 |
|
| Science Forum Index » Mathematics Forum » Extending the reals |
|
Page 1 of 4 Goto page 1, 2, 3, 4 Next |
|
| Author |
Message |
| David R Tribble |
 Posted: Wed Dec 27, 2006 5:24 pm |
|
|
|
Guest
|
Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
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.
[quote:2a664f256d]From there, an entire set H of h-numbers is defined as containing other
h-numbers, being sums and products of reals and n1.[/quote:2a664f256d]
The article goes on to explore arithmetic for the h-numbers and ends
up defining an entire hierarchy of such numbers (n1, n2, etc.) and
their multiplicative inverses (e1 = 1/n1, e2, etc.).
The article (which requires a browser capable of rendering certain
mathematical HTML characters) is at:
http://david.tribble.com/text/hnumbers.html
Comments and suggestions are welcome. I'm curious to know if
something like this has been done before, or whether it's
mathematically inconsistent.
-drt
| Rev 1.0, 2006-12-08
| iQA/AwUBRYQ+eHS9RCOKzj55EQJFrQCeMuZ+B+D9s0iLYs9j80SIjxNzAsoAoJWG
| /I5J051vnLpnmDctzmYe1dy6
| =iGOQ |
|
|
| Back to top |
|
|
|
| Guest |
| Posted: Wed Dec 27, 2006 7:23 pm |
|
|
|
|
David R Tribble wrote:
[quote:acea181b04]Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
[/quote:acea181b04]
<snip>
[quote:acea181b04]The article (which requires a browser capable of rendering certain
mathematical HTML characters) is at:
http://david.tribble.com/text/hnumbers.html
Comments and suggestions are welcome. I'm curious to know if
something like this has been done before, or whether it's
mathematically inconsistent.
[/quote:acea181b04]
Nice job. In conversations with Tony Orlow, I also thought of a similar
system (essentially, an ordered field extension of R using a polynomial
basis over some "unit infinity" B). The unanswered question (to me) in
your "Loose ends" section is defining a topology on the h-numbers so
that limits can be expressed. Will think about this on the drive home
from Xmas Festivities...
(Holiday) Cheers - Chas |
|
|
| Back to top |
|
|
|
| David R Tribble |
| Posted: Thu Dec 28, 2006 4:49 pm |
|
|
|
Guest
|
David R Tribble wrote:
[quote:0d62b3ce1f]Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
The article (which requires a browser capable of rendering certain
mathematical HTML characters) is at:
http://david.tribble.com/text/hnumbers.html
[/quote:0d62b3ce1f]
Chas Brown wrote:
[quote:0d62b3ce1f]Nice job. In conversations with Tony Orlow, I also thought of a similar
system (essentially, an ordered field extension of R using a polynomial
basis over some "unit infinity" B).
[/quote:0d62b3ce1f]
David R Tribble wrote:
[quote:0d62b3ce1f]I was hoping this conversation would go quite a bit further
before Tony's name was mentioned. His "unit infinity" is an
extermely flawed and inconsistent concept from the get go.
Oh well.
For the record, I was inspired by some concepts of NSA for
creating my h-numbers.
[/quote:0d62b3ce1f]
Ross A. Finlayson wrote:
[quote:0d62b3ce1f]Ha ha ha ha ha ha. Troll! Heh.
I applaud your notion, to be honest.
Your "seed", or eta_1 or what have you of your h-numbers, does look
quite similar to Tony's H-riffics' "unit infinity", or Yaroslav
Sergeyev's grossone or (1), or the unit scalar infinity. However, it
appears they are more similar to the Robinso(h)nian hyperreals. Tony's
rule would seem to apply, or not.
[/quote:0d62b3ce1f]
They have nothing in common. My n1 (eta_1) is in almost all respects
another kind of real, obeying relations such as 1+n1>n1. Infinite
numbers don't act that way, and I make it quite clear that n1 is no
such thing. They are similar to Robinson's illimited numbers, and
I'd like to know just how closely related.
[quote:0d62b3ce1f]The blackboard bold H is used for quaternions, the set and division
algebra. You might want to select a different symbol to avoid
ambiguity, although it is clear in context.
[/quote:0d62b3ce1f]
Such details can be worked out later. "H" would appear to reflect the
name "Hamiltonian". My "H" is more properly pronounced "Eta".
I originally started to call them "T-numbers", but I thought "eta"
looked like a better choice for "number". But these are minor details
that can be adjusted later.
