Dividing and subtracting infinite numbers

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A man I despise, William Lane Craig, has said something weasely like "the inverse functions of division and subtraction are prohibited in the transfinite calculus."

But thank goodness for John Horton Conway.

I don't understand the proof very well, but here's the jist: Conway invented/discovered an ordered field of numbers, which he calls "surreal numbers," defined as the conjunction of two sets we'll call the 'left' and the 'right' set. Each set can contain any combination of surreal numbers as long as no member of the left set is greater than any member of the right set (or vice-versa, as long as you're consistent).

These numbers yield all the real numbers, rational and irrational, as well as ordinal (infinite) and infinitesimal numbers, and probably other magic. What's interesting is that the addition of these numbers leads to mathematical meaningful results - and of course subtraction, multiplication and division do as well.

So I've been giddily exasperated for an entire day that I have this new example of Mr. Craig being absolutely wrong about the hardest science there is as a result of overstepping his bounds as a so-called philosopher, thereisaiditthanksforlistening. Wikipedia has a lovely article on the "Surreal number", but I can't post the URI yet.

Edit: I can't believe I can't even write "you-are-'ell."

I'm afraid that Mr Craig is right. There is no nice notion of subtraction of ordinals, if we take addition of ordinals to mean the usual thing. And that "addition" isn't even a nice addition, since it isn't commutative!

The surreals include all of the ordinals as sets, but the notion of addition which is placed on the surreals is not the same notion of "addition" which is used for ordinals when considering them as ordinals in the transfinite calculus. In other words, it is a sort of semantic overload.

As an example of how this sort of confusion might arise, consider the statement

This might seem like nonsense, but in fact in the usual axiomatisation of arithmetic in ZFC, it is literally true, because we interpret "2" and "3" to be certain sets which have this property. But we would never write this, since once we have the notion of numbers we no longer have to remember that they are "really" sets. In the same way, the surreal field uses the ordinals as "raw materials", but does not deal with them as ordinals.

Took a while to find out where I knew Conway from, but it's obviously his version of Game of Life.

That aside, is the proof that there exists an inverse of subtraction and division, or actually what that inverse is?

Have a look here:

http://en.wikipedia.org/wiki/Ordinal_arithmetic


There is no ordinal division. Only addition, multiplication and exponentiation.

ruveyn

Shorttail, he was referring to subtraction and division themselves as inverse functions, which is his own fault.

That's quite interesting, Declension. Is it that ω+3 and ω/2 aren't, per se, transfinite in quantity, then?

It might interest you to know that Mr. Craig (I do thank you for not calling him "doctor") was using this argument to assert that infinite physical quantities like space or time were prohibited in a debate with actual physicist Lawrence Krauss, who reminded us that we can't prove there are no such things, but perhaps I'll post that topic on a different forum.

Not true. Ordinal division does exist, but it is not well defined; especially when the divisor is an infinite ordinal. The left quotient can be computed using the generalized Euclidean Algorithm. The right quotient isn't defined for infinite divisors because of the axiom of foundation which forbids infinite descending chains.

You're playing a bit fast and loose with words there, aren't you? First of all, "division exists" is not the same as "left-division exists". And secondly, it isn't even true that "left-division exists"! "Left-division exists" in this context surely means which isn't true. What you mean is that "left-division with remainder exists", in other words that
which is not what we were talking about.

You've brought up an interesting point, but you shouldn't have said that ruveyn was wrong, since he isn't. It's as if he said "There are no apples in this bowl!" and you said, "Not true! There is a pear in the bowl!"

Um, you are correct that what I meant is that left division with remainder exists. Right division is not defined because that would require a mulitplicative inverse for each ordinal, and standard ZFC does not allow infinitesimal ordinals. ruveyn didn't properly define division when he mentioned that there is no ordinal division.


For example, the statement:

For every nonzero ordinals alpha and beta, there exists a unique nonzero ordinal gamma such that alpha = gamma*beta

is a closed formula in the language of set theory(it has no free variables). One can use the axiom schema of separation to construct this ordinal gamma(which is a set itself, as the ordinals form a proper class). The trouble you run into is that with infinite ordinals, this can lead to infinitesimals which aren't allowed in ZFC by the axiom of foundation. But here's an example:

Let Omega be the 1st infinite ordinal(which is countable) defined by Omega = [1, 2, 3, 4,............,n,......] for every natural number n.
Now given that for each natural number n, we can construct a multiplicative inverse (1/n) such that n*(1/n)=(1/n)*n = 1. Then let's define Omega^-1 = [1, 1/2, 1/3,1/4,.........................,1/n,..........] for every natural number n. Than Omega^-1 is a limit ordinal. Now say
we define the new element epsilon as the limit of the sequence {1/n} as n approaches Omega. Clearly for any n, epsilon < (1/n) as well as being smaller than ANY real number. Also we define the sum of countable many epsilon = 1. This is a nonstandard model of arithmetic that we have just *created*.

So now we define Omega^-1 = [1,1/2,1/3,1/4,.............................1/n,..............................epsilon]. Now we have Omega * Omega^-1 = Omega^-1 * Omega = [ 1/1, 2/2,..................,n/n,.......] = [1,1,....................] which is order isomorphic to 1, and voila! Now we have just constructed an inverse of an infinite ordinal.

