HDTenEightyP/NetNews huggingface dataset:
From: lady@uhunix.uhcc.Hawaii.Edu (Lee Lady)
Subject: Re: consequences of the Axiom of Choice
Followup-To: sci.math
Summary: About those different kinds of noetherianity...
Keywords:  noetherian   axiom of choice
Organization: University of Hawaii (Mathematics Dept)
Date: Mon, 5 Oct 1992 09:09:04 GMT

In article <26913.Oct200.15.5792@virtualnews.nyu.edu> brnstnd@nyu.edu (D. J. Bernstein) writes:
>
>All metamathematical issues aside, there is a sound _pedagogical_ reason
>to explicitly identify each use of the axiom of choice. Most statements
>people prove do not depend on the axiom of choice, and those which do
>don't always appear that way at first glance. So the typical reader
>   ... 
>
>For the same reason, it helps the reader to say whether a Noetherian
>condition is uniform (any collection of submodules has a maximal
>element), or strong (any submodule is finitely generated), or weak (any
>ascending chain of submodules stabilizes). By adding one adjective
>before the word ``Noetherian'' you make it clear exactly which
>assumptions you _really_ want to make.

Now you've got me curious.  I never knew anyone to make that distinction.
It seems to me that  uniform [noetherian] => strong => weak  and it seems
plausible that none of these implications is reversible without AC.  
Can you actually show that?  

And what would be some examples of standard theorems about noetherian
rings (or modules) which require uniform (a.k.a. the maximum principle)
and can't be proved with merely strong or weak?

And what would be some examples of rings which are weakly noetherian but
don't satisfy the maximum principle?  I guess the question of the
nullstellensatz arises here.  It seems to me (without checking) that
the most popular proof shows that if R is strongly noetherian then so
is R[X].  And it's probably not too hard to show that if R is weakly
noetherian then so is R[X].  The uniform case might be harder, though.

--
It is a poor sort of skepticism which merely delights in challenging
those claims which conflict with one's own belief system.  
                                                          --Bogus quote 
lady@uhunix.uhcc.hawaii.edu         lady@uhunix.bitnet
From: n9041888@henson.cc.wwu.edu (Jenny Bridge R)
Newsgroups: sci.math
Subject: Barbie
Summary: new doll badmouths math
Keywords: barbie, math, doll
Date: 28 Sep 92 21:40:18 GMT
Organization: Western Washington University

Mattelle has come out with a new talking teen Barbie.
One of the four things it says is, "Math class is hard."
If you would like to comment to Mattelle, the phone number
is 1-800-421-2887
Thanks.
From: epstein@rene.ma.utexas.edu (Paul Epstein)
Newsgroups: sci.math
Subject: primes
Date: 30 Sep 92 01:55:22 GMT
Distribution: sci
Organization: University of Texas at Austin Mathematics

Is (p-1)!+1 ever a power of p?  This problem is given in a book by Apostol
on elementary number theory but I am unable to solve it.
My e-mail address is epstein@math.utexas.edu
Newsgroups: sci.math,sci.logic
From: frank@Cookie.secapl.com (Frank Adams)
Subject: Re: ZFC etc. (was Re: Report on Philosophies of Physicists)
Date: Wed, 30 Sep 1992 15:19:32 GMT
Organization: Security APL, Inc.

In article <TORKEL.92Sep29232927@bast.sics.se> torkel@sics.se (Torkel Franzen) writes:
>Now, I've all but forgotten what was supposed to be the point of this.
>Let's see. I said that a disproof of "every formula has a disjunctive
>normal form" would show that our minds are seriously flawed, every bit
>as much as a proof that propositional logic is inconsistent.  Is there
>anything in your present reflections that tells against this?

No, but you go on to argue that there is no more point to talking about our
minds being seriously flawed in one case than the other.  This is the point
I was arguing against.

