HDTenEightyP/NetNews huggingface dataset:
From: ilan@leland.Stanford.EDU (ilan vardi)
Subject: Re: e-mail address in published papers
Organization: DSG, Stanford University, CA 94305, USA
Date: Mon, 31 Aug 92 20:52:27 GMT

In article <BtHpGM.FoG@watserv1.uwaterloo.ca> shallit@graceland.uwaterloo.ca (Jeffrey Shallit) writes:
>I just got the proofs for a paper that will appear in the journal
>Graphs and Combinatorics, and they have removed the e-mail address
>I included in my typeset draft.
>
>This has happened on more than one occasion.  What is the rationale
>behind this?  How can we convince editors to stop doing this?
>
>Jeff Shallit
>University of Waterloo

If this keeps happening I suggest getting your name legally changed to
shallit@graceland.uwaterloo.ca. This might pose some other problems
though. For example, it's unclear if the DMV has the ``@'' symbol to
put on your driver's license.

-ilan
Newsgroups: sci.math
From: steiner@andy.bgsu.edu (Ray Steiner)
Subject: Re: Evaluation of an integral
Keywords: exponential
Organization: Bowling Green State University B.G., Oh.
Date: Sun, 23 Aug 1992 20:49:56 GMT


There was an article about expressing integrals in terms of
elementary functions in the American Mathematical Monthly
(November, 1972). It dealt with a kind of differential
Galois theory for solving problems of this sort. 
In this article one can find a proof that the integral
of exp(x)/x is not elementary. To get the integral of
exp(exp(x) simply let u= exp(x), x= ln(u), dx= du/u. Then
the integral of exp(exp(x))= integral of exp(u)/u and thus
is not elementary either.
Ray Steiner
-- 
steiner@andy.bgsu.edu
Newsgroups: sci.math
From: shallit@graceland.uwaterloo.ca (Jeffrey Shallit)
Subject: Einstein's Erdos number
Organization: University of Waterloo
Date: Tue, 1 Sep 1992 17:12:52 GMT

The Erdos number of a person X, as most readers of this
newsgroup probably know, is defined to be the least number
of links (via joint papers) needed to connect X
with Erdos.  Erdos has Erdos number 0, co-authors of
Erdos have Erdos number 1, etc.

In the preface to "A Tribute to Paul Erdos", Cambridge
University Press, 1990, it is stated that Einstein's
Erdos number is 2.  That means there exists a person
who has written a paper with Einstein, and a paper with
Erdos.  Anybody know who it is?

Jeff Shallit
University of Waterloo
Newsgroups: sci.math
From: hporopudas@tnclus.tele.nokia.fi
Subject: Re: Proof of God's Existence
Followup-To: sci.math
Organization: Nokia Telecommunications.
Date: Tue, 1 Sep 1992 12:52:06 GMT


In article <pmanne-010992093107@ma-mac29.uio.no.>, pmanne@math.uio.no
(Per Manne) writes:
> In article <17ui6kINNsft@matt.ksu.ksu.edu>, bubai@matt.ksu.ksu.edu
> (P.Chatterjee) wrote:
>>
>> I was told by somebody that there is a mathematical proof of God's
>> existence; was wondering if somebody could shed some light on the same.
>>
>> Thanks. Looking forward to the response.
>
> The following proof is found in Seven Years of Manifold (1968-1980),
> edited by Ian Stewart and John Jaworski.
>
> THEOREM (due to Anselm, Aquinas, and others.)
> The Axiom of Choice is equivalent to the existence of a unique God.
>
> PROOF:
> =>   (Assuming the equivalence of the Axiom of Choice and Zorn's Lemma.)
> Partially order the set of subsets of the set of all properties of
> objects by inclusion. This set has maximal elements. God is by
> definition (according to Anselm) one of these maximal elements. Now
>
>     God  c  God u {existence}        [c - subset,  u - union]
>
> hence God = God u {existence}. Therefore God exists.
>   To prove uniqueness, let God and God' be two gods. Then
>
>     God u God'  c  God
>
> (according to Aquinas), therefore God' c God. Similarly God c God'.
> Hence God = God'.
>
> <=   Given a set {A_i : i \in I} of sets, let the unique God pick
> x_i \in A_i for each i \in I. (He can do so by omnipotence, proved as for
> existence above.) Then [in TeX notation, or it will look too awkward!]
>
>     (x_i)_{i\in I}  \in  \prod_{i\in I} A_i
>
> as required. QED


