| Topic: |
Science > Physics |
| User: |
"Jonathan Hoyle" |
| Date: |
23 May 2007 11:12:14 AM |
| Object: |
Re: Infinitesimal Arithmetic |
On May 23, 9:33 am, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
IST, Nelson's Internal Set Theory, which is Robinson's NSA's
hyperreals in set theory, has been proven _coconsistent_ with ZFC,
Zermelo-Fraenkel set theory with choice, not consistent by itself. So
that is not entirely correct.
Yes, when I said "consistently defined", I should have said "co-
consistent with ZFC", as there cannot be an absolute proof of NSA's
consistency (unless of course NSA were inconsistent).
How about Conway surreal infinitesimals? Are they not infinitesimals,
some of the non-zero ones?
Good point, I had neglected to consider Conway numbers. Also, I am
unaware of any proofs of Conway's relative consistency.
If there is the constant infinitesimal then Vitali type results about
the existence of non-measurable sets in the reals don't hold in the
same way. Structure is preserved in that way. Consider where the
infinite and infinitesimal terms are linearly dependent.
What do you mean by "constant infinitesimal"? Extending the reals to
include infinitesimals does not remove the existence of non-measurable
sets, at least using the usual definitions of the terms.
I wonder, in what ways would you consider IST and NSA any different
from each other? Basically IST posits a field tree which NSA uses to
contains non-real halos, except halos also are everywhere in a way
contiguous, or (thus) their points are. IST has the fluents and
fluxions.
What do you mean "IST has the fluents and fluxions"? IST (Internal
Set theory) is the simply the set-theoretic foundation to Robinson's
NSA. Nothing in IST makes any reference to the archaic terms "fluent"
and "fluxion". As for your question as to how IST and NSA differ,
IST is simply the axiomatic basis for a set theory, whilst NSA is the
analysis rest on top of it. It is analogous to standard Real Analysis
being founded upon ZFC.
No, nothing I wrote there is incorrect. You're just using the
standard definition of diverge and converge. Consider the definition
of the derivative in terms of an infinitesimal from two posts ago.
That's standard. The limit of the derivative for the free variable of
change is not finite and not zero. For example: rise over run,
slope, the derivative.
Even if you allow i to be infinitesimal, the integral still diverges
for non-zero i. This is provable from the Transfer Principle.
Partitioning the unit interval into infinitely many equal sized
partitions...
<snip>
This is provably impossible, both in standard and Non-Standard
Analysis. The rub comes with your term "equal sized". You can of
course partition an interval into finitely many equal-sized
partitions. But once you go to infinitely many, these partitions must
be either non-measurable (and thus incomparable) or of measure 0.
It gets worse. If your partitions are of measure 0, you must have
uncountably many of them. If you want only countably many infinite
partitions, they must be non-measurable. Moreover, these "partitions"
are disconnected scattered collections of points, nothing like an
interval.
They're degenerate intervals, because, they're intervals that only
contain the one number, for example as we were discussing about well-
ordering the real numbers. Yet, in terms of finite approximations as
are actually used in the applications, the logical number of
partitions is the Nyquist frequency of twice the product of the
frequencies of the events. That way, the least cross cancellative
terms are preserved, anti-principal component analysis on the large
systems. Any sample of the real numbers is probably _not_ the number,
they're the continuum, it's all quite simple.
Unfortunately, it is this "simplicity" that is hiding its self-
contradictions. To perform this calculation, you cannot have more
than countably many intervals, as only countable addivity is defined
for this (uncountable addivity is inconsistent with this process).
But to have countably infinitely many partitions, they cannot be
intervals, nor are these partitions even measurable.
Your process fails due to simple theorems proven in introductory
Measure Theory.
Really the point of that is the approximation correctors with the
polydimensional points, for example in these sensical non-real
functions that I describe. In physics terms, Heaviside was called
unorthodox, in for example the application of the step and impulse
function as indicator series in digital logic, where the actual N-body
component simulations readily synthesize.
In that way, real numbers seem consistent. They are quite consistent,
how could they not be?
Real numbers are consistent (at least not proven to be inconsistent).
The inconsistency arises from your faulty assumptions and invalid
leaps of reasoning.
I strongly encourage you to pick up an introduction on Measure
Theory. Its first chapter will explain all this to you and make quite
apparent why you cannot do what you hope to (at least with the
construction you have right now).
Regards,
Jonathan Hoyle
Eastman Kodak
.
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| User: "Ross A. Finlayson" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 08:37:19 PM |
|
|
On May 23, 9:12 am, Jonathan Hoyle <jonho...@mac.com> wrote:
On May 23, 9:33 am, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
IST, Nelson's Internal Set Theory, which is Robinson's NSA's
hyperreals in set theory, has been proven _coconsistent_ with ZFC,
Zermelo-Fraenkel set theory with choice, not consistent by itself. So
that is not entirely correct.
Yes, when I said "consistently defined", I should have said "co-
consistent with ZFC", as there cannot be an absolute proof of NSA's
consistency (unless of course NSA were inconsistent).
How about Conway surreal infinitesimals? Are they not infinitesimals,
some of the non-zero ones?
Good point, I had neglected to consider Conway numbers. Also, I am
unaware of any proofs of Conway's relative consistency.
If there is the constant infinitesimal then Vitali type results about
the existence of non-measurable sets in the reals don't hold in the
same way. Structure is preserved in that way. Consider where the
infinite and infinitesimal terms are linearly dependent.
What do you mean by "constant infinitesimal"? Extending the reals to
include infinitesimals does not remove the existence of non-measurable
sets, at least using the usual definitions of the terms.
I wonder, in what ways would you consider IST and NSA any different
from each other? Basically IST posits a field tree which NSA uses to
contains non-real halos, except halos also are everywhere in a way
contiguous, or (thus) their points are. IST has the fluents and
fluxions.
What do you mean "IST has the fluents and fluxions"? IST (Internal
Set theory) is the simply the set-theoretic foundation to Robinson's
NSA. Nothing in IST makes any reference to the archaic terms "fluent"
and "fluxion". As for your question as to how IST and NSA differ,
IST is simply the axiomatic basis for a set theory, whilst NSA is the
analysis rest on top of it. It is analogous to standard Real Analysis
being founded upon ZFC.
No, nothing I wrote there is incorrect. You're just using the
standard definition of diverge and converge. Consider the definition
of the derivative in terms of an infinitesimal from two posts ago.
That's standard. The limit of the derivative for the free variable of
change is not finite and not zero. For example: rise over run,
slope, the derivative.
Even if you allow i to be infinitesimal, the integral still diverges
for non-zero i. This is provable from the Transfer Principle.
Partitioning the unit interval into infinitely many equal sized
partitions...
<snip>
This is provably impossible, both in standard and Non-Standard
Analysis. The rub comes with your term "equal sized". You can of
course partition an interval into finitely many equal-sized
partitions. But once you go to infinitely many, these partitions must
be either non-measurable (and thus incomparable) or of measure 0.
It gets worse. If your partitions are of measure 0, you must have
uncountably many of them. If you want only countably many infinite
partitions, they must be non-measurable. Moreover, these "partitions"
are disconnected scattered collections of points, nothing like an
interval.
They're degenerate intervals, because, they're intervals that only
contain the one number, for example as we were discussing about well-
ordering the real numbers. Yet, in terms of finite approximations as
are actually used in the applications, the logical number of
partitions is the Nyquist frequency of twice the product of the
frequencies of the events. That way, the least cross cancellative
terms are preserved, anti-principal component analysis on the large
systems. Any sample of the real numbers is probably _not_ the number,
they're the continuum, it's all quite simple.
Unfortunately, it is this "simplicity" that is hiding its self-
contradictions. To perform this calculation, you cannot have more
than countably many intervals, as only countable addivity is defined
for this (uncountable addivity is inconsistent with this process).
But to have countably infinitely many partitions, they cannot be
intervals, nor are these partitions even measurable.
Your process fails due to simple theorems proven in introductory
Measure Theory.
Really the point of that is the approximation correctors with the
polydimensional points, for example in these sensical non-real
functions that I describe. In physics terms, Heaviside was called
unorthodox, in for example the application of the step and impulse
function as indicator series in digital logic, where the actual N-body
component simulations readily synthesize.
In that way, real numbers seem consistent. They are quite consistent,
how could they not be?
Real numbers are consistent (at least not proven to be inconsistent).
The inconsistency arises from your faulty assumptions and invalid
leaps of reasoning.
