HDTenEightyP/NetNews huggingface dataset:
From: dmsilev@athena.mit.edu (Daniel M Silevitch)
Subject: Re: TIME HAS INERTIA--INDEPENDENT VALIDATION OF ABIAN
Organization: Massachusetts Institute of Technology
Date: Tue, 15 Dec 1992 13:52:54 GMT

In article <234323@tamsun.edu>, echinter@flash.tamsun.edu (Eric Chinter) writes:
|> I was extremely SCEPTICAL that any of ABIAN's claims were TRUE, but
|> after I E-MAILED him for a more thorough DESCRIPTION of his THEORY, I
|> was able to determine that it is CONSISTENT of itself and EXPLAINS
|> adaquately many WELL-KNOWN and ACKNOWLEDGED PHENOMINONS.  ALL of his
|> EQUATIONS follow neatly and cleverly from his several stated AXIOMS
   ^^^^^^^^^
We haven't seen any equations (well, one equation), and it is not derived
from the axioms, it is simply stated as fact.

|> and POSTULATES.  FURTHERMORE, his THEORY makes several PROFOUND
|> STATEMENTS that apparently have never before come to the attention of
|> OTHER SCIENTISTS!!!  They are in fact so PROFOUND and EARTH-SHATTERNIG
|> in their IMPLICATIONS that I will not reveal them except in private
|> e-mail, but SEVERAL LABS have been notified and are setting up

Newsflash: Abian's results released, world scientific community collapses!
They can't be _that_ bad, can they. Besides, Abian has already pestered us
enough with his theory, his life might be easier if some verification was
given. There would then be a non-zero (but infitesimal) chance that people
would believe him.

|> EXPERIMENTS to test WHETHER the UNFORSEEN IMPLICATIONS are indeed
|> TRUE!!!  Preliminary results from one of the labs support the brilliant
|> THEORY of DOCTOR ABIAN, but of course only time will VINDICATE him
|> COMPLETELY.
|> 
|> Anyone who desires to see the document of INDEPENDENT VERIFICATION whihc
|> I will have prepared within ONE (1) week should send me PRIVATE E-MAIL.
|> 
|> Thank you, and GOD-WILLING we'll set this mess straight!
|> 
|> Eric CHINTER, PhD Berekeley

Why don't you post it for all of us to look at and marvel. I'm sure that
we'll all be astounded by the verification of this theory, which until this
point, has not even been able to prove that bodies of unequal mass fall at the
same rate in a uniform gravitational field, let alone explain why the universe
is the way it is.

Also, DON'T SHOUT, it's not polite, and we can read lower case very well.

BTW, I've removed sci.chem and sci.bio from the propagation of this article,
as they are not appropriate to this discussion (not that sci.math and sci.physics
are, it belongs in alt.physics.new-theories).

Daniel Silevitch                               dmsilev@athena.mit.edu
Massachusetts Institute of Technology
Newsgroups: sci.math,rec.puzzles
From: bcash@crchh410.BNR.CA (Brian Cash)
Subject: Re: Marilyn Vos Savant's error?
Reply-To: bcash@crchh410.BNR.CA (Brian Cash)
Organization: Bell-Northern Research Ltd.
Date: Wed, 16 Dec 1992 07:29:22 GMT

