| Topic: |
Science > Physics |
| User: |
"Lester Zick" |
| Date: |
08 May 2007 11:28:48 AM |
| Object: |
Infinitesimal Arithmetic |
Infinitesimal Arithmetic
~v~~
It's curious that for any finite r, infinitesimal i, and transfinite
I, the same people who understand I+r=I have such difficulty
understanding r+i=r.
~v~~
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| User: "PD" |
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| Title: Re: Infinitesimal Arithmetic |
30 May 2007 03:31:19 PM |
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On May 30, 1:32 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 29 May 2007 15:28:31 -0700, PD <TheDraperFam...@gmail.com> wrote:
On May 29, 5:23 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 29 May 2007 12:50:59 -0700, PD <TheDraperFam...@gmail.com> wrote:
Well empirics are considerably better at forcible tellings I daresay
than at mechanical explanations for what they forcibly tell. At least
I can explain the origin of quantum particle spin properties and
Planck's constant better than you
Do that.
Been there, done that.
Methinks thou liest.
And methinks thou dost protest too much.
Not if thou liest.
J'Accuse!
Yeah, yeah, Isis. The accusation just seems a trifle vague to support
a charge of high treason against science. High Truth maybe. At least
Zola had some specifics. As I've said before, let me give you a clue:
if you want people to pay attention to you, you have to say something
worth paying attention to.
What's vague about it? You claimed that you could explain the origin
of quantum particle spin properties and Planck's constant, and that in
fact you'd already done it. I accused you of lying about that. Doesn't
get more concrete. I'm not interested in others paying attention to me
-- I'm interested in whether you're lying or not.
There is no high treason against science involved, just your
shamelessly self-aggrandizing prevarications. That's a simple
misdemeanor, but one that will deliver you a well-deserved spit-gob or
two.
PD
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
30 May 2007 05:52:46 PM |
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On 30 May 2007 13:31:19 -0700, PD <TheDraperFamily@gmail.com> wrote:
On May 30, 1:32 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 29 May 2007 15:28:31 -0700, PD <TheDraperFam...@gmail.com> wrote:
On May 29, 5:23 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 29 May 2007 12:50:59 -0700, PD <TheDraperFam...@gmail.com> wrote:
Well empirics are considerably better at forcible tellings I daresay
than at mechanical explanations for what they forcibly tell. At least
I can explain the origin of quantum particle spin properties and
Planck's constant better than you
Do that.
Been there, done that.
Methinks thou liest.
And methinks thou dost protest too much.
Not if thou liest.
J'Accuse!
Yeah, yeah, Isis. The accusation just seems a trifle vague to support
a charge of high treason against science. High Truth maybe. At least
Zola had some specifics. As I've said before, let me give you a clue:
if you want people to pay attention to you, you have to say something
worth paying attention to.
What's vague about it?
What's vague about it is you say nothing specific.
You claimed that you could explain the origin
of quantum particle spin properties and Planck's constant, and that in
fact you'd already done it. I accused you of lying about that.
Well technically, Isis, you didn't say any such of a thing. You just
said you accuse. You didn't specify of what.
Doesn't
get more concrete. I'm not interested in others paying attention to me
-- I'm interested in whether you're lying or not.
As am I. When you get to figurin what I'm lying about get back to me.
There is no high treason against science involved, just your
shamelessly self-aggrandizing prevarications.
Are there any other kind?
That's a simple
misdemeanor, but one that will deliver you a well-deserved spit-gob or
two.
Yeah, yeah, whatever, princess.
~v~~
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| User: "PD" |
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| Title: Re: Infinitesimal Arithmetic |
30 May 2007 09:22:02 PM |
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On May 30, 5:52 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 30 May 2007 13:31:19 -0700, PD <TheDraperFam...@gmail.com> wrote:
On May 30, 1:32 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 29 May 2007 15:28:31 -0700, PD <TheDraperFam...@gmail.com> wrote:
On May 29, 5:23 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 29 May 2007 12:50:59 -0700, PD <TheDraperFam...@gmail.com> wrote:
Well empirics are considerably better at forcible tellings I daresay
than at mechanical explanations for what they forcibly tell. At least
I can explain the origin of quantum particle spin properties and
Planck's constant better than you
Do that.
Been there, done that.
Methinks thou liest.
And methinks thou dost protest too much.
Not if thou liest.
J'Accuse!
Yeah, yeah, Isis. The accusation just seems a trifle vague to support
a charge of high treason against science. High Truth maybe. At least
Zola had some specifics. As I've said before, let me give you a clue:
if you want people to pay attention to you, you have to say something
worth paying attention to.
What's vague about it?
What's vague about it is you say nothing specific.