I strongly encourage you to pick up an introduction on Measure
Theory. Its first chapter will explain all this to you and make quite
apparent why you cannot do what you hope to (at least with the
construction you have right now).
Regards,
Jonathan Hoyle
Eastman Kodak
It seems the key is to recognize that there already is an
"infinitesimal arithmetic" of sorts, in the "infinitesimal analysis"
that most these days call the integral calculus, or just calculus.
The definition of the derivative of a (non-vector-valued) function (at
a point) is in terms of the difference in rise for an infinitesimal
difference in the run. For a straight line, any two points can be
compared and the derivative is a constant between those two points.
For any continuous function, two points can be compared and the
function has as a value of its derivative for at least one point in
the interval the same as the line through those points. Yet, without
analyzing the function to the infinitesimal, without geometrical or
other arguments about the function, that point is unknown.
So, then in the consideration of the derivative, it is similar in
notion to the post recently made about the degenerate intervals and
how they contain only one point. Here, instead there is the
expectation of a sort that there will be two points, basically
adjacent, that describe a line, the slope of which is tangent and the
derivative of the function of that line through the point and its left
or right or both.
Consider all the other types of calculus there are, it's great.
Calculus of finite differences, lambda calculus (types), pi calculus
(processes), ....
http://en.wikipedia.org/wiki/Calculus
There is much background in infinitesimals in the integral calculus,
questions arise as how to reasonably extend mathematics of the
infinitesimal to extend the body of knowledge of physics, in terms of
theories with predictive content. The infinitesimal analysis already
works very well in the predictive modelling in the physics, for
example in the classical.
Of course, the opposite of an infinitesimal is an infinity. There are
a variety of systems in the physics that make do in approximation with
the notion of infinite quantities, by the same token there are some
systems where, for example, the value of a quantity that is generally
non-negative and finite is infinite, for example the center of a black
hole, which is matter with infinite mass.
Calculus, with the definition of the derivative being via the limit,
that is the limit as generally x goes to 0, lim x ->0+, is not a
method of finite differences. It's also not a method of zero
difference. The integral calculus is not a method of finite
differences, the resummation of the areas, after they are variously
divided, partitioned with minimal information, into infinitely many
rows or columns must be over infinitely many of them.
Given the infirmity of ZF I can prove a variety of things in it. I'm
interested to hear about this volume on questions of universal
quantification. I borrowed Barrow's "Theories of Everything". It is
not really explaining much while it is excellent. Consider for
example mentioning the diagonal argument, then saying Cohen proved CH
independent of ZF(C) without explaining how. Do you _really_ think
Cohen proved CH independent of ZF(C)? Very few people have not just
been told that. So, explain in detail how Cohen uses universal
quantification, or not, in finding a maximal ordinal.
Some say light is a wave. Points are wave collisions, light is an
electromagnetic wave.
Ross
--
Finlayson Consulting
.
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
25 May 2007 09:00:13 AM |
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It seems the key is to recognize that there already is an
"infinitesimal arithmetic" of sorts, in the "infinitesimal analysis"
that most these days call the integral calculus, or just calculus.
That is precisely what Calculus is NOT. Calculus involves limits, not
infinitesimals. In fact, there are no references at all to any
infinitesimals anywhere in modern Calculus.
It is true that the original view of Calculus in the 18th century
involved some inconsistent objects called "infinitesimals", but they
riddled the field with contradictions and paradoxes. In the 19th
century, Calculus was cleaned up of all these paradoxes, and Calculus
taught can be taught without the need of those inconsistent
infinitesimals.
In the 20th Century, as you know, Abraham Robinson was able to re-
introduce infinitesimals in a consistent way (although these bare
little resemblance to the inconssistent objects of Newton). So you
can now prove the theorems of calculus with infinitesimals again
(although it is almost never taught that way in Freshman Calculus
classes).
The definition of the derivative of a (non-vector-valued) function (at
a point) is in terms of the difference in rise for an infinitesimal
difference in the run.
That is incorrect. The definition of the derivative is in terms of
limits.
For a straight line, any two points can be compared and the
derivative is a constant between those two points. For any
continuous function, two points can be compared and the
function has as a value of its derivative for at least one point in
the interval the same as the line through those points.
That is incorrect. For example, for the function f(x) = x^2, there
are no two distinct points x1 and x2 such that the derivative is the
same on f(x). This is of course easy to prove: f'(x) = 2x, and so if
f'(x1)=f'(x2) then 2x1=2x2 which implies x1=x2. So it is impossible
for two different points to have the same derivative. This is true
even in Non-Standard Calculus with x1 and x2 being infinitesimally
close to each other.
Yet, without analyzing the function to the infinitesimal, without
geometrical or other arguments about the function, that point is
unknown.
Untrue. That point can be easily proven using limits and without
infinitesimals.
So, then in the consideration of the derivative, it is similar in
notion to the post recently made about the degenerate intervals and
how they contain only one point. Here, instead there is the
expectation of a sort that there will be two points, basically
adjacent, that describe a line, the slope of which is tangent and the
derivative of the function of that line through the point and its left
or right or both.
That is incorrect. No two points are ever adjacent, not even in
Newton's inconsistent infinitesimal approach. Even Newton knew that
between any two points (even those infinitesimally close to each
other), there are an infinite number of points between them.
Consider all the other types of calculus there are, it's great.
Calculus of finite differences, lambda calculus (types), pi calculus
(processes), ....
http://en.wikipedia.org/wiki/Calculus
Yes, it is great, I agree. I'm pleased you are familiar with that
site.
There is much background in infinitesimals in the integral calculus,
questions arise as how to reasonably extend mathematics of the
infinitesimal to extend the body of knowledge of physics, in terms of
theories with predictive content. The infinitesimal analysis already
works very well in the predictive modelling in the physics, for
example in the classical.
I agree that infinitesimal analysis can certainly work as well as
standard analysis (perhaps even better) for physical modelling.
However, it would be the infinitesimal analysis of today, not the
inconsistent variety of the 18th century, that would be the workable
one.
Of course, the opposite of an infinitesimal is an infinity. There are
a variety of systems in the physics that make do in approximation with
the notion of infinite quantities, by the same token there are some
systems where, for example, the value of a quantity that is generally
non-negative and finite is infinite, for example the center of a black
hole, which is matter with infinite mass.
Actually, black holes do not have infinite mass. They have infinite
density at its point of singularity, but the mass is always finite.
Perhaps that is what you meant to say?
Calculus, with the definition of the derivative being via the limit,
that is the limit as generally x goes to 0, lim x ->0+, is not a
method of finite differences. It's also not a method of zero
difference. The integral calculus is not a method of finite
differences, the resummation of the areas, after they are variously
divided, partitioned with minimal information, into infinitely many
rows or columns must be over infinitely many of them.
I don't understand your point here. Could you rephrase?
Given the infirmity of ZF I can prove a variety of things in it. I'm
interested to hear about this volume on questions of universal
quantification. I borrowed Barrow's "Theories of Everything". It is
not really explaining much while it is excellent. Consider for
example mentioning the diagonal argument, then saying Cohen
proved CH independent of ZF(C) without explaining how. Do you
_really_ think Cohen proved CH independent of ZF(C)? Very few
people have not just been told that. So, explain in detail how
Cohen uses universal quantification, or not, in finding a maximal
ordinal.
Of course Cohen explained how. It is in his proof. You can read it
for yourself at http://links.jstor.org/sici?sici=0027-8424%2819631215%2950%3A6%3C1143%3ATIOTCH%3E2.0.CO%3B2-5
, but I recommend boning up on your Mathematical Logic before trying
to read it, as it is pretty advanced.
Regards,
Jonathan Hoyle
Eastman Kodak
.
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| User: "Bob Kolker" |
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| Title: Re: Infinitesimal Arithmetic |
25 May 2007 11:22:06 AM |
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Jonathan Hoyle wrote:
That is precisely what Calculus is NOT. Calculus involves limits, not
infinitesimals. In fact, there are no references at all to any
infinitesimals anywhere in modern Calculus.
Not true any more. See Non Standard Analysis and Hyperreals. Robinson
has made infinitesimals logically sound.
Bob Kolker
.
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
25 May 2007 02:42:02 PM |
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That is precisely what Calculus is NOT. Calculus involves limits, not
infinitesimals. In fact, there are no references at all to any
infinitesimals anywhere in modern Calculus.
Not true any more. See Non Standard Analysis and Hyperreals. Robinson
has made infinitesimals logically sound.
This is certainly true, as I mention in a later portion of my posts.
What I was arguing against is the presumption that infinitesimals are
a required for Calculus. As you would agree, they certainly aren't
required.
