HDTenEightyP/NetNews huggingface dataset:
From: torkel@sics.se (Torkel Franzen)
Subject: Re: You know, the integers (was: Re: Stupid question about FLT)
Organization: Swedish Institute of Computer Science, Kista
	<1992Jul21.150931.22637@sics.se> <1992Jul22.100121.1@amherst.edu>
Date: Wed, 22 Jul 1992 15:38:57 GMT

In article <1992Jul22.100121.1@amherst.edu> djvelleman@amherst.edu writes:

   >I find the cumulative hierarch of sets very *unclear*--especially in
   >comparison to the integers.  My reaction to the statement, "You know, THE
   >INTEGERS" would be "Oh yes, of course, I know those."  I'm afraid my
   >reaction to "You know, THE SETS" is "No, I don't know them.

  Certainly the cumulative hierarchy of sets is unclear if you compare it with
the integers! Who ever claimed otherwise? But the answer to the question
"what sets?" which I mentioned was not "you know, THE SETS", but the
very different answer given by Zermelo.
From: brnstnd@nyu.edu (Dan Bernstein)
Newsgroups: sci.math
Subject: Re: You know, the integers (was: Re: Stupid question about FLT)
Date: 22 Jul 92 23:54:26 GMT
Organization: IR

In article <1992Jul21.183305.16522@galois.mit.edu> tycchow@riesz.mit.edu (Timothy Y. Chow) writes:
> Is it what mathematicians have done?

Yes. Oh, sure, you might assign semantics to mathematical concepts, and
when you see how amazingly well math applies to the real world you might
start believing (as I do) that it's all consistent... but when push
comes to shove you demonstrate *everything* syntactically. That's how
modern mathematics works. That's how it's worked since Euclid.

How many mathematicians care whether ZF has a model within some larger
system? Or whether ``the sets'' described by ZF are unique? Very few,
I'd say. All that matters is that the axioms which make up ZF are
useful for producing interesting statements. What a ``set'' is, beyond
the properties assigned to it by ZF, is absolutely, totally irrelevant.

> Texts on set theory will begin with a
> bunch of axioms for sets.  In general, the term "set" is not defined, much
> less defined to be a syntactic entity.

Of course it's not ``defined to be a syntactic entity.'' It *is* a
syntactic entity. It's a word. ``Set.'' It has a precise syntactic
meaning (which will vary from text to text, of course): i.e., you're
given certain axioms for set theory which tell you how you're allowed to
push the word ``set'' (together with ``element'') around the page,
within the framework of some appropriate logical system.

> But you go further, concluding from this fact that a
> set IS a syntactic entity,

You might think of sets as having semantics beyond the syntax, but you
can't convince any other mathematician of those semantics without
reducing them to syntax. I'm just trying to make a practical observation
about how math works.

> There is no need to
> say that we don't know what a mathematical object "is" just because
> its properties cannot be exhausted by a first-order axiomatization.

What I am saying is that anyone who talks about a mathematical object
and can't write down axioms for it is a crank. As far as I'm concerned
such objects aren't mathematical.

---Dan
Newsgroups: sci.math
From: jnjohann@immd4.informatik.uni-erlangen.de (Jan Johannsen)
Subject: Re: You know, the integers (was: Re: Stupid question about FLT)
Organization: CSD., University of Erlangen
Date: Thu, 23 Jul 1992 12:00:29 GMT

djvelleman@amherst.edu writes:

>In article <BrquHt.E5p@immd4.informatik.uni-erlangen.de>, jnjohann@immd4.informatik.uni-erlangen.de (Jan Johannsen) writes:
>> 
>> Following this thread I always wondered why noone ever mentioned the following
>> possible answer:
>> 
>> Take the axioms for the upper half of a discretely ordered ring (about 15 axioms)
>> and add the (infinite) \omega-rule of induction:
>> 
>>   A(0)   A(s0)  A(ss0)  A(sss0) ....
>> --------------------------------------
>>              \forall x A(x)
>> 
>> Then every true (first order) sentence about the integers is provable 
>> with this machinery, and vice versa.
>> Thus we have characterized the integers up to elementary equivalence.

>  And what exactly does the "..." in the rule above refer to?  Your
>"characterization" of the integers has the integers hidden in it.