Here, let me help you recapture the thread, despite your short
attention span.
http://groups.google.com/group/sci.physics/msg/86246f75b813ccd9
where the conversation went like this:
Zick: At least I can explain the origin of quantum particle spin
properties and Planck's constant better than you...
PD: Do that.
Zick: Been there, done that.
Then following this,
http://groups.google.com/group/sci.physics/msg/e83bdb3335b4d1a9
where the conversation continued:
PD: Methinks thou liest.
Zick: And methinks thou dost protest too much.
PD: Not if thou liest. J'Accuse!
There. Does that help you remember what you said about 24 hours ago?
PD
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
31 May 2007 12:13:40 PM |
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On 30 May 2007 19:22:02 -0700, PD <TheDraperFamily@gmail.com> wrote:
On May 30, 5:52 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 30 May 2007 13:31:19 -0700, PD <TheDraperFam...@gmail.com> wrote:
On May 30, 1:32 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 29 May 2007 15:28:31 -0700, PD <TheDraperFam...@gmail.com> wrote:
On May 29, 5:23 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 29 May 2007 12:50:59 -0700, PD <TheDraperFam...@gmail.com> wrote:
Well empirics are considerably better at forcible tellings I daresay
than at mechanical explanations for what they forcibly tell. At least
I can explain the origin of quantum particle spin properties and
Planck's constant better than you
Do that.
Been there, done that.
Methinks thou liest.
And methinks thou dost protest too much.
Not if thou liest.
J'Accuse!
Yeah, yeah, Isis. The accusation just seems a trifle vague to support
a charge of high treason against science. High Truth maybe. At least
Zola had some specifics. As I've said before, let me give you a clue:
if you want people to pay attention to you, you have to say something
worth paying attention to.
What's vague about it?
What's vague about it is you say nothing specific.
Here, let me help you recapture the thread, despite your short
attention span.
http://groups.google.com/group/sci.physics/msg/86246f75b813ccd9
where the conversation went like this:
Zick: At least I can explain the origin of quantum particle spin
properties and Planck's constant better than you...
PD: Do that.
Zick: Been there, done that.
Then following this,
http://groups.google.com/group/sci.physics/msg/e83bdb3335b4d1a9
where the conversation continued:
PD: Methinks thou liest.
Zick: And methinks thou dost protest too much.
PD: Not if thou liest. J'Accuse!
There. Does that help you remember what you said about 24 hours ago?
Yeah, Isis. I remembered the conversation to begin with only too well.
So what? I also remember you claimed I lie - not that I lied about
this or that or the other. Don't expect me to reconstruct your meaning
for you just because you're lazy and want to lie in the jacuzzi.
~v~~
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| User: "PD" |
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| Title: Re: Infinitesimal Arithmetic |
31 May 2007 12:27:40 PM |
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On May 31, 12:13 pm, Lester Zick <dontbot...@nowhere.net> wrote:
Here, let me help you recapture the thread, despite your short
attention span.
http://groups.google.com/group/sci.physics/msg/86246f75b813ccd9
where the conversation went like this:
Zick: At least I can explain the origin of quantum particle spin
properties and Planck's constant better than you...
PD: Do that.
Zick: Been there, done that.
Then following this,
http://groups.google.com/group/sci.physics/msg/e83bdb3335b4d1a9
where the conversation continued:
PD: Methinks thou liest.
Zick: And methinks thou dost protest too much.
PD: Not if thou liest. J'Accuse!
There. Does that help you remember what you said about 24 hours ago?
Yeah, Isis. I remembered the conversation to begin with only too well.
So what? I also remember you claimed I lie - not that I lied about
this or that or the other. Don't expect me to reconstruct your meaning
for you just because you're lazy and want to lie in the jacuzzi.
Ah, so you don't mind being stupid, as long as you talk prettily while
doing it.
Allow me to be absolutely, specifically clear, combining several short
sentences into one sentence so that punctuation doesn't interrupt your
comprehension:
I claim you are lying when you say that you have explained the origin
of quantum particle spin properties and Planck's constant at all let
alone better than someone else on this group.
You are free to demonstrate that you are not lying.
Or you are free to dance stupidly.
PD
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
31 May 2007 06:58:42 PM |
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On 31 May 2007 10:27:40 -0700, PD <TheDraperFamily@gmail.com> wrote:
On May 31, 12:13 pm, Lester Zick <dontbot...@nowhere.net> wrote:
Here, let me help you recapture the thread, despite your short
attention span.
http://groups.google.com/group/sci.physics/msg/86246f75b813ccd9
where the conversation went like this:
Zick: At least I can explain the origin of quantum particle spin
properties and Planck's constant better than you...
PD: Do that.
Zick: Been there, done that.