However, in many ways NSA is far more intuitive an approach. For
example, the classical definition for a function f being continuous at
some point x0 is:
For every e>0 there exists a d>0 such that for all x that satisfies |x-
x0|<d then |f(x)-f(x0)|<e.
This cumbersome definition lacks an certain ease of understanding...at
least not until you've read a number of times and gotten used to it.
In NSA, however, the definition is:
For all x infinitesimally close to x0, f(x) is infinitesimally close
to f(x0).
This is far more intuitive and captures the heart of continuity very
nicely.
Regards,
Jonathan Hoyle
Eastman Kodak
.
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| User: "Tony Orlow" |
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| Title: Re: Infinitesimal Arithmetic |
25 May 2007 09:58:04 AM |
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Jonathan Hoyle wrote:
It seems the key is to recognize that there already is an
"infinitesimal arithmetic" of sorts, in the "infinitesimal analysis"
that most these days call the integral calculus, or just calculus.
That is precisely what Calculus is NOT. Calculus involves limits, not
infinitesimals. In fact, there are no references at all to any
infinitesimals anywhere in modern Calculus.
It is true that the original view of Calculus in the 18th century
involved some inconsistent objects called "infinitesimals", but they
riddled the field with contradictions and paradoxes. In the 19th
century, Calculus was cleaned up of all these paradoxes, and Calculus
taught can be taught without the need of those inconsistent
infinitesimals.
In the 20th Century, as you know, Abraham Robinson was able to re-
introduce infinitesimals in a consistent way (although these bare
little resemblance to the inconssistent objects of Newton). So you
can now prove the theorems of calculus with infinitesimals again
(although it is almost never taught that way in Freshman Calculus
classes).
I am curious as to what the exact inconsistencies are which arise from
the primitive notion of infinitesimals. I mean, there are certainly at
least two different ways to view them, which are mutually inconsistent,
at least at first glance, but why the notion of an infinitesimal is
inherently inconsistent is not apparent to me.
The definition of the derivative of a (non-vector-valued) function (at
a point) is in terms of the difference in rise for an infinitesimal
difference in the run.
That is incorrect. The definition of the derivative is in terms of
limits.
Yes, but essentially what the derivative at a given point indicates is
the slope of the function at that point.
For a straight line, any two points can be compared and the
derivative is a constant between those two points. For any
continuous function, two points can be compared and the
function has as a value of its derivative for at least one point in
the interval the same as the line through those points.
That is incorrect. For example, for the function f(x) = x^2, there
are no two distinct points x1 and x2 such that the derivative is the
same on f(x). This is of course easy to prove: f'(x) = 2x, and so if
f'(x1)=f'(x2) then 2x1=2x2 which implies x1=x2. So it is impossible
for two different points to have the same derivative. This is true
even in Non-Standard Calculus with x1 and x2 being infinitesimally
close to each other.
You misunderstand Ross. He is saying that, given two points on a curve,
they determine a line with a slope that is equal to the slope of the
curve at at least one point between those original two. That's not
incorrect, really, is it?
Yet, without analyzing the function to the infinitesimal, without
geometrical or other arguments about the function, that point is
unknown.
Untrue. That point can be easily proven using limits and without
infinitesimals.
Well, it requires calculus, in whatever form. That point was able to
determined before calculus was rigorized, so determining that point does
not REQUIRE the use of limits as it's done today. There are many roads
leading to Rome. :)
So, then in the consideration of the derivative, it is similar in
notion to the post recently made about the degenerate intervals and
how they contain only one point. Here, instead there is the
expectation of a sort that there will be two points, basically
adjacent, that describe a line, the slope of which is tangent and the
derivative of the function of that line through the point and its left
or right or both.
That is incorrect. No two points are ever adjacent, not even in
Newton's inconsistent infinitesimal approach. Even Newton knew that
between any two points (even those infinitesimally close to each
other), there are an infinite number of points between them.
Well, that goes back to the two or more ways of viewing infinitesimals.
There may be considered points between, or those uncountably numerous
objects between infinitesimally different points may be considered
sub-points, on a yet-more-infinitesimal level.
Consider all the other types of calculus there are, it's great.
Calculus of finite differences, lambda calculus (types), pi calculus
(processes), ....
http://en.wikipedia.org/wiki/Calculus
Yes, it is great, I agree. I'm pleased you are familiar with that
site.
There is much background in infinitesimals in the integral calculus,
questions arise as how to reasonably extend mathematics of the
infinitesimal to extend the body of knowledge of physics, in terms of
theories with predictive content. The infinitesimal analysis already
works very well in the predictive modelling in the physics, for
example in the classical.
I agree that infinitesimal analysis can certainly work as well as
standard analysis (perhaps even better) for physical modelling.
However, it would be the infinitesimal analysis of today, not the
inconsistent variety of the 18th century, that would be the workable
one.
Of course, the opposite of an infinitesimal is an infinity. There are
a variety of systems in the physics that make do in approximation with
the notion of infinite quantities, by the same token there are some
systems where, for example, the value of a quantity that is generally
non-negative and finite is infinite, for example the center of a black
hole, which is matter with infinite mass.
Actually, black holes do not have infinite mass. They have infinite
density at its point of singularity, but the mass is always finite.
Perhaps that is what you meant to say?
True, they don't have infinite mass, or they would have infinite
gravitational force over any finite distance. They also do not
necessarily have infinite density, but only sufficient mass density to
overcome light's ability to escape. That, of course, may not apply to
all frequencies of light, as lower frequencies have lower effective mass
per photon, but I'm no expert in that area.
Calculus, with the definition of the derivative being via the limit,
that is the limit as generally x goes to 0, lim x ->0+, is not a
method of finite differences. It's also not a method of zero
difference. The integral calculus is not a method of finite
differences, the resummation of the areas, after they are variously
divided, partitioned with minimal information, into infinitely many
rows or columns must be over infinitely many of them.
I don't understand your point here. Could you rephrase?
That might be good, but I think what Ross is pointing at is the notion
that the sum of an infinite number of infinitesimals, each considered
zero in standard analysis, can actually be a finite value, or even an
infinite value. Similarly, infinite series may be considered not only to
diverge rather than converge, but to have a specific infinite value. At
least, that's my take on it.
Given the infirmity of ZF I can prove a variety of things in it. I'm
interested to hear about this volume on questions of universal
quantification. I borrowed Barrow's "Theories of Everything". It is
not really explaining much while it is excellent. Consider for
example mentioning the diagonal argument, then saying Cohen
proved CH independent of ZF(C) without explaining how. Do you
_really_ think Cohen proved CH independent of ZF(C)? Very few
people have not just been told that. So, explain in detail how
Cohen uses universal quantification, or not, in finding a maximal
ordinal.
Of course Cohen explained how. It is in his proof. You can read it
for yourself at http://links.jstor.org/sici?sici=0027-8424%2819631215%2950%3A6%3C1143%3ATIOTCH%3E2.0.CO%3B2-5
, but I recommend boning up on your Mathematical Logic before trying
to read it, as it is pretty advanced.
Regards,
Jonathan Hoyle
Eastman Kodak
Tony
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
25 May 2007 11:05:46 AM |
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I am curious as to what the exact inconsistencies are which arise from
the primitive notion of infinitesimals. I mean, there are certainly at
least two different ways to view them, which are mutually inconsistent,
at least at first glance, but why the notion of an infinitesimal is
inherently inconsistent is not apparent to me.
The notion is not inherently inconsistent; what was inconsistent was
Newton's use and definition of them (if you can call what Newton had
was a 'definition"). Specifically, he inconsistently used them as
both zero and non-zero quantities. There was of course no rule
determine when it should be considered a zero and when it should not
be, other than when it was convenient for his equations. Bishop
Berkeley was one of the most proficient at discovering these
problems. See: http://links.jstor.org/sici?sici=0007-0882(195305)4:13%3C22:BCOTI%3E2.0.CO;2-K
..
Modern uses of infinitesimals are more rigorous and do not suffer from
Newtonian inconsistencies.
For a straight line, any two points can be compared and the
derivative is a constant between those two points. For any
continuous function, two points can be compared and the
function has as a value of its derivative for at least one point in
the interval the same as the line through those points.
That is incorrect. For example, for the function f(x) = x^2, there
are no two distinct points x1 and x2 such that the derivative is the
same on f(x). This is of course easy to prove: f'(x) = 2x, and so if
f'(x1)=f'(x2) then 2x1=2x2 which implies x1=x2. So it is impossible
for two different points to have the same derivative. This is true
even in Non-Standard Calculus with x1 and x2 being infinitesimally
close to each other.