   You have not defined "existence", which you used above and definition of
 of God is not based on the Holy Bible (second book of Moses 3:14-15 and
 gospel of Johannes 8:28-29), where God says that He is "Existence Himself".
   There may also exist several unknown dimensions in very small scale about
 which we don't know yet anything which are related to "existence". This is
 why I think that definition of true "existence" is and always will be
 impossible for us humans.

            Hannu.

    "Don't you like your body as 'neutrino cage' for your 'neutrino soul'. ?
     If you don't, then simply ask God to release you."


Newsgroups: sci.math
From: platt@watson.ibm.com (Daniel E. Platt)
Subject: Re: Godel and the US constitution
Date: Wed, 02 Sep 1992 13:41:07 GMT
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: IBM T.J. Watson Research Center

In article <1992Sep1.203402.4257@magnus.acs.ohio-state.edu>, regeorge@magnus.acs.ohio-state.edu (Robert E George) writes:
|> In article <1992Sep1.175844.16825@nas.nasa.gov> asimov@wk223.nas.nasa.gov (Dani
|> el A. Asimov) writes:
|> [deletions]
|> >
|> >Out of curiosity, does anyone know how Godel envisioned a
|> >consitutional dictatorship possibly arising in the U.S. ???
|> [deletions]
|> 
|> I have no idea how Godel envisioned a dictatorship, but it is not very
|> difficult to see how it could happen: any of the Bill of Rights can be
|> repealed, just as Prohibition was repealed; terms could be extended
|> indefinitely, the Supreme Court could be abolished, elections could be
|> abolished, the veto repealed, et cetera. All that
|> would be required would be a series of constitutional amendments, of which
|> we have already had many. Each would require a 2/3 majority in both the
|> Senate and the House, and control of 3/4 of the State legislatures. (Another
|> route could be via the state convention method, or even the constitutional
|> convention.) Gaining such majorities would not be easy, but it is not
|> impossible either: the only action prohibited by the Constitution appears
|> to be the abolition of equal representation in the Senate -- but note
|> that an amendment could strip the Senate of any and all powers.
|> 
|> 
|> So, have a nice day!
|> 
|> Robert George
|> (speaking only for myself)
|> 
|> "The very essence of individual freedom is equal justice under a rule of
|> law, a law to which every man shall be subject and which no executive
|> can modify."
|>     Senator Robert Taft , March 11, 1944

Taft was wrong.  Abraham Lincoln, under the state of emergency, revoked the
writ of habeas corpus, effectively making it possible to imprison people
indefinitely only upon an accusation.  The powers associated with a state of
emergency might be adequate to keep a sufficiently popular president in
office effectively as a dictator.  It was through the state-of-emergency 
mechanisms in the German government that Hitler obtained his power.


Dan
From: dbc@ecs.soton.ac.uk (Bryan Carpenter)
Newsgroups: sci.math
Subject: What kind of algebra is this?
Date: 28 Aug 92 16:28:08 GMT


     In formal theories of processes (like ``Communicating Sequential
Processes'') two operators are generally introduced.  Call them + and *.
(They're related to internal and external non-determinism, but
that's not important here).  They distribute over one another, and
they're both idempotent:

     a * (b + c) = (a * b) + (a * c)

     a + (b * c) = (a + b) * (a + c)

     a * a = a + a = a.

Also + and * are both commutative and associative.
This makes the algebra a bit like Boolean arithmetic, except
that the `not' operation is missing.

     Mathematicians must have a name for this kind of structure.
Does someone know what it is?

     (An interesting exercise is to work out a
decision procedure to determine whether two expressions involving
variables combined with + and * are equal according to the laws
above).