The premiss of the \omega-rule should perhaps be more accurately written as:

    A(t)    for every term t built up from the constant symbol 0 and the
            unary function symbol s
  ------------------------------------------------------------------------

This is what my dots referred to, a purely syntactic definition. 
Where do you see integers here ?


>  Dan Velleman
>  Dept. of Mathematics & Computer Science
>  Amherst College

Jan Johannsen        jnjohann@immd1.informatik.uni-erlangen.de

Newsgroups: sci.math
From: bardo@ief-paris-sud.fr (Bardo Muller)
Newsgroups: sci.math
Subject: Re: POLYNOMIAL DIVISION
Date: 24 Jul 92 05:54:01 GMT
Reply-To: bardo@ief-paris-sud.fr
Organization: Universite de Paris

Dear ???

Hope you're helped with this piece of flatex.
Salut,
Bardo
___________________________________________________________________________
| Bardo MULLER                          | phone  : [33] [1] 69 41 78 50    |
| Institut d'Electronique Fondamentale  | fax    : [33] [1] 60 19 25 93    |
| Bat. 220 Universite Paris Sud         | e-mail : bardo@ief-paris-sud.fr  |
| 91405 ORSAY CEDEX FRANCE              |             THE NET WORKS !      |
|_______________________________________|__________________________________|


Pour une fonction de transfert repre'sente'e par une fraction rationnelle comme
l'e'quation (\ref{eq:etoile}), dont l'expression ge'ne'rale est
\begin{equation}\label{eq:transfert}
	Y(s)\, =\, \frac{N_0+N_1s+N_2s^2+\ldots+N_ms^m}
			{1+D_1s+D_2s^2+\ldots+D_ns^n}
\end{equation}
on peut obtenir les coefficients d'un de'veloppent en se'rie de Laurent 
\begin{equation}
	Y(s)\, =\, \sum_{i=o}^\infty\, C_is^i
\end{equation}
par division polynomiale \cite{LAL74}.
\begin{equation}\label{eq:tabula}
	G(s)\, =\, A_{21}-A_{31}s+A_{41}s^2-A_{51}s^3+\ldots
\end{equation}
avec la forme tabulaire suivante:
% direct division lal/mittra
\begin{equation}\label{eq:tabulb}
\matrix{
	1		\hfill&A_{12}=D_1	\hfill&A_{13}=D_2	\hfill&\ldots	\cr
	A_{21}=N_0 	\hfill&A_{22}=N_1	\hfill&A_{23}=N_2	\hfill&\ldots	\cr
	A_{31} 		\hfill&A_{32}		\hfill&A_{33}		\hfill&\ldots	\cr
	\phantom{A}\vdots\hfill&\phantom{A}\vdots\hfill	&\phantom{A}\vdots\hfill&
}
\end{equation}
Comme indique', les deux premie`res lignes ci-dessus sont les constantes de
la fonction de transfert (\ref{eq:transfert}) et les autres e'le'ments
sont obtenues par la relation
\begin{equation}\label{eq:tabulc}
	A_{k,l} \, =\,  A_{k-1,1}A_{1,l+1}-A_{k-1,l+1}
\end{equation}


\def\AC{IEEE Trans.\ Automatic Control}


@article{LAL74,
	author =	{M.~Lal {et} R.~Mitra},
	title =		{Simplification of Large System Dynamics Using a
			Moment Evaluation Algorithm},
	pages =		{602--603},
	journal =	{\AC},
	label =		{Lal74},
	month =		{oct},
	year =		{1974}
}
Xref: sparky sci.math:9707 sci.physics:12095
From: johan@blade.stack.urc.tue.nl (Johan Wevers)
Newsgroups: sci.math,sci.physics
Subject: Re: Correction.
Date: 31 Jul 92 07:58:19 GMT
Followup-To: sci.math
Organization: MCGV Stack, Eindhoven University of Technology, the Netherlands.

> In article <4895@tuegate.tue.nl> I write:
> 
>But because generally the speed of sound becomes higher as the density rho
>of the rod becomes higher (v=sqrt(E/rho), if this formule still holds under
>these circumstances), a rod in which te speed of sound has a value comparable 
>with the speed of licht must have the density of a nuetron star or black hole.

OOoops...... I forgot to say that usually E becomes more than lineary
larger with increasing rho. So, if rho gets larger, E/rho gets also larger.