You misunderstand Ross. He is saying that, given two points on a curve,
they determine a line with a slope that is equal to the slope of the
curve at at least one point between those original two. That's not
incorrect, really, is it?
If that is what he meant, then he is correct, and I misunderstood. I
thought that he was saying that there were two points infinitesimally
close to each other with the same derivative. This is almost exactly
the error Newton made (although Newton never made the absurd
assumption that these points were consecutive).
So, then in the consideration of the derivative, it is similar in
notion to the post recently made about the degenerate intervals and
how they contain only one point. Here, instead there is the
expectation of a sort that there will be two points, basically
adjacent, that describe a line, the slope of which is tangent and the
derivative of the function of that line through the point and its left
or right or both.
That is incorrect. No two points are ever adjacent, not even in
Newton's inconsistent infinitesimal approach. Even Newton knew that
between any two points (even those infinitesimally close to each
other), there are an infinite number of points between them.
Well, that goes back to the two or more ways of viewing infinitesimals.
There may be considered points between, or those uncountably numerous
objects between infinitesimally different points may be considered
sub-points, on a yet-more-infinitesimal level.
The main point being that you never have the case in which there are
two points (no matter how close) with no points in between them.
Of course, the opposite of an infinitesimal is an infinity. There are
a variety of systems in the physics that make do in approximation with
the notion of infinite quantities, by the same token there are some
systems where, for example, the value of a quantity that is generally
non-negative and finite is infinite, for example the center of a black
hole, which is matter with infinite mass.
Actually, black holes do not have infinite mass. They have infinite
density at its point of singularity, but the mass is always finite.
Perhaps that is what you meant to say?
True, they don't have infinite mass, or they would have infinite
gravitational force over any finite distance. They also do not
necessarily have infinite density, but only sufficient mass density to
overcome light's ability to escape. That, of course, may not apply to
all frequencies of light, as lower frequencies have lower effective mass
per photon, but I'm no expert in that area.
The infinite density is not throughout the black hole, but only at its
point of singularity, where all the mass lives. The remaining portion
of the black hole is defined as that area from which the potential
escape velocity exceeds the speed of light (and hence no escape
velocity).
Calculus, with the definition of the derivative being via the limit,
that is the limit as generally x goes to 0, lim x ->0+, is not a
method of finite differences. It's also not a method of zero
difference. The integral calculus is not a method of finite
differences, the resummation of the areas, after they are variously
divided, partitioned with minimal information, into infinitely many
rows or columns must be over infinitely many of them.
I don't understand your point here. Could you rephrase?
That might be good, but I think what Ross is pointing at is the notion
that the sum of an infinite number of infinitesimals, each considered
zero in standard analysis, can actually be a finite value, or even an
infinite value. Similarly, infinite series may be considered not only to
diverge rather than converge, but to have a specific infinite value. At
least, that's my take on it.
I'm not sure what you mean by "sum". Do you mean the combined
measures of sets of points, or are you referring to the arithmetic of
hyper-real values, or perhaps something else entirely? I ask because
they have different parameters, and I historically have a poor record
of guessing what you or Ross mean. :-)
Regards,
Jonathan Hoyle
Eastman Kodak
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| User: "Tony Orlow" |
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| Title: Re: Infinitesimal Arithmetic |
26 May 2007 10:40:29 AM |
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Jonathan Hoyle wrote:
I am curious as to what the exact inconsistencies are which arise from
the primitive notion of infinitesimals. I mean, there are certainly at
least two different ways to view them, which are mutually inconsistent,
at least at first glance, but why the notion of an infinitesimal is
inherently inconsistent is not apparent to me.
The notion is not inherently inconsistent; what was inconsistent was
Newton's use and definition of them (if you can call what Newton had
was a 'definition"). Specifically, he inconsistently used them as
both zero and non-zero quantities. There was of course no rule
determine when it should be considered a zero and when it should not
be, other than when it was convenient for his equations. Bishop
Berkeley was one of the most proficient at discovering these
problems. See: http://links.jstor.org/sici?sici=0007-0882(195305)4:13%3C22:BCOTI%3E2.0.CO;2-K
.
Modern uses of infinitesimals are more rigorous and do not suffer from
Newtonian inconsistencies.
I tried to take a look, but it appears I don't have JSTOR access at
home. I'll cut and paste the url and try at the library or Cornell.
Thanks for the reference.
For a straight line, any two points can be compared and the
derivative is a constant between those two points. For any
continuous function, two points can be compared and the
function has as a value of its derivative for at least one point in
the interval the same as the line through those points.
That is incorrect. For example, for the function f(x) = x^2, there
are no two distinct points x1 and x2 such that the derivative is the
same on f(x). This is of course easy to prove: f'(x) = 2x, and so if
f'(x1)=f'(x2) then 2x1=2x2 which implies x1=x2. So it is impossible
for two different points to have the same derivative. This is true
even in Non-Standard Calculus with x1 and x2 being infinitesimally
close to each other.
You misunderstand Ross. He is saying that, given two points on a curve,
they determine a line with a slope that is equal to the slope of the
curve at at least one point between those original two. That's not
incorrect, really, is it?
If that is what he meant, then he is correct, and I misunderstood. I
thought that he was saying that there were two points infinitesimally
close to each other with the same derivative. This is almost exactly
the error Newton made (although Newton never made the absurd
assumption that these points were consecutive).
Yes, well, he did say that, as we take the limit as the endpoints
approach each other, it's kind of like we end up at that limit with two
infinitesimally distant points forming the line segment with the desired
slope, or derivative. Do those two points "exist"? If one can imagine
them, I suppose, as much as any points "exist".
As regards the notion of consecutive points, I am sure it sounds absurd,
but how different is a point on a line from an infinitesimal segment on
that line, viewed from the finite scale? If there can be presumed to be
some infinite number of reals per unit interval, then can't the
reciprocal of that infinite number represent the space occupied on the
line by each real? I don't think that's entirely absurd.
Now, I know what you'll probably say. There are smaller numbers still,
than these infinitesimals, if there is any difference between
consecutive "real" numbers. That's true. But, since two "consecutive
reals" are indistinguishable on the finite scale, they are the same
standard real, and anything between them is also equal on that scale. On
the infinitesimal scale, sure, you can have half an iota, or any finite
multiple of an iota, and still have an infinitesimal. There is use of
the idea that the square of an infinitesimal can be considered equal to
zero, and there is some sense to that, even though one can extend levels
of infinity and infinitesimality to any arbitrary number of levels.
Oh well. You probably still think it's absurd, but I'm always happy to
read your comments.
So, then in the consideration of the derivative, it is similar in
notion to the post recently made about the degenerate intervals and
how they contain only one point. Here, instead there is the
expectation of a sort that there will be two points, basically
adjacent, that describe a line, the slope of which is tangent and the
derivative of the function of that line through the point and its left
or right or both.
That is incorrect. No two points are ever adjacent, not even in
Newton's inconsistent infinitesimal approach. Even Newton knew that
between any two points (even those infinitesimally close to each
other), there are an infinite number of points between them.
Well, that goes back to the two or more ways of viewing infinitesimals.
There may be considered points between, or those uncountably numerous
objects between infinitesimally different points may be considered
sub-points, on a yet-more-infinitesimal level.
The main point being that you never have the case in which there are
two points (no matter how close) with no points in between them.
True. If there is any measurable difference between them, then that
difference can always be subdivided, on the scale of the difference
between them.
Of course, the opposite of an infinitesimal is an infinity. There are
a variety of systems in the physics that make do in approximation with
the notion of infinite quantities, by the same token there are some
systems where, for example, the value of a quantity that is generally
non-negative and finite is infinite, for example the center of a black
hole, which is matter with infinite mass.
Actually, black holes do not have infinite mass. They have infinite
density at its point of singularity, but the mass is always finite.
Perhaps that is what you meant to say?
True, they don't have infinite mass, or they would have infinite
gravitational force over any finite distance. They also do not
necessarily have infinite density, but only sufficient mass density to
overcome light's ability to escape. That, of course, may not apply to
all frequencies of light, as lower frequencies have lower effective mass
per photon, but I'm no expert in that area.
The infinite density is not throughout the black hole, but only at its
point of singularity, where all the mass lives. The remaining portion
of the black hole is defined as that area from which the potential
escape velocity exceeds the speed of light (and hence no escape
velocity).