Then following this,
http://groups.google.com/group/sci.physics/msg/e83bdb3335b4d1a9
where the conversation continued:
PD: Methinks thou liest.
Zick: And methinks thou dost protest too much.
PD: Not if thou liest. J'Accuse!
There. Does that help you remember what you said about 24 hours ago?
Yeah, Isis. I remembered the conversation to begin with only too well.
So what? I also remember you claimed I lie - not that I lied about
this or that or the other. Don't expect me to reconstruct your meaning
for you just because you're lazy and want to lie in the jacuzzi.
Ah, so you don't mind being stupid, as long as you talk prettily while
doing it.
I certainly don't mind your being stupid.
Allow me to be absolutely, specifically clear, combining several short
sentences into one sentence so that punctuation doesn't interrupt your
comprehension:
That would be a first.
I claim you are lying when you say that you have explained the origin
of quantum particle spin properties and Planck's constant at all let
alone better than someone else on this group.
I don't quite see that anyone else on this group has explained
Planck's constant at all let alone better than me.
You are free to demonstrate that you are not lying.
Just as you are free to demonstrate that I am lying. So far all you've
done is claim claims whose claim to truth is ambiguous. But by all
means feel free to demonstrate that you're not.
Or you are free to dance stupidly.
Better than you? Not much hope of that.
~v~~
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
22 May 2007 12:27:16 PM |
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On 22 May 2007 07:11:54 -0700, PD <TheDraperFamily@gmail.com> wrote:
On May 21, 6:59 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 21 May 2007 12:41:43 -0700, Jonathan Hoyle <jonho...@mac.com>
wrote:
On May 21, 1:21 pm, Lester Zick <dontbot...@nowhere.net> wrote:
Nonsense. r+i=r exactly "says" what i actually is: i is zero, in
symbols: i = 0. Got it?!
Well not exactly. If I+i=I does that mean i=0? Not hardly.
That's precisely what it means.
No. Any finite added to a transfinite produces the same transfinite
result. That's why the arithmetic is transfinite to begin with. Unless
you're contending that the term "arithmetic" only applies to finites.
Which is a position I can understand and sympathize with. But when the
term "arithmetic" is applied to any combination of transfinites,
infinitesimals, and finites you're not doing finite arithmetic anymore
and the rules are different.
Take a close look at finite r in circular rotation. Finite r is not
just a rigid rod of some kind. It's mechanized by a combination of
tangential dr and centripetal dr.
Gotta love the "tangential dr" part.
So there's no tangential dr in circular rotation? So there's no
tangential velocity in circular rotation? Or tangential velocity
doesn't equal dr/dt? Or the definite integral of v=dr/dt between 0 and
dt doesn't equal dr? Mystery of mysteries. Then however does circular
rotation rotate, I wonder? Does god just attach a string to particles
and whirl them around the way Stringfellow thinks?
Zick swears he's seen some calculus someplace, sometime, and this
looks something like that.
This kind of babble would be safe if Zick could be assured that
everyone is at the same level of incompetence as his. It apparently
hasn't occurred to him that a substantial number of his readers can
recognize a pretense when they see it.
At least I'm not pretentious enough to post the same comments three
times. So how do you really feel about it? Care to explain your own
theory of particle spin?
Tangential dr is mechanized by
instantaneous definite integral of tangential velocity from 0 to dt
and centripetal dr is mechanized through the definite instananeous
integral of centripetal acceleration between 0 and dt to produce
finite centripetal velocity and the definite instantaneous integral of
finite centripetal velocity between 0 and dt to produce dr.
There is no other way to produce circular rotation. In fact even in
the case of a rigid rod the forces which keep it rigid are balanced
internally and centripetal acceleration is still required to produce
curvilinear rotation which is still subject to the same mechanics. A
finite number of infinitesimals just doesn't produce a finite change.
~v~~
~v~~
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| User: "PD" |
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| Title: Re: Infinitesimal Arithmetic |
22 May 2007 11:02:34 AM |
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On May 21, 6:59 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 21 May 2007 12:41:43 -0700, Jonathan Hoyle <jonho...@mac.com>
wrote:
On May 21, 1:21 pm, Lester Zick <dontbot...@nowhere.net> wrote:
Nonsense. r+i=r exactly "says" what i actually is: i is zero, in
symbols: i = 0. Got it?!
Well not exactly. If I+i=I does that mean i=0? Not hardly.
That's precisely what it means.
No. Any finite added to a transfinite produces the same transfinite
result. That's why the arithmetic is transfinite to begin with. Unless
you're contending that the term "arithmetic" only applies to finites.
Which is a position I can understand and sympathize with. But when the
term "arithmetic" is applied to any combination of transfinites,
infinitesimals, and finites you're not doing finite arithmetic anymore
and the rules are different.