***********************************************************************
* J.C.A. Wevers          *       _\       /e+  _\   _   /e+           *
* johan@stack.urc.tue.nl *    1  u \  X  /     d \  X  /           2  *
**************************  P=-|    >---<    +    >---<   + ......|   *
* Physics is really all  *    N    /     \       /     \              *
* there exists.  Me.     *        /d     u\     /u    d \             *
***********************************************************************

Newsgroups: sci.math,rec.puzzles,comp.theory
From: sethb@fid.morgan.com (Seth Breidbart)
Subject: Re: Ten equals 9.9999....
Organization: my opinions only
Date: Thu, 30 Jul 1992 23:26:41 GMT

In article <1992Jul29.165039.15193@CSD-NewsHost.Stanford.EDU>
bhayes@Xenon.Stanford.EDU (Barry Hayes) writes:

>For example, I will define the following sequence:
>P[n] is the nth digit in the decimal expansion of Pi: P[1] = 1, P[2] = 4. . .
>A[n] is 0 if P[n] through P[100+n] are all ones.
>A[n] is 1/10^n if P[n] through P[n+100] are all twos.
>A[n] is P[n]/10^n
>
>Note that since any P[n] can be computed, any A[n] can be computed.  
>
>Now define A as the infinite sum of all the A[n]'s.  If you believe in
>Pi, you have to believe in A.  Well, what's the order between A and
>Pi?  Depends a lot on Pi, doesn't it?

A is a computable number.  I don't know if A < pi, A = pi, or A > pi.
If A != pi, then we'll eventually find out (when the program runs long
enough).  If A = pi, we might or might not eventually find out.

Define B as the 10^10^100'th digit in the decimal expansion of pi.
Is B > 5?

Seth		sethb@fid.morgan.com
Newsgroups: sci.math
From: tycchow@riesz.mit.edu (Timothy Y. Chow)
Subject: What more do you want?
Organization: None.  This saves me from writing a disclaimer.
Date: Tue, 21 Jul 92 19:02:43 GMT

For a change of pace, here's a dramatization of the frustration that
existence proofs can sometimes cause.

A [rushing in excitedly]: I've done it!  I've proved that Goldbach's
   conjecture holds for all sufficiently large even numbers!
B: Congratulations!  So now you can write a computer program to check
   all the remaining even numbers and Goldbach's conjecture will be
   completely settled, right?
A: Well, not quite...I proved that there exists a constant C beyond
   which Goldbach's conjecture holds, but I don't have any idea how
   big C is.  On the other hand, that's not a big problem, because
   C is computable.
B: You mean there's an algorithm that will give you the value of C
   explicitly?
A: Exactly.
B: So what are we waiting for?  Let's run the algorithm and compute C!
A: Well, there's a slight problem here...I just managed to prove that
   such an algorithm exists, and I don't have any idea what the algorithm
   is like.
B: Hmmm...well, maybe it's not worth running the algorithm anyway.  Do
   you have any idea how long it would take to compute C?
A: Not really, but the running time T of the algorithm is computable...
-- 
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
From: gerry@macadam.mpce.mq.edu.au (Gerry Myerson)
Subject: Re: Integer Sequence Problem
Organization: ceNTRe for Number Theory Research
Date: Thu, 30 Jul 92 03:07:19 GMT

In article <1992Jul28.225605.18476@pasteur.Berkeley.EDU>, luzeaux@bellini.berkeley.edu (Dominique Luzeaux) writes:
> 
> In article <1992Jul28.173452.24364@news.tu-graz.ac.at>,
> hhassler@iaik.tu-graz.ac.at (Hannes Hassler) writes:
> |> For S an infinite, increasing sequence of positive integers (1=a_1 < a_2 <
> |> a_3 < ... < a_i < a_{i+1} < ...), we define the VALUE val(S) of S 
> |> 
> |>         val(S)  :=    sup_{n>1}  \sum_{i=1}^n a_i   / a_{n-1}

> |> PROBLEM: Find the minimum (infimum) of val(S) over all possible sequences S.

[analysis deleted]

> Furthermore, we can say that val(S) takes as only finite values q^2/(q-1) 
> where q is an integer >=2, hence 4 is the minimum of val(S) over all
> sequences S.

