HDTenEightyP/NetNews huggingface dataset:
From: scott@techbook.com (Scott Huddleston)
Newsgroups: sci.math,sci.psychology
Subject: Re: mathematician's marrying ages.
Summary: it's a joke, son
Date: 2 Sep 92 23:17:15 GMT
Article-I.D.: techbook.1992Sep2.231715.3836
Organization: TECHbooks --- Public Access UNIX --- (503) 644-8135

After reading the comment below, I couldn't resist posting this old
chestnut.

A lawyer, a doctor, and a mathematician were comparing the virtues of
a wife versus a mistress.  The doctor preferred a wife, because it was
better for the health.  The lawyer preferred a mistress, on grounds it
was better for the pocketbook.

The mathematician recommended having both a wife and a mistress,
because it gave him more time.  When he left one of them, the wife
thought he was with the mistress, the mistress thought he was with the
wife, and he was free to do mathematics.

In article <1992Aug24.134508.27438@cc.tut.fi> jk87377@cc.tut.fi (Juhana Kouhia) writes:
>Well, we don't have details so, it would be silly to argue on this
>anymore. I think it is better to choose the one which causes
>less harm:
>-if you don't marry, then you don't have a wife and will have mental
>problems
>-if you marry, then you have wife and very fun  :-)
-- 
Scott Huddleston  	scott@techbook.com
503-524-9442
From: fearnley@ERE.UMontreal.CA (Fearnley Anne)
Newsgroups: sci.math
Subject: Re: What kind of algebra is this?
Date: 1 Sep 92 17:10:24 GMT
Organization: Universite de Montreal

In article <12701@ecs.soton.ac.uk> dbc@ecs.soton.ac.uk (Bryan Carpenter) writes:
>
>     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?
>

If the operations also follow the following rules:
         a * (a + b) = a       and      a + (a * b) = a     (absorption)
then the algebra you describe is a distributive lattice.

Anne
From: bubai@matt.ksu.ksu.edu (P.Chatterjee)
Newsgroups: sci.math
Subject: A Math Induction Problem
Date: 29 Aug 92 18:45:34 GMT
Article-I.D.: matt.17pjpeINNouf
Organization: Kansas State University
NNTP-Posting-Host: matt.ksu.ksu.edu

I have solved/near-solved a math induction problem. Was wondering if somebody could point out whether it's flawed or not.

I am trying to prove that n^3 + 5n is divisible by 6.

This is how I proceeded:

Assume k^3 + 5k = 6p.

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?

Thanks for any kind of help.
Newsgroups: sci.math
From: polani@Informatik.Mathematik.Uni-Mainz.DE (Daniel Polani)
Subject: C(++) test suite for checking random number generator?
Organization: Computer Science, Johannes Gutenberg-Universitaet
Date: Thu, 27 Aug 1992 13:17:39 GMT

Hi,

as already said in the subject line, I'm looking for a C or C++ test suite 
to check the quality of random number generators. Does anyone have such a suite?
Information for ftp-sites is welcome as well.

Many aTdHvAaNnKcSe

Daniel Polani   				    


							 











                            
Newsgroups: sci.math
From: jesperse@nas.nasa.gov (Dennis Jespersen)
Subject: Location of electronic journal Solstice
Organization: Computational Fluid Dynamics Branch, NASA Ames Research Center
Date: 4 Sep 92 06:55:01

The August-September issue of the American Mathematical Monthly
contains a telegraphic review (on p. 702) of "SOLSTICE: An Electronic
Journal of Geography and Mathematics".  The review says this is one of
the world's first electronic journals, in TeX, distributed both on
paper (for a fee) and electronically (for free).  How can one obtain
this journal electronically?
--
Dennis Jespersen                Voice: (415) 604-6742                      
MS 202A-1		        FAX:   (415) 604-1095                      
NASA Ames Research Center       email: jesperse@rft28.nas.nasa.gov         
Moffett Field, CA 94035-1000    We have met the enemy and he is us. -- Pogo
Newsgroups: sci.math
From: hstitan@neumann.uwaterloo.ca (Hari S. Titan)
Subject: Concordia's prof killer's message
Organization: University of Waterloo
Date: Fri, 4 Sep 1992 19:06:16 GMT


Hi everyone,

I heard that the prof. killer at Concordia had sent an e-mail message to 
many mathematicians complaining about problems he had with his job a week
before he began the killings.  My understanding is that he meant this to 
be public letter and wouldn't mind more people reading it.  If someone 
still has his message could they e-mail me a copy.  