[quote:0d62b3ce1f]Do you see any applications of your system?
[/quote:0d62b3ce1f]
A strange question to ask of abstract mathematics.
[quote:0d62b3ce1f]Are you able to formalize otherwise inaccessible or what are
deemed paradoxical true results using your system?
[/quote:0d62b3ce1f]
Such as?
[quote:0d62b3ce1f]If so, do they apply to the real numbers and thus
demand that all numbers on the real number line are real numbers,
demanding reevaluation of sufficiency and necessity in representing the
real numbers?
[/quote:0d62b3ce1f]
As you can see at the end of the article, the reals R are a subset of
the suprareals H, designated H0. The entire hierarchy of H is a field.
I also make it quite clear that the h-numbers are an extension to the
reals, requiring additional axioms for their existence. So what is
true of the reals is not necessarily true of the h-numbers.
[quote:0d62b3ce1f]What's the point?
[/quote:0d62b3ce1f]
Good question. What IS the point of the real number line? |
|
|
| Back to top |
|
|
|
| Ross A. Finlayson |
| Posted: Fri Dec 29, 2006 3:17 am |
|
|
|
Guest
|
David R Tribble wrote:
[quote:050dc95582]David R Tribble wrote:
Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
The article (which requires a browser capable of rendering certain
mathematical HTML characters) is at:
http://david.tribble.com/text/hnumbers.html
Chas Brown wrote:
Nice job. In conversations with Tony Orlow, I also thought of a similar
system (essentially, an ordered field extension of R using a polynomial
basis over some "unit infinity" B).
David R Tribble wrote:
I was hoping this conversation would go quite a bit further
before Tony's name was mentioned. His "unit infinity" is an
extermely flawed and inconsistent concept from the get go.
Oh well.
For the record, I was inspired by some concepts of NSA for
creating my h-numbers.
Ross A. Finlayson wrote:
Ha ha ha ha ha ha. Troll! Heh.
I applaud your notion, to be honest.
Your "seed", or eta_1 or what have you of your h-numbers, does look
quite similar to Tony's H-riffics' "unit infinity", or Yaroslav
Sergeyev's grossone or (1), or the unit scalar infinity. However, it
appears they are more similar to the Robinso(h)nian hyperreals. Tony's
rule would seem to apply, or not.
They have nothing in common. My n1 (eta_1) is in almost all respects
another kind of real, obeying relations such as 1+n1>n1. Infinite
numbers don't act that way, and I make it quite clear that n1 is no
such thing. They are similar to Robinson's illimited numbers, and
I'd like to know just how closely related.
The blackboard bold H is used for quaternions, the set and division
algebra. You might want to select a different symbol to avoid
ambiguity, although it is clear in context.
Such details can be worked out later. "H" would appear to reflect the
name "Hamiltonian". My "H" is more properly pronounced "Eta".
I originally started to call them "T-numbers", but I thought "eta"
looked like a better choice for "number". But these are minor details
that can be adjusted later.
Do you see any applications of your system?
A strange question to ask of abstract mathematics.
Are you able to formalize otherwise inaccessible or what are
deemed paradoxical true results using your system?
Such as?
If so, do they apply to the real numbers and thus
demand that all numbers on the real number line are real numbers,
demanding reevaluation of sufficiency and necessity in representing the
real numbers?
As you can see at the end of the article, the reals R are a subset of
the suprareals H, designated H0. The entire hierarchy of H is a field.
I also make it quite clear that the h-numbers are an extension to the
reals, requiring additional axioms for their existence. So what is
true of the reals is not necessarily true of the h-numbers.
What's the point?
Good question. What IS the point of the real number line?
[/quote:050dc95582]
Well, I wonder, can you partition the unit interval into eta_1 many
equal-sized intervals?
About application, and where your numbers are a field that makes them
very similar to the hyperreals, where via the transfer principle the
large part of true statements about the hyperreals is true about the
standard reals and vice versa, and thus they're indistinguishable at a
distance, and interchangeable in practice, I wonder if there are any
what you would see as true facts about the real numbers that you can
verify via properties of the h-numbers or what have you that you don't
see verifiable from the standard properties of the "standard" real
numbers.
I'm glad you interpreted the question about the point in that way.
What's the point? Do you find it conceivable that the points and
variously all the lines in some ultimate space don't need to be defined
but that the existence of objects with those properties is a
consequence of the existence of space? How about space as a
consequence of existence of anything, even nothing?
