HDTenEightyP/NetNews huggingface dataset:
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Retarded log question
Date: 11 Sep 92 14:51:35 GMT
Organization: U of Cambridge, England

In article <1992Sep10.225134.21189@hellgate.utah.edu>, tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:

> If you are given a number, such as 2^X, how can you determine what
> the same number would be with a different base, such as 10?
> 
> For example, 2^100,000,000 = 10^ ??

     2^100000000 = 10^x
 <=> log_10(2^100000000) = log_10(10^x) = x

          and now you need to know:
            log_b(x^y) = y.log_b(x)        (1)
          which gives

 <=> 100000000.log_10(2) = x

          and now you need to know:
            log_b(x) = log_c(x)/log_c(b) for any b,c [proof in a moment]
          which gives

 <=> 100000000.log(2)/log(10) = x

where "log" is to any base you want. This is your answer.

  Proof of the base-conversion thing above:
  it's the same as log_b(x).log_c(b) = log_c(x);
  well, x=b^log_b(x) so
  log_c(x) = log_c(b^log_b(x))
           = log_b(x).log_c(b)   by (1).

I might as well prove (1) too while I'm at it.

  log_b(x^y) is defined by b^log_b(x^y) = x^y
  but we know b^(y.log_b(x)) = (b^log_b(x))^y = x^y.

Here I've used x^yz = (x^y)^z, which I am *not* going to prove.

-- 
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@cus.cam.ac.uk  Cambridge University, England.    [Research student]
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Couple of questions
Date: 10 Sep 92 00:48:17 GMT
Organization: U of Cambridge, England

In article <1992Sep9.102457.15049@news.columbia.edu>, pvl2@cunixb.cc.columbia.edu (Priscilla V Loanzon) writes:

> Could someone please explain to me a few basic things:
> 
> What is the logic used to answer questions of the below type?
> 
> 1) If the finite group G contains a subgroup of order 7 but no element
> (other than the identity) is its own inverse then the order of group G
> could be (a)27   (b)28   (c)35   (d)37   (e)42.
> 
> It says that the correct anwer is (c). I know why we can eliminate (a)
> and (d) but don't know how to proceed further.

If the group has no element equal to its own inverse, then it can be
partitioned into {1} and {x,inverse(x)} for various x. Hence its order
is odd. That eliminates b,e.

-- 
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@cus.cam.ac.uk  Cambridge University, England.    [Research student]
From: tim@maths.tcd.ie (Timothy Murphy)
Newsgroups: sci.math
Subject: Re: what are quadturnians?
Keywords: coordinate transformations
Date: 8 Sep 92 23:47:57 GMT
Organization: Dept. of Maths, Trinity College, Dublin, Ireland.

astrod@kentaurus.dftsrv.gsfc.nasa.gov (AstroD Developer) writes:

>Would someone please give me a reference and a basic description of  
>quadturnians. I understand them to be a compact representation for 3  
>dimensional coordinate transformations(?). 

Next year (1993) is the 150th anniversary
of the discovery of quaternions
by William Rowan Hamilton,
Professor of Mathematics at Trinity College Dublin.

The quaternions are a non-commutative extension of the complex numbers.
A quaternion is expressible in the form

	q = t + xi + yj + zk

where t, x, y, z are real numbers.
Multiplication is defined by the rules:

	i*i = j*j = k*k = -1,
	i*j = k = - j*i,
	j*k = i = - k*j,
	k*i = j = - i*k.

Multiplication is non-commutative but associative.

The quaternions form a division-algebra (or skew-field):
that is, every non-zero quaternion has an inverse.
In fact, the quaternions are the only finite-dimensional division-algebra
over the reals, apart from the reals themselves and the complex numbers.


-- 
Timothy Murphy  
e-mail: tim@maths.tcd.ie
tel: +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Newsgroups: sci.math
From: holmes@opal.idbsu.edu (Randall Holmes)
Subject: Re: What's going on here ???
Organization: Boise State University Math Dept.
Date: Fri, 11 Sep 1992 15:29:45 GMT

In article <1992Sep11.214701.783@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>The following rather silly problem cropped up in our department just now.
>
>Evaluate    S = 0! - 1! + 2! - 3! + 4! - ....
>
>(yes, I know it's divergent, but blame Euler, not me.)

