HDTenEightyP/NetNews huggingface dataset:
From: bhonsle@bhonsle.iss.nus.sg (Shailendra K Bhonsle)
Newsgroups: sci.math
Subject: Re: u(v^n)w prime puzzle - HINTS
Date: 20 Aug 92 04:32:32 GMT
Organization: Institute of Systems Science, NUS, Singapore

In article <1992Aug19.194149.24353@wri.com>, roach@bikini.wri.com (Kelly Roach) writes:
|> 
|> 
|> 
|>       Prove or disprove:  There are three non-empty
|>       strings of digits u,v,w such that all the
|>       numbers in
|>            L = {u(v^n)w | n is a natural number}
|>              = {uw, uvw, uvvw, uvvvw, uvvvvw, ...}
|>       are prime numbers.
|> 
|> 
|> 			HINT TIME!!!
|> 
|>      This is the puzzle I posted Monday.  Today I'll
|> give out some hints.  On Friday, I'll explain the
|> solution if I can't get anyone to solve the problem
|> before then.
|> 
|>      HINTS:
|>       
|>      (1) Fermat's Little Theorem says
|>      
|>      c^(p-1) = 1 (mod p)
|> 
|> if p is prime and (c,p)=1.
|>      (2) Let a=|v|=length of v, b=|w|=length of w.
|> Then
|> 
|>      u(v^n)w = u*10^(a*n+b) + v*Sum[10^(a*i+b),{i,0,n-1}] + w
|> 
|> Consider the two pieces
|> 
|>      u*10^(a*n+b) + w
|>      
|>      v*Sum[10^(a*i+b),{i,0,n-1}]
|> 
|> separately.
|>      (3) If prime number p does not divide 10, then
|>      
|>      10^(a*i+b) (mod p)
|> 
|> is a periodic function of i.
|>      (4) If uw is not a prime, there is not much to prove.
|> Might as well assume uw is a prime number.
|> 
|> 
|> 

-- 
As I mentioned in my last article let us take prime p="uw".

Let's call the first piece as I and the second piece as J.

step 1: p=u*10^(b)+w
        I=u*10^(b)*10^(a*n) + w
     ==> I= u*10^(b) +w +u*10^(b)*{10^(an) -1}
     ==> I= p + u*10^(b)*{10^(an) -1} 

step 2: Either 10^(a)==1(mod p) or choose n such that n=p-1
   ==> p divides I
       and  10^(an) == 1(mod p)    { Fermat's Little Theorem }



step 3:
       J=  v*Sum[10^(a*i+b),{i,0,n-1}]
      ==> J= v*10^b [ 1 + 10^a + 10^(2a) + ... + 10^((n-1)a)]
      ==> J= [v*10^b {10^(an) -1}]/[10^a - 1]
       Since from step 2 10^(an) -1 is divisible by p
      ==> If 10^a-1 is not divisible by p then J is divisible by p


So we have proved so far that if p does not divide 10^a - 1 then pdivides J.
And in this case it is not possible to have such triples {u,v,w}.

Now we show that p cannot divide 10^a -1 to complete the proof.



step 4:
        Assume p divides 10^a - 1
       p= u*10^b +w


    case 1: b > a
           let b=l* a + k   k<b
            ==> 10^b== 10^k (mod p)
            ==> p divides u*10^k +w
           But this is not possible since u*10^k + w < u*10^b + w =p
           So in this case p does not divide 10^a - 1
    
     case 2: b <= a
             SIMPLE, Can you do it ?

So we complete the proof.




Shailendra
From: magdi@uxh.cso.uiuc.edu (Magdi N Azer)
Newsgroups: sci.math
Subject: Evaluation of an integral
Keywords: exponential
Date: 20 Aug 92 14:09:14 GMT
Article-I.D.: news.BtACnF.CxK
Organization: University of Illinois at Urbana

Can anyone tell me if there is an analytical expression one can obtain
]for exp[exp(x)].  I've done a Taylor Series expansion to fourth order,
but I don't know if I am missing something obvious about this. 