In article <1992Dec15.091823.10100@nmt.edu>, jduncan@nmt.edu (John
Duncan) writes:
|>In article <1992Dec15.063213.11742@bnr.ca> bcash@crchh410.BNR.CA
(Brian Cash) writes:
|>>In article <1992Dec15.052211.24395@CSD-NewsHost.Stanford.EDU>,
|>>snewman@Xenon.Stanford.EDU (Steven Newman) writes:
|>>|>
|>>|>   1) You meet a man on the street and ask him how many children he has.
|>>|>      He replies "two, and one is a boy."  What is the probability that
|>>|>      his other child is also a boy?
|>>|>
|>>|>   2) You meet a man on the street and ask him how many children he has.
|>>|>      He replies "two, and the older one is a boy."  What is the
|>>|>      probability that his other child is also a boy?
|>>|>
|>>|>The answer to problem 1 is 1/3, while the answer to problem 2 is 1/2.
|>>|>(One must be careful in the interpretation of the statement in problem
|>>|>1; it means "I have two children, and it is not the case that both of
|>>|>them are girls.")
|>>|>
|>>|> - Steve Newman (snewman@cs.stanford.edu)
|>>
|>>Sorry!  You made the same mistake as the first poster!
|>
|>actually, no.  it is you that repeated the misteak :)
|>
|>>The odds of the other child being a boy is 1/2 in both cases.
|>>Observe:
|>>
|>>Case 1)  Child = boy   Other = ?
|>>Yes  ->  a) Child = boy , Other = boy
|>>No   ->  b) Child = boy , Other = girl
|>>         c) Child = girl, Other = boy
|>>         d) Child = girl, Other = girl
|>>         Obviously, c and d are invalid.
|>
|>WRONG !!! c is not invalid
|>one is still a boy so the chances are, in fact, 1/3 that the other is a boy.
|>
|>Regards
|>		--John P. Duncan (jduncan@prism.nmt.edu)
                                      
Uh, I'm sorry, but I don't follow the logic in that.
The man said "one is a boy".   The "Child" in case 1 is
the child the man refered to when he said "one". 
c states that the child he is talking about is a girl,
and the one he doesn't mention is a boy.  This is
invalid. 

ex.
"I have two coins that add to $.35 and one is not a quarter, how is this so?"

One = dime   Other = quarter
NOT One = quarter Other = dime

What am I doing wrong?

I think you might be dividing the children in another way:

a) child A = boy   child B = boy
b) child A = boy   child B = girl
c) child A = girl  child B = boy
d) child A = girl  child B = girl

Given this and "one is a boy" only d is invalid, but it also
makes b and c the *same* (one boy, one girl...order is irrelevant).
So now you have two events: a and b/c.  Odds = 1/2
I flip a coin 100 times.  99 of those times it came up heads,
what are the odds that the other flip came up heads also?
1/2.  Which coin is the "mystery flip" is irrelevant to the problem.

The only cases where you can say {A = x, B = y} and {A = y, B = x}
are two separate probabilities is when A and B are not interchangable
(ie. Younger and Older are not interchangable but child A and child B 
are).

Does anyone follow what I am saying?

Brian /-|-\ 
Xref: sparky sci.math:16821 comp.theory:2695
From: jerry@cs.mcgill.ca (Gerald (Jerry) KUCH)
Newsgroups: sci.math,comp.theory
Subject: Re: Apotheosis of textbookhood: summary
Date: 14 Dec 92 01:18:20 GMT
Organization: SOCS - McGill University, Montreal, Canada

In article <1992Dec10.135602.20997@sophia.smith.edu> orourke@sophia.smith.edu (Joseph O'Rourke) writes:
>Thanks to all who responded to my query for excellent undergraduate
>textbooks.  Only a few books were mentioned by more than one respondent,
>demonstrating a lack of agreement that is to be expected from such a 
>diverse group.  But I found the variety of well-loved textbooks 
>enheartening:  there are more very good textbooks out there than
>is at first apparent.


Some of these replies are very interesting...

>Halmos 			Naive Set Theory
>Halmos 			Naive Set Theory
>Halmos 			Naive Set Theory

This is a book that I'm sometimes afraid to mention having read.  I didn't 
mind it, although I didn't like it nearly as much as "Finite Dimensional
Vector Spaces."  The reactions I've received from professors on this title
though have been extremely bipolar.  It's either "a pleasant and clever
treatment of the subject" or "an ill-conceived, abominable waste of 
effort," to quote two people I've talked to about it.

>Lewis & Papadimitriou	Elements of the Theory of Computation

Very nice book---especially considering some of the lame texts on this topic
that have dribbled out in the last few years.  This really ought to be used
in a first year (possibly second semester) course for computer science majors.
Theory, including formal languages, usually gets ditched until late in the
curriculum when everyone's been turned into a Pascal-puking grind or left
the subject looking for more interesting pastures...

>Spivak			Calculus

Great book.  I don't know why the standard plethora of inferior paraphrasings
of the Thomas and Finney "Calculus and Analytic Geometry" aren't starting to
give way to this book yet...

