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| Hans Aberg... |
 Posted: Sun Oct 11, 2009 3:05 pm |
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Brian M. Scott wrote:
[quote:754cc3a7e6]In ordinary working mathematics, outside of formal set
theory? Don't be silly.
Yes, of course.
If you mean 'Yes, of course, the generalized associative and
commutative rules for addition must be shown at the level of
working mathematics', you merely demonstrate your ignorance
of mathematical practice.
[/quote:754cc3a7e6]
It seems you demonstrate the ignorance of how to build up a theory from
scratch.
[quote:754cc3a7e6]You're wrong. At the normal level of mathematical rigor it
is entirely legitimate to argue as follows:
Let S = sum_{k=0}^{n-1} (a + kd), and let
b = a + (n - 1)d. For arbitrary k we have a + kd =
a + [(n - 1) - (n - 1 - k)]d = b - (n - 1 - k)d, so
S = sum_{k=0}^{n-1} [b - (n - 1 - k)d] =
sum_{k=0}^{n-1} (b - kd). Hence 2S =
Here you use induction.
No. I use known facts about the field of real numbers. How
those are justified at a more foundational level is
irrelevant.
[/quote:754cc3a7e6]
That is of course the core.
[quote:754cc3a7e6]Right, though there are exceptions in category theory.
Category theory isn't working mathematics any more than
abstract set theory is.
[/quote:754cc3a7e6]
It's needed in some types of homological algebra, which then may depend
on the axiomatic set theory.
[quote:754cc3a7e6]I compare it with the interfaces in computing.
This is fine in naive set theory as well, normally used in
working math.
The point is that in *working* mathematics the basic
facts about the real field are taken for granted.
The way learned it was deriving it from the supremum
axiom.
It isn't an axiom: it's a consequence of the topological
completeness of the real numbers, which in turn is a
provable consequence of their construction from the
rationals, be it by Dedekind cuts or by equivalence classes
of Cauchy sequences.
[/quote:754cc3a7e6]
Those are just equivalent forms of that axiom.
[quote:754cc3a7e6]Of course, one can take it as an axiom
and avoid the hard work of constructing the real numbers,
but in that case one might as well take as basic properties
the generalized commutative and associative laws of addition
and multiplication.
[/quote:754cc3a7e6]
There are different ways to build it.
[quote:754cc3a7e6]This is called analysis. You perhaps learned
calculus.
Supercilious jackass. I've *taught* real analysis. And set
theory. And topology. And algebra. My doctorate is in
pure mathematics, with an emphasis in topology and set
theory.
[/quote:754cc3a7e6]
So how come you you do not know how to introduce a summation symbol, and
prove its properties?
[quote:754cc3a7e6]A generalization of (countable) mathematical induction is
the axiom of choice.
Not true. The generalization is transfinite induction,
which does not require the axiom of choice. AC merely
extends the range of settings in which induction can be
used.
?? - They are equivalent.
Absolutely not.
You can pretty much choose a limit cardinal up to which
they are valid.
Transfinite induction is valid up to any ordinal, limit or
successor, in ZF; AC is not needed to prove this. (I'm a
set-theoretic topologist; this is an area that I know well.)
You probably mean that if you have a well-ordered set, you
can do transfinite induction,
That is exactly what is meant by the principle of
transfinite induction.
but in order to show that the set has a well-ordering, you
will need the AC.
That depends on the set. The ordinals, finite and infinite,
are well-ordered sets whose existence does not in any way
depend on AC.
[/quote:754cc3a7e6]
Hm, so then I can only wish you good luck building you working math
theory on that axiom of infinity set.
Hans |
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| Brian M. Scott... |
| Posted: Sun Oct 11, 2009 3:32 pm |
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On Sun, 11 Oct 2009 23:05:25 +0200, Hans Aberg
<haberg_20080406 at (no spam) math.su.se> wrote in
<news:hathbj$eqi$1 at (no spam) news.eternal-september.org> in sci.lang:
[quote:59fec2d686]Brian M. Scott wrote:
In ordinary working mathematics, outside of formal set
theory? Don't be silly.
Yes, of course.
If you mean 'Yes, of course, the generalized associative and
commutative rules for addition must be shown at the level of
working mathematics', you merely demonstrate your ignorance
of mathematical practice.
It seems you demonstrate the ignorance of how to build up
a theory from scratch.
[/quote:59fec2d686]
Building a theory from scratch is not the subject under
consideration, so obviously I have demonstrated nothing at
all about my knowledge of how to do so.
