HDTenEightyP/NetNews huggingface dataset:
From: schow@bqneh23.bnr.ca (Stanley T.H. Chow)
Subject: Re: Godel and the US constitution
Followup-To: misc.legal
Organization: Bell Northern Research Ltd, Ottawa
Date: Thu, 3 Sep 1992 17:58:44 GMT

[note followup directed to misc.legal]

In article <1992Sep3.005923.22443@cbnewsm.cb.att.com> mls@cbnewsm.cb.att.com (mike.siemon) writes:
>> Fortunately there is the second amendment of the Bill of Rights.

Unfortunately, Lincoln ignored the bill anyway.

>Utterly useless if the dictator comes to power with quasi-fascist
>gun-toters as his storm troopers.  I suppose one could hold out the
>"hope" that the urban mob might be opposed to a rural-right takeover,
>but that is hardly what I regard as decent checks & balances.

As a foreign observer of the US, I see that the DEA and various
police agencies acquiring a lot of power. It is debatable whether
the German SS had more power then (legal or otherwise). Read 
misc.legal for a while to see the powers that are being granted
to the police type agencies.

--
Stanley Chow            InterNet: schow@BNR.CA
Bell Northern Research  UUCP:     ..!uunet!bnrgate!bqneh3!schow
(613) 763-2831
Me? Represent other people? Don't make them laugh so hard.
Newsgroups: sci.math
From: chalcraft@uk.tele.nokia.fi (Adam Chalcraft)
Subject: Re: A Math Induction Problem
Organization: cpd
Date: Mon, 31 Aug 1992 08:32:42 GMT

You could even prove that 2 divides k^2 + k in the induction way you
already understand, if you like.

Assume k^2+k = 2p. Then (k+1)^2+(k+1) = k^2+3k+2 = 2p+2k+2 = 2[p+k+1].

Oh, by the way, you ought to observe that 0^3+5.0=6.0 and 0^2+0=2.0 :-)
--
  ____________________________________________________________________________
 /  _Name: Adam|Names are linguistic constructs expressed in some language._  \
|\_|/|Chalcraft|They correspond to objects in some universe of discourse. |\|_/|
|`   |         |The corresponence between names (in the language) and     |`   |
|`   |Opinions:|objects (in the universe of discourse) is the  relation of|`   |
|`   |     Mine|identifying. A name identifies the object to which it is  |`   |
|`   |_________|bound.__________________________ISO_7498-3_:_1989_(E)_5.1_|`   |
 \__/                                                                      \__/
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 4 Sep 92 16:37:26 GMT
Article-I.D.: netnews.87960
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)

Al Adler missed my point.  A proof is valid or not, regardless of whether
the masses hold by your definition.
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
Newsgroups: sci.math
From: John_Jenkins@taligent.com (John H. Jenkins)
Subject: Re: Proof of God's Existence
Followup-To: sci.math
Organization: Taligent, Inc.
Date: Wed, 2 Sep 1992 05:56:28 GMT

In article <1992Sep1.235230.8805@massey.ac.nz>, news@massey.ac.nz
(USENET News System) wrote:
> 
> > 
> > In article <Btw9nH.BIE@world.std.com> rsilver@world.std.com (Richard
Silver) writes:
> > >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.
> > >
> I sent an e-mail reply, but since no-one else has brought this up, it
might
> be of interest here.
> 
> I forget the details, but a philospher who had quite a reputation at
the Russian
> court offered to debate the existence of God. It happened that a well
known
> French mathematician was visiting at the time (possibly Laplace - I'm
not
> sure). He claimed to have a mathematical proof.
> He said something like "3x+2 = 7 donc Dieu exist".
> Unfortunately the philosopher did not know enough maths to understand
> that he was being had and left discomfitted.
> 

This is from E.T. Bell's "Men of Mathematics":

"Invited by Catherine the Great to visit her Court, [Denis] Diderot [the
French encyclopedist and atheist] earned his keep by trying to convert
the courtiers to atheism.  Fed up, Catherine commissioned Euler to
muzzle the windy philosopher.  This was easy because all mathematics was
Chinese to Diderot.  De Morgan tells what happened (in his classic
_Budget of Paradoxes_, 1872):  'Diderot was informed that a learned
mathematician was in possession of an algebraical demonstration of the
existence of God, and would give it before all the Court, if he desired
to hear it.  Diderot gladly consented....Euler advanced toward Diderot,
and said gravely, and in a tone of perfection conviction:  "Sir, (a + b
** n) / n = x, hence God exists; reply!"