I won't argue with the analysis (which I've deleted), nor with min val(S)=4, 
but it's not true that val(S) can only be q^2/(q-1) with q an integer not 
less than 2. The Fibonacci sequence has val(S)=q^2/(q-1) with q the golden 
ratio ((1+sqrt5)/2). More generally, given real q>1, and any geometric 
series G with common ratio q, if S is a sequence of integers close enough to 
G (like, s_n=integer part of g_n), then val(S)=q^2/(q-1). Of course, the 
minimum value of q^2/(q-1) is 4, attained at q=2. 

> 	Dominique Luzeaux

Gerry Myerson
Newsgroups: sci.math
From: armando@synapse.ee.ufl.edu (Armando Barreto)
Subject: HELP. TEST FOR EXTREMA IN f(x,y,z)
Keywords: Partial derivatives, Extrema, function of 3 variables.
Reply-To: armando@synapse.ee.ufl.edu
Date: Wed, 29 Jul 92 01:05:42 GMT


Hello,

Most Calculus books include a theorem like this to test extrema of a function  
of two variables (This one taken from Calculus ..  by Swokowsky):

Let f(x,y) be a function of two variables which has continuous second partial  
derivatives on a rectangular region Q, and let

		g(x,y) = fxx(x,y)fyy(x,y) - [fxy(x,y)]^2
		
for all (x,y) in Q. If (a,b) is in Q and fx(a,b)=0, fy(a,b) =0, then the  
following statements hold.

(i) f(a,b) is a local maximum if g(a,b)>0 and fxx(a,b) < 0.
(ii) f(a,b) is a local maximum if g(a,b)>0 and fxx(a,b) > 0.
(iii) f(a,b) is not an extremum of f if g(a,b)<0.

<<< NOTATION: fx(x,y) is the first partioal w.r.t. x, fxx(x,y) is the second  
partial w.r.t. x and fxy(x,y) is the second mixed partial >>> 

My question is :

IS THERE SUCH A TEST FOR EXTREMA IN A FUNCTION OF 3 VARIABLES :f(x,y,z) ?

I will appreciate any answers to the question or bibliographical references  
that can guide me in finding the test myself.

Also, I tend to interpret the first term of g(x,y) as the amount of concavity  
or convexity of f(x,y) at the test point (simultaneous curvature in x and y),  
but, is there a :"graphical interpretration" for the second term of g(x,y) ?

Thanks in advance.

Armando Barreto     armando@synapse.ee.ufl.edu
From: rose@fsu1.cc.fsu.edu (Kermit Rose)
Newsgroups: sci.math
Subject: Re: Stupid question about FLT
Date: 23 Jul 92 18:14:46 GMT
Reply-To: rose@fsu1.cc.fsu.edu
Organization: Florida State University
News-Software: VAX/VMS VNEWS 1.3-4

In article <1992Jul18.212618.15509@mixcom.com>, ttyytt@mixcom.com (Adam Costello) writes...
>In article <1992Jul18.005818.8468@infodev.cam.ac.uk> gjm11@cus.cam.ac.uk (G.J. McCaughan) writes:
>>The mistake is this:
>>
>>"True in all models is the same as provable. That's the completeness theorem."
>>
>>There isn't a completeness theorem for any of the systems M that anyone uses.
>>So FLT might be true in all models but not provable.
> 
>That was no mistake.  There is a completeness theorem.  A sentence is
>provable from M iff it is true in all models of M.  The problem here is
>that FLT is a theorem about the natural numbers, and there is no first-
>order axiomatization M of the natural numbers.  Any recursive set of
>axioms in a first-order language that purports to describe the natural
>numbers will always admit some models which aren't exactly like the
>natural numbers.
> 
>In particular, there are always models containing natural numbers which
>are not represented by any terms of the language.  In such a model FLT
>might be false and still not disprovable because there would be no way to
>cite the counterexample.
> 
>AMC

How does this relate to the proof that the integral domain represented by 
the integers is categorical?  I had understood that this meant that any 
model that satisfied the given axioms was isomorphic to the integers.  Is 
not the natural numbers isomorphic to a proper subset of this categorical 
system?