Also, what's the point of your h-numbers, as in, why have you extended
effort to describe what you see as abstract? What perceived
insufficiency in simpler language about the real numbers do you see as
justifying the effort in inventing a notation and description of these
things? If you can't prove statements about the real numbers not
already proven by those of the standard real numbers, how is your
treatment of these objects any different than relabelling the
properties of the real numbers?
If the reals are complete in their total ordering there aren't any
numbers in that ordering not in it.
I thought it was funny to write "Troll!", maybe you can understand that
many perceive something in terms of numbers and number theories beyond
the standard, as you do, not all who troll are trolls.
Why bother addending to the standard real numbers?
Do these h-numbers have any geometric meaning?
For the record, I was inspired by the numbers.
The point is polydimensional.
Ross |
|
|
| Back to top |
|
|
|
| hagman |
| Posted: Fri Dec 29, 2006 6:43 am |
|
|
|
Guest
|
David R Tribble schrieb:
[quote:895c7b8949]Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
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.
The article goes on to explore arithmetic for the h-numbers and ends
up defining an entire hierarchy of such numbers (n1, n2, etc.) and
their multiplicative inverses (e1 = 1/n1, e2, etc.).
The article (which requires a browser capable of rendering certain
mathematical HTML characters) is at:
http://david.tribble.com/text/hnumbers.html
Comments and suggestions are welcome. I'm curious to know if
something like this has been done before, or whether it's
mathematically inconsistent.
-drt
| Rev 1.0, 2006-12-08
| iQA/AwUBRYQ+eHS9RCOKzj55EQJFrQCeMuZ+B+D9s0iLYs9j80SIjxNzAsoAoJWG
| /I5J051vnLpnmDctzmYe1dy6
| =iGOQ
[/quote:895c7b8949]
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}.
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!
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. |
|
|
| Back to top |
|
|
|
| Guest |
| Posted: Fri Dec 29, 2006 8:52 pm |
|
|
|
|
hagman wrote:
[quote:4b16c02ae9]David R Tribble schrieb:
Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
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.
The article goes on to explore arithmetic for the h-numbers and ends
up defining an entire hierarchy of such numbers (n1, n2, etc.) and
their multiplicative inverses (e1 = 1/n1, e2, etc.).
The article (which requires a browser capable of rendering certain
mathematical HTML characters) is at:
http://david.tribble.com/text/hnumbers.html
Comments and suggestions are welcome. I'm curious to know if
something like this has been done before, or whether it's
mathematically inconsistent.
-drt
| Rev 1.0, 2006-12-08
| iQA/AwUBRYQ+eHS9RCOKzj55EQJFrQCeMuZ+B+D9s0iLYs9j80SIjxNzAsoAoJWG
| /I5J051vnLpnmDctzmYe1dy6
| =iGOQ
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}.
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!
[/quote:4b16c02ae9]
Also, for example, 18a claims that u + v in H u R u L; but given h in
H, h + 1/h is not in any of H, R, or L.
[quote:4b16c02ae9]
Instead, You should have used R(X), the field of rational functions in
one variable.
[/quote:4b16c02ae9]
This is actually what I /thought/ the OP had defined (so much for my
careful reading!).
[quote:4b16c02ae9]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.
[/quote:4b16c02ae9]
Which leads us back to the question of what topology to use for these
numbers, and how/whether limits are to be defined.
Using open intervals seems the most "natural" basis for a topology for
this space; the sub-space topology on R is then the usual topology.
This gives us a Hausdorff space (I think, regular Hausdorff?).
Then, for example, the set of "finite" suprareals is an open set as is
the set of "infinite" suprareals, so we get the lack of connectedness
implied in the drawings the OP made. (Many other
examples of the disconnectedness of this system can be found.) Thus,
this space is not normal (the above two sets are also closed and
disjoint, but are not separated).
Let us denote the suprareals as H^. We can define lim (x->c) {f(x)} = L
as:
L in H^ is the limit of f(x) as x approaches c if, for every 0 <
epsilon in H^, there exists a 0 < delta in H^ such that for all x such
that |x-c| < delta, |f(x) -L| < epsilon.
It doesn't seem like this yields a complete space; but I can't
construct any counterexamples off the top of my head.