It converges in the p-adic numbers, does it not?

>
>------------------------
>Crazy solution (1)
>
>    oo           oo     oo            oo     oo          oo
>   \~~   n      \~~   n /  n -x       /  -x \~~  n       /   -x 1
>S = > (-1) n!  = > (-1) | x e  dx  =  | e    >(-x) dx  = |  e  --- dx
>   /__          /__    /             /      /__         /      1+x
>   n=0          n=0    0             0      n=0         0
>
>  = .5963..
>
>-----------------------
>Crazy solution (2)
>
>                                      2      3      4
>S = f(1)   where  f(x) = 0! - x 1! + x 2! - x 3! + x 4! - ...
>
>                              2      3
>So   (x f(x))' = 1! - x 2! + x 3! - x 4! + ...
>
>               = (1 - f(x))/x
>        2                                                     1        1
>Thus   x f' + (1+x)f = 1         this has integrating factor  - exp(- ---)
>                                                              x        x
>
>Thus  [ x exp(-1/x) f(x) ]' = exp(-1/x)/x
>
>                               x
>                              /
>thus      x exp(-x) f(x)    = | exp(-1/t)/t dt      now put x=1
>                              /
>                             0
>                               1
>                              /
>thus              f(1)/e  =   | exp(-1/t)/t dt      and put u=1/t
>                              /
>                             0
>                               oo
>                              /
>                          = e | exp(-u)/u du    =  .5963..
>                              /
>                             1
>
>being essentially the same integral as in crazy solution (1).
>-----------------------
>
>So what's going on here ????
>
>Two crazy solutions to a crazy problem giving the same answer.   Why ?
>
>--------------------------------------------------------------------------
>            Bill Taylor              wft@math.canterbury.ac.nz
>--------------------------------------------------------------------------
> MATH: necessary consequences of arbitrary axioms about meaningless things.
>--------------------------------------------------------------------------
>

Maybe it has something to do with the fact noted above.  Algebraic
manipulations of formal series often do give sensible results, which
might _really_ work in an unintended interpretation.  After all,
...1111 (base 2) _is_ -1 in the 2-adics.
-- 
The opinions expressed		|     --Sincerely,
above are not the "official"	|     M. Randall Holmes
opinions of any person		|     Math. Dept., Boise State Univ.
or institution.			|     holmes@opal.idbsu.edu
From: sichase@csa3.lbl.gov (SCOTT I CHASE)
Newsgroups: sci.math
Subject: Re: Lebesgue integral (was: Couple of questions
Date: 11 Sep 92 19:05:18 GMT
Organization: Lawrence Berkeley Laboratory - Berkeley, CA, USA
Distribution: na
Reply-To: sichase@csa3.lbl.gov
NNTP-Posting-Host: 128.3.254.198
News-Software: VAX/VMS VNEWS 1.3-4   

In article <1992Sep11.130033.26063@watson.ibm.com>, platt@watson.ibm.com (Daniel E. Platt) writes...
> 
>I think you would end up flat on your back.  The problem is that many (most)
>of the techniques revolving around Fourier series and integrals, completeness
>of a basis, etc, ultimately involve being able to evaluate 'improper' integrals
>as a limit of an integral of a sequence of functions.  They often look like
>Dirac-delta functions (the word 'function' is a misnomer, it is more like
>a limit of a family of functions) which just hides the complexity under
>some notation so that physicists don't have to worry about L^2(R^3).

Physicists (not including mathematical physicists) deal with this issue in 
a handwaving kind of way.  For example, take me.  I studied the theory of 
distributions a little bit a while back.  I know that the tempered
distributions are an interesting class of objects for studying QM, because
they are in a well-defined sense the largest class of "potential wave-functions"
 on which the Fourier transform is well-defined.  This means, to me, that the
other distributions cannot be used for wave-functions because they so unphysical
as to have no normal way to define the momentum-space representation. But more
than this, I do not know. 