"'It sounded like sense to Diderot.  Humiliated by the unrestrained
laughter which greeted his embarassed silence, the poor man asked
Catherine's permission to return at once to France.  She graciously gave
it.'

"Not content with this masterpiece, Euler in all seriousness painted his
lily with solemn proofs, in deadly earnest, that God exists and that the
soul is not a material substance.  It is reported that both proofs
passed into the treatises on theology of his day.  These are probably
the choicest flowers of the mathematically unpractical side of his
genius."

I love the story -- but since Bell tells it, I suppose it's probably
apocryphal (even if de Morgan *is* his source).  :-(

(Besides, I thought the Babel fish was the ultimate clincher in proving
God's existence/non-existence -- oops.  Wrong newsgroup.)

John H. Jenkins
John_Jenkins@taligent.com
Newsgroups: sci.math
From: rose@fsu1.cc.fsu.edu (Kermit Rose)
Subject: not unique factorization
News-Software: VAX/VMS VNEWS 1.3-4   
Reply-To: rose@fsu1.cc.fsu.edu
Organization: Florida State University
Distribution:  world
Date: 26 AUG 92 23:10:07    

Hello anyone.

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

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

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

6 = a * b * c * d where

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

This gives  

b = 2/a

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

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

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

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

rose@fsu1.cc.fsu.edu          To be sure I see your response, use e-mail.
-----------------------------------------------------------------------
You may post, repost, or publish ANY communication received from me.
To mend an undesirable barrier is to destroy the barrier.
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: Re: Beloved Books + Request
Date: 2 Sep 92 06:20:09 GMT
Organization: M.I.T. Artificial Intelligence Lab.


I have found Shoenfield's book useful sometimes, but also hard to use for
reference. The reason is that he assumes you remember where and when he
introduced his notation for something, even if it is a hundred pages later,
and has no index of notation.

One of my favorite books is Sierpinski's Hypothese du continu, which studies
the continuum hypothesis and its consequences and equivalents in analysis.
For example, the continuum hypothesis is equivalent to the following
assertion:

There exists a decomposition of the plane into two disjoint subsets A and B
such that the intersection of A with any horizontal line is at most
countable and the intersection of B with any vertical line is at
most countable.

Another equivalent is the conjunction of the following two assertions:
(1) every set of cardinality less than the continuum has measure zero
(2) there exists a set of cardinality the continuum whose intersection
    with every set of measure zero is at most countable.

Another result in the book, which depends on the continuum hypothesis is 
that there is a permutation of the reals which carries the class of sets of
measure zero bijectively onto the class of sets of first category.
This implies that in the equivalence of CH to (1) and (2) above, one
could replace "measure zero" by "first category". The resulting statement
is also equivalent to CH and the existence of the permutation explains it.
But the existence assumes CH.

One of the things people do now is to try to see what happens if one
negates the continuum hypothesis and assumes Martin's axiom. There was
a paper about 15 years ago, I think, which had a table of the results
in Sierpinski's book and summarized the status of each of them if one
replaces CH by -CH+MA.There is a book on Martin's axiom but I don't
know if it deals with this stuff.

Although not set theory, the book of Abraham Robinson, Introduction to Model
Theory and the Metamathematics of Algebra, is another favorite of mine. It
is quite sensitive and philosophical, and it is wonderful mathematics.

Another one I like very much is Jech's book The Axiom of Choice.

The long paper of Scott and Solovay on the independence of the axiom of
choice and the continuum hypothesis using Boolean valued models was also
quite nice. Unfortunately, it never appeared in print and I have been unable
to obtain a copy of it for myself in recent years. Anyone want to send me one?
Yes, I know that other books do it. I want this one.

Allan Adler
ara@altdorf.ai.mit.edu
Newsgroups: sci.math
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Subject: Re: Polygon point enclosure
Organization: U of Cambridge, England
Date: Wed, 26 Aug 1992 23:57:00 GMT

In article <7122@bigbird.hri.com.hri.com>, 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.