[...]
[quote:59fec2d686]The point is that in *working* mathematics the basic
facts about the real field are taken for granted.
The way learned it was deriving it from the supremum
axiom.
It isn't an axiom: it's a consequence of the topological
completeness of the real numbers, which in turn is a
provable consequence of their construction from the
rationals, be it by Dedekind cuts or by equivalence classes
of Cauchy sequences.
Those are just equivalent forms of that axiom.
[/quote:59fec2d686]
Don't be silly: they aren't axioms at all. They are
constructions of the real numbers from the rationals; having
performed either construction, one must then show that the
object so constructed has the desired properties, one of
which is order-completeness.
You're making it increasingly obvious that you're completely
out of your depth here.
[...]
[quote:59fec2d686]This is called analysis. You perhaps learned
calculus.
Supercilious jackass. I've *taught* real analysis. And set
theory. And topology. And algebra. My doctorate is in
pure mathematics, with an emphasis in topology and set
theory.
So how come you you do not know how to introduce a
summation symbol, and prove its properties?
[/quote:59fec2d686]
I've said nothing to justify either inference on your part.
Of course I know how to prove these things; I also know when
to prove them -- and when *not* to do so.
[quote:59fec2d686]A generalization of (countable) mathematical induction is
the axiom of choice.
Not true. The generalization is transfinite induction,
which does not require the axiom of choice. AC merely
extends the range of settings in which induction can be
used.
?? - They are equivalent.
Absolutely not.
You can pretty much choose a limit cardinal up to which
they are valid.
Transfinite induction is valid up to any ordinal, limit or
successor, in ZF; AC is not needed to prove this. (I'm a
set-theoretic topologist; this is an area that I know well.)
You probably mean that if you have a well-ordered set, you
can do transfinite induction,
That is exactly what is meant by the principle of
transfinite induction.
but in order to show that the set has a well-ordering, you
will need the AC.
That depends on the set. The ordinals, finite and infinite,
are well-ordered sets whose existence does not in any way
depend on AC.
Hm, so then I can only wish you good luck building you
working math theory on that axiom of infinity set.
[/quote:59fec2d686]
I suggest that you learn some set theory before you make an
even greater fool of yourself. Karel Hrbacek and Thomas
Jech, _Introduction to Set Theory_, is a solid introductory
text at the first-year graduate level; so is Judy Roitman's
_Introduction to Modern Set Theory_, which can be had for
free at <http://www.math.ku.edu/~roitman/>. |
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| Hans Aberg... |
| Posted: Sun Oct 11, 2009 3:52 pm |
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Brian M. Scott wrote:
[quote:9a1f686b5d]It seems you demonstrate the ignorance of how to build up
a theory from scratch.
Building a theory from scratch is not the subject under
consideration, so obviously I have demonstrated nothing at
all about my knowledge of how to do so.
[/quote:9a1f686b5d]
The topic was to demonstrate a simple formula, so it seemed important to
see what math is involved.
[quote:9a1f686b5d]It isn't an axiom: it's a consequence of the topological
completeness of the real numbers, which in turn is a
provable consequence of their construction from the
rationals, be it by Dedekind cuts or by equivalence classes
of Cauchy sequences.
Those are just equivalent forms of that axiom.
Don't be silly: they aren't axioms at all. They are
constructions of the real numbers from the rationals; having
performed either construction, one must then show that the
object so constructed has the desired properties, one of
which is order-completeness.
[/quote:9a1f686b5d]
With set theory in hand, those can be demonstrated, but that is not
necessarily the way it is done.
[quote:9a1f686b5d]You're making it increasingly obvious that you're completely
out of your depth here.
[/quote:9a1f686b5d]
You might benefit from looking up some math history.
[quote:9a1f686b5d]That depends on the set. The ordinals, finite and infinite,
are well-ordered sets whose existence does not in any way
depend on AC.
Hm, so then I can only wish you good luck building you
working math theory on that axiom of infinity set.
I suggest that you learn some set theory before you make an
even greater fool of yourself. Karel Hrbacek and Thomas
Jech, _Introduction to Set Theory_, is a solid introductory
text at the first-year graduate level; so is Judy Roitman's
_Introduction to Modern Set Theory_, which can be had for
free at <http://www.math.ku.edu/~roitman/>.
[/quote:9a1f686b5d]
So what exiting working math ordinals do you get without the axiom of
choice?