Second, can this be integrated(other than numerically)
The limits of the integral are 0 and 0.5
thank you

Magdi N Azer
magdi@uxh.cso.uiuc.edu
From: sec@otter.hpl.hp.com (Simon Crouch)
Date: Thu, 13 Aug 1992 05:57:02 GMT
Subject: Re: mathematician's marrying ages.
Organization: Hewlett-Packard Laboratories, Bristol, UK.
Newsgroups: sci.math


> My friend (yu@math.harvard.edu) keep advocating that marrying
> early is good for mathematician.  He gave several examples
> ( most of them are number theorist, according to his statemnet)
> Examples he gave are Delinge, Gauss, Milnor, Serre, Zariski.
> He told me that he got these information from bibliography
> (maybe from introduction to authors...etc)

> Any counterexample? ( Marry late and have quite an achievement )


Well...I'm still not married I haven't yet proved the Riemann conjecture....

;-)

Simon.

Newsgroups: sci.math
From: bs@faron.mitre.org (Robert D. Silverman)
Subject: Re: Primes
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Date: Mon, 17 Aug 1992 11:02:34 GMT

In article <1992Aug17.160252.10145@waikato.ac.nz> bill@waikato.ac.nz writes:
:How many primes are there less than 2 to the power of p?
:
2^p/(p log 2)   [approximately]

:Is there a way of calculating this, or does anyone know of where I can get my
 
How big is p? If p isn't too large it can be calculated exactly using
methods developed by Lagarias, Odlyzko, and Coppersmith. These are
improvements on the Meisel-Lehmer method.

:hands on a list of primes so I can find out how many primes there are
:less than 2^16 or 2^32 for example?
 
For p THIS small you can do it directly by sieving. 2^16 will take
milliseconds. How long 2^32 will take is memory dependent, but should
only be a few hours at most.

:Is there some way of determining (or estimating) the expected magnitude of the
:nth prime? Or of the growth of the difference between the nth and (n+1)th
 
the n'th prime is approximately n log n.

the growth of the difference is not well understood. studying it involves
some deep methods in analytic number theory. For example, if one has
a 'random' sequence of integers going to infinity, whose normal order
difference is log n, then one has lim sup (n-->infinity) p_n/log n = 2,
whereas for primes, the lim sup is infinite. Further, Rankin has shown
that the difference is

C log n  loglog n  loglogloglog n
---------------------------------

	logloglog n ^2

infinitely often.

I suggest you start by reading a book on number theory. Start with
Hardy & Wright.
--
Bob Silverman
These are my opinions and not MITRE's.
Mitre Corporation, Bedford, MA 01730
"You can lead a horse's ass to knowledge, but you can't make him think"
Newsgroups: sci.math,sci.math.symbolic
From: king@physics.su.OZ.AU (Peter King)
Subject: Solving Rate Equations
Organization: School of Physics, University of Sydney, Australia
Date: Thu, 20 Aug 1992 22:46:23 GMT

I'm interested in solving the following equations (in Mathematica format)
for a cw optically pumped laser:

eqns = {
        wv N3 - wp (N0 - g0 / g2 N2) == 0
        B21 n (N2 - g2 / g1 N1) - wr N1 == 0
        wp (N0 - g0 / g2 N2) - wr N2 - B21 n (N2 - g2 / g1 N1) == 0
        N0 + N1 + N2 + N3 == N
        B21 n (N2 - g2 / g1 N1) - 2 Pi y n == 0
       }

simultaneously.  In Mathematica I naively tried
        Solve[eqns, {N0, N1, N2, N3, n}]
and ran out of memory on a 40 Mb RAM workstation.  I've tried eliminating n
from the last equation to reduce the system to 4 equations but then I can't
find the solution to n without running out of memory.

        Does anyone know of any tricks/techniques/simplications I can make
to solve this system?  Alternatively does anyone know of simple references
to Groebner bases?

                                thanks in advance

                                     Peter

--
********************************************************************************
Peter King                               *         The home of TORTUS.
Plasma Physics Department                *   T
University of Sydney, NSW 2006           *   O
AUSTRALIA                                *   Roid of
    voice: 61 2 692 3371                 *   The 
      FAX: 61 2 660 2903                 *   University of
    email: king@physics.su.OZ.AU         *   Sydney
********************************************************************************
Newsgroups: sci.math
From: roach@bikini.wri.com (Kelly Roach)
Subject: u(v^n)w prime puzzle SOLUTION
Organization: Wolfram Research, Inc.
Date: Fri, 21 Aug 1992 22:07:19 GMT




      Prove or disprove:  There are three non-empty
      strings of digits u,v,w such that all the
      numbers in
           L = {u(v^n)w | n is a natural number}
             = {uw, uvw, uvvw, uvvvw, uvvvvw, ...}
      are prime numbers.


