Dividing and subtracting infinite numbers

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Please, do post it here.

I'm aware that ordinal addition and multiplication are not piecewise. So transfinite ordinal(right)division involves replacing every entry in a well-ordered set by a copy of the multiplicative inverse of the set you're dividing by! THAT is the definition of ordinal division I'm using here. Since any infinite set has a 1-to-1 correspondence with a proper subset of itself, transfinite ordinal subtraction involves removing a well ordered proper subset from a well-ordered set. I hope that makes more sense.

But there is no such thing as the "multiplicative inverse" of a set when the elements of the set are not guaranteed to be numbers! I really can't tell if you are a genius or a crank, because everything you are saying is nonsensical to me.

Let's go back to basics. Let me know if you disagree with anything I say.

A well-ordered set is an ordered pair (S, R) where R is a well-ordering on S.

If (S,R) and (S',R') are two well-ordered sets, then they are isomorphic if there is a bijection f: S -> S' which preserves the ordering.

An ordinal is an equivalence class of well-ordered sets under the equivalence relation of isomorphism.

If (S,R) and (S',R') are well-ordered sets, then we define the sum of the ordinals which they represent to be the ordinal which is represented by the well-ordered set which is formed by sticking (S,R) and (S',R') end-to-end.

If (S,R) and (S',R') are well-ordered sets, then we define the product of the ordinals which they represent to be the ordinal which is represented by the well-ordered set which is formed by taking (S',R') and replacing every entry by a copy of (S,R).

Given these definitions, it is perfectly clear that there is no ordinal alpha such that alpha + 3 = 0, where "3" refers to the ordinal represented by the well-ordered set ({1,2,3},<=) and "0" refers to the ordinal represented by the unique empty well-ordered set . Why? Well, because the sum of two ordinals is always at least as big as each of them!

You( and Mr Craig) are correct that subtraction is prohibited in transfinite calculus given the definition of transfinite ordinal addition. You cannot simply chop off infinitely many elements from the larger end of an ordinal so subtraction would have to be defined quite differently than addition(it would not be equivalent to the addition of an additive inverse since there is no such thing for a transfinite number).

But as I showed you earlier, it is possible to construct an infinitesimal ordinal such that ordinal multiplication of an infinitesimal by an infinite yields a FINITE result. If you substitute omega^-1 into omega as follows:

omega = [1,2,3,..........,n,......]
omega^-1 = [1,1/2,1/3,..........,1/n,.......]

and omega^-1 * omega = [ omega^-1, 2* omega^-1,................,n*omega^-1,.............], the limit of this product as n->omega is (first finite ordina)number 1.

Words mean things!

Ordinals are precisely what I said that they are. They are equivalence classes of well-ordered sets under the equivalence relation of isomorphism. You can't just construct some object and call it an ordinal!

Are the following statements true or false?





I claim that they are true. In fact, I can prove it.

Since they are true, it follows that there is no nice notion of subtraction or division for ordinals.

There are things which are sort of like subtraction and division for ordinals: there is left-subtraction for pairs "alpha, beta" where "beta <= alpha", and there is left-division with remainder. But these are not truly subtraction and division.

By the same token, there is no nice notion of division for integers. There is only division with remainder.

Look, the most charitable interpretation of your posts that I can think of is that you have shown that the semiring of ordinals can be considered to sit inside a larger semiring in which a nice property holds that does not hold in the semiring of ordinals. You need to do the following four things:

Words mean things, eh? I never woulda guessed.


Now with regard to the definition of an Ordinal Number and the set 1/omega as I defined it earlier ITT, (1/omega) IS well ordered under the reverse ordering of the natural numbers, and does indeed describe an order type: That order type is that of an infinite descending chain with a greatest element.

However, the set (1/omega)+omega under ordinal addition results in a set that is not well ordered under the forward ordering nor the reverse ordering as it has no least or greatest element. Hence, there is no additive closure for the sum of infinitesimal and infinite ordinals but there is closure under multiplication.

There is a set defined by Mr Conway known as the Surreal Numbers which include infinitesimals like I described. This allows for right division in transfinite calculus.

The pair ( {1,1/2,1/3,...,1/n,...} , <=) is not an ordinal, since the set has not a least element, and hence <= is not a well-ordering of it.

You are doing arithmetic in a superclass of the ordinals.