Bryan
From: kevin.jessup@mixcom.mixcom.com (kevin.jessup)
Newsgroups: sci.math
Subject: Re: Calendar (NEW)
Date: 27 Aug 92 16:57:17 GMT
Article-I.D.: mixcom.1992Aug27.165717.21567
Distribution: usa
Organization: Milwaukee Internet Xchange BBS, Milwaukee, WI U.S.A.

In <VICTOR.92Aug25094827@terse4.watson.ibm.com> victor@watson.ibm.com (Victor Miller) writes:

>For a nice treatment of all of these issues I would recommend looking
>at the paper "Calendrical Calculations" by Ed Reingold in Software
>Practice and Experience (sometime in the last year, I don't remember
>the exact issue).

Another good source of info, that goes directly to the mathematical
manipulations without getting overly invlolved in the theory behind
it all, is "Practical Astronomy With Your Calculator" by Peter Duffet
Smith.  Published by Cambridge University Press.

-- 
Kevin Jessup, bbwwbb@mixcom.mixcom.com
"You've got to know when to code 'em, know when to load 'em, know when to
emulate, know when to run.  You never count your money, when your sittin' at
the keyboard: there'll be time enough for countin', when the software's done."
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: Abel's proof of the insolubility of the quintic
Date: 3 Sep 92 15:28:32 GMT
Article-I.D.: netnews.87834
Reply-To: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Organization: The Wistar Institute of Anatomy and Biology
In-reply-to: martin@lyra.cis.umassd.edu (Gary Martin)

In article <MARTIN.92Sep2212731@lyra.cis.umassd.edu>, martin@lyra (Gary Martin) writes:
>I don't know a good reference, but as I recall, you need to know a
>little bit about field extensions, how groups permute roots of
>polynomials, the relation between solvability of groups and solvability
>of equations, and the fact that the alternating group of degree 5 is
>simple (and hence that the corresponding symmetric group is not
>solvable).

Actually, it's neat to prove non-solvability of A_n, n>=5 directly: if
we compute right-to-left, (abc)(cde)(cba)(edc)=(adc).  So the commutator
subgroup contains all three cycles, hence all of A_n.  For the proposed
unravel-the-Galois-proof project, this may allow a dispensation with
any explicit group theory.  [This proof is from Artin's little book.]

I spent my undergraduate days under the misconception that the proof of
the simplicity of A_5 was overly complicated and uninformative.  Then I
came across the proof in Rotman: identify the conjugacy classes of A_5,
and notice the impossibility of any of them forming a partition for a
size that non-trivially divides 60.  This too might mean something in
the context of the proposed project.
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
Newsgroups: sci.math
From: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
Subject: n doesn't  divide .......
Organization: Rose-Hulman Institute of Technology
Date: Fri, 28 Aug 1992 23:00:54 GMT


Allow me to nurse my ego a bit.  I have a Ph.D, in number theory. I  
have taken on a project this summer, which is to produce a solution  
manual for someone else's number theory book.  The book has 1235  
problems.  I have worked 1214 of them so far, and TeXed them up.  I  
posted one of the remaining 21 problems (to prove that n doesn't divide  
2^n-1) which I spent about 2 hours on and thought someone might help me  
out a bit.  Thanks to Adam Logan and Keith Ramsay who gave concise  
proofs. It's a big help for someone is has worked number theory  
problems until he is nearly cross-eyed.  I intend to post a couple more  
of these problems during the next week or so as I finish this project.   
When I do, I would like to request that ALL ARROGANT ASSHOLES KEEP  
THEIR BULLSHIT OUT OF MY MAILBOX.  
Here a nice solution, appropriate to the level of the problem, given by
Adam Logan:
>If n is even, the result is trivial.

>Otherwise, letn be an exception, and let  p be the least prime  
>dividing n. Since n > 1, such a prime exists.  By hypothesis, n | 2^n  
>- 1, so p | 2^n - 1, i.e. 2^n == 1 (mod p).  Fermat proved that  
>2^(p-1) == 1 (mod p). If gcd(n, p-1) = 1, we can choose a and b with  
>an + b(p-1) = 1; then, raising the first equation to the power a, and  
>the second to the power b, and multiplying, we get 2 == 1 (mod p),  
>which is absurd.  Hence gcd(n, p-1) > 1, contradicting the minimality  
>of p. (== means congruent to.)