rose@fsu1.cc.fsu.edu          To be sure I see your response, use e-mail.
-----------------------------------------------------------------------
Be of good cheer, for it is much more fun than being depressed.
Newsgroups: sci.math
Subject: Re: HELP. TEST FOR EXTREMA IN f(x,y,z)
From: news@massey.ac.nz (USENET News System)
Date: Thu, 30 Jul 92 22:12:30 GMT
Organization: Massey University

In article <1992Jul30.144053.28402@news.cs.brandeis.edu>, palais@binah.cc.brandeis.edu writes:
> 
> Armando Barre asked                                                     
> 
> >My question is :
> 
> >IS THERE SUCH A TEST FOR EXTREMA IN A FUNCTION OF 3 VARIABLES :f(x,y,z) ?
> 
> Actually there is; someone else will have to provide a reference, but:
> 
> (i) f(a,b) is a local maximum if the Hessian is negative definite
> (ii) f(a,b) is a local minimum if the Hessian is positive definite
> (iii) f(a,b) is a local minimum if the Hessian is indefinite.
> 
> Arthur Rubin answered: 
> 
>    I think Arthur meant to say saddle point for case (iii) (but one must
> assume that the Hessian matrix is non-singular). The same result is true 
> in any dimension (and even in Hilbert space). The best reference is to any 
> proof of the "Morse-Lemma", which says that after a change of variables the 
> function locally is equal to a constant (its value at the critical point) 
> plus the quadratic function given bythe Hessian. (See Lang's "Differentiable 
> Manifolds" for the Hilbert space proof.)
> 
Of course there is a fourth case. When the Hessian is zero the stationary
point might be any of those types. This  also applies in case (iii) if the
Hessian does not have full rank.
From: stallcup@stsci.edu (Scott Stallcup)
Newsgroups: sci.math
Subject: Solution to a system of nonlinear inequalities
Keywords: nonlinear systems
Date: 28 Jul 92 22:02:01 GMT
Organization: Space Telescope Science Institute


 Can anyone point me in the direction to a possible solution for 
 the following system of nonlinear inequalities ?

  Given that P, T, Q, and N are 3D real cartesian vectors
  and r, a, and b are real scalers.  N is a unit vector.

      N . Q < r cos(a)
     
       T - P       Q - P  
     --------- . ---------  < cos(b)
     //T - P//   //Q - P//

     r = //Q//


     r < //T//
    
     r < //P//

     0 < a < Pi/2

     0 < b < Pi/2

  The vector Q is the only unknown (but the length (r) of Q is known).

  I don't need an actual solution for Q.  I only need to know if any
  solutions exist (a boolean). 

  References to canned software, books, or papers on solving systems
  of nonlinear inequalities would be helpful.


 ---------------------------------------------------
 Scott Stallcup (AURA),  stallcup@stsci.edu

 Space Telescope Science Institute
 Johns Hopkins University
From: ttyytt@mixcom.com (Adam Costello)
Newsgroups: sci.math
Subject: Re: You know, the integers (was: Re: Stupid question about FLT)
Date: 23 Jul 92 04:22:50 GMT
Article-I.D.: mixcom.1992Jul23.042250.10738
Organization: Milwaukee Internet Xchange BBS, Milwaukee, WI U.S.A.

In article <1992Jul20.173716.6310@galois.mit.edu> tycchow@riesz.mit.edu (Timothy Y. Chow) writes:
>
>But this is precisely the point!  Consider the "REAL" integers:  every
>set of axioms and syntactic rules that the "REAL" integers satisfies
>also admits nonstandard models.  How then do you propose to "define" the
>REAL integers with a fixed set of axioms and syntactic rules?

I'm sorry if I begin to sound like a broken record, but please remember
that this is only true if you restrict your axioms to first-order
languages.  The REAL integers _can_ be defined (to within isomorphism)
in a second-order language.  Of course, this still doesn't get you a
complete number theory.  Also, using a second-order language requires
that one already knows what sets are.  If these still have to be defined,
then we're back to the same problem.  It's inescapable.  Set theory is
built on logic, but logic is defined in terms of sets.  People obviously
knew and understood both concepts well, before either was formalized.