Nevertheless, I regularly stick distributions of various kinds under integral
signs and come up with correct answers.  I have never studied Lesbegue
integration, though I am not proud of that fact.  (You can't study everything - 
I once got a lecture on Riemann-Steiltjes integrals, which is more
than many physicists ever see.) 

-Scott  
--------------------
Scott I. Chase			"The question seems to be of such a character
SICHASE@CSA2.LBL.GOV		that if I should come to life after my death
				and some mathematician were to tell me that it
				had been definitely settled, I think I would
				immediately drop dead again."      - Vandiver
Newsgroups: sci.math
From: jbaez@riesz.mit.edu (John C. Baez)
Subject: Re: Proof of God's Existence
Keywords: . . . and that's all what I am!
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Wed, 16 Sep 92 01:29:04 GMT

In article <8967@amsaa-cleo.brl.mil> matt@amsaa-cleo.brl.mil (Matt Rosenblatt) writes:
>For those who have never heard of him, Popeye is a fictional American
>cartoon and comic-strip character.  He is the one who gains superhuman
>strengh whenever he eats a can of spinach.  In his own dialect, his
>slogan is, 
>
>           "I YAM WHAT I YAM, AND THAT'S ALL WHAT I YAM,
>            I'M POPEYE, THE SAILOR MAN."
>
>With all that, no matter how many times he asserts, "I AM WHAT I AM,"
>he is still a fictional character.  Therefore, one may not prove the
>existence of a being solely on the basis that he is quoted as having
>said, "I AM WHAT I AM."  Let Popeye appear in thunder and lightning
>and smoke to six hundred thousand men with their wives and children
>and tell them, "I AM," and then I'll believe he exists.

Thanks, I was waiting for someone to say that - but I don't think Hannu
will cease his space potato reasoning due to this clarification.  After
all, Popeye isn't omnipotent the way God is, so he can't prove his own
existence just by claiming he exists.  :-)  Maybe some more
spinach might do it?

I'm surprised, actually, that nobody has brought up the most venerable
proof of God's existence - proof by repetition.  "What I tell you three
times is true," said the Butcher.
Newsgroups: sci.math
Subject: Re: Computability of the universe
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 15 Sep 92 21:20:43 GMT
 <1992Sep10.183603.2357@sei.cmu.edu> <1992Sep11.013141.28378@galois.mit.edu> <1992Sep15.090238.186@nntp.nta.no>

In <1992Sep15.090238.186@nntp.nta.no> klaus@hal.nta.no (Klaus Gaarder FNI) writes:

>Classical Physics:
>------------------
>	The so-called "three body problem", computing the trajectory of three
>(yes only 3) bodies under mutual gravitational (Newton-style) interaction, is not
>computable.

What do you mean, "not computable".  Do you mean not stable?
--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
My interaction with our news system is unstable; please mail anything important.
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Newsgroups: sci.math
Subject: Re: Solitaire games
Date: 4 Sep 92 21:59:53 GMT
Distribution: na

In <26102@dog.ee.lbl.gov> sichase@csa1.lbl.gov (SCOTT I CHASE) writes:

>In article <NICKH.92Sep3143716@VOILA.VENARI.CS.CMU.EDU> nickh@CS.CMU.EDU (Nick Haines) writes:
>> I play a game with triplets of digits (such as one finds on 99% of
>> British numberplates),

>This reminds me of the "annual puzzle" which appears each January in the 
>alumnus section of Technology Review, in a math/games column which appears
>in each issue.  (MIT grads get a special insert with class news, MIT 
>propaganda, etc.)

>Problem:  Construct all the numbers from 1 to 100 using all four digits
>of the year (in our case: 1,9,9,2),preferably in order, by interposing
>apropriate operators.  For example,

>1  = 1**992
>2  = 1**99 * 2 
>3  = 1**99 + 2
>7  = (1**9)*9 - 2 
>16 = 1*(9+9) - 2



>and so forth.  If you've never tried it before, give it a whack.  Warning:
>the best solution does *not* include all 100 numbers in most (if not all) years.

