HDTenEightyP/NetNews huggingface dataset:
From: chalcraft@uk.tele.nokia.fi (Adam Chalcraft)
Subject: Re: Another induction problem
Organization: cpd
Date: Mon, 31 Aug 1992 08:50:02 GMT

Any chance of explaining how an induction can get that complicated?

Ordinary induction works because w (read omega, the ordinal) is well-ordered,
so there must be a minimal counter-example.
"Two deep" induction (of the form: Induce over pairs (x,y), where a pair
(x,y) < (x',y') iff x<x' or (x=x' and y<y'), i.e. so long as x goes down, I
don't care how much y goes up) works because w^2 is well-ordered.
My thesis goes up to w^3.
The proof that Conway's Sylver Coinage game (see Winning Ways) terminates
works because w^w is well-ordered.

There are other well-ordered sets, of course, but the very definition of
ordinals means that every well-ordered set is order-equivalent to an ordinal.

Maybe I've missed something. Can induction exist without relying on a well-
ordered set? Are we talking about transfinite induction? But it seems to me
that the complexity of any induction scheme can be described by quoting the
ordinal it relies on.

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?

Thanks.
--
  ____________________________________________________________________________
 /  _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_|`   |
 \__/                                                                      \__/

Newsgroups: sci.math
From: bhonsle@bhonsle.iss.nus.sg (Shailendra K Bhonsle)
Newsgroups: sci.math
Subject: Re: Can you prove it?
Date: 24 Aug 92 03:03:09 GMT
Organization: Institute of Systems Science, NUS, Singapore

In article <1992Aug20.210400.1773@wri.com>, roach@bikini.wri.com (Kelly Roach) writes:

|> 
|>       We need q prime to get (4).  We didn't really
|> need p to be prime nor q<p for the proof to work.  We
|> just need p to be odd (for the trivial case q=2).
|> 
|> 
|> 
|> 
This simple proof is also correct, nice to see the observation that p needn't be prime as well as q needn't be less than p.


Thanks.
Shailendra

--
From: amnell@klaava.Helsinki.FI (Marko Amnell)
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 1 Sep 92 17:30:25 GMT
Organization: University of Helsinki

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.

'With pleasure.  At least two versions of the ontological proof come to
mind.  First, consider this proof from 'positive properties' attributed
to Goedel:

Axiom 1.     (Dichotomy) A property is positive iff its negation is negative.
Axiom 2.     (Closure) A property is positive if it necessarily contains a
             positive property.
Theorem 1.   A positive property is logically consistent (ie. possibly it
             has some instance).
Definition.  Something is God-like iff it possesses all positive
             properties.
Axiom 3.     Being God-like is a positive property.
Axiom 4.     Being a positive property is (logical, hence) necessary.
Definition.  A property P is the essence of x iff x has P and P is
             necessarily minimal.
Theorem 2.   If x is God-like, then being God-like is the essence of x.
Definition.  NE(x): x necessarily exists if it has an essential
             property.
Axiom 5.     Being NE is God-like.
Theorem 3.   Necessarily there is some x such that x is God-like.

Now, gape at this short modal version by Alvin Plantinga:
Define 'unsurpassable greatness' to be equivalent to _maximal greatness
in every possible world_.  Then

(1)  There is a possible world in which unsurpassable greatness is
     exemplified.
(2)  The proposition <a thing has unsurpassable greatness if and only if
     it has maximal excellence in every possible world> is necessarily
     true.
(3)  The proposition <whatever has maximal excellence is omnipotent,
     omniscient, and morally perfect> is necessarily true.

Try refuting these babies!

Marko Amnell
Newsgroups: sci.math
From: bdm@cs.anu.edu.au (Brendan McKay)
Subject: Re: Proof of God's Existence
Organization: Australian National University
Date: Tue, 1 Sep 1992 12:04:13 GMT

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 will take it that you are referring to a serious proof, rather 
than a tongue-in-cheek proof.  