Take a close look at finite r in circular rotation. Finite r is not
just a rigid rod of some kind. It's mechanized by a combination of
tangential dr and centripetal dr.
Gotta love the "tangential dr" part.
Zick swears he's seen some calculus someplace, sometime, and this
looks something like that.
This kind of babble would be safe if Zick could be assured that
everyone is at the same level of incompetence as his. It apparently
hasn't occurred to him that a substantial number of his readers can
recognize a pretense when they see it.
Tangential dr is mechanized by
instantaneous definite integral of tangential velocity from 0 to dt
and centripetal dr is mechanized through the definite instananeous
integral of centripetal acceleration between 0 and dt to produce
finite centripetal velocity and the definite instantaneous integral of
finite centripetal velocity between 0 and dt to produce dr.
There is no other way to produce circular rotation. In fact even in
the case of a rigid rod the forces which keep it rigid are balanced
internally and centripetal acceleration is still required to produce
curvilinear rotation which is still subject to the same mechanics. A
finite number of infinitesimals just doesn't produce a finite change.
~v~~
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| User: "PD" |
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| Title: Re: Infinitesimal Arithmetic |
22 May 2007 11:19:01 AM |
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On May 21, 6:59 pm, Lester Zick <dontbot...@nowhere.net> wrote:
On 21 May 2007 12:41:43 -0700, Jonathan Hoyle <jonho...@mac.com>
wrote:
On May 21, 1:21 pm, Lester Zick <dontbot...@nowhere.net> wrote:
Nonsense. r+i=r exactly "says" what i actually is: i is zero, in
symbols: i = 0. Got it?!
Well not exactly. If I+i=I does that mean i=0? Not hardly.
That's precisely what it means.
No. Any finite added to a transfinite produces the same transfinite
result. That's why the arithmetic is transfinite to begin with. Unless
you're contending that the term "arithmetic" only applies to finites.
Which is a position I can understand and sympathize with. But when the
term "arithmetic" is applied to any combination of transfinites,
infinitesimals, and finites you're not doing finite arithmetic anymore
and the rules are different.
Take a close look at finite r in circular rotation. Finite r is not
just a rigid rod of some kind. It's mechanized by a combination of
tangential dr and centripetal dr.
Gotta love the "tangential dr" part.
Zick swears he's seen some calculus someplace, sometime, and this
looks something like that.
This kind of babble would be safe if Zick could be assured that
everyone is at the same level of incompetence as his. It apparently
hasn't occurred to him that a substantial number of his readers can
recognize a pretense when they see it.
Tangential dr is mechanized by
instantaneous definite integral of tangential velocity from 0 to dt
and centripetal dr is mechanized through the definite instananeous
integral of centripetal acceleration between 0 and dt to produce
finite centripetal velocity and the definite instantaneous integral of
finite centripetal velocity between 0 and dt to produce dr.
There is no other way to produce circular rotation. In fact even in
the case of a rigid rod the forces which keep it rigid are balanced
internally and centripetal acceleration is still required to produce
curvilinear rotation which is still subject to the same mechanics. A
finite number of infinitesimals just doesn't produce a finite change.
~v~~
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
21 May 2007 02:43:04 PM |
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On Mon, 21 May 2007 10:21:49 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
In the usual treatment of infinitesimals, r+i is *never* equal
to r except in the case i=0. So that's not a good definition of
infinitesimal.
Well, Daryl, it's true that r+i=r is not a completely satisfactory
definition, because it doesn't say what i actually is.
Nonsense. r+i=r exactly "says" what i actually is: i is zero, in
symbols: i = 0. Got it?!
Well not exactly.
Huh?
I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
Not hardly.
Yes, exactly.
F.
--
E-mail: info<at>simple-line<dot>de
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
21 May 2007 06:23:28 PM |
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On Mon, 21 May 2007 21:43:04 +0200, G. Frege <nomail@invalid> wrote:
On Mon, 21 May 2007 10:21:49 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
In the usual treatment of infinitesimals, r+i is *never* equal
to r except in the case i=0. So that's not a good definition of
infinitesimal.
Well, Daryl, it's true that r+i=r is not a completely satisfactory
definition, because it doesn't say what i actually is.
Nonsense. r+i=r exactly "says" what i actually is: i is zero, in
symbols: i = 0. Got it?!
Well not exactly.
Huh?
I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
Not hardly.
Yes, exactly.
Finitely yes infinitesimally no.
~v~~
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
22 May 2007 12:58:47 PM |
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I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
Not hardly.
Yes, exactly.
Finitely yes infinitesimally no.