I understand that, except that the black hole has a finite mass, and in
order to have an infinite density, would have to actually have zero
volume, which I don't believe is ever the case. The matter is squashed
probably beyond our comprehension, but I don't believe any theory holds
that the mass inside the black hole is actually squashed into a
singularity without volume. It still maintains some physical structure,
which requires space. It's just that the space is smaller than the event
horizon, not zero. Are you sure I'm wrong about that? I coulda sworn...
Calculus, with the definition of the derivative being via the limit,
that is the limit as generally x goes to 0, lim x ->0+, is not a
method of finite differences. It's also not a method of zero
difference. The integral calculus is not a method of finite
differences, the resummation of the areas, after they are variously
divided, partitioned with minimal information, into infinitely many
rows or columns must be over infinitely many of them.
I don't understand your point here. Could you rephrase?
That might be good, but I think what Ross is pointing at is the notion
that the sum of an infinite number of infinitesimals, each considered
zero in standard analysis, can actually be a finite value, or even an
infinite value. Similarly, infinite series may be considered not only to
diverge rather than converge, but to have a specific infinite value. At
least, that's my take on it.
I'm not sure what you mean by "sum". Do you mean the combined
measures of sets of points, or are you referring to the arithmetic of
hyper-real values, or perhaps something else entirely? I ask because
they have different parameters, and I historically have a poor record
of guessing what you or Ross mean. :-)
Regards,
Jonathan Hoyle
Eastman Kodak
Haha! That's understandable. I don't think it's your fault. :)
It's certainly not the sum of the measures of points, since points have
zero measure, and any sum is presumed to be zero. I used to make the
mistake of saying 0*oo=1, but this is only true if 0 is replaced by a
specific infinitesimal which is the reciprocal of a specific oo. That's
the motivation behind some of Ross' and my ideas. There's a specific
infinite number of reals in the unit interval, as a unit on an infinite
scale. Each occupies an infinitesimal space in that interval. I'm not
really clear on how Robinson treats the squares of infinitesimal values,
but I think they can be ignored or preserved, depending on the
situation. That probably goes back the Berkeley's criticism of Newton?
Tony Orlow
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
26 May 2007 03:33:05 PM |
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Modern uses of infinitesimals are more rigorous and do not suffer from
Newtonian inconsistencies.
I tried to take a look, but it appears I don't have JSTOR access at
home. I'll cut and paste the url and try at the library or Cornell.
Thanks for the reference.
Okay, sorry. The Newtonian infinitesimal i has this strange
property that f(x+i) is not equal to f(x) when f is a fluent (what we
today call a function), but f'(x+i)=f'(x) when f' is a fluxion (what
we today we could call the derivative).
Let me try to explain it this way using a concrete example. Let f(x)
= x^2, and we wish to find its derivative, f'(x). Loosely, these are
the two approaches (I say loosely because I am dropping the usual
rigor to make this easier to follow):
NEWTONIAN APPROACH:
Let i be an infinitesimal. The derivative f'(x) = [f(x+i)-f(x)]/i.
At this point of the calculation, we want to treat i as non-zero (to
avoid a division by 0 problem). So we follow the calculation, f'(x) =
[f(x+i) - f(x)] / i = [(x+i)^2 - x^2] / i = [(x^2 + 2xi + i^2) - x^2]/
i = (2xi + i^2)/i = 2x+i
Now Newton decided that at this end of the equation to set i=0, so
that the derivative f'(x) is 2x, not 2x plus an infinitesimal. So he
believed in non-zero infinitesimals on the left-hand side of the
calculation, but lost faith by the time he got to the right-hand side.
Essentially, Newton's infinitesimals had the property of being both
zero and non-zero at the same time, swapping identities based upon his
convenience.
STANDARD APPROACH:
The derivative is defined in terms of limits, not infinitesimals:
f'(x) = lim t->x [f(t) - f(x)] / (t-x) = lim t->x [t^2 - x^2] / (t-x)
= lim t->x (t+x)(t-x)/(t-x)
= lim t->x (t+x) = (x+x) = 2x.
Note that no infinitesimals were assumed to exist in this proof. The
calculation is just as easy, and it is more obvious what the intention
is.
I will respond to the rest of your response in another post.
Jonathan Hoyle
Eastman Kodak
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| User: "Androcles" |
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| Title: Re: Infinitesimal Arithmetic |
26 May 2007 04:13:15 PM |
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"Jonathan Hoyle" <jonhoyle@mac.com> wrote in message
news:1180211585.923395.131670@g4g2000hsf.googlegroups.com...
:> > Modern uses of infinitesimals are more rigorous and do not suffer from
: > > Newtonian inconsistencies.
: >
: > I tried to take a look, but it appears I don't have JSTOR access at
: > home. I'll cut and paste the url and try at the library or Cornell.
: > Thanks for the reference.
:
: Okay, sorry. The Newtonian infinitesimal i has this strange
: property that f(x+i) is not equal to f(x) when f is a fluent (what we
: today call a function), but f'(x+i)=f'(x) when f' is a fluxion (what
: we today we could call the derivative).
:
: Let me try to explain it this way using a concrete example. Let f(x)
: = x^2, and we wish to find its derivative, f'(x). Loosely, these are
: the two approaches (I say loosely because I am dropping the usual
: rigor to make this easier to follow):
:
: NEWTONIAN APPROACH:
: Let i be an infinitesimal. The derivative f'(x) = [f(x+i)-f(x)]/i.
: At this point of the calculation, we want to treat i as non-zero (to
: avoid a division by 0 problem). So we follow the calculation, f'(x) =
:
: [f(x+i) - f(x)] / i = [(x+i)^2 - x^2] / i = [(x^2 + 2xi + i^2) - x^2]/
: i = (2xi + i^2)/i = 2x+i
:
: Now Newton decided that at this end of the equation to set i=0, so
: that the derivative f'(x) is 2x, not 2x plus an infinitesimal. So he
: believed in non-zero infinitesimals on the left-hand side of the
: calculation, but lost faith by the time he got to the right-hand side.
:
: Essentially, Newton's infinitesimals had the property of being both
: zero and non-zero at the same time, swapping identities based upon his
: convenience.
:
:
: STANDARD APPROACH:
: The derivative is defined in terms of limits, not infinitesimals:
:
: f'(x) = lim t->x [f(t) - f(x)] / (t-x) = lim t->x [t^2 - x^2] / (t-x)
: = lim t->x (t+x)(t-x)/(t-x)
: = lim t->x (t+x) = (x+x) = 2x.
:
: Note that no infinitesimals were assumed to exist in this proof. The
: calculation is just as easy, and it is more obvious what the intention
: is.
:
: I will respond to the rest of your response in another post.
:
: Jonathan Hoyle
: Eastman Kodak
i = x-t when lim t->x is an infinitesimal of miniscule magnitude > 0
[f(x+i) - f(x)] / i is conceptually the same as lim t->x [f(t) - f(x)] /
(t-x),
surely?
I see a notational difference but not a conceptual one.
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
26 May 2007 05:20:20 PM |
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i = x-t when lim t->x is an infinitesimal of miniscule magnitude > 0
Incorrect. lim t->x (x-t) = 0, not "an infinitesimal of miniscule
magnitude > 0".
[f(x+i) - f(x)] / i is conceptually the same as lim t->x [f(t) - f(x)] /
(t-x), surely?
I see a notational difference but not a conceptual one.
Incorrect. The former is a calculation of division involving an
infinitesimal, whilst the second is makes no assumptions about the
existence of infinitesimals at all.
The "lim t->x" notation is not just some loose language being thrown
around that is open to interpretation. It has a specific meaning.
More formally, the text
"lim x->a f(x) = L"
is merely shorthand notation for:
"For every e>0 there exists a d>0 such that 0<|x-a|<d implies |f(x)-L|
< e".
Note that e,d,x,a,L are real numbers (not infinitesimals) in this
example.
Jonathan Hoyle
Eastman Kodak
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| User: "Androcles" |
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| Title: Re: Infinitesimal Arithmetic |
26 May 2007 06:01:23 PM |
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"Jonathan Hoyle" <jonhoyle@mac.com> wrote in message
news:1180218020.796772.260960@h2g2000hsg.googlegroups.com...
: Incorrect.
Incorrect.
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
26 May 2007 09:04:35 PM |
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On May 26, 7:01 pm, "Androcles" <Engin...@hogwarts.physics> wrote:
"Jonathan Hoyle" <jonho...@mac.com> wrote in message
news:1180218020.796772.260960@h2g2000hsg.googlegroups.com...
: Incorrect.
Incorrect.