Well, here's one approach. Imagine a ray (=half-line) coming off from the point
in some direction. Count how many of the sides of the polygon it goes through.
(If it goes *along* any of those sides you're in trouble and should try another
direction.) If it's an odd number the point is inside the polygon; if it's an
even number the point is outside the polygon.
Newsgroups: sci.math
From: israel@unixg.ubc.ca (Robert B. Israel)
Subject: Re: x 2^x = 2^n ?
Organization: University of British Columbia, Vancouver, B.C., Canada
Date: Fri, 28 Aug 1992 07:14:09 GMT

In <israel.714951978@unixg.ubc.ca> israel@unixg.ubc.ca (Robert B. Israel) writes:

>In <17gnuqINNr7k@darkstar.UCSC.EDU> soeren@cse.ucsc.edu (Soren Soe) writes:

>>Can anyone tell me what `x' is in this equation:

>>	x 2^x = 2^n

>>Is there a general solution, or do I have to approximate `x' to something like
>>x = n - lg n + a_little_more ?
>>In fact I can do with an approximation to x, iff 
>>		floor(x) <= app(x) <= x

>>I need the "floor" of `x' to solve the following summation, where k 
>>would be equal to floor(x):

>>	i=n-1
>>	SUM   min(2^i, 2^2^(n-i))
>>	i=0

>>	i=k	     i=n-1
>>    =   SUM  2^i  +  SUM   2^2^(n-i)
>>	i=0	     i=k+1

>>--soeren

>>Soren Soe,
>>soeren@ce.ucsc.edu

>I assume n is a positive integer.  For n = 1, x = 1, so suppose n >= 2.

>Write n = 2^m k where m is a positive integer and 1 <= k < 2, i.e. 
>m = floor(lg n).

>Let x1 = n - m - 1.  Then x1 2^x1 < n 2^x1 = k 2^(n-1) < 2^n.  So x1 < x.

>Let x2 = n - m + 1.  Then x2 2^x2 = 2^(n+1) k x2/n >= 2^(n+1) (1 - (m - 1)/n)
>                                 >= 2^(n+1) (1 - (m-1) 2^(-m)) > 2^n

>since  m - 1 < 2^(m-1) for all m.  So x > x2.  Thus floor(x) is either
>x3 = n - m or x1 = n - m - 1.

>Now x3 2^x3 = (n-m) 2^(n-m) = (n-m)k/n 2^n.  To have floor(x) = x3 we need
>n <= (n-m) k = (n-m) n 2^(-m), i.e. n >= 2^m + m.  Thus we conclude that

Oops! Reverse those inequalities, we need n <= 2^m + m.

>floor(x) = n - m - 1 if 2^m <= n < 2^m + m for some m
>floor(x) = n - m     if 2^m + m <= n < 2^(m+1).

So it's really:
floor(x) = n - m if 2^m <= n <= 2^m + m
floor(x) = n - m - 1 if 2^m + m < n < 2^(m+1).

>  
>-- 
>Robert Israel                            israel@math.ubc.ca
>Department of Mathematics             or israel@unixg.ubc.ca
>University of British Columbia
>Vancouver, BC, Canada V6T 1Y4
-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             or israel@unixg.ubc.ca
University of British Columbia
Vancouver, BC, Canada V6T 1Y4
From: Antoine@osd.ulaval.ca (Antoine Gautier)
Newsgroups: sci.math
Subject: Linear Programming on NeXT
Date: 4 Sep 92 13:18:13 GMT
Reply-To: Antoine@osd.ulaval.ca    (Antoine Gautier)
Organization: Universite Laval

Does anyone know about a LINDO-look alike, preferably shareware, that will run on  
NeXT?
thks
From: landsbur@troi.cc.rochester.edu (S. Landsburg)
Newsgroups: sci.math
Subject: Unix denies the axiom of choice
Date: 1 Sep 92 17:45:32 GMT
Organization: University of Rochester (Rochester, NY)

In article <pmanne-010992093107@ma-mac29.uio.no.> pmanne@math.uio.no (Per Manne) writes:

>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:

    (Remainder of article deleted.)

I tried to save this posting to a file called "god".  I got back the message
that "god does not exist".


Steve Landsburg
landsbur@troi.cc.rochester.edu
Newsgroups: sci.math
From: rickertj@athena.cas.vanderbilt.edu (John Rickert)
Subject: Re: Godel and the US constitution
Organization: Vanderbilt University Mathematics Department
Date: Thu, 3 Sep 92 23:45:49 GMT

In article <1992Sep1.160045.15281@husc3.harvard.edu> elkies@ramanujan.harvard.edu (Noam Elkies) writes:
>In article <1992Sep1.175844.16825@nas.nasa.gov>
>asimov@wk223.nas.nasa.gov (Daniel A. Asimov) writes:
>>Out of curiosity, does anyone know how Godel envisioned a
>>consitutional dictatorship possibly arising in the U.S. ???
>
>It would seem that all you need is 2/3 of the Senators and
>Representatives to agree on an Amendment that would make
>(say) Madonna the absolute ruler of the United States...
>
>--Noam D. Elkies (elkies@zariski.harvard.edu)
>  Dept. of Mathematics, Harvard University


   Now _that's_ what I'd call "power of the purse"!