Perhaps you are referring to the arithmetical patterns which 
supposedly permeate the text of the Bible.  Jewish and Christian
writers have written about such patterns for millenia, but the
study was taken to new heights by a Russo-Canadian named Ivan Panin,
who devoted his entire life to the subject.  The "proof" runs
roughly along the lines of "these patterns are so intricate and
remarkable that it is quite impossible for a human to have produced 
them on purpose, or for mere chance to explain them; therefore God 
exists".  There are quite a few books on the subject, complete with 
copious tables, probability calculations, and pious analysis.
It all looks quite impressive at first glance.

Unfortunately for Mr. Panin (who somehow became a "famous 
mathematician" by getting a degree in languages) the patterns 
are neither remarkable nor unlikely.  They occur in any text at
all---you just need a little patience to find them.  Moreover,
after some practice it is easy to write text which displays much
better patterns than the Bible does.  (I proved this by doing it
myself.  Hey, maybe I'm God!)  As for the probability calculations,
pick up a large phone book (say one million numbers) and choose
six numbers completely at random.  Do you know that the chances
of picking exacly those numbers in exactly that order was only
one in 100000000000000000000000000000000000?  Remarkable!

In summary, if you want to believe in God, find some other reason.
Incidentally, I could quote some of it if there is sufficient interest.

Brendan.
From: hporopudas@tnclus.tele.nokia.fi
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 1 Sep 92 07:00:11 GMT
Organization: Nokia Telecommunications.


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 think that God's existence is an axiom. He says in the Holy Bible (Second
 book of Moses 3:14-15 in Old Testament and also in Johannes 8:28-29 in New
 Testament) that He is "Existence Himself".
   If someone is arguing that "Existence Himself" does not exist, I think that
 at the same time he/she is arguing that he/she himself/herself does not exist.

              Hannu.
Organization: Mathematics, Carnegie Mellon, Pittsburgh, PA
Newsgroups: sci.math
Date: Wed,  2 Sep 1992 21:10:05 -0400 
From: Oswald Wyler <ow0a+@andrew.cmu.edu>
Subject: Circular logic (was: Re: p prime, p|ab ...)

Get real, all you cicular reasoners.  Observe:
Theorem.  An integral domain D has unique factorization into irreducibles
iff D satisfies:
(i) Every non-zero element of D is a product of irreducibles.
(ii) If p is irreducible and p|ab, then p|a or p|d.
p irreducible means: p not zero, not divisor of 1, and if p = uv, then u
or v divides 1.
So how do you prove unique factorization before the p|ab business?
From: gore@yoko.rutgers.edu (Bittu)
Newsgroups: sci.math
Subject: Re: n doesnt divide 2^n-1
Date: 3 Sep 92 20:29:43 GMT
Organization: Hill Center

From kedlaya@b124.super.org (Kiran Sridhara Kedlaya) Fri Aug 28 16:31:44 1992

> 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.

But what if d = n? I don't see the contradiction in that case. Am I
missing something? How does one show that d < n? There is probably a
theorem that says that the order of 2 w.r.t an odd integer n is always
less than n.

-Bittu
From: cet1@cl.cam.ac.uk (C.E. Thompson)
Newsgroups: sci.math
Subject: Re: Calendar
Date: 2 Sep 92 16:24:40 GMT
Reply-To: cet1@cl.cam.ac.uk (C.E. Thompson)
Organization: U of Cambridge Computer Lab, UK

In article <1992Sep1.203418.17246@leland.Stanford.EDU>,
ilan@leland.Stanford.EDU (ilan vardi) writes:
|> There is a high probability of a lunar eclipse during Passover.  In
|> fact, given that non Israeli Jews celebrate Passover on two successive
|> evenings, the probability that some passover celebration in the world
|> occurs during a lunar eclipse must be quite high (perhaps > 1/10).
|> This does not hold for Easter since the computation uses a pseudo
|> lunar computation (the Paschal moon).

Apologies for continuing this thread in sci.math, in which it is hardly
appropriate, but I couldn't let that last remark pass.

Both the Gregorian calendar (w.r.t. the date of Easter) and the Jewish
calendar use fictitious moons, and there is no significant difference
in the accuracy with which they simulate the real one. The reason that 
Easter is less likely to coincide with the full moon (and hence an
eclipse) is that is postponed from a particular point in the lunar
cycle to the next Sunday. The corresponding variation in the Jewish
calendar is only the 1 or 2 days by which the start of the year
sometimes has to be shifted to avoid certain religious festivals
falling on particular days of the week.