		     THE SOLUTION

     I've seen some good solutions to my puzzle already,
but as promised, I now present the solution I said I would
post on Friday (today).
     Let a=|v|=length of v, b=|w|=length of w.  Then

     u(v^n)w = u*10^(a*n+b) + v*Sum[10^(a*i+b),{i,0,n-1}] + w

Fermat's Little Theorem says
     
     c^(p-1) = 1 (mod p)

if p is prime and (c,p)=1.  If uw is not a prime, there
is not much to prove.  So, let p=uw be prime.  If
(p-1) | n, then by F.L.T.

     u*10^(a*n+b) + w = u*10^b + w = uw = 0 (mod p)

All that remains is to arrange things so that

     Sum[10^(a*i+b),{i,0,n-1}] = 0 (mod p)

Prime number p does not divide 10 because uw has at least
2 digits.  Hence

     10^(a*i+b) (mod p)

is a periodic function of i.  The period divides p-1
and that means that if p(p-1) | n, then each residue
appearing in the list

     10^b, 10^(a+b), ..., 10^(a*(n-1)+b)     (mod p)

appears a multiple of p times.  Consequently, it is
sufficient to let n=p(p-1) to get p | u(v^n)w.  Obviously
u(v^n)w > p=uw.

			  QED


     It is possible to make n smaller in my proof
by using p*ord(10,p) instead.  Also, Arthur Rubin
suggests using p=uvw and n=p.  Then

     u*10^(a*n+b) + vw = u*10^(a+b) + vw = uvw = 0 (mod p)

     Sum[10^(a*i+b),{i,1,n-1}]
     = (10^(a*(p-1))-1)/(10^a-1) * 10^(a+b)	

The Sum is divisible by p since p > 10^a-1 and F.L.T.
applies to 10^(a*(p-1))-1.
     There were two other nice solutions posted which
reduced the problem to an instance of the more general
problem of whether

     x_{n+1} = a*x_n + b

could always be prime and then solved that problem.
     About the u(v^n)w puzzle itself:  It arises from
the problem of investigating whether any regular language
can generate only prime numbers.  There is a pumping
lemma for the regular languages which basically reduces
the problem down to the u(v^n)w puzzle which I posted
to sci.math.  So, everyone who solved my puzzle also
just proved that no finite state machine can accept
an infinite language consisting of only prime numbers.
If you generalize the proof a bit, you can also prove
that u(v^n)w(x^n)z with |vx| != 0 cannot generate all
prime numbers either.  This shows that no infinite
context-free language can contain only prime numbers.
I'm pretty sure these are old results.



Newsgroups: sci.math
From: guckes@math.fu-berlin.de (Sven Guckes)
Subject: Re: graph problem
Originator: guckes@medusa
Organization: Free University of Berlin, Germany
Date: Wed, 12 Aug 1992 23:49:21 GMT

tfrey@rp.CSIRO.AU (Thierry Frey) writes:

>Keywords: 
>If someone could help me with this, it would be fantastic!
>
>Given a connected graph of n nodes, where each link between
>nodes is given a positive weights, first find a path between
>nodes A and B given, then find the shortest ( or 'lightest')
>path between the two.

Dijkstra.

>I am looking for a program in C capable of such achievements.

Knuth.
From: orpheus@reed.edu (P. Hawthorne)
Newsgroups: sci.math
Subject: A Graph Theory Newsgroup, Perhaps?
Date: 18 Aug 92 08:49:52 GMT
Article-I.D.: reed.1992Aug18.084952.27270
Organization: Reed College, Portland OR


  Graph theory is at the heart of so many fields that it is somewhat
obscured by its own versatility.

  I feel that there ought to be a graph theoretic newsgroup, if only for
the insights that such a multidisciplinary newsgroup could offer.

  I'd be interested in hearing from others, to see if a more formal
approach to creating such a newsgroup is warranted.


  Theus (Prometheus Hawthorne - Jones, orpheus@reed.edu)
Newsgroups: sci.math
From: s9054960@rkw-risc.cs.up.ac.za ()
Subject: Sequence
Date: 13 Aug 92 14:58:04 GMT
Organization: University of Pretoria, South Africa

What is the next number in the sequence, and how did you get it?