Yes, that's what I said. To avoid future errors, I recommend you pick
up any Calculus text book written with the last 100 years.
Regards,
Jonathan Hoyle
Eastman Kodak
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| User: "Androcles" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 12:34:40 AM |
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"Jonathan Hoyle" <jonhoyle@mac.com> wrote in message
news:1180231475.499841.296020@k79g2000hse.googlegroups.com...
: On May 26, 7:01 pm, "Androcles" <Engin...@hogwarts.physics> wrote:
: > "Jonathan Hoyle" <jonho...@mac.com> wrote in message
: >
: > news:1180218020.796772.260960@h2g2000hsg.googlegroups.com...
: >
: > : Incorrect.
: >
: > Incorrect.
:
: Yes, that's what I said.
Correct, you were incorrect.
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| User: "Tony Orlow" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 10:37:22 AM |
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Jonathan Hoyle wrote:
Modern uses of infinitesimals are more rigorous and do not suffer from
Newtonian inconsistencies.
I tried to take a look, but it appears I don't have JSTOR access at
home. I'll cut and paste the url and try at the library or Cornell.
Thanks for the reference.
Okay, sorry. The Newtonian infinitesimal i has this strange
property that f(x+i) is not equal to f(x) when f is a fluent (what we
today call a function), but f'(x+i)=f'(x) when f' is a fluxion (what
we today we could call the derivative).
Ah! Let me rub a couple neurons together, then read on...that might make
some sense...
Let me try to explain it this way using a concrete example. Let f(x)
= x^2, and we wish to find its derivative, f'(x). Loosely, these are
the two approaches (I say loosely because I am dropping the usual
rigor to make this easier to follow):
NEWTONIAN APPROACH:
Let i be an infinitesimal. The derivative f'(x) = [f(x+i)-f(x)]/i.
At this point of the calculation, we want to treat i as non-zero (to
avoid a division by 0 problem). So we follow the calculation, f'(x) =
[f(x+i) - f(x)] / i = [(x+i)^2 - x^2] / i = [(x^2 + 2xi + i^2) - x^2]/
i = (2xi + i^2)/i = 2x+i
Okay. Looks good....
Now Newton decided that at this end of the equation to set i=0, so
that the derivative f'(x) is 2x, not 2x plus an infinitesimal. So he
believed in non-zero infinitesimals on the left-hand side of the
calculation, but lost faith by the time he got to the right-hand side.
Um, well, that makes some sense to me, actually. The use of the
infinitesimal in the calculation, and then dismissal of it in the final
result doesn't seem like a contradiction, exactly, to me. In fact, I
came up with a similar result in "
Essentially, Newton's infinitesimals had the property of being both
zero and non-zero at the same time, swapping identities based upon his
convenience.
Um, I would say that it depends on what scale you are measuring the
curve. The use of an infinitesimal for measuring the state of a "point"
seems reasonable, while the residual infinitesimal portions of the
result are not worth including on the finite scale. Adding some finite
number of points to the end of a line segment doesn't increase its
finite measure. So, some may see that as a contradiction. I don't have a
problem with it. But then, I'm not trying to save Catholicism from the
ravages of "natural philosophy" ;)
STANDARD APPROACH:
The derivative is defined in terms of limits, not infinitesimals:
f'(x) = lim t->x [f(t) - f(x)] / (t-x) = lim t->x [t^2 - x^2] / (t-x)
= lim t->x (t+x)(t-x)/(t-x)
= lim t->x (t+x) = (x+x) = 2x.
Note that no infinitesimals were assumed to exist in this proof. The
calculation is just as easy, and it is more obvious what the intention
is.
Oh yes, limits work very well for calculus, and they eliminate the need
to refer to any infinitesimals, but I certainly don't think Newton's use
of infinitesimals is "wrong". Perhaps the principles of elimination of
infinitesimal values from the final result were just never well explained.
I will respond to the rest of your response in another post.
Jonathan Hoyle
Eastman Kodak
Okay, thanks.
Tony Orlow
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 11:25:34 AM |
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On May 27, 11:37 am, Tony Orlow <t...@lightlink.com> wrote:
Jonathan Hoyle wrote:
Now Newton decided that at this end of the equation to set i=0, so
that the derivative f'(x) is 2x, not 2x plus an infinitesimal. So he
believed in non-zero infinitesimals on the left-hand side of the
calculation, but lost faith by the time he got to the right-hand side.
Um, well, that makes some sense to me, actually. The use of the
infinitesimal in the calculation, and then dismissal of it in the final
result doesn't seem like a contradiction, exactly, to me. In fact, I
came up with a similar result in "
The contradiction come from the fact that i is either zero or it's
not. If it's is zero, then the equation [f(x+i)-f(x)]/i is an illegal
division by 0 operation. if it is not zero, then 2x+i is never equal
to 2x, even when i is an infinitesimal.
I am not denying that the use of infinitesimals can be very
instructive in calculus. it certainly can. What I am saying is that
Newton's approach is invalid. Robinson's hyper-real approach is
valid, uses infinitesimals, and suffers from none of these drawbacks.
Newton's basic contradiction is this: He wants f'(x+i)=f'(x) but not
have f(x+i)=f(x) for any infinitesimal i. This was just simply bad
mathematics, even in his day. He forgiven this lapse for the same
reason Einstein was forgiven for his cosmological constant: these
small defects (which in each case got repaired later) are minor when
compared with the world changing introduction of Calculus (and
Relativity).
Regards,
Jonathan Hoyle
Eastman Kodak
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 11:36:34 AM |
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On 27 May 2007 09:25:34 -0700, Jonathan Hoyle <jonhoyle@mac.com>
wrote:
Um, well, that makes some sense to me, actually. The use of the
infinitesimal in the calculation, and then dismissal of it in the final
result doesn't seem like a contradiction, exactly, to me. In fact, I
came up with a similar result in "
The contradiction come from the fact that i is either zero or it's
not. If it's is zero, then the equation [f(x+i)-f(x)]/i is an illegal
division by 0 operation. if it is not zero, then 2x+i is never equal
to 2x, even when i is an infinitesimal.
How about the following approach?
f'(x) = real(f(x+i)-f(x)]/i) = ... = real(2x+i) = 2x.
Where i is an infinitesimal, and real(.) just gives the "real part"
of a real number + an infinitesimal? (I'm sorry, I didn't read any
introduction to Nonstandard Analysis so far. Hence my knowledg
econcerning Nonstandard Analysis is infinitesimal, and I'm only
"guessing". But at least to me this "approach" would make sense.)
He forgiven this lapse for the same reason Einstein was forgiven for
his cosmological constant: these small defects ...
Huh? :-)
"This latest finding is consistent with Einstein's explanation for
what dark energy is, the researchers noted. Einstein's
'cosmological constant' idea, which he called his biggest blunder
and later rejected, turned out to be the same thing that scientist
now see as the repulsive form of gravity called dark energy."
Source:
http://www.space.com/scienceastronomy/061116_darkenergy_infantuniverse.html
F.
--
E-mail: info<at>simple-line<dot>de
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| User: "Tony Orlow" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 12:06:34 PM |
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G. Frege wrote:
On 27 May 2007 09:25:34 -0700, Jonathan Hoyle <jonhoyle@mac.com>
wrote:
Um, well, that makes some sense to me, actually. The use of the
infinitesimal in the calculation, and then dismissal of it in the final
result doesn't seem like a contradiction, exactly, to me. In fact, I
came up with a similar result in "
The contradiction come from the fact that i is either zero or it's
not. If it's is zero, then the equation [f(x+i)-f(x)]/i is an illegal
division by 0 operation. if it is not zero, then 2x+i is never equal
to 2x, even when i is an infinitesimal.
How about the following approach?
f'(x) = real(f(x+i)-f(x)]/i) = ... = real(2x+i) = 2x.
Where i is an infinitesimal, and real(.) just gives the "real part"
of a real number + an infinitesimal? (I'm sorry, I didn't read any
introduction to Nonstandard Analysis so far. Hence my knowledg
econcerning Nonstandard Analysis is infinitesimal, and I'm only
"guessing". But at least to me this "approach" would make sense.)
That's sort of what I was saying. In the final standard result, the
nonstandard infinitesimal can be dropped. I'm not sure why that's a
problem. Standard(2x+i)=2x.
He forgiven this lapse for the same reason Einstein was forgiven for
his cosmological constant: these small defects ...
Huh? :-)
"This latest finding is consistent with Einstein's explanation for
what dark energy is, the researchers noted. Einstein's
'cosmological constant' idea, which he called his biggest blunder
and later rejected, turned out to be the same thing that scientist
now see as the repulsive form of gravity called dark energy."