Cheers - Chas |
|
|
| Back to top |
|
|
|
| David R Tribble |
| Posted: Fri Dec 29, 2006 9:59 pm |
|
|
|
Guest
|
David R Tribble schrieb:
[quote:a440746648]Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
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.
[/quote:a440746648]
hagman wrote:
[quote:a440746648]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}.
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!
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.
[/quote:a440746648]
Thank you very much for the feedback. I confess that, as an
amateur, this is out of my depth. But I'm willing to learn. Any
good books to recommend that cover R(X), R[X], etc.?
-drt |
|
|
| Back to top |
|
|
|
| Guest |
| Posted: Sun Dec 31, 2006 6:53 pm |
|
|
|
|
David R Tribble wrote:
[quote:0fe90a9706]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?
[/quote:0fe90a9706]
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 1 , 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..
[quote:0fe90a9706]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?)
[/quote:0fe90a9706]
That's what he means; the "leading coefficient" is the coefficient of
the largest power of n1 which is non-zero.
[quote:0fe90a9706]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
[/quote:0fe90a9706]
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 |
|
|
| Back to top |
|
|
|
| Ross A. Finlayson |
| Posted: Sun Dec 31, 2006 8:26 pm |
|
|
|
Guest
|
cbrown@cbrownsystems.com wrote:
[quote:2a11b40892]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.
....
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
[/quote:2a11b40892]
Hello,
Browsing arXiv the other day I noticed there is a recent paper about
Non-Dedekindian numbers. The author claims in the abstract that all
types of hyperreals are the soi-disant "non-Dedekindian" numbers.
http://arxiv.org/abs/math.GM/0612590
The complete ordered field, is the complete ordered field. Where there
is only one of those, hyperreals are them. If there are infinitesimals
in the reals, they're reals. The real number are already the compete
ordered field.
Schmieden and Laugwitz' are not hyperreals.
Built on axiomatics of integers, each of the above systems (hyper-,
super-, supra-, partially ordered ring of with rather restricted
transfer principle, reals) is incomplete.
Ross |
|
|
| Back to top |
|
|
|
| Tony Orlow |
| Posted: Mon Jan 01, 2007 6:24 pm |
|
|
|
Guest
|
cbrown@cbrownsystems.com wrote:
[quote:00b314015b]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").
[/quote:00b314015b]
I'm sorry Chas, but that doesn't make sense to me. In the finite case it
does not hold. Say we have n1=4 and e1=1/4. 1/3 does not equal 1/4 with
a remainder of 1/4. 1/(n1-1)=((n1-1)+1)/(n1-1)-(n1-1)/(n1-1)= n1/(n1-1)-1.
Sorry, David, but I have to put this in terms of T-riffics.
Say we use base 2 T-riffics, and n1=1:000...000.000...000
n1-1=0:111...111.000...000. 1/(n-1)=0.000...001:000...001::::. That is,
in T-riffics, 1/(Big'un-1)= sum(n=-1->-oo: Big'un^n). This sum actually
holds for all bases.
[quote:00b314015b]
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.
[/quote:00b314015b]
Actually that sounds like the result with the T-riffics: an unending sum
of subsequent infinitesimal levels.
[quote:00b314015b]
(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 1, 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..
[/quote:00b314015b]
Uh, yeah, the set of ih numbers needs to to be embedded in each real
unit interval, and likewise with each standard real "halo" about each
h-number. Any combination of the three should suffice as an element of
this universe.
[quote:00b314015b]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.
[/quote:00b314015b]
Lexicographically!! :)
[quote:00b314015b]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
[/quote:00b314015b]
Cheers |
|
|
| Back to top |
|
|
|
| David R Tribble |
| Posted: Mon Jan 01, 2007 11:12 pm |
|
|
|
Guest
|
hagman wrote:
[quote:252c5ccffe]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!
[/quote:252c5ccffe]
David R Tribble wrote:
[quote:252c5ccffe]I don't see why they are not equivalent. And why is there no inverse?
[/quote:252c5ccffe]
Chas Brown wrote:
[quote:252c5ccffe]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).
[/quote:252c5ccffe]
Yes, I had suspected something like that was the case, since 1/h for
an arbitrary suprareal h, being a polynomial, is a non-terminating
polynomial (a la Newton's expansion of 1/P(x), if I remember
correctly). I didn't follow up on it, though.
Limiting the suprareals to finite-power polynomials makes the
handling of their inverses tricky.