You might also look in the Mensa Bulletin for related puzzles.
--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
My interaction with our news system is unstable; please mail anything important.
Newsgroups: sci.math
From: martin@lyra.cis.umassd.edu (Gary Martin)
Subject: Re: Couple of questions
Organization: University of Massachusetts Dartmouth
Date: Wed, 9 Sep 1992 20:21:39 GMT

In article <1992Sep9.102457.15049@news.columbia.edu> pvl2@cunixb.cc.columbia.edu (Priscilla V Loanzon) writes:

   Could someone please explain to me a few basic things:

   What is the logic used to answer questions of the below type?

   1) If the finite group G contains a subgroup of order 7 but no element
   (other than the identity) is its own inverse then the order of group G
   could be (a)27   (b)28   (c)35   (d)37   (e)42.

   It says that the correct anwer is (c). I know why we can eliminate (a)
   and (d) but don't know how to proceed further.

Cauchy's Theorem: if p|o(G) then G has an element of order p.

Studying for GRE's?



--
Gary A. Martin, Assistant Professor of Mathematics, UMass Dartmouth
Martin@cis.umassd.edu
From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math
Subject: Re: D^k f^n ?
Keywords: differentiation, combinatorics
Date: 13 Sep 92 23:38:56 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <1992Sep10.172604.6358@jarvis.csri.toronto.edu> frank@csri.toronto.edu (Frank Kschischang) writes:
>
>Let f(x) denote some function of x.  Let D=d/dx denote the
>operation of differentiation with respect to x.
>
>Question:
>Has anybody ever worked out a formula for D^k f^n, the
>k'th derivative of the n'th power of f(x) with respect to x?

It's Called Faa di Bruno's formula. (Yes that's not a typo...) You can
find one use for it in the work of M. A. Dostal and R. Macchia on the
differentiability of roots of smooth functions. This problem grew out
of Glaeser's wierdo example...

Now the only place I can remember which is a reference for this formula off
the top of my head is the book on Fractional Calculus in the Academic Press
series on Mathematics in Science and Engineering. (The red ones with the
gold letters.)

No doubt now that we have the name of it, someone else might have a better
reference.

Later,
Andrew Mullhaupt
Newsgroups: sci.math
From: jbaez@riesz.mit.edu (John C. Baez)
Subject: Re: measures of the `size' of infinite sets
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Wed, 9 Sep 92 04:23:45 GMT

In article <1992Sep8.134624.11005@newstand.syr.edu> hgraber@lynx.cat.syr.edu (Harry Graber) writes:

>     An associate of mine here tells me that there are other measures that
>have been devised for the size of an infinite set.  Some of these measures
>give a different result, and agree with our `intuitive' position that the set
>of multiples of 29 is in some definite sense smaller than the set of numbers
>that are not multiples of 29.  However, this is an area neither of us has
>specialized in.  I never knew that these other measures even existed, and he
>had heard of them but is not able to tell me what any of them is.  

Well, there are various ways of keeping track of the "size" of infinite
sets, depending on the context.  One branch of math that specializes
this is called measure theory.  The simplest example of that is the fact
that a line segment which is twice as long as another has twice the
"measure," even though by Cantor's definition they have the same number
of points.  But for your problem the best answer involves not measure
but "density".  Given a set of integers, we calculate its density as
follows.  Figure out how many integers in your set lie between -n and n
- say that m do.  Take the ratio n/2m ... this just tells us what
fraction of the integers between -n and n lie in your set.  Now take the
limit as n goes to infinity!  If the limit exists and equals d (some
number between 0 and 1), we say your set has density d.

The set of integers that are multiples of 29 has density 1/29.
The set of integers that aren't has density 28/29.

Exercise to see if you get it: figure out, or guess, the density of the
prime numbers.