~v~~
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
22 May 2007 06:05:38 PM |
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On 22 May 2007 10:58:47 -0700, Jonathan Hoyle <jonhoyle@mac.com>
wrote:
I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
Not hardly.
Yes, exactly.
Finitely yes infinitesimally no.
Did you have something to add to the discussion?
~v~~
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| User: "Jonathan Hoyle" |
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| Title: Re: Infinitesimal Arithmetic |
22 May 2007 01:00:05 PM |
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I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
Not hardly.
Yes, exactly.
Finitely yes infinitesimally no.
Finitely yes, infinitesimally yes.
Jonathan Hoyle
Eastman Kodak
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
22 May 2007 06:08:35 PM |
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On 22 May 2007 11:00:05 -0700, Jonathan Hoyle <jonhoyle@mac.com>
wrote:
I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
Not hardly.
Yes, exactly.
Finitely yes infinitesimally no.
Finitely yes, infinitesimally yes.
So a finite plus an infinitesimal produces a finite change? Surpassing
strange but no one can quite seem to find a finite change to constant
r in circular rotation. Probably haven't looked hard enough I expect.
~v~~
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
22 May 2007 06:46:28 PM |
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On Tue, 22 May 2007 16:08:35 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
So a finite plus an infinitesimal produces a finite change?
Huh? What the hell are you talking about, man?!
Look, #%&$%&, a finite number r plus an infinitesimal i is NOT
identical to the (original) finite number r, but to the finite
number r _plus_ an infinitesimal i.
With other words,
r + i =/= r
(but r + i = r + i, of course).
In addition, we have that
r + i = r
implies
i = 0.
Since if i were not zero (i.e. i =/= 0) -infinitesimal or not- we
would have
r + i =/= r.
(Hint: A finite number + SOMETHING cannot be identical with the
original finite number, except that SOMETHING actually were
NOTHING, i.e. = 0)
Reference:
Robert Kaplan, The Nothing that Is: A Natural History of Zero, OUP,
2000.
--
E-mail: info<at>simple-line<dot>de
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
23 May 2007 01:52:09 PM |
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On Wed, 23 May 2007 01:46:28 +0200, G. Frege <nomail@invalid> wrote:
On Tue, 22 May 2007 16:08:35 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
So a finite plus an infinitesimal produces a finite change?
Huh? What the hell are you talking about, man?!
I thought the question was rather straightforward myself.
Look, #%&$%&, a finite number r plus an infinitesimal i is NOT
identical to the (original) finite number r, but to the finite
number r _plus_ an infinitesimal i.
Well, gee, that's kinda obvious since the problem itself was
originally stated that way. The question is whether there is any
finite difference between r and r+i. And the answer is no.
With other words,
r + i =/= r
(but r + i = r + i, of course).
In addition, we have that
r + i = r
implies
i = 0.
Since if i were not zero (i.e. i =/= 0) -infinitesimal or not- we
would have
r + i =/= r.
(Hint: A finite number + SOMETHING cannot be identical with the
original finite number, except that SOMETHING actually were
NOTHING, i.e. = 0)
Which is really why I raised the subject of infinitesimal arithmetic
to begin with. You're trying to articulate and apply the same concept
here that applies to transfinite arithmetic where I+r=I. When you add
a finite number to "many" you're still left with "many". And when you
add any finite number of infinitesimals to a finite r you continue to
have the same finite r.
The only mechanically definitive method to achieve finite magnitudes
from infinitesimals is definite integration over finite intervals. But
when dealing with definite integration over infinitesimal intervals as
we are here there is no finite change to r.
The same idea applies to imaginary numbers only the mechanics are
different. Would you say that r plus the square root of -1 is greater
than r just because you claim "(Hint: A finite number + SOMETHING
cannot be identical with the original finite number, except that
SOMETHING actually were NOTHING, i.e. = 0)"? Same principle.
Reference:
Robert Kaplan, The Nothing that Is: A Natural History of Zero, OUP,
2000.
~v~~
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
23 May 2007 04:53:52 PM |
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On Wed, 23 May 2007 11:52:09 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
[bla bla bla]
The question is whether there is any finite difference between
r and r+i.
No, that wasn't the question. The "question" was if
r + i = r
implies
i = 0.
The answer was/is _yes_.
Concerning the question if there is a finite difference between r
and r+i for an infinitesimal i, the answer obviously is no, since
the difference just is
|(r+i) - r| = |i| ,
which obviously is infinitesimal for an infinitesimal i.
Still...
...we have that
r + i = r
implies
i = 0.
Since if i were not zero (i.e. i =/= 0) -infinitesimal or not- we
would have
r + i =/= r.
(Hint: A finite number + SOMETHING cannot be identical with the
original finite number, except that SOMETHING actually were
NOTHING, i.e. = 0)
Which is really why I raised the subject of infinitesimal arithmetic
to begin with.