Thanks very much.

H. Titan 
(hstitan@neumann.uwaterloo.ca)
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 3 Sep 92 18:58:36 GMT
Reply-To: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Organization: The Wistar Institute of Anatomy and Biology
In-reply-to: ara@zurich.ai.mit.edu (Allan Adler)

In article <ARA.92Sep3133928@camelot.ai.mit.edu>, ara@zurich (Allan Adler) writes:
>I don't see why everyone is trying to prove the existence of God. I
>think the people of the world are virtually unanimous in the opinion
>that God does not exist. To see why this is so, let us merely assert,
>in addition to the mere use of the word "God", some property of God
>and ask if people believe that God exists and has that property and
>overwhelmingly the response will be no. [...]

This is clearly a proof on par with all horses are purple (by induction).
The same technique will prove that the people of the world do not believe
in the existence of "clear communication".  Go around and ask, and watch
all the blank stares you get from the vast majority.  So much for "clear
communication".

Of course, if you're not trying to be mathematically silly, and I managed
to completely misunderstood your point, please, restrict such postings in
the future to talk.religion.misc where the context will tell us you're
actually being serious.
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
Newsgroups: sci.math
From: jasonp@cunix4.Prime.COM (Jason R. Pascucci)
Subject: Re: Problematic smurfs!
Keywords: smurfs
Organization: Prime Computer, Inc.
Date: Sun, 30 Aug 1992 21:01:43 GMT

Fairly easy, methinks:

1000 smurfs rise. 
The color of the hats of the smurfs that rise are blue or red. 

(The assumption is that 'not apart of' should be 'not a part of'
If you meant 'not apart from', then it's only 999)

The smurfs, being logical little vlue devils, noted that the 
mayor didn't say 'Only smurfs wearing red hats stand up.' (i.e.
exclusive of the smurfs wearing blue hats), he told
the red hatted smurfs to stand up, so they did. 

-- 
Jason R. Pascucci
jasonp@primerd.prime.com
Newsgroups: sci.math
From: arnold@cs.toronto.edu (Arnold Rosenbloom)
Subject: Help finding a book
Organization: Department of Computer Science, University of Toronto
Distribution: na
Date: 1 Sep 92 00:58:47 GMT

I was reading the book The Theory of Numbers by McCoy and have
decided to purchase a copy. Unfortunately the publisher says
its out of print. If you have a copy you are willing to part with
or know where I can find the author please let me know.

Thanks
Arnold Rosenbloom
PS Other suggested number theory books would be appreciated.


Newsgroups: sci.math
Newsgroups: sci.math
From: pedmonds@cs.toronto.edu ("Philip G. Edmonds")
Subject: Re: Solitaire games
Organization: Department of Computer Science, University of Toronto
Date: 2 Sep 92 21:17:38 GMT

In article <1992Sep2.204931.29320@wdl.loral.com> mab@wdl39.wdl.loral.com (Mark A Biggar) writes:
>In article <Btyo5z.D8q@cs.psu.edu> morgan@math.psu.edu (Ilene Morgan) writes:
>>My game with license plates is to try to find a word that contains the
>>letters of the license plate in order, not necessarily consecutively.
>>For example, XPV -> expressive.  My license plate used to have XPF; I never
>>did think of one for that.
>
>expeditefull
>excerptfull
>exemplify
>exemplification

ex post facto

(from Webster's)
Phil.
From: cjh@tinton.ccur.com (Christopher J. Henrich)
Newsgroups: sci.math
Subject: Re: Another induction problem
Date: 2 Sep 92 21:33:12 GMT
Organization: Concurrent Computer Corp., Tinton Falls, NJ

In article <CHALCRAFT.92Aug31105002@laurel.uk.tele.nokia.fi> chalcraft@uk.tele.nokia.fi (Adam Chalcraft) writes:
>Any chance of explaining how an induction can get that complicated?
>
...
>
>Maybe I've missed something. Can induction exist without relying on a well-
>ordered set? 
>
>Of course, writing down an ordinal can be tricky (Challenge for the naive
>reader: Devise a general scheme for writing down any ordinal :-) :-) :-)),
>but is that really the major conceptual difficulty here?
>
Induction can also be done over a partially ordered set, such that
every subset has a minimal element.  In a really gnarled proof, the
structure of the partial order over which you are trying to induct
may be irregular, or may depend on variable properties of the
structures under consideration.  