Hans |
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| Joachim Pense... |
| Posted: Sun Oct 11, 2009 4:47 pm |
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Brian M. Scott (in sci.lang):
[quote:3d5f8c6f30]On Sun, 11 Oct 2009 14:35:00 +0100, Chuck Riggs
chriggs at (no spam) eircom.net> wrote in
news:b7n3d5lugbj4tqopg88bjk5jv5t3oec4um at (no spam) 4ax.com> in
sci.lang:
On Sat, 10 Oct 2009 15:21:26 +0200, Joachim Pense
snob at (no spam) pense-mainz.eu> wrote:
[...]
Sure, it's not a trick but a formula that is taught at
schools <http://en.wikipedia.org/wiki/Arithmetic_progress
ion>.
You're confusing the result with the trick by which it's
easily obtained.
[/quote:3d5f8c6f30]
you're right.
Joachim |
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| Brian M. Scott... |
| Posted: Sun Oct 11, 2009 4:58 pm |
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On Sun, 11 Oct 2009 23:52:40 +0200, Hans Aberg
<haberg_20080406 at (no spam) math.su.se> wrote in
<news:hatk3n$52o$1 at (no spam) news.eternal-september.org> in sci.lang:
[quote:4f2f2eb9a0]Brian M. Scott wrote:
It seems you demonstrate the ignorance of how to build up
a theory from scratch.
Building a theory from scratch is not the subject under
consideration, so obviously I have demonstrated nothing at
all about my knowledge of how to do so.
The topic was to demonstrate a simple formula, [...]
[/quote:4f2f2eb9a0]
Not really, no.
[quote:4f2f2eb9a0]It isn't an axiom: it's a consequence of the topological
completeness of the real numbers, which in turn is a
provable consequence of their construction from the
rationals, be it by Dedekind cuts or by equivalence classes
of Cauchy sequences.
Those are just equivalent forms of that axiom.
Don't be silly: they aren't axioms at all. They are
constructions of the real numbers from the rationals; having
performed either construction, one must then show that the
object so constructed has the desired properties, one of
which is order-completeness.
With set theory in hand, those can be demonstrated, but
that is not necessarily the way it is done.
You're making it increasingly obvious that you're completely
out of your depth here.
You might benefit from looking up some math history.
[/quote:4f2f2eb9a0]
The history of mathematics has nothing to do with the
mathematical error on your part that I corrected above.
(And in all likelihood I'm more familiar with the relevant
history than you anyway.)
[...]
[quote:4f2f2eb9a0]So what exiting working math ordinals do you get without
the axiom of choice?
[/quote:4f2f2eb9a0]
Any that you want, of course. To be sure, most
mathematicians have no need for ordinals beyond \omega_0,
but they're all available for the few who do. |
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| Hans Aberg... |
| Posted: Sun Oct 11, 2009 5:38 pm |
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Brian M. Scott wrote:
[quote:05b8f1357f]The history of mathematics has nothing to do with the
mathematical error on your part that I corrected above.
[/quote:05b8f1357f]
Those things were done before axiomatic set theory was developed, and
definitions are needed for the reals.
[quote:05b8f1357f]So what exiting working math ordinals do you get without
the axiom of choice?
Any that you want, of course.
[/quote:05b8f1357f]
That does not work, because you need to show that the sets you are
working with are well ordered.
[quote:05b8f1357f]To be sure, most
mathematicians have no need for ordinals beyond \omega_0,
but they're all available for the few who do.
[/quote:05b8f1357f]
Too low, if that is the first uncountable ordinal.
Hans |
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| Chuck Riggs... |
| Posted: Mon Oct 12, 2009 6:29 am |
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On Sun, 11 Oct 2009 16:36:09 +0100, Nick
<3-nospam at (no spam) temporary-address.org.uk> wrote:
[quote:b98caef178]Chuck Riggs <chriggs at (no spam) eircom.net> writes:
On Sat, 10 Oct 2009 11:37:30 +0100, Nick
3-nospam at (no spam) temporary-address.org.uk> wrote:
Chuck Riggs <chriggs at (no spam) eircom.net> writes:
Since it is highly probable that neither Mr Brown nor Mr Green have
anything to do with President Obama's life, the puzzle can not be
solved. Shame on you, Ron, for asking that we try to.
Mr Brown would like to. Mr Obama appears to share many of our opinions
of him.
When a known character appears in a recreational puzzle and we're
asked to compare some aspect of that character with those of phoney
characters, no valid conclusions can be drawn.
That, I think, is a better wording of what I wrote yesterday.