>Similar but harder problem: prove that n^2 | 2^n + 1 iff n = 1 or 3.
>(The description of numbers n s.t. n | 2^n + 1 is not so simple.)


Here's a sample of the drivel I've been getting:

>     Sure.  Proof by extended example:
>
>          n=2,  2^n-1=3  -> n obviously does not divide 2^n-1 for n=2
>          n=3,  2^n-1=8  -> n obviously does not divide 2^n-1 for n=3
>          n=4,  2^n-1=15 -> n obviously does not divide 2^n-1 for n=4
>          ...
>     QED.
>
>     I'm sure that you can fill in the dots with all the time you have
>     from not doing homework...

(har har har...You seem to have plenty of free time.)

>    Look up "Fermat's theorem" and the "Chinese remainder theorem"
>    in any book on number theory.

(Big help.  Actually, one would have to strain oneself to use the CRT  
on this problem.)

I have a head cold just now, and may regret my tirade later when I'm  
not so grouchy, but I doubt it. 

Very, very sincerely,
Bart

P.S.  If you meant your comments all in fun, and are surprised I took  
offence, I sorry, I'm just not in the mood.  Further, keep any  
apologies to yourself, I don't want to read it, just consider it  
accepted.
Newsgroups: sci.math
From: root@cc.mcgill.ca (Steeve)
Subject: Re: Proof of God's Existence
Organization: McGill University
X-Newsreader: Tin 1.1 PL5
Date: Tue, 1 Sep 1992 16:10:30 GMT

Richard Silver (rsilver@world.std.com) wrote:
: In article <17ui6kINNsft@matt.ksu.ksu.edu> bubai@matt.ksu.ksu.edu (P.Chatterjee) writes:
: >I was told by somebody that there is a mathematical proof of God's existence;
: >was wondering if somebody could shed some light on the same. 
: >Thanks. Looking forward to the response.
: 
: You do not need proof. Just ask those who die in his name. 
: 
: And please find other newsgroups for such lame queries. 
: 

	Time for your valium!
Newsgroups: sci.math
From: victor@watson.ibm.com (Victor Miller)
Subject: Re: not unique factorization
Date: Thu, 27 Aug 1992 15:11:42 GMT
Reply-To: victor@watson.ibm.com
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: IBM, T.J. Watson Research Center

>>>>> On 26 AUG 92 23:10:07    , rose@fsu1.cc.fsu.edu (Kermit Rose) said:

Kermit> Hello anyone.

Kermit> I had read that the integers extended by the sqrt(-5) did not
Kermit> have unique factorization into primes.  The example was given

Kermit> 6 = 2 * 3 = (1+sqrt(-5)) * (1 - sqrt(-5))

Kermit> I considered whether or not this ring could be fixed up by additional 
Kermit> extension.   Perhaps in a larger ring we could have

Kermit> 6 = a * b * c * d where

Kermit> a * b = 2
Kermit> c * d = 3
Kermit> a * c = 1 + sqrt(-5)       and
Kermit> b * d = 1 - sqrt(-5).

Kermit> This gives  

Kermit> b = 2/a

Kermit> c = (1 + sqrt(-5))/a

Kermit> d = (1 - sqrt(-5))/b = (1 - sqrt(-5)) * (a/2)

Kermit> One way to have a,b,c,d to be elements of the ring is to
Kermit> define a = 1/2 to be an element of the ring.  Then all powers
Kermit> of 2 becomes units, and the factorization 6 = 2 * 3 does not
Kermit> count since 2 is not prime.

Kermit> Now for my question.  Does this fix solve all the cases of non-unique 
Kermit> factorization in the ring of integers extended by sqrt(-5)?

Kermit, You are correct about this.  There are basically two ways to
fix this up: 1) Localize at a finite set of primes, which generate the
ideal class group (this is a group which measures in a precise way how
unique factorization fails).  In your case of sqrt(-5), the ideal
class group is of order two and generated by the prime 2.  So you have
done exactly that.  2) Go to an extension field: the Hilbert class
field has the property that every ideal downstairs, "capitulates":
i.e. becomes principal.