Regards,
Chris Henrich
Newsgroups: sci.math
From: kenb@amc.com (Ken Birdwell)
Subject: Triangular systems of equations.
Organization: Applied Microsystems Corporation
Date: Fri, 28 Aug 1992 00:25:56 GMT

I'm looking for a method to help me solve large systems of 
simultanious equations (10,000+) that are related to one another in 
the form of an irregular triangular mesh.  My problem seems to be in 
not being able to order the points into a tight enough diagonal 
sparse matrix.  If you know of any references on solving systems 
like this, please forward them.


The problem:
    Given an irregular non-overlapping 2D open triangular mesh of a N 
    points, with known coordinates for all edge points,  calculate "good" 
    coordinates for all the interior points while maintaining the feature of 
    non-overlapping triangles.  

My solution so far:
    Cycle through all the interior points, calculating for each one a new 
    coordinate by averaging the coordinates of all of its 'mesh' neighbors.
    Repeat until the points stop moving (much).
    
    I've also added the feature that all the points are weighted by 
    how well their new triangles match the surface area of their 
    original triangles.  This kinda forces a equal area distortion, which 
    is desirable.

    The problem is that it takes N*sqr(N) time for the interitive approach
    to settle down into a solution. When N > 10k, this is a problem, and 
    can take over a 1/2 hour and a 486/33.

What I tried to do:
    Well, I tried to build this into a very large system of system of 
    simultanions equations, and interate through it one or two times until 
    I could calculate what weights to use.  
    
    I wrote some software that does sparse matrices (zero's take up no 
    memory), and got this solution to work fine for systems with around 
    2,000 points. It also has very nice linear computational cost (about 30 
    seconds per 1000 points).   With systems larger then about 10,000, the 
    sparse matrix starts to fill up (I'm using Gaussian elimination with 
    back subsitution). Systems any larger start to take up alot more memory, 
    and pretty soon fill up my 32MB of RAM.

    The problem, I think, is that I don't have a good way to order the 
    points in the matrix so that connected points keep a tight enough 
    diagonal.  What I do now is to pick a single starting point, and 
    interate outward from that point, numbering each point as they get 
    farther and farther away from the starting point.  This creates (in the 
    matrix) what look like a central band with two other parallel bands, one 
    on each side of the central band, that bow out in the middle. This 
    bowing in the middle ends up causing the matrix to fill up with non-zero 
    values, and also killing performance.

    Any method that will order the points better, or a way to swap rows and 
    columns to minimize the memory used would be much appreciated.
    

--
From: loader@vax.oxford.ac.uk
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 1 Sep 92 21:40:45 GMT
Organization: Oxford University VAX 6620

In article <1992Sep1.215331.89956@vaxc.cc.monash.edu.au>, kevin@vaxc.cc.monash.edu.au writes:
> If one reads through philosophy, one will find many attempts at mathematical
> proofs of God's existence. All these 'ontological proofs' have very obvious
> and silly errors. It is quite clear that mathematics is not capable of making
> a statement on the ontological status of God. Sure, maths can make lots of 
> comments about the hyperbolic plane, the Zariski Topology, the lattice of r.e
> sets and so forth. But God? No way!
> Asides from limitations in the expressive power of mathematics, the following
> will show any proof of the ontic status of God futile.
> 1. It is possible to imagine a world, completely agreeing with the current one
> on all sensory data, in which God exists. 
> 2. It is possible to imagine a world, completely agreeing with the current one
> on all sensory data, in which God doesn't exists.
> Call a world satisfying condition 1 M, and a world satisfying 2 M'.
> Then the existence of these epistemically indistinguishable worlds proves that
> any ontological statement about God is INDEPENDENT of any epistemically 

It is possible to imagine a world completely agreeing with the current
one on all sensory data, in which Fermat's last theorem is false.
It is possible to imagine a world completely agreeing with the current
one on all sensory data, in which Fermat's last theorem is true.
Then by the fact that mathematical truths are necessary truths and true
in all possible worlds, FLT is true and false.

Kabloom
(The world disappears in a puff of logic.)