For the last time: it is inconsistencies in our reasoning process which
lead us to doubt the perfection of our minds.  A disproof of "every formula
has a disjunctive normal form" would be a problem only in that, in
conjunction with its proof, it constitutes an inconsistency.  A proof of the
inconsistency of propositional logic is a more direct attack on the
consistency of our reasoning process, thus considerations of possible mental
flaws are more reasonable when discussed in the context of that theorem.
From: prakash@cs.mcgill.ca (Prakash Panangaden)
Newsgroups: sci.math
Subject: Re: Topos theory for mere mortals (?)
Date: 29 Sep 92 19:38:48 GMT
Organization: SOCS, McGill University, Montreal, Canada

In article <1992Sep29.181206.26236@galois.mit.edu> jbaez@riesz.mit.edu (John C. Baez) writes:

>It's a good point: one can indeed completely get rid of sets
>and do topos theory from scratch in first-order logic.  I really
>recommend MacLane and whatshisname's new book, which treats
>this.  (Sorry, I always forget that name.)

Ieke Moerdijk.

I endorse this recommendation.

Prakash
Newsgroups: sci.math,geometry.research
From: orourke@sophia.smith.edu (Joseph O'Rourke)
Subject: Re: Platonic solids: history?
Organization: Smith College, Northampton, MA, US
Date: Wed, 30 Sep 1992 02:16:44 GMT

I would like to thank those who responded to my query, suggesting
as sources Coxeter's "Regular Polytopes" and Heath's translation of 
Euclid.  Both were useful.  In addition, I'd like to bring a more
obscure source to the attention of those interested in these
matters:
	@incollection{Malkevitch
	, title =	"Milestones in the history of polyhedra"
	, author =	"J. Malkevitch"
	, booktitle =	"Shaping Space: A Polyhedral Approach"
	, pages = 	"80--92"
	, publisher =	"Birkh{\"{a}}user"
	, location =	"Boston"
	, year =	1988
	, editor =	"M. Senechal and G. Fleck"
	}
Newsgroups: sci.math
From: matmcinn@nuscc.nus.sg (Mcinnes B T (Dr))
Subject: Re: Mattel says "Math is tough."
Organization: National University of Singapore
Date: Mon, 5 Oct 1992 08:40:02 GMT

iltiss@owlnet.rice.edu (Steven Patrick Iltis) writes:
: I don't normally read this news group, so I apologize if this is not
: the appropriate place to post this. But I believe this is relevant.
: 
:         Mattel has just released a new product. A Barbie doll that says
: "Math is tough".

What would you have her say?:
From: cxm7@po.CWRU.Edu (Colin Mclarty)
Newsgroups: sci.math
Subject: Re: Topos theory for mere mortals (?)
Date: 28 Sep 92 15:41:47 GMT
Article-I.D.: usenet.1a793rINNsgu
Reply-To: cxm7@po.CWRU.Edu (Colin Mclarty)
Organization: Case Western Reserve University, Cleveland, OH (USA)
NNTP-Posting-Host: slc12.ins.cwru.edu


In a previous article, zeleny@husc9.harvard.edu (Michael Zeleny) says:

>In article <19qoafINNiqf@usenet.INS.CWRU.Edu> 
>cxm7@po.CWRU.Edu (Colin Mclarty) writes:
>
>>	There is a difference though for Platonists.  Set theoretic
>>realists generally think there is some one universe of all sets, AC 
>>is true in it or is not, and that universe is really the only place 
>>to work.  So the question of what to "postulate" may be hard but
>>it has some right answer--they are not "postulates" at all in the
>>sense of "postulate theorists" like Huntingdon (of the many 
>>axiomatizations of Boolean algebra).  Zeleny has made this point 
>>well in another context.  Category theorists, even if realists, tend 
>>to think it is worth working in all kinds of various ambient 
>>categories for various purposes.  So that you always do need to say 
>>what you postulate about your present ambient.
>
>I have hesitated for a long time before following up on this claim,
>especially since I don't recollect too well just what point it is that I
>made in another context.  However, I wish to go on record as stating that,
>though it probably is a sociological fact that most realists who bother
>thinking about it, think that there is exactly one universe of all sets,
>and that, conversely, some people have gone on record announcing their
>conversion to formalism on the grounds of their loss of belief in the
>unique existence of such a universe, I do not myself believe that a realist
>philosophy of mathematics cannot accomodate a multiplicity of set-theoretic
>(or type-theoretic) universes possessed of different size and structural
>properties.  There is, after all, no good reason to suppose that everything
>in the mathematical universe is a set, or equivalently, that the
>development of a particular set theory is bound to yield privileged
>insights into the nature of all mathematical objects.  So your contrast
>between set theory and category theory may be right, but only for
>accidental reasons.
>
>cordially,
>mikhail zeleny@husc.harvard.edu
>"Un de mes plus grands plaisirs est de jurer Dieu quand je bande."
>
	Yes, I think it is just a sociological remark.  To me, the
point is that a set theorist could be quite realist but also believe
the term "set" is more or less ambiguous, so that various alternative
"universes of all sets" really exist depending on which sense you mean.