As an intersting aside, fields with class number 2 (one of which is
given by your example), are characterized by the property that they
don't have unique factorization, but that every factorization into
prime ideals has the same NUMBER of factors.

Kermit> rose@fsu1.cc.fsu.edu          To be sure I see your response, use e-mail.
Kermit> -----------------------------------------------------------------------
Kermit> You may post, repost, or publish ANY communication received from me.
Kermit> To mend an undesirable barrier is to destroy the barrier.
--
		Victor S. Miller
		Bitnet: VICTOR at WATSON
		Internet: victor@watson.ibm.com
		IBM, TJ Watson Research Center
		"Great artists steal; lesser artists borrow" Igor Stravinsky
From: SXW103@psuvm.psu.edu (S. Wukovitz)
Newsgroups: sci.math
Subject: Re: A Math Induction Problem
Date: 30 Aug 92 05:07:04 GMT
Organization: Penn State University

In article <17pjpeINNouf@matt.ksu.ksu.edu>, bubai@matt.ksu.ksu.edu
(P.Chatterjee) says:
>Then, (k+1)^3 + 5(k+1) = k^3 + 5k + 3k^2 + 3k + 6
>                       =6p + 6[(1/2)k^2 + (1/2)k + 1]
>Does that make sense? Or, are 1/2 's disallowed?

Whether k is even or odd, k^2 + k is even, so there is
no problem here.

Hope it helps.

-Stephanie Wukovitz
steph@math.cornell.edu
Newsgroups: sci.math
From: tim@maths.tcd.ie (Timothy Murphy)
Subject: Re: Polygon point enclosure
Organization: Dept. of Maths, Trinity College, Dublin, Ireland.
Date: Wed, 26 Aug 1992 23:09:09 GMT

mps@sparc68.hri.com (Mark Stockley) writes:

>I'm looking for a method to determine if a point is enclosed
>by a polygon. Any help would be greatly appreciated.

If P is the point,
you could work out the angle PX goes through
as X runs round the polygon.
Should be 0 if P is outside, 2\pi or -2\pi if inside.

-- 
Timothy Murphy  
e-mail: tim@maths.tcd.ie
tel: +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Newsgroups: sci.math
From: shallit@graceland.uwaterloo.ca (Jeffrey Shallit)
Subject: Re: Einstein's Erdos number
Organization: University of Waterloo
Date: Wed, 2 Sep 1992 18:46:36 GMT

In article <1992Sep2.053332.11070@newshost.anu.edu.au> gerry@macadam.mpce.mq.edu.au (Gerry Myerson) writes:
>In article <Btwt5H.A6n@watserv1.uwaterloo.ca>, shallit@graceland.uwaterloo.ca (Jeffrey Shallit) writes:
>[...]
>> In the preface to "A Tribute to Paul Erdos", Cambridge
>> University Press, 1990, it is stated that Einstein's
>> Erdos number is 2.  That means there exists a person
>> who has written a paper with Einstein, and a paper with
>> Erdos.  Anybody know who it is?
>
>Just a guess, but Ernst Straus was an assistant to Einstein in the late 40s, 
>and surely a co-author of Erdos'. 
>
>> Jeff Shallit
>> University of Waterloo
>
>Gerry Myerson

Ernst G. Straus is the correct answer; he co-wrote papers with
Einstein in 1945-6 (e.g. Ann. Math. 47 (1946), 731-741) and 
many with Erdos (e.g. Pacific J. Math. 3 (1953), 689-694).

Jeff
From: marwk@levels.unisa.edu.au
Newsgroups: sci.math
Subject: Re: Combinatorial Problem
Date: 29 Aug 92 16:58:45 GMT
Organization: University of South Australia