2 3 3 5 10 13 ...

Reply to: 9054960@rkw-lan.cs.up.ac.za

Thanks,
Gregory Knott.
Newsgroups: sci.math
From: tycchow@riesz.mit.edu (Timothy Y. Chow)
Subject: Re: The Converse of Kaplansky's Therorem
Organization: None.  This saves me from writing a disclaimer.
Date: Wed, 12 Aug 92 22:24:05 GMT

In article <1992Aug12.165234.11088@thunder.mcrcim.mcgill.edu> boshuck@triples.math.mcgill.ca (William Boshuck) writes:
<In article <1349@newsserver.cs.uwindsor.ca> tarokh@server.uwindsor.ca (TAROKH VAHID             ) writes:
<>Would somebody please tell me if the converse of Kaplansky's Theorem
<>is true or not? That is if R is a ring with identity such that all the
<>projective R modules are free, then is R necessarily Local? If we assume
<>that R does not have identity what could be said?
<
<The answer to the first question is NO. Look at the ring of integers.
<I haven't thought about the second question since I have absolutely
<no experience with rings without identity.

More generally, consider any principal ideal domain.  Any submodule of a
free module over a principal ideal domain is free.  Projective modules are
direct summands (hence submodules) of free modules.  Any non-local PID is a
counterexample.

I also have no wisdom to offer about rings without unit.
-- 
Tim Chow     tycchow@math.mit.edu
Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs
30 tons, computers in the future may have only 1,000 vacuum tubes and weigh
only 1 1/2 tons.                               ---Popular Mechanics, March 1949
Newsgroups: sci.math
Subject: Re: A Word-Problem
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Date: 13 Aug 92 11:36:15 EDT
Organization: Harvard Math Department

In article <1992Aug13.135704.2280@unibi.uni-bielefeld.de>
umatf071@unibi.uni-bielefeld.de (0105) writes:
"This is the problem the word-problem comes from:
"
"                   1
"                  1 1
"                 2 2 3
"                3 2 3 3
"               3 3 1 1 2
"              1 2 2 1 2 2
"             1 1 2 3 3 1 1
"            2 3 3 1 3 2 1 3
"           2 2 3 1 1 2 2 3 3
"                                                            o
"This figure shows that you can tile a triangle T9 with T2: o o
"
"QUESTION: which T(n) can be tiled with T2?
[...]

So it was a tiling problem, but not the one I had imagined...

"  Noam Elkies is right, that the homomorphism of David Sibley shows that
"  it is impossible to tile an odd rectangle with the L-triomino (only
"  180-degree rotation allowed) too. In this case you have to show that
"  a^3 b <> b a^3.

I omitted a step here: the commutators [a^3,b^2] and [b^2,a^3] vanish
(aaabb=a(aabb)=a(baba)=(abab)a=(bbaa)a=bbaaa, and likewise aabbb=bbbaa;
in ffect I'm tiling 2x3 and 3x2 rectangles with the allowed L-triominos),
and aaabbb=bababa iff a^3 and b^3 commute since b(abab)a=b(bbaa)a;
so the group-theory identity [a^m,b^n]=1 that would result from a tiling
of any odd mxn rectangle by allowed L-triominos is equivalent to any of
[aaa,b]=1, [aaa,bbb]=1, or aaabbb=bababa.

--Noam D. Elkies (elkies@zariski.harvard.edu)
  Dept. of Mathematics, Harvard University
From: cjh@tinton.ccur.com (Christopher J. Henrich)
Newsgroups: sci.math
Subject: Re: Binomial like recurrence
Date: 13 Aug 92 23:02:45 GMT
Organization: Concurrent Computer Corp., Tinton Falls, NJ

In article <1992Aug12.171516.24574@menudo.uh.edu> rmverma@cs.uh.edu (Dr. R.M. Verma) writes:
>In article <1992Aug12.100933.13400@aifh.ed.ac.uk> tw@aisb.ed.ac.uk (Toby Walsh) writes:
>>
>>The binomial recurrence,
>>
>>    c(n,k) = c(n-1,k-1) + c(n-1,k)
>>
>>is well known. Has the related recurrence,
>>
>>    u(n,k) = 2.u(n-1,k-1) + u(n-1,k)
>>
>>been studied? Does it have a closed form?
>>What is its asymptotic behaviour?
>>
>>And more generally, how about,
>>
>>    v(n,k) = a.v(n-1,k-1) + b.v(n-1,k)
>>
>>for integer a and b?
>>
>>toby
>
>Yes, they have been studied and there are answers to some of your
>questions in the literature. 
>
>See the paper by Monier in Journal of ALgorithms 1980, and
>Ch. 8 of The Analysis of Algorithms by Purdom and Brown (more references
>can be found here).