Chris Thompson
JANET:    cet1@uk.ac.cam.phx
Internet: cet1@phx.cam.ac.uk
Newsgroups: sci.math
From: mareg@warwick.ac.uk (Dr D F Holt)
Subject: Re: Calendar (NEW)
Organization: Computing Services, Warwick University, UK
Date: Wed, 26 Aug 1992 12:40:42 GMT

In article <2906@ucl-cs.uucp> P.Samet@cs.ucl.ac.uk (Paul Samet) writes:
>Actually Easter is the first Sunday following the first full moon which occurs
>on or after March 21 (Spring equinox), not March 20, as claimed.
>

Surely 'on or after March 21' is the same as 'after March 20'. But am I
correct in thinking that if the full moon occurs on a Sunday then one has
to wait for the following Sunday? So the earliest possible date for Easter
is presumably March 22.

Derek Holt.
Newsgroups: sci.math
From: israel@unixg.ubc.ca (Robert B. Israel)
Subject: Re: dividing 100 digit nos.?
Organization: University of British Columbia, Vancouver, B.C., Canada
Date: Wed, 26 Aug 1992 17:57:45 GMT

In <1992Aug26.145600.9346@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:

>In article <17g3f1INN220@hilbert.math.ksu.edu> bennett@math.ksu.edu (Andy Bennett) writes:
>>comptec91065@camins.camosun.bc.ca writes:
>>
>>>	Can you help me locate the easiest method of dividing
>>>	integers of at least 100 digits (possibly up to 4300 digits) ?
>>
>>Use UBASIC. This is a public domain basic developed by Yuji Kida in Japan
>>with full support for integers up to about 2500 digits. It also comes with
>>state of the art number theory algorithms already coded (for primality
> 
>With all due respect: [this is not a flame]

>I do not believe that you answered the question that was asked, although
>you may have answered what the poster *meant to ask*.

>What was asked about was the easiest *method* for doing division on 100
>digit integers. That information can be found in Knuth Vol. 2
>The poster may have wanted to know how it is done, rather than wanted
>a black box to do it.

>I may be wrong, in that the poster was really asking where he could
>find software to perform what he wanted, but in absence of other information
>I answer what is asked, rather than try to fathom what is *meant*.
>
What is easier: programming an algorithm from Knuth, or using software
that already exists?  Without any other clues from the poster, I'd assume
that the second method was acceptable.  Someone who wanted to be directed
to Knuth could have written "algorithm" rather than "method", or
"an efficient" rather than "the easiest".  And I suspect that someone
who is really capable of implementing arbitrary-precision arithmetic
would already know about Knuth. 
-- 
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: ap@levangerhs.no (Andrei Prasolov)
Subject: Elementary solution needed: harmonic bi-sequences
Organization: Hogskolen i Levanger
Date: Thu, 27 Aug 92 17:00:48 GMT

   There is a problem which seems to have a very simple solution.
I remember that I heard one in my youth. That solution used
the epsilon-delta language only. However, I cannot recall that.

DEF. Let`s call a bi-sequence (a(i,j), i,j from Z) HARMONIC if for all i,j

(*)       a(i,j) = 1/4 (a(i,j+1) + a(i,j-1) + a(i+1,j) + a(i-1,j)).

PROBLEM. Prove that any bounded (|a(i,j)| < C) harmonic bi-sequence is constant.

   I know the proof which uses Fourier-transform of generalized functions
(distributions):

Let f(x,y) = Sum_i,j(delta(x-i,y-j)).
Than f is a generalized function of polynomial (of degree zero) growth,
so its Fourier-transform F(x,y) has also polynomial growth and is 2pi periodical.
(*) becomes:

(**)   (2cos(x) + 2cos(y) - 4) F(x,y) = 0.

It follows that F(x,y)=0 unless x=2k pi and y=2l pi, k, l from Z. So F, being
periodical, can be represented by

     F(x,y) = Sum_i,j(ksi(x - 2 i pi, y - 2 j pi))

where ksi is a finite sum of delta-function and its derivatives,
therefore a(i,j) is a polynomial of i,j and, being bounded, is constant.