>Strang 			Linear algebra and its applications

Another book about which opinions seem to neatly fall into "love" and "hate"
camps.  I really liked it...  it certainly shouldn't be an undergraduate's
only exposure to linear algebra but its treatment of many of the applications
is a lot of fun (and useful), and parts of it seem great for building 
intuition.  Having a two headed (possibly year long) undergraduate introductory
linear algebra course that approaches the subject from two angles, using 
this book, and maybe M.L. Curtis's "Abstract Linear Algebra" (Springer-Verlag,
1990, not to be confused with the Charles Curtis linear text) together
might be interesting.


-- 
 Jerry Kuch (jerry@cs.mcgill.ca) | Wishing I were an emaciated chihuahua. 
 "I was wrong to play God.  Life is precious, not a thing to be toyed with.
  Now take out that brain and flush it down the toilet."
                  --- M. Burns  "Treehouse of Horror II"
From: bruck@mtha.usc.edu (Ronald Bruck)
Newsgroups: sci.math
Subject: Re: Rounding Rules
Date: 20 Dec 1992 08:06:55 -0800
Organization: University of Southern California, Los Angeles, CA
NNTP-Posting-Host: mtha.usc.edu

In article <1992Dec20.003018.14325@sol.ctr.columbia.edu> shaw@toadflax.UCDavis.EDU (Rob Shaw) writes:
>
>What is the rationale behind the following rounding rule?
>
>when dealing with 5's followed by all zero's, check the
>next digit to the left. If it's even, round down; odd,
>round up.
>
>For example both 1.13500 and 1.14500 are 1.14 to 3 places.
>What is the advantage of having the interval closed at both
>ends around even digits, and open at both ends around odds?
>

The purpose is to make addition more accurate.  Assuming there
is no systematic error, i.e. in half the cases the 5 is preceded
by an even and in half the cases by an odd, then this rule will
cancel the errors.  "Round to even" sounds as if it discriminates
against odds, but if you add 10^9 terms following some other
rule, your purported sum may be way off.

--Ron Bruck
bruck@mtha.usc.edu preferred



Newsgroups: sci.math
From: nissim@mary.fordham.edu (Leonard J. Nissim)
Subject: Re: Need some help with Topology
Organization: Fordham University
Date: Mon, 14 Dec 1992 19:54:00 GMT
News-Software: VAX/VMS VNEWS 1.41    
Distribution: usa
X-Posted-From: mary.fordham.edu
NNTP-Posting-Host: sol.ctr.columbia.edu

In article <1gc73iINNl4g@matt.ksu.ksu.edu>, 
bubai@matt.ksu.ksu.edu (P. Chatterjee) writes...
>I had a few questions (none homework, though!) and would appreciate any   
>kind of help from the math-knowledgeables on the net.
> 
> 
>a) What does it mean to say that a set A is 'infinite'? 'Finiteness', by 
>definition, implies that A is equivalent to a portion of the set of 
>positive integers. Can this definition of 'finiteness' be used to 
>motivate one for 'infiniteness'?

The usual definition (from set theory, not necessarily with a topology on
the set):  A set S is infinite iff there exists a *proper* subset of S, call
it T, and a *bijection* from S onto T.  

> 
>b) Show that A is OPEN <==> X \ A is closed. 
>Isn't this a definition or can it be proven?
> 
In some books it is the definition of a closed set, and then you prove what 
properties closed sets have.  In other texts, you define a closed a closed set
by properties, and prove that a set is closed iff its complement is an open 
set.

-------------------------------------------------------------------------------
Leonard J. Nissim  (nissim@mary.fordham.edu)   
Disclaimer:  "I speak only for myself."
-------------------------------------------------------------------------------
Newsgroups: sci.math,rec.puzzles
From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
Subject: Re: Marilyn Vos Savant's error?
Keywords: savant
Organization: Computer Science Department,  Stanford University.
Date: Thu, 17 Dec 1992 07:13:40 GMT

In article <1992Dec17.041022.29031@thunder.mcrcim.mcgill.edu> mouse@thunder.mcrcim.mcgill.edu (der Mouse) writes:
>In article <1992Dec15.055832.26324@galois.mit.edu>, jbaez@riesz.mit.edu (John C. Baez) writes:
>
>> 1)  You draw 4 cards from a well-shuffled standard deck.  You turn
>>     one over and it's an ace.  What's the probability that they are
>>     all aces?
>
>There are 4! C(52,4) = 52!/48! = 6497400 sets of 4 cards you could
>draw, counting different orders as different.  In one case in 13 you'll
>turn up an ace, for a total of 499800, while in only 24 cases are they
>all aces.  The answer to this question is therefore 24/499800, which is
>1 in 20825, or, to 20 places from dc, .00004801920768307322.