Source:
http://www.space.com/scienceastronomy/061116_darkenergy_infantuniverse.html
F.
I don't know what to make of "dark energy". Somehow, I think the
acceleration of the universe is an illusion, perhaps based on some
assumptions of astral physics that acted somewhat differently billions
of years ago, but I guess that'll be figured out eventually. I'm no
expert on that....
Tony
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| User: "Ross A. Finlayson" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 01:34:40 PM |
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On May 27, 9:25 am, Jonathan Hoyle <jonho...@mac.com> wrote:
On May 27, 11:37 am, Tony Orlow <t...@lightlink.com> wrote:
Jonathan Hoyle wrote:
Now Newton decided that at this end of the equation to set i=0, so
that the derivative f'(x) is 2x, not 2x plus an infinitesimal. So he
believed in non-zero infinitesimals on the left-hand side of the
calculation, but lost faith by the time he got to the right-hand side.
Um, well, that makes some sense to me, actually. The use of the
infinitesimal in the calculation, and then dismissal of it in the final
result doesn't seem like a contradiction, exactly, to me. In fact, I
came up with a similar result in "
The contradiction come from the fact that i is either zero or it's
not. If it's is zero, then the equation [f(x+i)-f(x)]/i is an illegal
division by 0 operation. if it is not zero, then 2x+i is never equal
to 2x, even when i is an infinitesimal.
I am not denying that the use of infinitesimals can be very
instructive in calculus. it certainly can. What I am saying is that
Newton's approach is invalid. Robinson's hyper-real approach is
valid, uses infinitesimals, and suffers from none of these drawbacks.
Newton's basic contradiction is this: He wants f'(x+i)=f'(x) but not
have f(x+i)=f(x) for any infinitesimal i. This was just simply bad
mathematics, even in his day. He forgiven this lapse for the same
reason Einstein was forgiven for his cosmological constant: these
small defects (which in each case got repaired later) are minor when
compared with the world changing introduction of Calculus (and
Relativity).
Regards,
Jonathan Hoyle
Eastman Kodak
No, Jonathan, again you make a true statement then go off the track.
(ZF is inconsistent.) The hyperreals aren't always valid.
The Finlayson reals, as far as I can tell the real numbers, have
infinitesimals in them. It's dually the complete ordered field and
partially ordered transfer ring, R, the vague fugue of the real
numbers, as I have explained. Natural languages are natural, and
quite expressive. To convince myself of that I carefully described it
to you.
Anyways, I could reply to your comments in detail. In discussing
these ways to formalize number systems and especially in the
application, there is much more use for large linear systems than
transfinite cardinals. I have studied mathematics before.
Would you not agree that the differential has properties of an
infinitesimal? It's very regular to describe continuum problems in
terms of infinities and infinitesimals. Analysis is all very standard
and correct.
With light speed being a constant, you might consider where Al says
that light speed is infinite, in the little table, c, 3.00 x 10^8 m/s,
2.998..., the speed of light, generally understood to be an absolute
barrier in the physical universe, the speed of light, except for
energy in the form of light.
Personally, I'm trying to understand this MEMS, for solid state
refined materials. As I now design MEMS, there seems much application
in multi-layer MEMS, microscale materials, hardened materials and so
on.
Consider, for example, very low power radioactive batteries, that, you
could sink in wells deep underwater. In the low power conversions,
ideal for salt water application, water could be source treated with
the offset light. Fresnel sub-effects in the optical line
conditioning allow simple optical interconnect, MEMS on the blade:
Source coated light ceramic optical microlensed network sensors.
Ross
--
Finlayson Consulting
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| User: "Aatu Koskensilta" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 01:51:35 PM |
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On 2007-05-27, in sci.logic, Ross A. Finlayson wrote:
The Finlayson reals, as far as I can tell the real numbers, have
infinitesimals in them. It's dually the complete ordered field and
partially ordered transfer ring, R, the vague fugue of the real
numbers, as I have explained. Natural languages are natural, and
quite expressive. To convince myself of that I carefully described it
to you.
What a delightful paragraph, a true gem! The vague fugue of the real numbers
is something to behold.
--
Aatu Koskensilta (aatu.koskensilta@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
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| User: "Aatu Koskensilta" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 01:50:03 PM |
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On 2007-05-27, in sci.logic, Ross A. Finlayson wrote:
The Finlayson reals, as far as I can tell the real numbers, have
infinitesimals in them. It's dually the complete ordered field and
partially ordered transfer ring, R, the vague fugue of the real
numbers, as I have explained. Natural languages are natural, and
quite expressive. To convince myself of that I carefully described it
to you.
What a deloghtful paragraph, a true gem! The vague fugue of the real
numbers is something to behold.
--
Aatu Koskensilta (aatu.koskensilta@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
.
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 08:19:27 PM |
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On May 27, 2:34 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
No, Jonathan, again you make a true statement then go off the track.
Would you be kind enough to show me specifically what you diagree
with?
(ZF is inconsistent.)
Which axiom(s) of ZF do you think introduces this inconsistency?
The hyperreals aren't always valid.
Non-Standard Analysis, from whence the hyper-reals come, is proven to
be equi-consistent with ZFC. So the validity of the hyper-reals is
stands, until you can demonstrate this inconsistency of ZF you allege.
The Finlayson reals, as far as I can tell the real numbers, have
infinitesimals in them.
Since Finlayson Set Theory is inconsistent, all these statements (as
well as their negations) are theorems in your analysis.
Anyways, I could reply to your comments in detail.
I wish you would. It is only then I can show you your errors.
In discussing these ways to formalize number systems and
especially in the application, there is much more use for large
linear systems than transfinite cardinals.
I am not sure what this sentence means. Can you please rephrase?
I have studied mathematics before.
Which math courses have you taken?
Would you not agree that the differential has properties of an
infinitesimal?
I would not agree.
It's very regular to describe continuum problems in terms of
infinities and infinitesimals. Analysis is all very standard
and correct.
I don't know what you mean by "very regular" but I can say that I have
never heard the Continuum Hypothesis being described in terms of
infinitesimals. And since I have read a great deal on the subject, I
think I can safely say that it is an irregular thing to do.
With light speed being a constant, you might consider where Al says
that light speed is infinite, in the little table, c, 3.00 x 10^8 m/s,
2.998..., the speed of light, generally understood to be an absolute
barrier in the physical universe, the speed of light, except for
energy in the form of light.
I don't know who Al is, nor have I ever heard anyone claim that light
speed is infinite. Certainly I would disagree with any such
characterization. Regardless, it seems to me to be an irrelevant
point to the subject at hand.
Personally, I'm trying to understand this MEMS, for solid state
refined materials. As I now design MEMS, there seems much application
in multi-layer MEMS, microscale materials, hardened materials and so
on.
Consider, for example, very low power radioactive batteries, that, you
could sink in wells deep underwater. In the low power conversions,
ideal for salt water application, water could be source treated with
the offset light. Fresnel sub-effects in the optical line
conditioning allow simple optical interconnect, MEMS on the blade:
Source coated light ceramic optical microlensed network sensors.
I am not faimilar with MEMS, so I cannot offer an opinion on this
portion of your post.
Regards,
Jonathan Hoyle
Eastman Kodak
.
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| User: "Tony Orlow" |
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| Title: Re: Infinitesimal Arithmetic |
27 May 2007 09:05:05 PM |
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Jonathan Hoyle wrote:
On May 27, 2:34 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
No, Jonathan, again you make a true statement then go off the track.
Would you be kind enough to show me specifically what you diagree
with?
(ZF is inconsistent.)
Which axiom(s) of ZF do you think introduces this inconsistency?
I believe he means externally inconsistent with other notions, not
internally inconsistent, as in producing self-contradictions.
The hyperreals aren't always valid.
Non-Standard Analysis, from whence the hyper-reals come, is proven to
be equi-consistent with ZFC. So the validity of the hyper-reals is
stands, until you can demonstrate this inconsistency of ZF you allege.
I think he means that one needn't always consider unending levels of
infinitesimality, not that the system is not logically valid in its
derivation. Maybe I'm wrong.
The Finlayson reals, as far as I can tell the real numbers, have
infinitesimals in them.
Since Finlayson Set Theory is inconsistent, all these statements (as
well as their negations) are theorems in your analysis.
Have you detected an actual inconsistency in what you've been able to
glean of Finlayson Theory?
Anyways, I could reply to your comments in detail.
I wish you would. It is only then I can show you your errors.