Two well known characters. I'm not sure who Mr Green is though.
[/quote:b98caef178]
Since I was describing what I feel is an invalid puzzle, character
names in the puzzle Ron presented are irrelevant, if you're following
me.
--
Regards,
Chuck Riggs,
who speaks AmE, lives near Dublin, Ireland,usually spells in BrE
and hasn't corrected his email address yet |
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| Brian M. Scott... |
| Posted: Mon Oct 12, 2009 10:51 am |
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On Mon, 12 Oct 2009 01:38:10 +0200, Hans Aberg
<haberg_20080406 at (no spam) math.su.se> wrote in
<news:hatqam$og8$1 at (no spam) news.eternal-september.org> in sci.lang:
[quote:02c6c3fa6f]Brian M. Scott wrote:
The history of mathematics has nothing to do with the
mathematical error on your part that I corrected above.
Those things were done before axiomatic set theory was
developed, and definitions are needed for the reals.
[/quote:02c6c3fa6f]
Non sequitur.
[quote:02c6c3fa6f]So what exiting working math ordinals do you get without
the axiom of choice?
Any that you want, of course.
That does not work, because you need to show that the sets
you are working with are well ordered.
[/quote:02c6c3fa6f]
If you're working with ordinals, you're automatically
working with well-ordered sets. This has nothing to do with
AC. Are you so ignorant as to think that one needs AC to
get uncountable well-ordered sets? The entire class of
ordinals exists in ZF without AC. I begin to think that you
don't really know what an ordinal is.
[...] |
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| Hans Aberg... |
| Posted: Mon Oct 12, 2009 11:38 am |
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Brian M. Scott wrote:
[quote:68f878780e]If you're working with ordinals, you're automatically
working with well-ordered sets. This has nothing to do with
AC. Are you so ignorant as to think that one needs AC to
get uncountable well-ordered sets? The entire class of
ordinals exists in ZF without AC. I begin to think that you
don't really know what an ordinal is.
[/quote:68f878780e]
The book you mentioned has a section "The axiom of choice" that
describes the relation to what it calls "standard mathematics".
Hans |
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| Evan Kirshenbaum... |
| Posted: Mon Oct 12, 2009 4:35 pm |
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msb at (no spam) vex.net (Mark Brader) writes:
[quote:e189b62f24]Peter Moylan:
My HP calculators are long gone. A year or so ago I tried
to buy a new calculator, with RPN as an essential
requirement. ...
Brian Scott:
Current HP scientific and graphing calculators offer both
modes. I just looked into the HP-50g, which is their
current top of the line graphing calculator...
I thought I'd heard about 5 years ago, when HP was undergoing one of
its bouts of internal turmoil and/or being bought and sold,
[/quote:e189b62f24]
Buying, never bought or sold, though there have been some spin-offs,
most notably Agilent.
[quote:e189b62f24]that it was getting out of the calculator business. Am I wrong, or
if not, what happened? Evan?
[/quote:e189b62f24]
We still make them[1] and sell them (about 15 of them). I don't
recall there being any serious talk of divesting that business.
[quote:e189b62f24](Me, I had an HP-29 for many years, and now an HP-32S II. RPN is
indeed essential.)
[/quote:e189b62f24]
Personally, I could never get it to feel natural to me. (And this is
coming from somebody who implemented both Forth and PostScript and who
did a lot of low-level PostScript programming.)
[1] I believe. I've never worked with the groups involved, so I
honestly have no idea how much of that business is in-house versus
built-to-our-design versus slap-a-label-on OEM these days. My
guess would be that at least the design and probably the
manufacture is in-house, but I really don't know.
--
Evan Kirshenbaum +------------------------------------
HP Laboratories |I need to get a new colander. My
1501 Page Mill Road, 1U, MS 1141 |old one has holes in it.
Palo Alto, CA 94304
kirshenbaum at (no spam) hpl.hp.com
(650)857-7572
http://www.kirshenbaum.net/ |
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| Mark Brader... |
| Posted: Mon Oct 12, 2009 11:26 pm |
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Mark Brader:
[quote:a77df133b7]I thought I'd heard ... that [HP] was getting out of the
calculator business. Am I wrong, or if not, what happened?
[/quote:a77df133b7]
Evan Kirshenbaum:
[quote:a77df133b7]We still make them[1] and sell them (about 15 of them). I don't
recall there being any serious talk of divesting that business.
[/quote:a77df133b7]
*Abandoning* it was what I thought I'd heard. Glad to hear I was wrong.