Newsgroups: sci.math,rec.puzzles,comp.theory
From: holmes@opal.idbsu.edu (Randall Holmes)
Subject: Re: more on .999... = 10.0
Organization: Boise State University Math Dept.
Date: Wed, 9 Sep 1992 15:22:34 GMT

In article <18dmplINNe0o@matt.ksu.ksu.edu> kodiak@matt.ksu.ksu.edu (Bryan D. Nehl) writes:
>Sorry for having taken so long to follow up on this subject.
> 
>Proof that 10.0 = 9.9999...
>using algebra:
> 
>        x =  9.999...
>  10x = 99.999...
>10x-x = 99.999... - 9.999...
>   9x = 90.0
>    x = 10
> 
>        10.0 = x = 9.999...
> 
>using geometric series:
> 
>x = 9 + 9/10 + 9/100 + ...
>x = 9/10^0 + 9/10^1 + 9/10^2 + ... + 9/10^n + ...
>x = 9 summation of 1/10^n from n = 0 to +infinity
> 
>this is a geometric series a + ar + ar^2 + ... + ar^n + ...
>a = 9 r = 1/10        
> 
>if -1<r<1 then the series = a/(1-r) 
>        9/(1-1/10) = 9/(9/10) = 9 * 10/9 = 10
>        
>someone emails me using the above algebraic method that 
>999... is equal to -1.
> 
>why is it that 999.... != -1?
> 
>...999 != 1 
>...999 is the series 9 + 9(10) + 9(10)^2 + ... + 9(10)^n + ...
>a = 9, r = 10
>iff -1 < r < 1 then the series converges
>since -1 < 10 !< 1 the series does not converge
>also recall that if the limit as n->+inf Sn != 0 then the summation of Sn
>diverges.
>Sn = 9(10)^n
>
>Bryan
>
>/* === Bryan Nehl ========== kodiak@Kodiakpc.Manhattan.KS.US =========+
>[            USDA-ARS-NPA-WERU          ][ bdn@chepil.weru.ksu.edu    ]
>[     913.532.6233 or 913.532.6495      ][ kodiak@matt.ksu.ksu.edu    ]

The point is that the algebraic method is invalid without reference to
issues of convergence.  In the "10-adic numbers", where the sequence
of positive powers of 10 converges to zero and the sequence of
negative powers of 10 does not, it is indeed the case that ...9999.0 =
-1.  The 10-adic numbers are not often considered, as the system
contains divisors of zero, but analogous results hold in the p-adic
numbers, p prime, which do make up a field.

10-adic numbers are represented by decimal expressions which may have
infinitely many digits to the LEFT of the decimal point but must have
finitely many to the RIGHT.  Experiment should convince one that one
can compute with them much as one computes with real numbers.
Division is not always successful.  The idea is that the _positive_
powers of 10 are "small", not the negative ones.


-- 
The opinions expressed		|     --Sincerely,
above are not the "official"	|     M. Randall Holmes
opinions of any person		|     Math. Dept., Boise State Univ.
or institution.			|     holmes@opal.idbsu.edu
Newsgroups: sci.math
From: holmes@opal.idbsu.edu (Randall Holmes)
Subject: Re: ZFC etc. (was Re: Report on Philosophies of Physicists)
Organization: Boise State University Math Dept.
Date: Tue, 15 Sep 1992 18:49:30 GMT

In article <716501145.10401@minster.york.ac.uk> cjhs@minster.york.ac.uk writes:
>I get a bit lost in some of the postings in the "Philosophies..." thread,
>but it does seem that some of the particpants might be able to help me
>a couple of things straight in my own mind. Maybe the questions have
>already been answered, and I just didn't see it...
>
>1) We all know that ZFC is consistent (or maybe just those of us who
>are sufficiently naive): I want to know why we have this confidence.

There is an intuitive picture of how the universe of ZFC is
"constructed" by taking iterated power sets of a base set, which seems
reasonably convincing.

>
>It just the case that lots of very intelligent people have failed
>to find an inconsistency? Or are there informal arguments (as there
>are for Church's thesis) to suggest consistency.