In discussing these ways to formalize number systems and
especially in the application, there is much more use for large
linear systems than transfinite cardinals.
I am not sure what this sentence means. Can you please rephrase?
I think he is saying something like I was saying, that it seems more
helpful to imagine infinite numbers as being colinear with the finite
real line, viewed as points infinitely far from the origin. At least,
that's what I was saying...
I have studied mathematics before.
Which math courses have you taken?
Would you not agree that the differential has properties of an
infinitesimal?
I would not agree.
It's very regular to describe continuum problems in terms of
infinities and infinitesimals. Analysis is all very standard
and correct.
I don't know what you mean by "very regular" but I can say that I have
never heard the Continuum Hypothesis being described in terms of
infinitesimals. And since I have read a great deal on the subject, I
think I can safely say that it is an irregular thing to do.
I read that at first as perhaps referring to CH, but no, Ross is
referring to analysis of continuous functions, the calculus.
With light speed being a constant, you might consider where Al says
that light speed is infinite, in the little table, c, 3.00 x 10^8 m/s,
2.998..., the speed of light, generally understood to be an absolute
barrier in the physical universe, the speed of light, except for
energy in the form of light.
I don't know who Al is, nor have I ever heard anyone claim that light
speed is infinite. Certainly I would disagree with any such
characterization. Regardless, it seems to me to be an irrelevant
point to the subject at hand.
Al Einstein said that matter with rest mass reaches a point of infinite
density and mass at c, which cannot be attained because it requires
infinite energy. But, he never said light "speed" is infinite. It was
already determined that it had a finite speed.
Personally, I'm trying to understand this MEMS, for solid state
refined materials. As I now design MEMS, there seems much application
in multi-layer MEMS, microscale materials, hardened materials and so
on.
Consider, for example, very low power radioactive batteries, that, you
could sink in wells deep underwater. In the low power conversions,
ideal for salt water application, water could be source treated with
the offset light. Fresnel sub-effects in the optical line
conditioning allow simple optical interconnect, MEMS on the blade:
Source coated light ceramic optical microlensed network sensors.
I am not faimilar with MEMS, so I cannot offer an opinion on this
portion of your post.
Regards,
Jonathan Hoyle
Eastman Kodak
Got me!
Tony
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
28 May 2007 12:08:41 AM |
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On May 27, 10:05 pm, Tony Orlow <t...@lightlink.com> wrote:
Jonathan Hoyle wrote:
Which axiom(s) of ZF do you think introduces this inconsistency?
I believe he means externally inconsistent with other notions, not
internally inconsistent, as in producing self-contradictions.
Exactly. It is not ZF which is inconsistent, but ZF+RA (the Ross
Assumptions). This is why I say that Finlaysoin Set Theory is
inconsistent. ZF, on the other hand, does not include RA in any of
its axioms.
Non-Standard Analysis, from whence the hyper-reals come, is proven to
be equi-consistent with ZFC. So the validity of the hyper-reals is
stands, until you can demonstrate this inconsistency of ZF you allege.
I think he means that one needn't always consider unending levels of
infinitesimality, not that the system is not logically valid in its
derivation. Maybe I'm wrong.
That's not the take I get from his posts, but you seem to have a
better understanding of Ross than I do, so I will withhold judgement.
Since Finlayson Set Theory is inconsistent, all these statements (as
well as their negations) are theorems in your analysis.
Have you detected an actual inconsistency in what you've been able to
glean of Finlayson Theory?
Yes, although many of these things might be weaseled away by his
changing definitions. One obvious inconssistency is his claim that
the Ross Reals are a complete, ordered field and yet contain
infinitesimals. Another is his claim that his theory is both
consistent and complete. Then there are higher-order proof theory
inconsistencies, such as claiming to create theorems without axioms.
In general, Ross's ideas are simply not well thought out. If he
sspent the time to rigorously define what he wishes to start out with,
create an axiomatic system based on his assumptions, and then see
where it takes him, he may discover some very interesting results.
But unfortunately, Ross has not seemed inclined to to that, so I can
only take what he says at face value and go with it.
It's very regular to describe continuum problems in terms of
infinities and infinitesimals. Analysis is all very standard
and correct.
I don't know what you mean by "very regular" but I can say that I have
never heard the Continuum Hypothesis being described in terms of
infinitesimals. And since I have read a great deal on the subject, I
think I can safely say that it is an irregular thing to do.
I read that at first as perhaps referring to CH, but no, Ross is
referring to analysis of continuous functions, the calculus.
Ah, I see. I think you are correct that I misread him there.
Describing continuity in terms of infinitesimals is certainly easier
to understand conceptually, although I think the standard delta/
epsilon approach is still far more common.
For example, the standard definition of a function f being continuous
at a point a is:
For every e>0, there exists d>0 such that |x-a|<d implies |f(x)-
f(a)|<e.
However, the Non-Standard Robinsonian definition is:
If x is infinitesimally close to a, then f(x) is infinitesimally
close to f(a).
This definition is better at conveying the concept of continuity much
better. Note also that Non-Standard Analysis does not suffer from any
Newtonian hand-waving and inconsistencies. Robinson can easily define
continuity, differentiability, integration, etc. with infinitesimals,
and do it with no concerns about an infinitesimal beginning as non-
zero and ending a s zero.
Regards,
Jonathan Hoyle
Eastman Kodak
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| User: "Tony Orlow" |
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| Title: Re: Infinitesimal Arithmetic |
29 May 2007 12:00:32 PM |
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Jonathan Hoyle wrote:
On May 27, 10:05 pm, Tony Orlow <t...@lightlink.com> wrote:
Jonathan Hoyle wrote:
Which axiom(s) of ZF do you think introduces this inconsistency?
I believe he means externally inconsistent with other notions, not
internally inconsistent, as in producing self-contradictions.
Exactly. It is not ZF which is inconsistent, but ZF+RA (the Ross
Assumptions). This is why I say that Finlaysoin Set Theory is
inconsistent. ZF, on the other hand, does not include RA in any of
its axioms.
If RA doesn't assume the axioms of ZF, then it may be internally
consistent while being mutually inconsistent with ZF. That doesn't make
it wrong any more than it makes ZF wrong. Right? By the way, I'm not
saying there aren't internal contradictions in RA, just that I haven't
noticed any glaring ones in what I've heard.
Non-Standard Analysis, from whence the hyper-reals come, is proven to
be equi-consistent with ZFC. So the validity of the hyper-reals is
stands, until you can demonstrate this inconsistency of ZF you allege.
I think he means that one needn't always consider unending levels of
infinitesimality, not that the system is not logically valid in its
derivation. Maybe I'm wrong.
That's not the take I get from his posts, but you seem to have a
better understanding of Ross than I do, so I will withhold judgement.
He and I have spoken plenty before. :)
Since Finlayson Set Theory is inconsistent, all these statements (as
well as their negations) are theorems in your analysis.
Have you detected an actual inconsistency in what you've been able to
glean of Finlayson Theory?
Yes, although many of these things might be weaseled away by his
changing definitions. One obvious inconssistency is his claim that
the Ross Reals are a complete, ordered field and yet contain
infinitesimals.
There probably is a contradiction there. I'm not sure.
Another is his claim that his theory is both consistent and complete.
He seems to have a interesting opinion of Godel which I don't understand.
Then there are higher-order proof theory
inconsistencies, such as claiming to create theorems without axioms.
Yes, there is that. I think he wants axioms to be, as Lester would say
it, "demonstrably true", or logically derived, rather than simply
assumed. I agree that axioms should be justifiable, but in order to work
them logically, you have to declare them as a starting point for proofs.
I know he's used the phrase "non-logical axioms" as a description of
his object of objection.
In general, Ross's ideas are simply not well thought out. If he
sspent the time to rigorously define what he wishes to start out with,
create an axiomatic system based on his assumptions, and then see
where it takes him, he may discover some very interesting results.
But unfortunately, Ross has not seemed inclined to to that, so I can
only take what he says at face value and go with it.
Well, it's not all that easy. I'm slowly working on developing some sort
of axiom set for my intuitive system, and it takes a lot of
consideration. How long did it take to get ZFC right? Oy! But. it's
certainly worth trying. It might be a good exercise for Ross, too. But
then, Lester will just call them assumptions of truth that are not
demonstrably true in universal terms of truth exhaustively defined in
mechanical terms of truth, and like that...
It's very regular to describe continuum problems in terms of
infinities and infinitesimals. Analysis is all very standard
and correct.
I don't know what you mean by "very regular" but I can say that | | | | | | | | | | | | | |