	The remark I refer to was the very nice point that axiomatization
and proof theoretic strength are mere technicalities to a realist, and
what matters is to know the truths about a subject.

Colin McLarty
"Un de mes voeux les plus sincere est de voir disparaitre ce sig qui
ne fait signe que de trop de catolicisme en guise de sacrilege"
Newsgroups: sci.math
From: cctimar@athena.cas.vanderbilt.edu (Cary Timar)
Subject: Roommate for Joint Meetings
Summary: Roommate needed
Organization: Vanderbilt U. Math Grad Student
Distribution: na
Date: Thu, 1 Oct 1992 03:42:20 GMT

I intend to go to the Joint Mathematics Meetings (of the AMS, MAA, &c.) in
San Antonio, Texas, on January 13-16.

I have noticed that even the student accomodation rates are prohibitive
(starting at $63/night).  I am looking for one or more other non-smokers
to share a room with.  

I will also look into accomodation at hotels not listed in the official
information.  Even there, it is cheaper to share.

Please reply via personal e-mail.
--
	-- Cary Timar, Mathematics Department, Vanderbilt University
Newsgroups: sci.math,rec.puzzles
From: ferenbaugh-charles@CS.YALE.EDU (Charles Ferenbaugh)
Subject: Re: EASY NUMBER SEQUENCE PUZZLE...
Organization: Yale University Math Dept.
Date: Fri, 2 Oct 1992 15:11:01 GMT

In article <aet.717945133@munagin> aet@mullian.ee.mu.OZ.AU (bert thompson) writes:
>25
>1729
>635318657
>???

You left out the first term...  There should be a 2 up at the top. 

:-)
Organisation: ECP-Paris France
Date: Monday, 5 Oct 1992 01:20:29 EDT
From: <COMTET@FRECP12.BITNET>
Newsgroups: sci.math
Subject: MARTINGALES & WIENER

I recently studied an example of Martingale in Stochastic processes. Let's imag
ine a cell with N elements : j elements A and N-j elements B. When the cell is
being duplicated first the number of A and B elements is multiplied by two and
then the cell divides in two cells with N elements for each one. If you describ
e the cell by the number of its A elements you get a Markovian process that is
indeed a Martingale. Couls somebody provide me with examples of Martingales tha
t do not use the hypergeometric law as in this process. I would like to get the
name of a book also only about Martingales.
Gilles Comtet-Varga
Dept of Applied Mathematics
Ecole Centrale Paris
Gde Voie des Vignes
92295 Chatenay-Malabry Cedex
FRANCE

COMTET at FRECP12.BITNET
From: jarleb@athena.uio.no (Jarle Brinchmann)
Subject: Re: Another Cute Math Joke  (explanation of)
Reply-To: jarleb@athena.uio.no (Jarle Brinchmann)
Organization: Institute for Theoretical Astrophysics.
Date: Thu, 1 Oct 1992 19:33:34 GMT