Oh dear, we're all still here. At least I am. Just goes to show I'm not part of
the real world after all.

> knowable axioms on the real world. So no proof either way exists. Bravo.
> Notes 1. Some people say world M' is impossible, because the beauty of the
> world we live in necessarily entails God's existence. There are many similar
> arguments, basically content isomorphic to this. In fact, this argument is
> content isomorphic to the trivial or empty argument, so not much need be said.
> I assume my response to such a person is implicitly made clear in the previous
> comments.
> 2. It will strike many as odd that even Godel believed he had a proof, using
> modal logic. If you can bother spending the time reading it (it is quite
> complex), I'm sure you'll spot the error. It's more sublte than most other
> ontological errors, but it's there, unavoidable, and unrect
> even great mathematicians can go loony every now and again.
> Well, there you have it. No further comment necessary.

Well, there you have it. No further comment possible.

> Kevin Davey
> Monash University 
> Australia.
Newsgroups: sci.math,misc.misc
From: patstew@oz.plymouth.edu (Pat Steward)
Subject: Re: value of pie
Reply-To: patstew@oz.plymouth.edu (Pat Steward)
Organization: Plymouth State College - Plymouth, N.H.
Distribution: na
Date: Fri, 4 Sep 92 13:42:46 GMT

They seem to cost around $2.99 here, though blueberry usually sells
for more than apple.

Hope this helps!
-- 
Pat Martin Steward                          patstew@oz.plymouth.edu
I myself have never been able to find out precisely what feminism is;
I only know that people call me a feminist whenever I express senti-
ments that differentiate me from a doormat.       Rebecca West, 1913
Xref: sparky sci.math:10696 rec.org.mensa:6441 rec.puzzles:5946
From: spacefox@acs.bu.edu (Godfrey Degamo)
Newsgroups: sci.math,rec.org.mensa,rec.puzzles
Subject: Problematic smurfs!
Keywords: smurfs
Date: 30 Aug 92 20:09:36 GMT
Followup-To: sci.math
Organization: Boston University Undergraduate Mathematics
Originator: spacefox@acs.bu.edu


     Help!  My friend gave me this problem.  Being quite lazy, I don't want to
think about it.  Anyways, we decided that this would be an interesting problem
to put on the net.  Anyways, if you decide to post anything about this problem,
please put the word "smurfs" somewhere in the subject header and keyword 
category.  We plan to record every post on it.  
Well, here goes, good luck!


-------------------------------------------------------------------------------
     A group of 1000 smurfs wear either a red or blue hat on their head.
They can only wear one color. The color hat on their head is known to all, but
the wearer.  The wearer of the hat has no means whatsoever for obtaining the
color of his hat.  It is never the case that all 1000 smurfs will wear
the same color hat.
     One day the mayor, not apart of the 1000 smurfs, decides to call a
town meeting.  All 1000 smurfs are gathered into the town hall.  The mayor
is not wearing a hat.  During the meeting, the mayor asks that all the
smurfs wearing red hats to stand up.
     There is a bit of commotion in the crowd, but no one stands up.
     The mayor requests the same demand a bit more sternly.
     Again, commotion, but no one stands up.
     The mayor, irate, demands the red hatted smurfs to stand up.
     Then, a certain amount of smurfs rise.

find:
     The exact amount of smurfs that rise.  
         (Hint: This number will be greater than one.)
     The color of the hats of the smurfs that rise.

more importantly,
     Explain how you deduced your answer.
-------------------------------------------------------------------------------

I hope that's clear enough for you.  If not, post, but all the necessary
information has been supplied.  I'll only answer questions about my poor
grammar and writing style.  Put on your smurf hats and think!


                                        -G. Degamo,
                                         spacefox@acs.bu.edu

PS: please, no Delft University students' input! ;)
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: sci.math
Subject: Re: Abel's proof of the insolubility of the quint
Date: 4 Sep 92 16:03:48 GMT
Article-I.D.: husc3.1992Sep4.120349.15365
Organization: Harvard Math Department

In article <87849@netnews.upenn.edu>
weemba@sagi.wistar.upenn.edu (Matthew P Wiener) writes:
:In article <1992Sep3.122450.15337@husc3.harvard.edu>
:elkies@ramanujan (Noam Elkies) writes:
:>>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 nice proof is reasonably well-known, but can one get a proof
:>of the simplicity of all A_n (n>4) from such ideas?
:
:I don't think so, although I don't know of a particular A_n where the
:partition is possible, of course without forming a subgroup.

