

As I said, it really depends on what sort of concept of number (and of infinity) you are working with.

edited 11th Sep '11 7:16:29 AM by Carciofus

That brings up the question: if A is infinite, then how can you even construct a set such that you have an ordered set A* composed of all the elements in A followed by all the elements in A. You'd never get to the second set of elements by definition of infinity.

Well, one formally defines an order as a pair (A, <=), where A is a nonempty set and <= is a well-founded linear ordering, that is, a relation such that

• For all a in A, a <= a;
• If a <= b and b <= a, then a = b;
• For all a, b and c, if a <= b and b <= c then a <= c;
• For all a and b, a <= b or b <= a;
• There is no infinite descending chain a1 >= a2 >= a3 >= ...

The last axiom is called the well-ordering axiom, and you need it in order to get only the ordinals and not also other stuff. However, note that it says nothing against a ascending chain - for example, the natural numbers with their usual order satisfies this requirement, as you never have an infinite descending chain, but the integers don't.

Also, of course, you can always define a < b as "a <= b and a != b" - it's just that it's technically a bit easier to take "less or equal than" instead of "less than" as the primitive notion.

All right?

Now, we say that two ordered sets (A, <=) and (B, <=') are the same ordinal if and only if there is an order-preserving bijection between them as I said before. Then, we define (A, <=) + (B, <=') as the ordered set

(C, <='')

where C is the disjoint union of A and B, and where

x <='' y if and only if

• x and y are in A and x <= y, or
• x and y are in B and x <=' y, or
• x is in A and y is in B.

With a little bit of work, you can show that (C, <='') is a well-order — the last axiom is the only one which requires a little bit of work, and the trick here is showing that an infinite descending sequence in C would contain an infinite descending sequence in A or one in B.

Note, by the way, that this sum is not commutative: for example, if omega is the set of the natural numbers then we have that 1 + omega = omega but omega + 1 != omega — yeah, ordinal arithmetic is freaky. Really useful in axiomatic set theory, though.

Finally, you can define 2 * (A, <=) simply as (A, <=) + (A, <=), and you are done.

edited 11th Sep '11 7:36:34 AM by Carciofus

I can't tell which is more surprising: the fact that there's a formal definition like that, or the fact that I understood more than 95% of it.

One of the most fun (at least for me) applications of this sort of thing is in infinite game theory — that is, the study of games with infinitely (for some ordinal-number kind of "infinitely") many moves.

Which is awesome in itself, I think, and also has some "useful" applications to formal logic (and in particular to descriptive complexity theory, which is related with the P-NP problem and that sort of stuff).

Not that I have any idea of what the P/NP problem is, but okay.

However, I do have a degree in economics, and I have taken a formal class in game theory, so I do share your excitement on infinite games. (See for instance how any finitely repeated instance of the prisoners' dilemma has a Nash equilibrium of always defect on every trial, but one that continues infinitely will lead to a solution of always cooperating.)

tl;dr version is, if P=NP then problems that can have solutions verified in asymptotically polynomial time can also be solved in asymptotically polynomial time. It's probably the most important standing problem in computer science.

edited 11th Sep '11 8:02:34 AM by Tzetze

Another fun use for ordinal numbers is infinitary logic, in which logical statements and proofs can be of infinite length. (Ordinal numbers are used because a proof — even an infinitely long one — is essentially an ordered list of statements, each of which follows from the axioms or the previous statements.) I don't know much about it, but it seems like a natural and interesting extension of the usual notions of statement and proof.

edited 11th Sep '11 8:50:23 AM by Enthryn

Well, if you divide one by three, you get 0.33333333333333333333... Thus, if you multiply 0.33333333333333333... by three, you get 0.999999999999999999999.... Since three times 0.3333... is one, than 0.99999999..... must be equal to one.

edited 12th Sep '11 11:15:14 AM by AlirozTheConfused