AMC
Newsgroups: sci.math
From: bs@gauss.mitre.org (Robert D. Silverman)
Subject: Re: Re; prime quadruplets ... partial apology
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Date: Wed, 29 Jul 1992 18:37:25 GMT

In article <keRf_Ru00iUy02UVMt@andrew.cmu.edu> ow0a+@andrew.cmu.edu (Oswald Wyler) writes:
:My memory was at fault, the problem whether there are infinitely many
:quadruplets 30n+11, 30n+13, 30n+17, 30n+19 of primes, as is the problem
:whether there are infinitely many twin primes p, p+2.  However, I cannot
:buy the "almost certainly" designation -- as little as I can buy Math
:by authority.  The question about quadruplet records still makes sense,
:however.


Your question is a special case of the prime k-tuples conjecture. This
conjecture is backed by enormous amounts of statistical evidence, as well
as many heuristic arguments which suggest why it is true. 

Read Halberstam & Richert's book "Sieve Methods". Then, if you would like
to discuss reasons why the conjecture might or might not be true, I will
be very glad to do so. 

I do not know what your math background is, so please don't take this as
a flame. It is not intended as such.

However, based upon your posts I believe that you simply don't know enough
about the mathematics at hand to either accept or reject the conjecture.
Thus, your statement "I cannot buy the almost certainly designation" is made
without sufficient knowledge for you to have a valid opinion. 

Go study the math behind the k-tuples conjecture. 

If a way can be found to solve the parity problem in sieve methods, it will
yield a proof of the twin prime conjecture. The same technique will apply
to the k-tuples conjecture.

You are not buying "math by authority". What you are buying is the reasoned
opinion of many different expert mathematicians who have studied this problem
from many different aspects. I know of no number theorist who doubts that it is
true.

--
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"
From: amlogan@phoenix.Princeton.EDU (Adam M. Logan)
Newsgroups: sci.math
Subject: Re: High Order Undulation Seeds and Binary Undulants
Date: 24 Jul 92 17:11:21 GMT
Organization: Princeton University
Originator: news@ernie.Princeton.EDU

In article <9207241316.AA14033@ucbvax.Berkeley.EDU> cliff@watson.ibm.com ("Cliff Pickover") writes:
>"Binary undulants" are numbers of the form 2**N that undulate
>(alternate) the digits 1 and 0 in their decimal expansion.
>
>For example, 2**[9]49 is the "highest quality" binary undulant so far
>discovered because it has the undulating sequence 101010 in it:
>
>2**949 = [...?]
>4758454107128905800953799994079681792420032645310062268978469949811010102913
[...]>Here N = 949 is called an undulation seed of order 6, since it
>gives rise to a 6-digit undulation pattern of adjacent 1's and 0's.
>
>Are undulation seeds of higher order easily found using either
>mathematical prowess or computation power?

Interesting problem...I'll try the "mathematical prowess".  In fact, I'll 
restrict myself to finding powers of 2 that start with adjacent,
alternating 1's and 0's.  Let a_n be the n-digit number 10101..., and let
b_n = log_10 (a_n), c_n = log_10 (a_n + 1).  Of course, a necessary and
sufficient condition that 2^k start with a_n is that the fractional part
of log_10 (2^k) be at least that of b_n and less than that of c_n.  But
log_10 (2^k) = k log_10 (2), and it is known (since log_10 (2) is
irrational) that this sequence is uniformly distributed mod 1, asymptotic-
ally the proportion of powers starting in a_n is c_n - b_n, which, in fact,
for decent-size a_n (say >= 10) is about 1/(a_n ln 10).  Of course, 
similar results hold for other bases, and thjis works for any
old digit string, not just a_n.  A more sophisticated analysis
could look for patterns within the number, of course, but this is good 
enough to show existence.  As for finding them, compute the logs to high 
enough precision, etc.

Adam
From: smith@gramian.harvard.edu (Steven Smith)
Newsgroups: sci.math
Subject: Re: Recreational Math Journals
Date: 24 Jul 92 17:01:05 GMT
Article-I.D.: gramian.SMITH.92Jul24120105
Organization: Harvard Robotics Lab, Harvard University

cliff@watson.ibm.com ("Cliff Pickover") writes:

> I'm interested in what mathematical journals exist that are interested
> in "light" math.
> . . . . . . . . .
> Are there others?

The Queen of MathLite: The Mathematical Intelligencer, published
monthly by Springer-Verlag.  Where else can you find postage stamps
featuring Dedekind with a factorization into prime ideals?

Steven Smith