DOES ANYBODY KNOW AN ELEMENTARY PROOF?

Thanks in advance.
From: kevin@vaxc.cc.monash.edu.au
Newsgroups: sci.math
Subject: Re: God's Existence
Date: 2 Sep 92 23:54:33 +1000
Organization: Computer Centre, Monash University, Australia

I can't seriously believe that someone can assert God's existence by
claiming it is an axiom - written in the bible.
Reason 1. This is a blatant account of circularity - for example, I could
write a book - THE BOOK OF KEVIN. This book consists of 1 line - the
following

	AXIOM : KEVIN IS GOD.

(END of book)
What conclusion is one to reach then? One cannot hope to prove to a 
non-believer that the existence of God is an axiom by quoting it from
the Bible. At first, when I read this in a posting I thought it was just
a joke. But it keeps on popping up? Am I missing the humor? 
Reason 2. It is not immediately clear what an axiom in the real world is,
or even more seriously, how we know we have one in our hands when we think
we do. Note that by an axiom I mean a NON-TAUTOLOGICAL sentence about the
world. I challenge someone to come up with such an 'axiom', that they can be 
sure is such an axiom. For example, Newton's law doesn't qualify, as I can't
be sure that tomorrow balls might fly upwards when dropped - the fact that
this probably won't happen is irrelevant. Once again, this stuff is all
well explored in the philosopy of science,
So it seems that we can't even start a deductive proof of the existence of
God, because the first steps - the axioms - aren't there.
Well, all this stuff has become too philosophical and non-mathematical, so
I'm not sure what it's future should be here.

P.S. To those who sent me a request for a reference for Godel's proof - 
I'm hunting down my photocopy, and will be in touch as soon as I have found
it. 
Regards,
Kevin Davey
Monash University
Australia.

Newsgroups: sci.math
From: asimov@wk223.nas.nasa.gov (Daniel A. Asimov)
Subject:  Triangulating a Region of R^3 Bounded by Triangulated Surfaces
Organization: NAS, NASA Ames Research Center, Moffett Field, CA
Date: Tue, 1 Sep 92 18:50:11 GMT

Some usual definitions:
	Let k+1 points in R^n be said to be in "general position" if 
	they lie on no (k-1)-dimensional plane.

	Let a subset S of R^n be called a "k-simplex" if it is the convex 
	hull of k+1 points in R^n in general position.  
	
	The k+1 points are the "vertices" of S. 

	A "face" of S is defined as the convex hull of a subset of the vertices.

	Define a compact subset X of R^n to be "triangulated" if X is expressed 
	as the union of finitely many simplices whose pairwise intersection 
	is a common face, or empty.
Now:
Let M be the closure of a bounded open subset of 3-space, and assume that 
the boundary of M consists of a finite number of connected components N(i).  

Assume that each of these N(i) is a compact triangulated surface 
(without boundary).  Call this the "boundary triangulation." 

QUESTION:  Is it necessarily true that M can be triangulated in such 
a way that the restriction of this triangulation to the boundary of M 
is the given boundary triangulation?  

It is easy to see that this is true if M is convex, but what if it isn't?

--Dan Asimov
asimov@nas.nasa.gov

Mail Stop T045-1
NASA Ames Research Center
Moffett Field, CA 94035-1000
(415) 604-4799
From: ct@dde.dk (Claus Tondering)
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 2 Sep 92 14:11:41 GMT
Organization: Dansk Data Elektronik A/S

jchen@tomahawk.me.gu.edu.au (Jinghong CHEN) writes:

>Proof of God's inexistence:
>God can do anything.
>So he can make a stone which he cannot lift.
>So he cannot lift that stone.
>So "God can do anything" is wrong.

To put your argument another way:

	Proof of God's inexistence:
	God can do anything.
	So he can make a snax fnyf muski.
	So he cannot luffe puqu mi.
	So "God can do anything" is wrong.


What I'm trying to say is this: Your statement "God can make a stone which
he cannot lift" is intrinsically nonsensical. It makes no more sense than
"snax fnyf muski". Your statement just happens to be syntactically well-
formed, whereas my statement does not.