Dr. Verma's answer sounds like a response to a harder problem than
the one Mr. Walsh proposed.  As far as I can see,

     u(n,k) = 2^k . c(n,k).

More generally,

     v(n,k) = a^k . b^(n-k) . c(n,k).

Regards,
Chris Henrich
From: oliveira@phnom-penh.berkeley.edu (Jorge-Nuno Oliveira Silva)
Newsgroups: sci.math
Subject: Prin. Math. Induction
Date: 21 Aug 1992 19:38:23 GMT
Organization: U.C. Berkeley Math. Department.
Distribution: world
NNTP-Posting-Host: phnom-penh.berkeley.edu


	I'm collecting stuff about the Principle of
Mathematical Induction (curious examples, applications,
fallacies, etc), if you have some please e-mail them
to me, I'll post a survey later.
	Thanx,
	jn






-- 
Jorge-Nuno Oliveira Silva				
Department of Mathematics		328 Wilson St.	      
University of California		Albany, CA 94710   USA
Berkeley CA 94720 USA               e-mail: oliveira@math.berkeley.edu
Newsgroups: sci.math
From: roach@bikini.wri.com (Kelly Roach)
Subject: Re: u(v^n)w prime puzzle
Organization: Wolfram Research, Inc.
Date: Wed, 19 Aug 1992 17:35:15 GMT

In article <1992Aug19.024106.25588@nuscc.nus.sg>  
bhonsle@bhonsle.iss.nus.sg (Shailendra K Bhonsle) writes:
> The approach is to take a integer x (let's say prime integer) and prove  
that by repeating 
> v one gets a string u(v^m)w (for positive integer m) which is divisible  
by x.
> 
> Let's take x=9 (because it has nice properties for base 10 numbers).


     OK, but x=9 is not prime.

 
> then for a number "uv"(ie. 10u+v) 
>      uv=u+v (mod 9)
> 
> 
> In our problem definition:
> 
> let u+w=m(mod 9)
> 
> Now it is possible to find a n for given v such that
> 
>                      v^n==n.v== 9-m (mod 9) ==> simple number theory 



     I see some of the techniques I'm expecting to see in a
correct proof.  But my puzzle is not solved so easily.  What
would happen to your proof if v is divisible by 3 and m is not
divisible by 3?  Solving n.v== 9-m (mod 9) for n will be
impossible.  As a particular case, consider

     u="56",v="0",w="39"
     5639, 56039, 560039, 5600039, 56000039, 560000039

Any example with v="0" cannot be handled by your proof.

				Kelly



Newsgroups: sci.math
From: mroussel@alchemy.chem.utoronto.ca (Marc Roussel)
Subject: Re: Calculus and Mathematica
Organization: Department of Chemistry, University of Toronto
Date: Wed, 19 Aug 1992 17:42:30 GMT

In article <1992Aug19.163805.3972@cs.rose-hulman.edu>
goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard) writes:
>Last year, I gave the  
>new class roughly the same list (but longer) and had them integrate  
>them on the machine.  (I had to check that the maching could do all of  
>them before I made the assignment.  If the machine can't, the problem  
>isn't a good example for this exercise.)

     Did you eventually show your students examples the computers
screwed up badly?  Integration is the one thing which computer algebra
systems all claim to do, but can't handle reliably.  If you use one of these
systems in your classes, you have to show the students just how
dangerously stupid these things are.