John Rickert
rickertj@athena.cas.vanderbilt.edu


Newsgroups: sci.math
From: chalmers@bronze.ucs.indiana.edu (David Chalmers)
Subject: Re: Godel and the US constitution
Organization: Indiana University
Date: Tue, 1 Sep 1992 21:47:43 GMT

In article <1992Sep1.175844.16825@nas.nasa.gov> asimov@wk223.nas.nasa.gov (Daniel A. Asimov) writes:

>Out of curiosity, does anyone know how Godel envisioned a
>consitutional dictatorship possibly arising in the U.S. ???

I believe that it had something to do with the amendment clause of
the US constitution (Article V), perhaps in conjunction with the
10th amendment, which can be viewed as an implicit amendment of the
amendment clause itself.  For more details, see Peter Suber's _The
Paradox of Self-Amendment_ (Peter Lang Publishing, 1990).

-- 
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."
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: Godel and the US constitution
Date: 3 Sep 92 15:01:28 GMT
Article-I.D.: netnews.87832
Reply-To: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Organization: The Wistar Institute of Anatomy and Biology
In-reply-to: radford@cs.toronto.edu (Radford Neal)

In article <92Sep2.183019edt.207@neuron.ai.toronto.edu>, radford@cs (Radford Neal) writes:
>Ahhh! But in Godel's time, there were also _territories_. Surely Alaska
>was plenty big enough for hundreds of states?

>For that matter, do states have to occupy any land? 

This reminds me of the suggestion that Gorbachev should have stayed on as
the one and only citizen of the USSR, for the sake of the UN vote/veto.
That would have been the real coup!
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
From: kevin@vaxc.cc.monash.edu.au
Newsgroups: sci.math
Subject: Re: Proof of God's existence
Date: 2 Sep 92 19:20:01 +1000
Organization: Computer Centre, Monash University, Australia

In defense of my undecidability thesis:
I claimed that God's proof was mathematically undecidable for two reasons,
1. Mathematics can talk about Hyperbolic space, Orthodox Semigroups, Banach
Algebras and Measurable Cardinals, but can't make statements about the real 
world.
(I know this will be misinterpreted whatever I say, so out of laziness I won't
elaborate)
2. It is easy to imagine epistemically indistinguishable worlds M and M' to our
real world, one in which God exists, one in which he doesn't.
(See my previous two postings.)
One critique of 2 is that is depends on the viewer. For example, someone who
claims to see a miracle or something contradictory to science may beleive
he has grounds to infer the necessary existence of a God.
(This, I gather, was the prime nature of one publicly posted response)
Hence my response:
The failure of what we believe to be a scientific law does not constitute 
evidence for God's existence. For, as any scientist must admit, science is
incomplete. Would Newton, upon conducting the Michelson-Morley experiment have
been able to conclude that necessarily, a God exists? No. Why? Because, in a
fairly subtle way, Newton's science was wrong. Aberrations occured, and so 
science was modified. 
In the same way, there is absolutely no (epistemic) reason to believe that
an unexplained phenomenon, (relative to the scientific knowledge of the time),
is sufficient evidence for inferring the existence of a God. Surely the Newton
example above shows this. In fact, regardless of what we ever do know, in
particular, even if we never do explain certain phenomenon, this is 
no reason to infer that no causal explanation exists. Quite simply, this is
confusing the issue of knowledge with existence, in that we are inferring
that we do not know of any explanation that there is no explantion. Such
questions of ontology must be kept quite distict from questions of 
epistemology.
Regardless of scientific misunderstandings, it seems undisputable that the
worlds M and M' do exists. Even if there was a chap who rose from the
dead and performed miracles, and visibly rose to the sky, even if one bucket
of fish fed thousands, even if two persons, Adam and eve suddenly materialized
and all evolution is wrong, it is still possible to imagine the world which
agrees with this but in which god doesnt't exist. Don't argue "surely this
is a direct contradiction of physics" or "but what physical law could account
for this?" . These questions have been resolved in the previous paragraphs.
Don't overestimate the importance of so-called laws - a little reading
in the philosophy of science will tell you this.
Anyway, this strikes me as a pretty good defence of my position. I invite
all comments, both e-mailed and public.
Kevin Davey.
Monash University.
Australia.