But what I will concede is that ordinal division as Declension described is not defined in ZFC set theory nor standard arithmetic. But as I have shown one can construct nonstandard models of set theory where division of infinite ordinals is possible/exists.

As for subtraction, it's important to remember that subtraction is essentially addition where the right operand is the additive inverse of the element. For example: For any 2 numbers a & b, a-b = a + (-b). But for real numbers, subtraction is non-commutative unless the operands are equal. a-b = -(b-a). The natural numbers can be extended by defining the new number -1 and the equation 1 + -1 = -1 + 1 = 0. And so for any natural number n, there is a negative integer -n such that n + -n = n - n = 0. In the case of the infinite ordinals, if we construct the first infinite ordinal(s) from the Integers, we have:

+Omega = [ 1, 2, 3,.............,n,...........]

-Omega = [-1,-2,-3,.............,-n,........]


And thus, -Omega can be used to construct the additive inverses of the infinite ordinals by means of subtraction and multiplication.

Hang on, "additive inverse" does not make sense for the ordinals. There is no ordinal alpha such that alpha + 3 = 0, for example.

Here is a summary, as far as I understand it.
Divison? No.
Left-division? No.
Right-division? No.
Division with remainder? No.
Left-division with remainder? Yes.
Right-division with remainder? No.
Subtraction? No.
Left-subtraction? No.
Right-subtraction? No.
Subtraction defined for "alpha, beta" where beta <= alpha? No.
Left-subtraction defined for "alpha, beta" where beta <= alpha? Yes.
Right-subtraction defined for "alpha, beta" where beta <= alpha? No.

If you look at that list, it seems perfectly reasonable to say that subtraction and division do not exist for the ordinals. You have to jump through all sorts of hoops to get something which is sort of like subtraction and division.



But that's what we were talking about! You are using completely different definitions. You might call your thing "ordinals", but it's not the same thing as what we were calling "ordinals".

Is it mathematically sensible to say, as Craig did, that these operations are "prohibited because they lead to contradictions"?

They violate the axiom of foundation, which prohibits infinitesimals(and infinite descending chains of belonging). I honestly think that this axiom should be slated for removal because there are models consistent with the other axioms of ZFC + the negation of the axiom of foundation.

I beg your pardon but there most certainly IS an alpha such that alpha +3 = 0. And that alpha = -3. Once you define -1 by the equation 1 + -1 = -1 + 1 = 0, you can construct alpha by alpha = -1 + -1 + -1.







An Ordinal Number is defined in set theory an equivalence class of well ordered sets that are order isomorphic to one another. In the case of -Omega, such a set CAN be well ordered under the reverse ordering of the natural numbers. Constructing additive inverses for the infinite ordinals does not violate the axiom of foundation(unlike constructing infinitesimals necessary for the multiplicative inverse).

But that's not relevant here!

Think of this as an analogy: does division exist in the world of integers? The answer is clearly no. Aha, you say, but we can consider the integers to sit inside a larger structure called the rationals, where division does exist! Well yes, we can. But that's not what we were talking about.

It is absolutely true that there is no ordinal alpha such that alpha + 3 = 0. You could be right when you claim that the ordinals can be considered to sit inside a larger structure in which 3 has a multiplicative left-inverse. But that larger structure is not what we are talking about!

Also, your attempt to construct the larger structure doesn't work. The things that you are calling +Omega and -Omega are the same object. Ordinals are equivalence classes of well-ordered sets under the equivalence relation of well-ordered set isomorphism. The well-ordered set ({1,2,...,n},<=) and the well-ordered set ({-1,-2,...,-n},>=) are isomorphic as well-ordered sets, so they belong to the same equivalence class, so +Omega = -Omega. Furthermore, when you add +Omega to -Omega using the definition of ordinal addition (which is the same thing as adding +Omega to +Omega, or adding -Omega to -Omega), you don't get the equivalence class of empty well-ordered sets (which is the additive identity, or "zero", in the world of ordinals), you actually get [({1,2,...,2n},<=)].

AspieRogue, is it possible that you have an entirely different notion of "ordinal arithmetic" than the standard one? I can't make any sense of your posts.

You seem to be saying that you add ordinals by adding them "piecewise". But that's not what addition of ordinals means! You add ordinals by connecting two well-ordered sets "end-to-end". You seem to be saying that you multiply ordinals by multiplying them "piecewise". But that's not what multiplication of ordinals means! You multiply ordinals by replacing every entry in a well-ordered set by a copy of another well-ordered set.

Furthermore, it doesn't even make sense to add or multiply ordinals "piecewise". For starters, well-ordered sets will almost always be of different sizes, so they won't "line up". If they do line up, then that means that they are isomorphic, which means that their equivalence classes are the same ordinal! Second, you can't define an operation on equivalence classes by referring to things which are just arbitrary artifacts of a particular member of the equivalence class, unless you can show that this definition works nicely. Third, the elements in a well-ordered set don't even have to be numbers! If they're not numbers, then how are you going to "add" or "muliply" them?

For the most part, transfinite ordinal arithmetic involves operations of ordinals of the same cardinality. But keep in mind that the natural numbers are ALSO ordinals. In the equation you mentioned: alpha + 3 = 0, that equation does not have a solution among infinite ordinals but has a solution among the integers.