The identity and simple transpositions together give a union of
conjugacy classes of size (n^2+n+2)/2, which divides |A_n|=n!/2
at least for n=11,18,27,37,38,46.

--Noam D. Elkies (elkies@zariski.harvard.edu)
  Dept. of Mathematics, Harvard University
Newsgroups: sci.math
From: kedlaya@b124.super.org (Kiran Sridhara Kedlaya)
Subject: Re: n doesnt divide 2^n-1
Organization: Supercomputing Research Center
Date: Fri, 28 Aug 1992 20:31:44 GMT

Sure. Let n be the smallest integer > 1 such that n divides 2^n - 1.
Let d be the smallest integer such that n divides 2^d - 1 (i.e. the order
of 2, with respect to n). Then d divides n (otherwise n = rd + s, where
0 < s < d, and since 2^n and 2^d are congruent to 1 mod n, so is
2^(n - rd) = 2^s, whence n divides 2^s - 1, which is impossible
by the choice of d). Also 2^d - 1 divides 2^n - 1, and d > 1 (otherwise
n divides 2^1 - 1 = 1). So d is an integer > 1 but smaller than n, such
that d divides 2^d - 1. Contradiction.

Kiran Kedlaya
Supercomputing Research Center
kedlaya@super.org (for the moment)
Newsgroups: sci.math
From: flax@frej.teknikum.uu.se (Jonas Flygare)
Subject: Real World Problem (handrail for stairs..)
Keywords: stumped
Reply-To: flax@mizar.docs.uu.se
Organization: Dept. of Control and Systems, Teknikum, Uppsala
Date: Fri, 4 Sep 1992 09:35:38 GMT

Here's a Real World (tm) problem for all you math types out in netland.

My girlfriends father is making spiraling stairs for a living, and at a
dinner a while ago aske d me about a problem he have.
When making the handrail, it is ordered from another company, and is made
in circle segments, from a circle with a radius larger than the radius
of the stair itself. (due to the height difference)
So, when ordering he need to specify the radius.
That is dependent on the radius of the staircase, the inclination (or
the height between each step)

I believe that the solution is somewhat dependent on the length of each 
segment, as they have to 'twisted' slightly in relation to each other
to fit. 

When I asked this at a local conference system I got several
answers that seemed to be well thought out. Unfortunately they
produce different results.. :-)

I'd really like to get a 'definite' answer to this so here goes:

Given the radius and inclination of a spiraling staircase (and 
eventually the length of each segment, or angle that the segment covers),
when approximating the handrail from circle segments, what radius should
that circle be?

If any of you think this is trivial, I don't mind you pointing that out, 
as long as you provide me with a solution. :-)


--------------------------------------------------------
Jonas Flygare, 		+ Wherever you go, there you are
V{ktargatan 32 F:621	+
754 22 Uppsala, Sweden	+
Newsgroups: sci.math
From: rairan@ecf.toronto.edu (RAI Ranjan)
Subject: Re: p prime, p divides ab => pdivides a or b
Keywords: Contrapositive
Organization: University of Toronto, Engineering Computing Facility
Date: Wed, 2 Sep 1992 18:12:37 GMT

In article <Btyo8q.E63@ux1.cso.uiuc.edu> ceblair@ux1.cso.uiuc.edu (Charles Blair) writes:
>
>   My recollection is that a number theory course I took presented
>this as a difficult result, only proved after doing some stuff
>with the Euclidean algorithm.  Is there a proof which avoids that?

  Contrapositive is a fancy way of saying reductio ad absurdum :)

  It basically means that for any propositional statements s and t:
                    s => t <=>  not(t) => not(s)
  So to prove p|(ab) => p|a or p|b, we just show that if p does not 
divide a AND p does not divide b, then p does not divide ab.  By the
Fundamental Theorem of Arithmetic this is obvious.  Euclid's algorithm is
not necessary.  Indeed, the proof of the extended Euclidean algorithm
depends on this result, I think.
-- 
---
Ranjan Rai                             <rairan@ecf.toronto.edu> 
Engineering Science 9T3                <rairan@eecg.toronto.edu>
University of Toronto                      
Newsgroups: sci.math