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David R Tribble wrote:
> David R Tribble wrote:
> >> I've written an article exploring the effects of adding a few axioms to
> >> standard arithmetic to extend the reals with what I call, for lack of
> >> a better term, "h-numbers". Briefly, an h-number has a magnitude
> >> greater than any real. The first axiom states that n1 (eta_1), a
> >> primitive h-number constant, exists, and that x < n1 for all x in R.
> >> >From there, an entire set H of h-numbers is defined as containing other
> >> h-numbers, being sums and products of reals and n1.
> >
>
> hagman wrote:
> >> The easiest, most obvious and literally universal method to extend the
> >> reals by adding new numbers is to look at R[X].
> >> In fact, that is what you did; additionally you defined an order
> >> relation on HuR = R[X] by declaring all non-zero-polynomials with
> >> positive leading coefficient as positive.
> >> It is trivially verified that for this set P we have P*P subset P, P+P
> >> subset P and R[X] is the disjoint union of P and -P and {0}.
> >
>
> I think I follow most of what you are saying from a more thorough
> reading of your post...
>
> >> Later you construct HuRuL in a way that essentially boils down to
> >> R[X,Y]/(XY-1) and claim that this is a field.
> >> However, Theorem 18c. fails to be true (at least with the first version
> >> of Axiom 8; the second version is not equivalent):
> >> In R[X,Y]/(XY-1), X+1 has no inverse!
> >
>
> I don't see why they are not equivalent. And why is there no inverse?
>
The second version of Axiom 8 states that there is an inverse for all h
in H; from which it follows immediately that there exists x such that
x*(n1 - 1) = 1.
The first version states only that there exists e1 such that e1*n1 = 1.
It must then be proven that, for example, there also exists x such that
x * (n1 - 1) = 1.
(I'll skip some rigor here). But if we try to find such an x by
(essentially) "long division", we get first that e1*(n1 - 1) = 1 - e1
(i.e., "1 divided by (n1 - 1) is e1 with a remainder of e1").
And (continuing our "long division"), then (e1 + e1^2)(n1 - 1) = 1 -
e1^2, and then that (e1 + e1^2 + e1^3)*(n1-1) = 1 - e1^3, and in
general (e1 + e1^2 + e1^3 + ... + e1^m)*(n1 - 1) = 1 - e1^m.
So we find that x = 1/(n1 - 1) = e1 + e1^2 + e1^3 + ... + e1^m + ... ;
but this is not a polynomial in e1 as it has an infinite number of
terms; so it is not in L.
(Side Bar: There are formulations of polynomials called "formal power
series" where such "infinite" polynomials are allowed. They come up in
combinatorics, amongst other places. Keywords: generating function,
formal power series).
Another (different) problem with your formulation is that (contrary to
theorem 18), H u R u L is not closed w.r.t to addition; because e1 + n1
is not in any of the sets H, R or L..
> >> Instead, You should have used R(X), the field of rational functions in
> >> one variable.
> >> The order on R[X] induces an order on R(X), i.e. f/g is positive if f
> >> and g are both positive or both negative.
> >> (Again, it is trivially verified that for this set P we have P*P subset
> >> P, P+P subset P and R[X] is the disjoint union of P and -P and {0}).
> >>
> >> To obtain nested hierarchies, repeat the step, i.e. use more variables:
> >> Let S be a totally ordered set (of variables, i.e. viewed as disjoint
> >> with R etc.; if you want to use S=R or the like, use some standard
> >> trick to obtain a disjoint copy)
> >> Then R[S] is a ring and R(S) is a field.
> >> To define an order relation on R[S] (and thus on R(S)), declare a
> >> non-zero element of R[S] as positive if the leading coefficient is a
> >> positive real.
> >> Note that each element of R[S] is in fact an element of R[X1,...,Xn]
> >> for a finite number of variables X1<...<Xn in S. Since
> >> R[X1,..,Xn]=R[X1,...,X{n-1}][Xn], the leading coefficient method can be
> >> applied step by step.
> >>
> >> This works for any totally ordered set S, be it finite, countable,
> >> continuum-sized or whatever.
> >
>
> Why does the leading coefficient determine the order (or perhaps I
> don't understand what "leading" means)? Given suprareals h and g,
> where
> h = 2 - 3n1^1 + 4n1^2
> g = 3 + 4n1^1 - 5n1^2
> then g < h because the highest power of n1 is -5n1^2 for g and
> is +4n1^2 for h, and because n1^p < n1^q for p < q.
>
> In other words, shouldn't it be that the coefficient for the largest
> power of n1 determines the order? (Or is that what you meant?)
>
That's what he means; the "leading coefficient" is the coefficient of
the largest power of n1 which is non-zero.
> Also bear in mind that for higher orders of suprareals, the
> polynomials over n2 (n3, n4, etc.) has coefficients that are not
> limited to reals but to all the suprareals of lower orders.
> For example, for h in H2,
> h = x0 + x1 n2^1 + x2 n2^2 + ... + xn n2^n
> where each coefficient x0, x1, x2, ..., xn can itself be a suprareal
> in H1, so that they are polynomials over n1, e.g.:
> x0 = y0 + y1 n1^1 + y2 n1^2 + ... + ym n1^m
Usually, when we speak of polynomial rings such as K[n1, n2] (where K
is some ring), we include such polynomials as n1 + n1*n2 + n2 (i.e.,
where n1 and n2 are "mixed"). This can be thought of also as the
polynomial n1 + (1 + n1)*n2. In the former case, we have a polynomial
in n1 and n2 with coefficients in K; in the latter (equivalent) case,
it's a polynomial in n2, with coefficients in K[n1].
For suprareals, the latter approach is useful for defining the ordering
needed to guarantee that each extension is indeed an ordered field; but
other than that, the two formulations are equivalent.
Cheers - Chas
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