Ha!  This is right and I was wrong.  I just took this to be the problem
where you look at your whole hand and observe at least one ace, and so
figured there were 48*47*46*45 = 4669920 ways to get no aces, hence
6497400-4669920 = 1827480 ways to get at least one ace, giving the
estimate as 24/1827480 = .000013.  But the problem clearly states you
just look at one card, for which the above calculation is correct.
Ditto for the remaining calculations.  Oh well, at least we now have
the numbers for both versions of the problem (look at whole hand/look
at one card).

-- 
Vaughan Pratt				There's safety in certain numbers.
From: dseal@armltd.co.uk (David Seal)
Newsgroups: sci.math
Subject: Re: Factor 69 digit integer
Date: 16 Dec 92 17:52:11 GMT
Distribution: sci
Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK

In article <1gb5bkINNm71@roundup.crhc.uiuc.edu> hougen@focus.csl.uiuc.edu
(Darrell Roy Hougen) writes:

>[attributions removed, since they were wrong]
>>>> Please help me.  I need the prime factors of 
>>>>  132686104398972053177608575506090561429353935989033525802891469459697
>
>Write your own infinite precision arithmetic program to solve the
>problem.  Here's a start:
>(1) Store each number in an array, e.g., A[1..69] where each element
>of the array is a digit, ie., 0,1,...,9.
>(2) Implement a divide and carry algorithm.  For each remainder,
>you'll have to try all dividends from 0 through 9 up the first that
>doesn't work.

Shouldn't you have put a large number of smileys after this? People might
take it seriously...

(For those who did:
  (a) implementing infinite precision arithmetic in *decimal* is distinctly
      unusual, not to say inefficient - a base like 2^16 or 2^32 would make
      more sense;
  (b) calculating a quotient digit by trying possibilities from 0 upwards
      until one doesn't work is highly inefficient, especially for
      reasonably large bases - binary chop would be better, some methods
      using the result of single precision divisions described in Knuth yet
      better;
  (c) if you've drawn the conclusion that you can then solve the problem by
      trial division, think again - if this number is at all difficult to
      factorise, you could run trial division on current hardware for the
      current age of the universe and still not have made any real inroads
      on the problem. Fortunately, there *are* better algorithms.
)

David Seal
dseal@armltd.co.uk

All opinions are mine only...
Xref: sparky sci.math:17045 sci.math.num-analysis:3606 alt.uu.math.misc:21 ucb.math:13
From: suleman@netcom.com (Dani Suleman)
Newsgroups: sci.math,sci.math.num-analysis,alt.uu.math.misc,ucb.math
Subject: Question about binary operators ??
Date: 16 Dec 92 19:32:14 GMT
Organization: Netcom - Online Communication Services  (408 241-9760 guest)

Hi, I have a binary question here. I need to find what operation(s) to do
such that when the source is (?) to the destination, if the source is all 1's,
the result will be the destination value, else if the source is not all 1's,
the result will be the source.
e.g.
    1111 ? 0111 = 0111  and   0101 ? 1110 = 0101
      ^      ^      ^--> result
      |      +--> destination
      +--> source

If it simplifies the question, you can also change the source to all 0's such
that when the source is all 0's and is (?) to the destination, the result
is the destination, and when the source is not all 0's, the result is the
source.
e.g.
    0000 ? 1001 = 1001   and    1100 ? 0100 = 1100

Any help/advice is DEEPLY appreciated !

p.s., pls respond to my account if possible, I'll post a summary if there's
any request.

--dani
Newsgroups: sci.math
From: ross@tarski.tmc.edu (David Ross)
Subject: Re: nonstandard analysis
Organization: University of Hawaii Mathematics Department
Date: Thu, 17 Dec 1992 22:07:58 GMT

In article <ARA.92Dec15153446@camelot.ai.mit.edu> ara@zurich.ai.mit.edu (Allan Adler) writes:
>
> I'm not entirely pleased that the
>emphasis in recent postings seems to be on Nelson's foundations IST.