Huh? :-o
When you add a finite number to "many" [bla bla bla]
Didn't know that /many/ is a number. Interesting concept!
And when you add any finite number of infinitesimals to a finite r
you continue to have the same finite r.
Huh?! I just TOLD YOU that
r + i =/= r
for any i =/= 0 (where i may be infinitesimal or not).
Are you really too dense to understand the simple truth that a
finite number + SOMETHING cannot be identical with the original
finite number? (Except that SOMETHING actually were NOTHING, i.e. =
0.)
The same is true for
r + n*i (for any n e N, n =/= 0).
With other words
r + n*i =/= r (for any n e N, n =/= 0).
The same idea applies to imaginary numbers only the mechanics are
different.
Oh yes, the "mechanics" are different. I guess the mechanics in
your brain are different too...
Would you say that r plus the square root of -1 is greater
than r just because you claim "(Hint: A finite number + SOMETHING
cannot be identical with the original finite number, except that
SOMETHING actually were NOTHING, i.e. = 0)"?
No, (since I'm not a complete idiot) *I* wouldn't claim that, but I
would "claim" that
r + i =/= r ,
(where i is the imaginary unit).
Hint: A finite number + SOMETHING cannot be identical with the
original finite number, except that SOMETHING actually were
NOTHING, i.e. = 0
Same principle.
Exactly.
Additional hint: If
r + x = r,
we can subtract r from both sides of the equation. This way we get
x = 0.
Got it?
Reference:
Robert Kaplan, The Nothing that Is: A Natural History of Zero, OUP,
2000.
--
E-mail: info<at>simple-line<dot>de
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
23 May 2007 07:11:07 PM |
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On Wed, 23 May 2007 23:53:52 +0200, G. Frege <nomail@invalid> wrote:
On Wed, 23 May 2007 11:52:09 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
I just told you that (with usual addition in a field)
r + i = r
implies
i = 0.
[bla bla bla]
The question is whether there is any finite difference between
r and r+i.
No, that wasn't the question. The "question" was if
No. The question was "So a finite plus an infinitesimal produces a
finite change?" I should know since I asked it. You're free to ask any
question you want but don't try to tell me what the question I asked
was.
r + i = r
implies
i = 0.
The answer was/is _yes_.
That's nice. Now go back to sleep.
Concerning the question if there is a finite difference between r
and r+i for an infinitesimal i, the answer obviously is no, since
the difference just is
|(r+i) - r| = |i| ,
which obviously is infinitesimal for an infinitesimal i.
Well bully for you.
Still...
...we have that
r + i = r
implies
i = 0.
Since if i were not zero (i.e. i =/= 0) -infinitesimal or not- we
would have
r + i =/= r.
(Hint: A finite number + SOMETHING cannot be identical with the
original finite number, except that SOMETHING actually were
NOTHING, i.e. = 0)
Which is really why I raised the subject of infinitesimal arithmetic
to begin with.
Huh? :-o
When you add a finite number to "many" [bla bla bla]
Didn't know that /many/ is a number. Interesting concept!
Yes, yes, I know. You haven't been paying attention. So what else is
new? The root post to my recent thread "One, Two, Three . . . Many"
spells it out clearly for those too cognitively challenged to note the
original.
And when you add any finite number of infinitesimals to a finite r
you continue to have the same finite r.
Huh?! I just TOLD YOU that
r + i =/= r
for any i =/= 0 (where i may be infinitesimal or not).
I know, I know. I really appreciate your telling what to think but I'm
so slow on the uptake I'll need to be told "many" (read "infinite")
number of times before what you tell me to think becomes what I think
instead of what you think which you can't demonstrate true in any
event so I doubt we really need to be concerned with what you think
because what you think and what is true are at considerable variance.
Are you really too dense to understand the simple truth that a
finite number + SOMETHING cannot be identical with the original
finite number? (Except that SOMETHING actually were NOTHING, i.e. =
0.)
Yes, yes I'm infinitely dense indeed. But I do apprecieate your
opinion on the subject nonetheless.
The same is true for
r + n*i (for any n e N, n =/= 0).
With other words
r + n*i =/= r (for any n e N, n =/= 0).
The same idea applies to imaginary numbers only the mechanics are
different.
Oh yes, the "mechanics" are different. I guess the mechanics in
your brain are different too...
Naturally unlike yourself I'm not to lazy or stupid to demonstrate the
truth of what I'm saying.
Would you say that r plus the square root of -1 is greater
than r just because you claim "(Hint: A finite number + SOMETHING
cannot be identical with the original finite number, except that
SOMETHING actually were NOTHING, i.e. = 0)"?