Newsgroups: sci.math
From: israel@unixg.ubc.ca (Robert B. Israel)
Subject: Re: x 2^x = 2^n ?
Organization: University of British Columbia, Vancouver, B.C., Canada
Date: Thu, 27 Aug 1992 21:46:18 GMT

In <17gnuqINNr7k@darkstar.UCSC.EDU> soeren@cse.ucsc.edu (Soren Soe) writes:

>Can anyone tell me what `x' is in this equation:

>	x 2^x = 2^n

>Is there a general solution, or do I have to approximate `x' to something like
>x = n - lg n + a_little_more ?
>In fact I can do with an approximation to x, iff 
>		floor(x) <= app(x) <= x

>I need the "floor" of `x' to solve the following summation, where k 
>would be equal to floor(x):

>	i=n-1
>	SUM   min(2^i, 2^2^(n-i))
>	i=0

>	i=k	     i=n-1
>    =   SUM  2^i  +  SUM   2^2^(n-i)
>	i=0	     i=k+1

>--soeren

>Soren Soe,
>soeren@ce.ucsc.edu

I assume n is a positive integer.  For n = 1, x = 1, so suppose n >= 2.

Write n = 2^m k where m is a positive integer and 1 <= k < 2, i.e. 
m = floor(lg n).

Let x1 = n - m - 1.  Then x1 2^x1 < n 2^x1 = k 2^(n-1) < 2^n.  So x1 < x.

Let x2 = n - m + 1.  Then x2 2^x2 = 2^(n+1) k x2/n >= 2^(n+1) (1 - (m - 1)/n)
                                 >= 2^(n+1) (1 - (m-1) 2^(-m)) > 2^n

since  m - 1 < 2^(m-1) for all m.  So x > x2.  Thus floor(x) is either
x3 = n - m or x1 = n - m - 1.

Now x3 2^x3 = (n-m) 2^(n-m) = (n-m)k/n 2^n.  To have floor(x) = x3 we need
n <= (n-m) k = (n-m) n 2^(-m), i.e. n >= 2^m + m.  Thus we conclude that
floor(x) = n - m - 1 if 2^m <= n < 2^m + m for some m
floor(x) = n - m     if 2^m + m <= n < 2^(m+1).
  
-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             or israel@unixg.ubc.ca
University of British Columbia
Vancouver, BC, Canada V6T 1Y4
Newsgroups: sci.math
From: michael@t13.lanl.gov (Michael I Angerman)
Subject: ftp sites for math preprints
Organization: Los Alamos National Laboratory, NM
Date: Mon, 31 Aug 1992 20:09:59 GMT


does anybody know of math preprints available from other places
besides stony brook.  Is there are list of sites that have preprints.

please respond to me directly via email

------------------------------------------------------------------------

math preprints are available at stony brook

If you have Internet access, these preprints are available via
anonymous ftp from math.sunysb.edu (129.49.31.57).

you can also receive preprints via email by

mail preprints@math.sunysb.edu
send info
Newsgroups: sci.math
From: torkel@sics.se (Torkel Franzen)
Subject: Re: Proof of God's Existence
Organization: Swedish Institute of Computer Science, Kista
	<1992Sep2.010215.2202@news.cs.brandeis.edu>
Date: Wed, 2 Sep 1992 08:42:57 GMT

In article <1992Sep2.010215.2202@news.cs.brandeis.edu> palais@binah.cc.
brandeis.edu writes:

   >I'm surprised that nobody has brought up the anecdote
   >of Euler's "proof" of the existence of God. 

  It's been brought up several times before, and according to the
orthodox theory, which nobody has refuted, it is a piece of fantasy
by De Morgan.
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math
Subject: Re: Organization for search
Date: 26 Aug 92 13:45:39 GMT
Organization: Purdue University Statistics Department

In article <1992Aug21.215538.26112@access.usask.ca> choy@skorpio.usask.ca writes:

>The information in libraries is so massive that I cannot read all
>of it. However, the information is so incomplete and deep that I
>may take me a long time to determine its existence or utility.
>Can we place the information in a computer to allow fast search
>and a chance to catch up on the pleasures of life? Is anyone doing
>this?

The problems are much greater than appear on the surface.  Even 
cataloguing the existence and location of the information is a massive
job, and is made worse by the fact that a paper, and even more so a book,
on one subject may have material even more relevant to something else,
and that this fact may not be apparent to the author or reviewer.

There are some attempts to do this for current literature.  There are
on-line indices of papers and books by title and author, and even somewhat
by subject.  But an index of concepts, definitions, theorems, proofs, and
methods, all of which are important, does not exist and never has existed.
Can it exist, in a usable form?  Consider the problem of entering, classifying,
and cross-indexing.  

I do not agree that it is not important.  One thing which I believe is quite
lacking is the information about material in one field relevant for another.
The so-called expository papers do very little of this.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@pop.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!pop.stat!hrubin(UUCP)
Newsgroups: sci.math