In article <1992Aug27.135518.8048@rhrk.uni-kl.de>, schwab@miranda.mathematik.uni-kl.de (Hartmut Schwab) writes:
>
> I have a Problem, which I think is very easy.
> But I don't know how to find the solution.
>
> Given an Integer K.
> Partition K into m Integers c[i], such that their sum is K.
> Condition: For all i :  0 <= c[i] <= k[i].
>
> How many different Solutions are there?
>
> Two solutions c0 and c1 are different, if there is an i such that
> c0[i] <> c1[i].
>
> Therefore in the case K = 3, m = 2 , 0 <= c[1] <= 2, 0 <= c[2] <= 3
> we get the solutions:  0+3, 1+2, 2+1. These are 3 different solutions.
>
> You could think of K bowls which are put into m boxes.
> The condition says, that box i is only able to take k[i] or less bowls.
>
> Without the condition I am able to solve it. But I have problems with
> the condition.
>
> I am interested in an easy to calculate term.
>
>
> Where can I get a solution to this problem?
>
> Thanks in advance.
>
>
> Hartmut Schwab
> schwab@mathematik.uni-kl.de

There was a book written in the 19th century, Choice and Chance, which
discusses several of these kinds of problems, and in particular solves
the problem mentioned - probably as the coefficients of a power of a poly
over another polynomial.  Exact results are then determined by
using an algebra package or your own computer program.

If you cannot find the book in your library I can provide the solution,
but this request occurs here about once every 6 months.

--
Raymond Kennington
LECTURER
University of South Australia   | Act in haste and repent at leisure
P.O. Box 1                      | Code too soon and debug forever
Ingle Farm                      | Knobs, knobs everywhere,
South Australia                 |              just vary a knob to think!
Newsgroups: sci.math,sci.philosophy.tech
From: chalmers@bronze.ucs.indiana.edu (David Chalmers)
Subject: Re: Proof of God's Existence
Organization: Indiana University
Date: Fri, 4 Sep 1992 00:51:17 GMT

In article <1992Sep3.215253.23999@ttinews.tti.com> jackson@soldev.tti.com (Dick Jackson) writes:

>I know that you prefaced this quote by saying it was not entirely serious,
>and that the main point of your article lies elsewhere. But if I (as a
>very amateur voyeur of philosophy) make a corresponding argument but
>substitute the string FCA (which might stand for first cause agent) for
>the string God, how would a defender of this argument rebut my claim that
>I have proved the existence of FCA but that FCA has no other properties of
>the entity commonly called by many, but not a majority of people, God.

That seems about right to me.  Here's what the paper has to say:

  So what is being claimed here, at least, is a sound argument to a
  true conclusion.  It is only reasonable to add the usual disclaimers.
  All that the argument seeks to establish is that there is a First
  Cause.  Readers will have to join the churches, mosques, temples, or
  cells of their choice to ascertain what other properties, if any,
  God might have.

-- 
Dave Chalmers                            (dave@cogsci.indiana.edu)      
Center for Research on Concepts and Cognition, Indiana University.
"It is not the least charm of a theory that it is refutable."
Newsgroups: sci.math
From: sshws@convx1.lerc.nasa.gov (Herb Schilling)
Subject: Re: Correction: Question on divergence of series given condition
Organization: NASA Lewis Research Center
Date: Fri, 4 Sep 1992 18:39:41 GMT

In my previous post given below, I failed to mention that the series is
complex. Please insert magnitudes where needed. Sorry about that.



In article <1992Sep4.174051.3223@eagle.lerc.nasa.gov> sshws@convx1.lerc.nasa.gov (Herb Schilling) writes:
>
>We think there is a theorem ( By Weierstrauss ? ) on this subject
>but we cannot find
>it. Can anyone out there tell us if the following is true? 
>
>=======================================================================
>
>  If the terms, a  , of a series have the following property:
>                 n
>
>         a                               A
>          n+1                1            n
>        ------ =    1   -  -----   +   -----    as n -> oo
>         a                   n           n
>          n
>
>
>
>    where A    ->   0 as n -> oo
>           n
>
>
>
>
>  then does sum of the a    diverge  ?
>                        n 
>
>
> Thanks in advance for the help.
>-- 
>Herb Schilling , NASA Lewis Research Center , 21000 Brookpark Road, MS 142-5
>Cleveland, Ohio 44135 . (216) 433-8955       sshws@convx1.lerc.nasa.gov


-- 
Herb Schilling , NASA Lewis Research Center , 21000 Brookpark Road, MS 142-5
Cleveland, Ohio 44135 . (216) 433-8955       sshws@convx1.lerc.nasa.gov