No, (since I'm not a complete idiot) *I* wouldn't claim that, but I
would "claim" that
Perhaps that's why you wrote it then?
r + i =/= r ,
(where i is the imaginary unit).
Hint: A finite number + SOMETHING cannot be identical with the
original finite number, except that SOMETHING actually were
NOTHING, i.e. = 0
Same principle.
Exactly.
Additional hint: If
r + x = r,
we can subtract r from both sides of the equation. This way we get
x = 0.
Got it?
Well somebody's definitely got it. What we can't say. But do try to
keep up. It becomes rather tedious trying to recapitulate the obvious
to the inept.
~v~~
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| User: "Phineas T Puddleduck" |
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| Title: Re: Infinitesimal Arithmetic |
23 May 2007 07:13:42 PM |
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In article <kmk953dcvkkmb93tta2oer7cadj4829mj2@4ax.com>,
Lester Zick <dontbother@nowhere.net> wrote:
Well somebody's definitely got it. What we can't say. But do try to
keep up. It becomes rather tedious trying to recapitulate the obvious
to the inept.
I guess thats why many of us no longer bother to talk to you.
--
COOSN-174-07-82116: Official Science Team mascot and alt.astronomy's favourite
poster (from a survey taken of the saucerhead high command).
Sacred keeper of the Hollow Sphere, and the space within the Coffee Boy
singularity.
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
24 May 2007 12:44:37 PM |
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On Thu, 24 May 2007 01:13:42 +0100, Phineas T Puddleduck
<phineaspuddleduck@gmail.com> wrote:
In article <kmk953dcvkkmb93tta2oer7cadj4829mj2@4ax.com>,
Lester Zick <dontbother@nowhere.net> wrote:
Well somebody's definitely got it. What we can't say. But do try to
keep up. It becomes rather tedious trying to recapitulate the obvious
to the inept.
I guess thats why many of us no longer bother to talk to you.
No. Actually the reason is that you have nothing to say. That and I'm
still trying to decide whether your surname is real or a pseudonym. If
real I'd be happy to entertain further notions of what you don't know.
~v~~
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
23 May 2007 07:58:21 PM |
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On Wed, 23 May 2007 17:11:07 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
Let's recapitulate some things.
Daryl McCullough:
In the usual treatment of infinitesimals, r+i is *never*
equal to r except in the case i=0. So that's not a good
definition of infinitesimal.
Lester Zick:
Well, Daryl, it's true that r+i=r is not a completely
satisfactory definition, because it doesn't say what i
actually is.
G. Frege:
Nonsense. r+i=r exactly "says" what i actually is: i is
zero, in symbols: i = 0. Got it?!
Lester Zick:
Well not exactly.
G. Frege:
Yes, exactly.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hint:
If
r + i = r,
(where r is some finite number) we can subtract r from both sides
of the equation. This way we get
i = 0.
Got it?
If not, this just means
r + i = r implies i = 0.
The question is whether there is any finite difference between
r and r+i.
No, that wasn't the question.
See quotes above.
Would you say that r plus the square root of -1 is greater
than r just because you claim "(Hint: A finite number + SOMETHING
cannot be identical with the original finite number, except that
SOMETHING actually were NOTHING, i.e. = 0)"?
No, (since I'm not a complete idiot) *I* wouldn't claim that.
Perhaps that's why you wrote it then?
I NEVER wrote that, idiot. I wrote:
Hint: A finite number + SOMETHING cannot be identical with
the original finite number, except that SOMETHING actually
were NOTHING, i.e. = 0
--
E-mail: info<at>simple-line<dot>de
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
23 May 2007 08:26:36 PM |
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On Thu, 24 May 2007 02:58:21 +0200, G. Frege <nomail@invalid>
wrote:
Lester Zick:
Well, Daryl, it's true that r+i=r is not a completely
satisfactory definition [bla]
No, it's not just "not a completely satisfactory definition" but a
_completely unsatisfactory_ definition.
Here's a better one, imho:
x is infinitesimal if 0 < |x| < 1/n for all n e N.
Rationale: An infinitesimal number is smaller than any finite
number - no matter how small.
F.
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
24 May 2007 01:58:42 PM |
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On Thu, 24 May 2007 03:26:36 +0200, G. Frege <nomail@invalid> wrote:
On Thu, 24 May 2007 02:58:21 +0200, G. Frege <nomail@invalid>
wrote:
Lester Zick:
Well, Daryl, it's true that r+i=r is not a completely
satisfactory definition [bla]
No, it's not just "not a completely satisfactory definition" but a
_completely unsatisfactory_ definition.
Thanks for the opine.
Here's a better one, imho:
Humility being your long suit no doubt and truth your short suit.
x is infinitesimal if 0 < |x| < 1/n for all n e N.