[quote:252c5ccffe]Another (different) problem with your formulation is that (contrary to
theorem 1, 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..
[/quote:252c5ccffe]
Yes, I got ahead of myself there. I was thinking of the total union
of all suprareal polynomials, including sums of h-numbers and
ih-numbers. This is one of the sections that needs reworking,
I suppose.
David R Tribble wrote:
[quote:252c5ccffe]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?)
[/quote:252c5ccffe]
Chas Brown wrote:
[quote:252c5ccffe]That's what he means; the "leading coefficient" is the coefficient of
the largest power of n1 which is non-zero.
[/quote:252c5ccffe]
Thanks, thought so.
David R Tribble wrote:
[quote:252c5ccffe]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
[/quote:252c5ccffe]
Chas Brown wrote:
[quote:252c5ccffe]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.
[/quote:252c5ccffe]
Yes, and it follows that when g is a suprareal in H2 having
coefficients in H1, the resulting terms are products of eta_1^n
and eta_2^m, as expected.
Thanks very much for the interest and the feedback. I hope to hammer
the article into better shape soon. |
|
|
| Back to top |
|
|
|
| David R Tribble |
| Posted: Mon Jan 01, 2007 11:21 pm |
|
|
|
Guest
|
David R Tribble wrote:
[quote:20465bf2dd]Please read the article. My suprareals are not infinite numbers.
[/quote:20465bf2dd]
Tony Orlow wrote:
[quote:20465bf2dd]I have now read it. Yes, you are very careful not to call them
"infinite", and even to point out that, even though you illustrate them
as residing colinear with the reals, they are not really in that
relationship. I didn't see the point in tiptoeing around that,
personally. Your h-numbers are "larger than any finite", meaning farther
along the line from 0.
[/quote:20465bf2dd]
Chas Brown wrote:
[quote:20465bf2dd]Except for the fact that the suprareals don't form a "line"; they are
toplogically disconnected. By "disconnected", I mean that the
suprareals are the union of two non-empty disjoint open sets.
The real number line is /not/ the union of two non-empty disjoint open
sets. Lines are connected; the suprareals are not connected; therefore
the suprareals are not a line.
[/quote:20465bf2dd]
Exactly. (I appreciate the answers, Chas.)
I thought my diagrams made this clear.
Tony Orlow wrote:
[quote:20465bf2dd]I had a few comments:
1. In the section "Even More Numbers," you say, "In fact it would appear
that every h-number can be represented as a polynomial over powers of
eta_1." Is that to say that one cannot have log_2(eta_1) and produce
another h-number using that function? How many bits are required to list
eta_1 elements?
[/quote:20465bf2dd]
Chas Brown wrote:
[quote:20465bf2dd](i) "2^eta_1" is not a polynomial in eta_1; so "2^eta_1" is not
meaningful as a suprareal. It follows that "log_2(eta_1)" is not
meaningful for suprareals.
(ii) eta_1 is not a cardinality; so it makes no sense to say "this set
has eta_1 elements"; just as it makes no sense to say "this set has
1/sqrt(2) elements".
One has at best "the cardinality of the suprareals <= eta_1 is the same
as the cardinality of the reals". Of course it is also true (as DT
proves) that "the cardinality of the suprareals > eta_1 is the same as
the cardinality of the reals"; and "the cardinality of the suprareals
= eta_1 is the same as the cardinality of the suprareals <=
1/sqrt(2)".
(iii) There are as many suprareals as there are real numbers.
Therefore, you cannot in any case "list" the elements of the
suprareals, just as you cannot list the real numbers.
[/quote:20465bf2dd]
Also, it makes no sense to talk about "how many bits" could be
used to encode a given suprareal, since eta_1 cannot be represented
in a binary notation, since it's not a real. It's kind of like asking
how many bits are required to encode i or w (omega).
Tony Orlow wrote:
[quote:20465bf2dd]4. "An Uncountable Hierarchy" struck me as odd, coming from you, David.
On the one hand, you are enumerating a sequence of sets, each defined as
being the elements larger than all elements in the previous set (rather
like limit ordinals) but then you suggest that each set may be numbered
with a real. Are you suggesting an uncountable sequence of sets, which
you previously proclaimed to be an idea without any sense?
[/quote:20465bf2dd]
Chas Brown wrote:
[quote:20465bf2dd](i) It's not an uncountable sequence; because a sequence is always, by
definition, countable (that's the part that is "an idea that doesn't
make any sense").