See above.  But the "intuition" remains informal.  There are possible
objections, on the grounds of impredicativity, for instance.

>
>2) We know that the continuum hypothesis is independent of ZFC. Godel
>provided the model (constructible sets) in which the CH is true, and
>Cohen provided the model (via forcing) in which the CH is false.
>
>Are the proofs by Cohen and Godel are formal proofs? (I would think
>so.) Can we identify the formal system in which the proofs were
>performed?

They are rigorous arguments.  No one ever does proofs in formal
systems, in practice.  The Godel proof is a construction in ZF (normal
set theory without choice).  The Cohen proof is a construction in ZFC
(Choice is used, I believe).

>
>Thanks for any informed comment.
>
>Felicitations -- Chris Ho-Stuart
>
>(PS this is fall-out from the alt.atheism thread, which has evolved to
>the point where sci.math is the better location)


-- 
The opinions expressed		|     --Sincerely,
above are not the "official"	|     M. Randall Holmes
opinions of any person		|     Math. Dept., Boise State Univ.
or institution.			|     holmes@opal.idbsu.edu
Newsgroups: sci.math
From: agl3@jane.uh.edu (Carlos Riveros)
Subject: Hermitian 'quasi' Toeplitz matrix eigensystem solution needed.
News-Software: VAX/VMS VNEWS 1.41    
Organization: University of Houston
Date: Fri, 11 Sep 1992 14:24:00 GMT

Hi,

	I'm looking for a _fast_ algorithm/subroutine to compute the
eigenvalues/eigenvectors of a complex Hermitian 'almost Toeplitz' matrix,
that is a matrix that is complex Hermitian Toeplitz everywhere except in
the main diagonal, where all numbers are (real and) different.  Adicionally,
some of the sub/super-diagonals may be zeros.

	I don't need to compute all eigval/eigvec pairs, just those who are
greater than a given (negative) number.  I've found 'ch' EISPACK (*all* eigval/
eigvec of a complex hermitian matrix), and there are other routines to compute
just *some* eigval/eigvec of a complex general matrix, but I wonder if 
somebody knows of something more appropiate.

	Any information will be welcome.  Please respond by e-mail.  
Thanks in advance.

--Carlos Riveros		| Allied Geophysical Labs.
salrcr@archie.agl.uh.edu	| University of Houston.
agl3@jane.uh.edu
From: jmarlan@titan.nmt.edu (Jon Marlan)
Newsgroups: sci.math
Subject: Re: Chess Problem
Date: 12 Sep 92 22:24:02 GMT
Article-I.D.: nmt.1992Sep12.222402.14408
Organization: New Mexico Tech


In article <BuGA8t.DL8@cmptrc.lonestar.org> carter@cmptrc.lonestar.org (Carter Bennett) writes:


>  FOLLOW-UP QUESTION FOR EXTRA CREDIT:
>
>  Determine the odds for doing the same with eight bishops!  ;-)

I'll give this one a try.  

Let B(i) = # of available "safe" places for i'th bishop

B(1) = 64 
B(2) = 64-8 = 56
B(3) = B(2)-8 = 48
B(4) = B(3)-7 = 41
B(5) = B(4)-7 = 34
B(6) = B(5)-6 = 28
B(7) = B(6)-6 = 22
B(8) = B(7)-5 = 17

Let b(i) = # of available places for i'th bishop

b(1) = 64
b(2) = 63
b(3) = 62
b(4) = 61
b(5) = 60
b(6) = 59
b(7) = 58
b(8) = 57

      B(1)*B(2)*B(3)*B(4)*B(5)*B(6)*B(7)*B(8)
P =   ---------------------------------------
      b(1)*b(2)*b(3)*b(4)*b(5)*b(6)*b(7)*b(8)

      64*56*48*41*34*28*22*17
P =   -----------------------
      64*63*62*61*60*59*58*57

P =  .0141


Jon

--
Jon Marlan
jmarlan@titan.nmt.edu    
--

P.S.  Anybody want to play a game of chess via e-mail?
Newsgroups: sci.math