In article <marc.717947769@vieta.math.uni-sb.de>, marc@vieta.math.uni-sb.de (Marc Conrad) writes:
|>From: marc@vieta.math.uni-sb.de (Marc Conrad)
|>Newsgroups: sci.math,sci.physics
|>Subject: Re: Another Cute Math Joke  (explanation of)
|>
|>dms@krang.vis.citri.edu.au (David Mcgregor SQUIRE) writes:
|>
|>>Q: What do you get if you cross an elephant with a grape?
|>
|>>A: |elephant||grape|sin(theta)
|>I'm stupid, but can someone explain this to me ??
|>Marc
|>
|>--
|>     \   /                     | Marc Conrad, Universitaet des Saarlandes 
|>      \ Luxemburg              | marc@math.uni-sb.de  phone: 0681 302-2297
|>France \|   Germany            | these opinions are not necessarily these   
|>        \x <---- you are here! | of the SIMATH-group (and maybe even not mine).
|>
--
Well, sorry if this disturbs you, but there isn't really an explanation. 
The original versions are :

Q: What do you get if you cross an elephant with a zebra.
A: Elephant zebra sin theta.

Q: What do you get if you cross an elephant with a mountain climber.
A: You can't do that.  A mountain climber is a scalar.

or 
"What do you get when you cross an elephant with a banana?

Elephant banana sine theta in a direction mutually perpendicular to the two
as determined by the right hand rule."
---------

Quite surrealistic really.


				Jarle.


      ---------------------------------------
       Internet: jarleb@medusa.uio.no         
       or        Jarle.Brinchmann@astro.uio.no
      ---------------------------------------


Newsgroups: sci.math
From: boerncke@kirk.fmi.uni-passau.de (Frank-Roland Boernke)
Subject: FAQ? - What is ZF?
Organization: University of Passau, Germany
Date: Tue, 6 Oct 1992 11:14:35 GMT

I often read the term "ZF" in this group. Can anybody give me an explanation
of what "ZF" means? I guess from the context, that it is kind of a formal
system, but I'm not quite sure.

Thanks in advance

Frank


 
-------------------------------------------------------------------------
Frank Boerncke                       ,,,   University of Passau - Germany
boerncke@kirk.fmi.uni-passau.de     (.~.)            phone: +49 0851 2267
--------------------------------oOO--(_)--OOo---------------------------

Newsgroups: sci.math
From: mrw9e@kelvin.seas.Virginia.EDU (Michael Robert Williams)
Subject: Re: Barbie
Summary: Don't flame me if I'm wrong
Organization: University of Virginia
Date: Thu, 1 Oct 1992 01:39:25 GMT

I don't know if it has anything to do with Barbie not liking math, but
there is going to be a Fox news segment on tommorrow morning called
"Bad mouthing Barbie." (6.30-9 a.m. I'll set my VCR, but I'm not
sacrificing sleep). Maybe there's a Benoit doll with a string that 
when pulled says "Barbie's a math-weenie." All appropriate :-)'s and
what-not.
In Real Life:Mike Williams     | Perpetual Grad Student
e-mail      :mrw9e@virginia.edu|   - It's not just a job, it's an indenture
---------------------------------------------------------------------------
The probability of something happening is inversely proportional
to its desirability
Xref: sparky sci.math:12674 sci.physics:15885
From: watson@clark.edu (Dennis Watson)
Newsgroups: sci.math,sci.physics
Subject: Re: Another Cute Math Joke  (explanation of)
Date: 4 Oct 92 20:51:37 GMT
Article-I.D.: clark.1992Oct4.205137.29695
Organization: Clark College, Vancouver, Wa.  USA

Sometime back someone posted the following  joke?   