[/quote:20465bf2dd]
Exactly.
[quote:20465bf2dd]Note that using this terminology, eta_b * eta_c is NOT EQUAL TO
eta_(b*c). It remains the polynomial expression eta_b * eta_c. The real
number labels a, b, c, etc. are only there to indicate an ordering.
[/quote:20465bf2dd]
Exactly.
Tony Orlow wrote:
[quote:20465bf2dd]What is the
set H_pi the set of, all elements larger than the element of H_x, where
x is the predecessor to pi in the natural order of the reals?
[/quote:20465bf2dd]
Chas Brown wrote:
[quote:20465bf2dd](ii) Your statement makes no sense; because there is no such
"predecessor to pi" in the usual ordering of the reals.
[/quote:20465bf2dd]
Exactly.
[quote:20465bf2dd](iii) It follows that if 0 < x < y where x and y are real numbers, then
H_x is a proper subset of H_y.
So what he has described is a total order on sets, which is /not/
synonymous with "a sequence of sets".
[/quote:20465bf2dd]
Exactly.
(Thanks, Chas.) |
|
|
| Back to top |
|
|
|
| David R Tribble |
| Posted: Thu Jan 04, 2007 7:11 pm |
|
|
|
Guest
|
David R Tribble wrote:
[quote:96814c7152]Yes. I have two choices:
a) allow suprareals to be polynomials with an infinite number of terms;
b) accept that H_1 U L_1 is not a field.
Option (b) appears to be the more acceptable at this point, because
it's simpler and because it allows me to keep the more general
suprareal construction as being a polynomial with both positive and
negative integer powers of eta_i.
h = sum{i= -n to +m} x_i eta_1, x_i in R
Obviously, I need to rework some of my article.
[/quote:96814c7152]
Oops, that should be:
h = sum{i= -n to +m} x_i eta_1^i, x_i in R
So a given suprareal h in H_1 U L_1 (in its corrected form) is
still a polynomial over integer powers of eta_1:
h = x + x1 eta_1^1 + x2 eta_1^2 + ... + xn eta_1^n,
where x, x1, x2, ..., xn in R and at least one x_i /= 0. |
|
|
| Back to top |
|
|
|
| Guest |
| Posted: Thu Jan 04, 2007 8:49 pm |
|
|
|
|
David R Tribble wrote:
[quote:e708ef647a]Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
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.
The article goes on to explore arithmetic for the h-numbers and ends
up defining an entire hierarchy of such numbers (n1, n2, etc.) and
their multiplicative inverses (e1 = 1/n1, e2, etc.).
The article (which requires a browser capable of rendering certain
mathematical HTML characters) is at:
http://david.tribble.com/text/hnumbers.html
Comments and suggestions are welcome. I'm curious to know if
something like this has been done before, or whether it's
mathematically inconsistent.
[/quote:e708ef647a]
I don't see how these numbers relate to the reals. It is all well and
good to build something with a certain property, but your numbers don't
seem to be defined in a way that allows you to actually show they are
larger than the reals -- you just declare them to be and that is that.
For instance, the standard model of the hyperreals is as infinite
sequences of real numbers {r_i} sorted into equivalence classes by a
free ultrafilter on the indices i. Then the reals are (roughly)
sequences whose members agree on an element of the ultrafilter, and the
infinite hyperreals are those sequences which have an unbounded
subsequence indexed by an element of the ultrafilter. With this
construction finite hyperreals (reals) and infinite hyperreals can be
compared, and have the same construction.
What is the common structure of the reals and your h numbers that
allows you to show that the h's are larger?
Baldin Lee Pramer |
|
|
| Back to top |
|
|
|
| Guest |
| Posted: Thu Jan 04, 2007 10:29 pm |
|
|
|
|
David R Tribble wrote:
[quote:2cd24613a1]Over the last few months I've been noodling around with the concept
of an extension to the reals, defining real-like numbers that are
larger than any regular real.
[/quote:2cd24613a1]
Just glancing over the article it seemed a lot like the ordered ring of
real polynomials in one variable. Just define x to be your n1. Then
making this a field is taking the ordered field of rational functions.
Read the following book
J. Bochnak, M. Coste, and MF. Roy. Real Algebraic Geometry. Springer,
1998.
Jiri |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Thu Nov 26, 2009 9:06 pm
|
|