What you have expressed is a paradox. On the same lines as "this sencetence
is false". It is easy to formulate paradoxes, but the possibility to
formulate a paradox does not prove that what the paradox talks about is
untrue.
-- 
Claus Tondering
Dansk Data Elektronik A/S, Herlev, Denmark
E-mail: ct@dde.dk
From: ilan@leland.Stanford.EDU (ilan vardi)
Subject: Re: Calendar
Organization: DSG, Stanford University, CA 94305, USA
Date: Tue, 1 Sep 92 20:34:18 GMT

There is a high probability of a lunar eclipse during Passover.  In
fact, given that non Israeli Jews celebrate Passover on two successive
evenings, the probability that some passover celebration in the world
occurs during a lunar eclipse must be quite high (perhaps > 1/10).
This does not hold for Easter since the computation uses a pseudo
lunar computation (the Paschal moon).  

Similarly for a solar eclipse on the Jewish new year. I would say
about 10 times more likely than on another given date of the Gregorian
calendar.

-ilan
Newsgroups: sci.math
From: rivin@SAIL.Stanford.EDU (Igor Rivin)
Subject: Re: Calendar (NEW)
Organization: /usr/lib/news/organizatoin
Date: Mon, 31 Aug 1992 17:04:42 GMT

In article <VICTOR.92Aug25094827@terse4.watson.ibm.com> victor@watson.ibm.com writes:
>>>>>> On 25 Aug 92 08:51:01 GMT, P.Samet@cs.ucl.ac.uk (Paul Samet) said:
>[...omitted...]
>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).
>--
>			Victor S. Miller
>			Vnet and Bitnet:  VICTOR at WATSON
>			Internet: victor@watson.ibm.com
>			IBM, TJ Watson Research Center

See also I. Vardi's "Computational Recreations" book.


Newsgroups: sci.math
From: thompson@atlas.socsci.umn.edu (T. Scott Thompson)
Subject: Re: Polygon point enclosure
Reply-To: thompson@atlas.socsci.umn.edu
Organization: Economics Department, University of Minnesota
Date: Thu, 27 Aug 1992 16:48:34 GMT

gjm11@cus.cam.ac.uk (G.J. McCaughan) writes:

>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.

This algorithm is in _Algorithms_ by Robert Sedgewick, Addison-Wesley
(1983), pp 315-318.  (I believe there is a second edition out, but I
don't have it.)  Sedgewick also gives some pseudo-code for the
algorithm that takes into account some other nasty special cases, as
well as the one where the ray includes one of the connecting line
segments.  For example, you must allow for the case where the ray
enters or leaves the polygon at a vertex properly in order for the
method to work.

The strategy of choosing a new direction when the ray goes along a
side is not efficient since there are ways of fixing up the count to
allow for this.  Sedgewick's algorithm provides the details.

--
T. Scott Thompson              email:  thompson@atlas.socsci.umn.edu
Department of Economics        phone:  (612) 625-0119
University of Minnesota        fax:    (612) 624-0209
From: gerardo@ferris.cray.com (Gerardo Cisneros)
Newsgroups: sci.math
Subject: Re: Problematic smurfs!
Keywords: smurfs
Date: 31 Aug 92 15:42:37 GMT
Organization: Cray Research, Inc.

In article <94952@bu.edu> spacefox@acs.bu.edu (Godfrey Degamo) writes:
>[...]
>     A group of 1000 smurfs wear either a red or blue hat on their head.
>[...]
>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.
>[...]

Several followups have come up with an answer of three red-hatted smurfs
standing up.  That can't be right, however, because everybody knows that
smurfs _all_ wear white hats, except Papa smurf, who wears a red hat.
:-) :-) :-)
-- 
Gerardo Cisneros		|Cray Research de M'exico, S.A. de C.V.
gerardo.cisneros@cray.com	|Camino a Sta. Teresa 480-A #302
(+52+5)622-8584 		|14020 Tlalpan, D.F., MEXICO
 Anything I said sound like an opinion? Then it's no one's but my own.
Newsgroups: sci.math