\epsilon < 0   

 I have been away, did anyone ever explain this joke??
watson@clark.edu
Newsgroups: sci.math
From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
Subject: Re: Topos theory for mere mortals (?)
Organization: Computer Science Department,  Stanford University.
Date: Sat, 26 Sep 1992 23:02:23 GMT

In article <1992Sep24.012308.4206@galois.mit.edu> jbaez@riesz.mit.edu (John C. Baez) writes:
>In article <19qoafINNiqf@usenet.INS.CWRU.Edu> cxm7@po.CWRU.Edu (Colin Mclarty) writes:
>
>Gian-Carlo Rota, who likes to be provocative, said to me something like
>"There is no such thing as equality!  There are only equivalence
>relations."
>

This is a reasonable point of view from the category perspective.  But
from the 2-category perspective (or better still, n-category) one sees
how provincial it is to deny equality, which is just 0-isomorphism,
i.e. isomorphism in a set or discrete category or 0-category, where the
only maps are identity maps.  In this numbering the usual notion of
isomorphism is 1-isomorphism, the categorical (= 1-categorical)
notion.  2-isomorphism is equivalence, in the sense of equivalence of
categories (e.g. equivalence of Finite Sets with the opposite of Finite
Boolean Algebras).  Here the 2-cells of the adjunction constituting the
equivalence are isomorphism.  And so on up.

-- 
=================================================== In short, I am floundering
Vaughan Pratt   pratt@cs.Stanford.EDU  415-494-2545 around and don't take what
=================================================== I say seriously.
Xref: sparky sci.math:12236 sci.math.symbolic:2539
From: brnstnd@nyu.edu (D. J. Bernstein)
Newsgroups: sci.math,sci.math.symbolic
Subject: Re: Integral Tables
Date: 26 Sep 92 07:19:30 GMT
Organization: IR

In article <1992Sep24.163552.12172@alchemy.chem.utoronto.ca> mroussel@alchemy.chem.utoronto.ca (Marc Roussel) writes:
> The problem is that in order
> to use the Risch algorithm it is normally necessary to do some symbolic
> manipulations first; that is where these programs (including Reduce and,
> I suspect, Axiom) normally fail.  For instance, how you should simplify
> sqrt((1-x)^2) depends on the range of the integral; abs(1-x) is always
> correct but causes the symbolic integrator fits.

Gee, was it just my imagination that Macsyma would pester me with ``is
1-x positive or negative?'' whenever I gave it such an integral? Sure,
Mathematica thinks that sqrt((-1)^2) = -1, but just because the new
technology produces wrong answers doesn't mean that the old technology
did.

---Dan
Newsgroups: sci.math
From: anthonyb@ecr.mu.OZ.AU (Anthony_Edward BROCKWELL)
Subject: prime twins
Keywords: prime twin conjecture sieve
Organization: Dept. Of Engineering Computer Resources, Uni. Of Melbourne
Date: Tue, 6 Oct 1992 00:23:53 GMT



If anyone out there is interested in prime twins, I have
a method (similar to Eratosthene's sieve)
of generating them (which I can prove as well),

	Start with the prime number 5.
	Take j=the closest int to 5/6,and then eliminate multiples
	of 5 plus and minus j, i.e. 4,6,9,11,14,16,etc.

	Now continue with the next prime (7), so
	we eliminate mults of 7 plus and minus 1, i.e.
	6,8,13,15,20,22,etc.

	Now repeat with the next prime (11), so
	we eliminate mults of 11 plus and minus 2 (11/6 is closest to 2),
	i.e. 9,13,20,24,etc.

	Keep going forever ...

The numbers remaining 

	i.e. 1,2,3,5,7,10,12,...

correspond to exactly the set of all prime twins in the following manner:

	Take n*6-1 and n*6+1 and they form a prime twin,

so 1 gives 5,7
   2 gives 11,13,
  12 gives 71,73,
  etc.

Note:  The only prime twin this sieve does not leave is 3 and 5.

I have thought about using this to prove the prime twin conjecture
but I never seem to have the time to sit down and really think about it,
since I am not really a mathematician.
A first step might be to say instead of going through eliminating mults of
primes plus and minus j, to use 6n plus minus one instead of the primes.
This would eliminate a superset of the numbers eliminated above, and
if it could be shown that there were infinitely many integers left
after eliminating this superset, the prime twin conjecture would
be solved.