IST seems to be more easily accessible to people with no background
in logic or logic-like mathematics.  I don't understand this myself, as
IST is more syntactic in basis than is 'normal' nonstandard analysis.

>
>I looked at Nelson's book in the Annals of Math studies and
>I found that he didn't take enough care to maintain a
>distinction between standard mathematical objects and nonstandard ones.

This is actually an important part of the ontology of work in IST; the
idea is that there *is* no difference between standard and nonstandard
entities, but rather a syntactic difference of what you can say about
them.  My friends who work in the IST framework act surprised if you
suggest that the 'real' reals *don't* contain infinitesimals.  I
find this kind of strange, myself.


- David


--
David Ross, Dept. of Math., Univ. of Hawaii at Manoa, Honolulu HI  96822
Internet: ross@math.hawaii.edu  -or- ross@tarski.math.hawaii.edu
Phone: 808-956-9949
Xref: sparky sci.math:17102 rec.games.abstract:640
From: dbush@csugrad.cs.vt.edu (David Bush)
Newsgroups: sci.math,rec.games.abstract
Subject: Re: Game of pentominos
Date: 17 Dec 92 19:37:26 GMT
Organization: Virginia Tech Computer Science Dept, Blacksburg, VA

nates@ll.mit.edu ( Nate Smith) writes:

>a LONG time ago, i was given a game called "Pan-Kai".  this is a
>variant of the above played with 2 sets of pentominoes in 2
>colors on a 10x10 board.  the objective was still the same - make
>the last move.  the difference was that now you could make a "box"
>in the shape of a piece you still had but your opponent didnt.  this
>would give you an extra move at the end.  i played that game quite
>a bit as a kid and found it quite intriguing.  i never got to play
>the 8x8 version because by the time i found out about it, nobody
>wanted to have anything to do with me & my pentominoes.  :-)  :-(

Oh, yeah? Very interesting. Back in the late sixties Parker Brothers came out
with "Universe," a pentomino game for 2-4 players. The rules were similar. For
2 players, the board is 10x10. In addition, there are four regions of
25 cells each appended to the 4 sides of the main square. The entire
board has 4-way rotational symmetry. For 3 players, use any two extra regions
in addition to the 10x10 grid. 4 players use the whole board. On your move, you
may not place a pentomino so as to "fence off" a region of < 5 cells connected
horizontally and/or vertically. (Of course only the active portion of the board
is considered.) If you have no move, you're out of the game.
The last player to move wins.

Trivia: the game box contained publicity stills from "2001:A Space Odyssey."
One shot shows an astronaut playing Hal a game of something that looks like
Universe. Apparently that scene never made it to the movie. :)
From: rmurphy@aludra.usc.edu (Bob Murphy)
Newsgroups: sci.math
Subject: Re: Pi in beatiful form --lost
Date: 16 Dec 1992 11:24:22 -0800
Organization: University of Southern California, Los Angeles, CA
NNTP-Posting-Host: aludra.usc.edu

In article <1992Dec16.144508.19371@nntpd.lkg.dec.com> edp@math.zk3.dec.com (Eric Postpischil) writes:
>In article <1gmvieINN7al@aludra.usc.edu>, rmurphy@aludra.usc.edu (Bob
>Murphy) writes:
>
>>(pi - 3)/4 = 1/(2*3*4) + 1/(4*5*6) + 1/(6*7*8) + ...
>
>The first term is 1/24, which is already greater than (pi-3)/4.  Should
>that have been an alternating series?
>

Thanks for pointing out that one of my formulas had an error.
I obtained this from "The Man Who Knew Infinity: A Life Of The
Genius Ramanujan".  The book included this typo.  Since I've
never seen a series like this before I don't know what the typo is.
That is, I don't know if it is supposed to be an alternating series
or if the left hand side of the equation is the one that needs to be 
changed.  The series is not attributed to Ramanujan so I'm not sure
where to look to find the correct series.  I'd love to know the
correction and a pointer to a proof if anyone has seen this before.