--
Mark Brader | "I wish to inform you now that the square peg is now
Toronto | in square whole and can be voguish for that your
msb at (no spam) vex.net | payment is being processed..." --seen in spam |
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| Peter T. Daniels... |
| Posted: Tue Oct 13, 2009 3:15 am |
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On Oct 13, 3:48 am, Evan Kirshenbaum <kirshenb... at (no spam) hpl.hp.com> wrote:
[quote:dbd4933ee9]"Peter T. Daniels" <gramma... at (no spam) verizon.net> writes:
On Oct 12, 6:22 pm, Evan Kirshenbaum <kirshenb... at (no spam) hpl.hp.com> wrote:
He was, of course. The class, "Computational Models for the Syntax of
Natural Language" (or something like that), was cross-listed between
Linguistics and Computer Science. Interestingly, although I had been
a linguistics undergrad for a couple of years by that point, it was
the first class that actually spent any time covering Transformational
Grammar. The Linguistics professors all apparently believed (and I
picked up their bias in this) that it was just too silly to take
seriously and waste time on.
And they were right. Were you there in the Greenberg era, or after?
I think it was right at the end. I had thought that he was already
(recently) emeritus when I started taking classes in '83, but his
obituary implies that he was full-time until '86. In any case, I
didn't take any classes from him and I only recall seeing him a few times..
[/quote:dbd4933ee9]
He certainly didn't slow down in his later years. I encountered him
only twice -- first when he was president of the LSA, in Chicago -- I
think 1977, when Hockett introduced me to Trager -- and at the 1991
NACAL in Berkeley, when Alan Kaye arranged a conclave of all the
senior Afroasiaticists. I have a magnificent snapshot by Alan of
Greenberg, Hodge, Rosenthal, and two others -- I think Gordon -- and I
don't know what the movers did with the frame that housed a group of
such photos and cards.
Twice he was supposed to speak at meetings I was at (1989 NACAL
Atlanta, 1993 ILA New York) but sent a paper for someone else to read
instead. |
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| Chuck Riggs... |
| Posted: Tue Oct 13, 2009 8:26 am |
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On Mon, 12 Oct 2009 15:22:42 -0700, Evan Kirshenbaum
<kirshenbaum at (no spam) hpl.hp.com> wrote:
[quote:9d4eb1f29b]Chuck Riggs <chriggs at (no spam) eircom.net> writes:
[/quote:9d4eb1f29b]
<snip>
[quote:9d4eb1f29b]...I've often seen him
riled. Am I wrong, Evan?
It happens, but I like to think that I *tend* to be even-tempered.
[/quote:9d4eb1f29b]
You tend to be fair, I would say. I hope that is also true of me, at
least most of the time.
[quote:9d4eb1f29b]I'll cop to "sarcastic", though.
[/quote:9d4eb1f29b]
Fair enough and I can be the same, but will sarcasm enhance our
chances of getting to allegorical heaven?
--
Regards,
Chuck Riggs,
who speaks AmE, lives near Dublin, Ireland,usually spells in BrE
and hasn't corrected his email address yet |
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| Adam Funk... |
| Posted: Tue Oct 13, 2009 12:58 pm |
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On 2009-10-09, Joe Fineman wrote:
[quote:658e38d388]W. H. Auden was delighted to discover that an article in _Scientific
American_ titled "Cleaning Shrimp" was not about how to clean shrimp,
but about shrimp that clean fish by eating parasites off them.
[/quote:658e38d388]
A bit like "doctor fish" (conceptually, not syntactically).
--
Civilization is a race between catastrophe and education.
[H G Wells] |
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| PaulJK... |
| Posted: Tue Oct 13, 2009 9:06 pm |
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Chuck Riggs wrote:
[quote:9a21af4de2]On Mon, 12 Oct 2009 15:22:42 -0700, Evan Kirshenbaum
kirshenbaum at (no spam) hpl.hp.com> wrote:
Chuck Riggs <chriggs at (no spam) eircom.net> writes:
snip
...I've often seen him
riled. Am I wrong, Evan?
It happens, but I like to think that I *tend* to be even-tempered.
You tend to be fair, I would say. I hope that is also true of me, at
least most of the time.
I'll cop to "sarcastic", though.
Fair enough and I can be the same, but will sarcasm enhance our
chances of getting to allegorical heaven?
[/quote:9a21af4de2]
Oh, yeah, sure it will. :-)
pjk |
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