If anyone can use this info. (or if the prime twin conjecture has
already been solved and no one has told me), please let me know.

			
				Anthony Brockwell.

Newsgroups: sci.math
From: thompson@atlas.socsci.umn.edu (T. Scott Thompson)
Subject: Re: Lebesgue measure in Fractal Dimension
Reply-To: thompson@atlas.socsci.umn.edu
Organization: Economics Department, University of Minnesota
Distribution: usa
Date: Tue, 6 Oct 1992 15:36:27 GMT

sassoj@rs6412.ecs.rpi.edu (John J. Sasso Jr.) writes:

>   In Real Analysis, I learned that the set {x : 0 < x < 1} has Lebesgue 
>measure 1 for the Lebesgue measure defined in R^1, but is 0 for the Lebesgue
>measure defined in R^2 (thus, the measure of a set is relative to what 
>space, i.e., R^1, R^2, etc..., the Lebesgue measure is defined in).

Technically {x : 0 < x < 1 } is _not_ a subset of R^2, R^3, etc., so
the Lebesgue measure of this set with respect to those dimensions is
undefined.  What you mean (for example) is that 
{(x,y): 0 < x < 1, y = y0} has Lebesgue measure zero for each fixed
y0.

>   Now, the Cantor set has Lebesgue measure 0 for the Lebesgue measure defined
>in spaces R^1, R^2, etc...  However, the Cantor does have a fractal dimension
>(or Hausdorff dimension) between 0 and 1 (I cannot recall the value), and
>I'll call this dimension Df.  Is it possible to define a Lebesgue measure in
>a space of dimension Df (R^Df) such that the measure of the Cantor set is a 
>value other than 0?

Sounds like you are looking for Hausdorff measure.  Look in Patrick
Billingsley, "Probability and Measure, 2nd Edition," (John Wiley
Press, 1986).  The idea is as follows (working from rather vague
memories here):

Fix a number H > 0.

Consider a set A embedded in R^n.  For any subset C of R^n let

               diam(C) =    sup    || x - y ||
                          x,y in C

Now for each epsilon > 0 let

           m(epsilon,A,H) =  const_H * inf sum  ( diam(C_i)^H )
                                            i

where the inf is taken over all countable coverings {C_i} of A.  The
normalizing constant const_H will be chosen later.

Now let   M(A,H) =   lim         m(epsilon,A,H)
                   epsilon -> 0

(The limit might be +infinity.)

This defines the Hausdorff (outer) measure of dimension H.

What I don't recall is what sigma-algebra is needed to make this a
proper measure.  If I remember correctly, all Borel measurable subsets
of R^n are Hausdorff measurable.  I also don't remember what the
normalizing constant is.  You want to pick the normalizing constants
const_H so that the Hausdorff measure coincides with Lebesgue measure
when H = n.

I think that you can see that M(A,H) = 0 if the Hausdorff dimension of
A is less than H, 0 < M(A,H) if the Hausdorff dimension of A is
exactly H, and M(A,H) is infinite if the Hausdorff dimension of A is
greater than H.  It is the natural restriction of Lebesgue measure to
lower-dimensional (than R^n) sets.

I don't know much about fractals.  Perhaps someone else knows what is
the Df-dimensional Hausdorff measure of the Cantor set.

Here's hoping the real mathematicians don't flame me too badly on
this.
--
T. Scott Thompson              email:  thompson@atlas.socsci.umn.edu
Department of Economics        phone:  (612) 625-0119
University of Minnesota        fax:    (612) 624-0209
From: bubai@matt.ksu.ksu.edu (P.Chatterjee)
Newsgroups: sci.math
Subject: Looking for a slick proof
Date: 7 Oct 92 01:34:44 GMT
Organization: Kansas State University
NNTP-Posting-Host: matt.ksu.ksu.edu

Hi,

How does one prove that 2^n divides n*(n+1)*(n+2)*......*(2n) without     
having to resort to the 'odd' and 'even' induction? Any ideas?