                                Bob Murphy (rmurphy@aludra.usc.edu)
Xref: sparky sci.math:17008 rec.puzzles:7961
From: jduncan@nmt.edu (John Duncan)
Newsgroups: sci.math,rec.puzzles
Subject: Re: Marilyn Vos Savant's error?
Date: 15 Dec 92 09:18:23 GMT
Article-I.D.: nmt.1992Dec15.091823.10100
Organization: New Mexico Tech

In article <1992Dec15.063213.11742@bnr.ca> bcash@crchh410.BNR.CA (Brian Cash) writes:
>In article <1992Dec15.052211.24395@CSD-NewsHost.Stanford.EDU>,
>snewman@Xenon.Stanford.EDU (Steven Newman) writes:
>|>
>|>   1) You meet a man on the street and ask him how many children he has.
>|>      He replies "two, and one is a boy."  What is the probability that
>|>      his other child is also a boy?
>|>
>|>   2) You meet a man on the street and ask him how many children he has.
>|>      He replies "two, and the older one is a boy."  What is the
>|>      probability that his other child is also a boy?
>|>
>|>The answer to problem 1 is 1/3, while the answer to problem 2 is 1/2.
>|>(One must be careful in the interpretation of the statement in problem
>|>1; it means "I have two children, and it is not the case that both of
>|>them are girls.")
>|>
>|> - Steve Newman (snewman@cs.stanford.edu)
>
>Sorry!  You made the same mistake as the first poster!

actually, no.  it is you that repeated the misteak :)

>The odds of the other child being a boy is 1/2 in both cases.
>Observe:
>
>Case 1)  Child = boy   Other = ?
>Yes  ->  a) Child = boy , Other = boy
>No   ->  b) Child = boy , Other = girl
>         c) Child = girl, Other = boy
>         d) Child = girl, Other = girl
>         Obviously, c and d are invalid.

WRONG !!! c is not invalid
one is still a boy so the chances are, in fact, 1/3 that the other is a boy.

Regards
		--John P. Duncan (jduncan@prism.nmt.edu)
Newsgroups: sci.math,rec.puzzles
From: snewman@Xenon.Stanford.EDU (Steven Newman)
Subject: Re: Marilyn Vos Savant's error?
Organization: Computer Science Department, Stanford University.
Date: Tue, 15 Dec 1992 05:22:11 GMT

> ... but while I'm on Bayes' theorem let me recall a problem similar
> to the above one.  
> 
> 1)  You draw 4 cards froma well-shuffled standard deck.  You turn one
> over and it's an ace.  What's the probability that they are all aces?
> 
> 2)   You draw 4 cards from a well-shuffled standard deck.  You turn one
> over and it's the ace of hearts.  What's the probability that they are
> all aces?
> 
> The answers to problems 1) and 2) are NOT THE SAME.
> 
> Probability theory, like quantum mechanics, takes a while to get used to.

Actually, both of these problems have answer 6/(49*50*51) = 1/20825.  There
are 52! possible orderings for a well-shuffled deck.  In 4! * 48! orderings
the first four cards are an ace; in 4 * 51! orderings the first card is an
ace; and the first set of orderings is a subset of the other.  Hence the
answer to problem 1 is (4! * 48!) / (4 * 51!) = 6 / (49*50*51).

Similarly, for problem 2, in 51! orderings the first card is the Ace of
Hearts, and in 6 * 48! of those orderings the next three cards are also
aces.  Hence the answer to this problem is (6 * 48!) / 51! = 6 / (49*50*51)
as well.

There are problems that sound very similar to these in which the answers
are not the same.  For example:

   1) You meet a man on the street and ask him how many children he has.
      He replies "two, and one is a boy."  What is the probability that
      his other child is also a boy?

   2) You meet a man on the street and ask him how many children he has.
      He replies "two, and the older one is a boy."  What is the
      probability that his other child is also a boy?

The answer to problem 1 is 1/3, while the answer to problem 2 is 1/2.
(One must be careful in the interpretation of the statement in problem
1; it means "I have two children, and it is not the case that both of
them are girls.")

 - Steve Newman (snewman@cs.stanford.edu)
Xref: sparky sci.math:17087 rec.games.abstract:638