If we would consider instead the ordinal O=({1,1/2,1/3,...,1/n,...} , >=) we would have O=omega. Simply by considering the bijection f: {1,2,3,...,n,...}->{1,1/2,1/3,...,1/n,...} defined by f(n)=1/n. We can check that n<=m implies f(n)>=f(m), so f preserves the ordering and gives an isomorphism between O and omega.
Hence O*omega=omega^2.

Um, I think you're confusing ordinals with cardinals. For any natural number n, f: omega -> O = (1/n) and while
n < n+1, f(n)>f(n+1). So this is not an order isomorphism, even though it is a bijection.

We have an ordinal omega=(S_1,R_1) with S_1={n}_n and R_1(x,y)=(x<=y).
And another ordinal O=(S_2,R_2) with S_2={1/n}_n and R_2(x,y)=(x>=y).
In both cases we have an element z in S such that R(z,x) for all x in S, as necessary to be R a well-order.

And the bijection f:S_1->S_2 defined as before satisfies R_1(x,y)<->R_2(f(x),f(y)).

Regarding the product of O and omega, the cartesian product of these 2 sets will consist of elements (n/m) for any natural numbers n and m. What this looks like is an omega by omega matrix of fractions. When n = m, (n/m)=1. The diagonal row consists of fractions of the form (n/n) which are countably many copies of the number 1. Since all elements of this diagonal are equal, we have the order type {1}. Keep in mind that O is a limit ordinal whose limit is infinitely smaller than the number 1. So for O*omega to be an infinite ordinal violates the rules of cardinal arithmetic(ordinals do have a corresponding cardinality FYI).

But we have proved that O=omega (by finding an isomorphism between the ordered sets used to define them), so O*omega must be omega^2 which is clearly different that 1.

Let us clear the product.
The product of two ordinals can be defined in the following form:
(S_1,R_1)*(S_2,R_2)=(S_3,R_3)
where S_3=S_1*S_2 (Cartesian product of sets)
and R_3 is the lexicographic product, this is, R_3(x,y) := R_1(x,y) or (x=y and R_2(x,y)).
Hence O*omega is
[1,2,3,...]*[1,1/2,1/3,...]=[(1,1),(1,1/2),(1,1/3),..., (2,1),(2,1/2),..., (3,1),(3,1/2),...., ....]
which is clearly isomorphic to omega^2=[(1,1),(1,2),(1,3),..., (2,1),(2,2),(2,3), ...., ....]

You seem to be using another product of sets, which is not a product of ordinals (since you get different results for O and omega).

No, *we* haven't. And here's why: Let f be the bijection from omega ->O than you defined earlier ITT, now for any natural numbers (n,m), n <= m -> f(n) >= f(m) because f (n) = 1/n. Thus f does not satisfy the definition of an order isomorphism because it reverses the ordering of omega rather than preserving it.

With the notation in which you reference, we have two posets:
The first poset P_omega:=(S,<=_s) with S:=Nat and (<=_s):=(<=)
And the second poset P_O:=(T,<=_T) with T:={ 1/k | k in Nat} and (<=_T):=(>=)
And the isomorphism f which I defined satisfies for all u,v in S that u<=_S v if and only if f(u)<=_T f(v). Which is exactly the definition of order isomorphism you are referencing.

Indeed, the first example of the Wikipedia is very similar:

For any natural numbers n and m, IF n < =m THEN (1/n) >= (1/m). The bijective mapping YOU defined from omega to O was f(n) = (1/n) for any natural number n(in omega). Also, much of your notation is extremely unclear to me( as in (_S v ) ). If f were an order ismorphism, we'd have

u <= v -> f(u) <= f(v) for all u, v in omega and f(u),f(v) in O. But the direction of the '<' is reversed when you map omega to O which fails to meet the criterion of an order isomorphism. Please use english or the appropriate mathematical symbols.

I have used the notation closer to wikipedia as possible under the restriction of the plaintext. Whenever I wrote <=_S, I meant the less or equal with subindex S used in wikipedia, and with <=_T the same but for the subindex T.
In LaTeX I could have written $\leq_S:=\leq$, $\leq_T:=\geq$, ending with $u\leq_S v \leftrightarrow f(u)\leq_T f(v)$.
Please, look at the first example of the wikipedia page, which I highlighted, in which two different orders are used, exactly like our case.