Well that only tells us when something is infinitesimal and not what
an infinitesimal actually is. So it's no more satisfactory than mine.
Rationale: An infinitesimal number is smaller than any finite
number - no matter how small.
Gee. Who'da thunk.
~v~~
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
24 May 2007 04:50:01 PM |
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On Thu, 24 May 2007 11:58:42 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
x is infinitesimal if 0 < |x| < 1/n for all n e N.
Well that [...] tells us when something is infinitesimal [...]
That's exactly what _definitions_ (like this) are for.
[but] not what an infinitesimal actually is.
That's right. So what? :-o (Nothing WRONG with that.)
So it's no more satisfactory than mine.
Nonsense. Of course it's "more satisfactory" than yours, since it
defines the notion of /infinitesimal/.
Btw, do you have a _better_ definition than that one? :-o
--- Tell me!
Rationale: An infinitesimal number is smaller than any finite
number - no matter how small.
F.
--
E-mail: info<at>simple-line<dot>de
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| User: "Lester Zick" |
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| Title: Re: Infinitesimal Arithmetic |
24 May 2007 06:29:29 PM |
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On Thu, 24 May 2007 23:50:01 +0200, G. Frege <nomail@invalid> wrote:
On Thu, 24 May 2007 11:58:42 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
x is infinitesimal if 0 < |x| < 1/n for all n e N.
Well that [...] tells us when something is infinitesimal [...]
That's exactly what _definitions_ (like this) are for.
As opposed to definitions which are not like this. Which would be what
exactly?
[but] not what an infinitesimal actually is.
That's right. So what? :-o (Nothing WRONG with that.)
Well hell I can do that. I already have. r+i=r. Big deal.
So it's no more satisfactory than mine.
Nonsense. Of course it's "more satisfactory" than yours, since it
defines the notion of /infinitesimal/.
Not actually. It only defines the notion of when something is
infinitesimal. It does nothing to define what an infinitesimal is.
Btw, do you have a _better_ definition than that one? :-o
--- Tell me!
Just did.
Rationale: An infinitesimal number is smaller than any finite
number - no matter how small.
~v~~
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
24 May 2007 06:37:16 PM |
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On Thu, 24 May 2007 16:29:29 -0700, Lester Zick
<dontbother@nowhere.net> wrote:
I already have. r+i=r.
Which implies that i = 0. So what? :-o
I asked for a definition of the notion /infinitesimal/.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
My definition is
x is infinitesimal if 0 < |x| < 1/n for all n e N ,
what's yours?
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| User: "G. Frege nomail@invalid" |
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| Title: Re: Infinitesimal Arithmetic |
24 May 2007 06:56:38 PM |
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On Fri, 25 May 2007 01:37:16 +0200, G. Frege <nomail@invalid>
wrote:
I already have. r+i=r.
Which implies that i = 0. So what? :-o
I asked for a definition of the notion /infinitesimal/.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Ok, I concede defeat. "r+i=r" IS a acceptable definition of
/infinitesimal/.
As a direct consequence from this definition we get that 0 is the
only infinitesimal (since r + i = r implies i = 0). Fine.
On the other hand, if you are not satisfied with this consequence
from your definition, you might consider an alternative one.*)
F.
*) For example
x is infinitesimal if 0 < |x| < 1/n for all n e N.
--
E-mail: info<at>simple-line<dot>de
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| User: "Glen Wheeler" |
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| Title: Re: Infinitesimal Arithmetic |
31 May 2007 03:40:44 AM |
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"G. Frege" <nomail@invalid> wrote in message
news:r69c53p126p53gu1c4du2768edbb9jj9vk@4ax.com...
x is infinitesimal if 0 < |x| < 1/n for all n e N.
You're saying that x is 0?
--
Glen
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| User: "Androcles" |
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| Title: Re: Infinitesimal Arithmetic |
31 May 2007 04:26:21 AM |
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"Glen Wheeler" <spam@gew75.com> wrote in message
news:465e8a5a@dnews.tpgi.com.au...
:
: "G. Frege" <nomail@invalid> wrote in message
: news:r69c53p126p53gu1c4du2768edbb9jj9vk@4ax.com...
: > x is infinitesimal if 0 < |x| < 1/n for all n e N.
: >
:
: You're saying that x is 0?
He's just said 0 < |x|, so no, he isn't saying x = 0, 0 is not less than 0.
Does notation bother you?
a < b -- a is less than b
a <= b -- a is less than or equal to b
a >= b -- a is greater than or equal to b
a >b -- a is greater than b
a > b <=> b <a -- a is greater than b if and only if b is less than a.
x = |-x| -- the absolute value of minus x is equal to x.
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