				Marc R. Roussel
                                mroussel@alchemy.chem.utoronto.ca
Newsgroups: sci.math
From: robert@triton.jpl.nasa.gov 
Subject: Polynomial least-squares curve fit functions???
Reply-To: robert@triton.Jpl.Nasa.Gov
Organization: JPL	
Date: Tue, 18 Aug 1992 22:46:35 GMT

hello,
	I'm in need of a C function that determines a polynomial
that fits a given discrete set of data.  Any help is appreciated.

thanks

-robert
Newsgroups: sci.math,sci.physics
From: frisinv@alleg.edu
Subject: tensors
Organization: Allegheny College
Date: Thu, 20 Aug 1992 19:00:41 GMT

  I was reading a book on general relativity and the author began talking  
about tensor analysis, something I'd never heard of. I get the feeling  
tensors are some sort of number, but I'm not sure. Could anyone enlighten  
me as to what they are and how to use them?
----
Vince
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Newsgroups: sci.math
Subject: Re: Help - non-integral power of a matrix?
Keywords: matrix
Date: 12 Aug 92 21:12:43 GMT

In <carle.713599691@vex> carle@vex.ugcs.caltech.edu (Matthew Thomas Carle) writes:

>gonzalez@clam.rutgers.edu (Ralph Gonzalez) writes:


>>Hi.  Does anyone know of an algorithm to find a non-integral
>>power of a matrix, e.g. A^.5 or A^1.3? Thus, A^2 is the same
>>as AxA and A^0 is the identity.

>>I imagine if such a thing is defined, then there are conditions
>>on A...

>(I've never actually seen this applied, but I don't know why it wouldn't work)
>Well, let's look at how we implement a^b, with real a and b
>c=a^b  ->   log c = b log a   ->   c=exp(b*log(a))
>For real b, but a matrix A, C=A^b=exp(b*log(A))
>log(I+A) = A + (A^2)/2 + (A^3)/3 + ... +(A^n)/n + ...
>b*log(A) = b * [(A-I) + ((A-I)^2)/2 + ((A-I)^3)/3 + ... + ((A-I)^n)/n + ...]
>where the powers of A are defined as usual; then stick that into the 
>power series expansion of the exponential of a matrix . . .
>C = I + [b*log(A)] + ([b*log(A)]^2)/2 + ... + ([b*log(A)]^n)/(n!) + ...
>I dunno know, looks kinda freaky to me.

>I would guess exp(A) converges for all A, but I don't know about the log.

log "obviously" converges if all eigenvalues of A are strickly within 1 of 1,
just like the complex case.
--
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.
Our news system is unstable; if you want to be sure I see a post, mail it.
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: Re: Collineations
Date: 13 Aug 92 23:07:15 GMT
Organization: M.I.T. Artificial Intelligence Lab.

In article <1992Aug12.194204.24356@uwm.edu> radcliff@csd4.csd.uwm.edu (David G Radcliffe) writes:

   Suppose f is a one-to-one function from the plane to itself
   which maps lines into lines, and suppose the image of f is 
   not contained in a line.  Must f be affine?

   I have found some partial results.  The restriction of f to
   Q^2 is equal to the restriction of a projective transformation.
   If one also assumes that f is surjective, or continuous, or that
   it preserves betweenness, then f must be affine. 

By the plane, I assume you mean R^2. In that case, yes, it must be an
affine transformation. By composing your function with a suitable
affine transformation, you can assume that it fixes the points (0,0),
(1,0) and (0,1). Since it must preserve parallel lines, there
are functions g(x) and h(y) such that f(x,y)=(g(x),h(y)). And then
the fact that the line y=x is mapped to itself, we have g=h.
Furthermore, g(0)=0 and g(1)=1. Therefore, f maps the line with
slope y through the origin (i.e. the line joining (0,0) and (1,y))
to the line with slope g(y) through the origin (i.e. the line joining
(0,0) and (1,g(y)). In particular, it maps (x,xy) to (g(x),g(x)g(y)),
which proves that g(xy)=g(x)g(y). Also, since f preserves parallograms,
we have g(x+y)=g(x)+g(y). Therefore, g is an automorphism of R.
It is easy to see that g fixes every rational. To prove that g fixes
every real, it is enough to show that it g is order preserving. But that
follows from the fact that a number is nonnegative if and only if it has
a square root. So g is the identity and therefore so is f.

If one relaces R by another field, the proof shows that f is the composition
of an affine transformation with an automorphism of the field applied to all
the coordinates. This works in n dimensional space for all n>1.

Allan Adler
ara@altdorf.ai.mit.edu
Newsgroups: sci.math