HDTenEightyP/NetNews huggingface dataset:
From: oliver@oak.math.ucla.edu (Mike Oliver)
Subject: Re: An elementary, "naturally occurring" problem
Organization: UCLA Mathematics Department
Date: Fri, 10 Apr 92 00:04:46 GMT

In article <1992Apr9.201215.1319@wdl.loral.com> mab@wdl39.wdl.loral.com (Mark A Biggar) writes:

>Question: is it possible to partition the interval [0,1] into a finite
>number of uncountable sets that are all dense on the interval [0,1]?
>If so, can you give a rule to determine partition membership?

Easily.  Given x \in [0,1], put x in the set A if x is rational and
less than 1/2 or irrational and greater than 1/2.  Put x in B if x
is irrational and less than 1/2 or rational and >= 1/2.  Then A and
B are both dense in [0,1] and [0,1] is the disjoint union of A and
B.

But of course this is sort of a trivial evasion taking advantage of the
fact that the question probably did not capture what was in the asker's
mind.  For example the intersection of A with [0,1/2] is countable,
as is the intersection of B with [1/2,1].  One might ask:  is
there such a partition such that the intersection of each set
with any open interval is uncountable?

Given the axiom of choice, there is (in fact we can make each
of the intersections have the cardinality of the continuum).
To see this, let S be the set of all open subintervals of [0,1],
and let h be a bijection from 2^{\Aleph_0} onto S X 2^{\Aleph_0}.
Also suppose given a wellordering of [0,1].
Now do a transfinite induction of length 2^{\Aleph_0} as
follows:  At stage \alpha, look at the interval given by the
first coordinate of h(\alpha).  Choose the smallest real
in this interval (that is, smallest in the given wellordering)
that hasn't yet been assigned to either A or B, and throw it into A.
Throw the next smallest into B.

You always have available reals in the appropriate interval to assign
to A and B, because you have always completed less than 2^{\Aleph_0}
steps whereas there are 2^{\Aleph_0} reals in each interval.

At the end, if any reals are left over, throw them into A.

Now A and B give a partition of [0,1], and each of them has
intersection with cardinality 2^{\Aleph_0} with every open
subinterval of [0,1].

I don't know whether it can be done without AC.  It seems like
the kind of thing that might be inconsistent with the Axiom of
Determinacy but I don't immediately see why.

--
------------------------------------------------------------------
Mike Oliver 
UUCP:...!{ucbvax,{hao!cepu}}!ucla-cs!math.ucla.edu!oliver
ARPA: oliver@math.ucla.edu
Newsgroups: sci.math
From: c3ar@zaphod.uchicago.edu (Walter C3arlip)
Subject: Re: nonexistance of counterexample -> existance of proof?
Organization: University of Chicago Computing Organizations
Distribution: na
Date: Fri, 10 Apr 1992 15:53:30 GMT

In article <1992Apr10.084947.6478@unixg.ubc.ca> israel@unixg.ubc.ca (Robert
B. Israel) writes: 
> [stuff deleted]
>    Actually, there IS a counterexample to your "theorem".  I assume that
>you have a precise definition of what it means for a "finite expression"
>to "represent" a real number (otherwise the "theorem" is ambiguous).
>Those "finite expressions" that represent real numbers can then be 
>enumerated (e.g. in order of the positive integers that we get when 
>representing ASCII strings as hexadecimal digits). 
>Let r(n) be the number represented by the n'th "finite expression"
>that represents a real number.  Let d(n) be the n'th decimal digit 
>(after the decimal point) of r(n) (with whatever convention you want
>for numbers with nonunique decimal expansion).  Let c(n) = 1 if d(n) = 2
>and c(n) = 2 otherwise.  My counterexample is 0.c(1)c(2)c(3)... .
>
>    Have I specified a unique real number?  Yes . . .

Yes, and your paragraph above is a finite expression that precisely
describes your counterexample (a number that cannot be expressed
via a finite expression) . . . oops. :-) :-) :-)

--Walter
_____________________________________________________________________________
Walter C3arlip 		        	**** carlip@ace.cs.ohiou.edu ****
(the "3" is silent)                  
_____________________________________________________________________________


_____________________________________________________________________________
Walter Carlip 		        	**** carlip@ace.cs.ohiou.edu ****
(the "3" is invisible)                  
_____________________________________________________________________________
Newsgroups: sci.math
From: martin@lyra.cis.umassd.edu (Gary Martin)
Subject: Re: factoid about new prime
Organization: University of Massachusetts Dartmouth
	<1992Apr9.111809.19221@linus.mitre.org>
	<MARTIN.92Apr10092553@lyra.cis.umassd.edu> <4068@crl.LABS.TEK.COM>
Date: Sat, 11 Apr 1992 04:56:08 GMT

In article <4068@crl.LABS.TEK.COM> scott@ferrari.LABS.TEK.COM (Scott Huddleston) writes:

   In article <MARTIN.92Apr10092553@lyra.cis.umassd.edu> martin@lyra.cis.umassd.edu (Gary Martin) writes:
   +In article <1992Apr9.111809.19221@linus.mitre.org> bs@faron.mitre.org (Robert D. Silverman) writes:
   +   :>In article <3987@crl.LABS.TEK.COM> scott@ferrari.LABS.TEK.COM (Scott Huddleston) writes:
   +   :>:It was mentioned on CNN that the new prime number discovered
   +   :>:recently is four times bigger then the previous record.
   +
   +   4 times as big as P does NOT equal 4P. Size is measured by number of
   +   bits. 4 times as big means 4 times as many bits. CNN got it RIGHT.
   +   Most people here are getting it wrong.
   +   --
   +What are the chances that CNN understood why what they said was right?
   +(Particularly considering that there are people here who don't.)
   +If one takes a skeptical view of the popular media's understanding
   +of math, it does make a good joke.

   And the joke works under either interpretation of size.  When the CNN joke
   came out, I cancelled my background job searching for primes of the form
   4*n.  After Bob clarified size for us, I had to cancel another background
   job searching for primes of the form n^4.
   --
But isn't the size of a number the integer part of the log?  In that case
lots of numbers are 4 times the size of n, not just  n^4.

--
Gary A. Martin, Assistant Professor of Mathematics, UMass Dartmouth
Martin@cis.umassd.edu
Newsgroups: sci.math
From: price@linus.serum.kodak.com (Kevin Price (x37128))
Subject: Re: Factoring an algebraic expression ---  HELP?
Organization: Clinical Products Division, Eastman Kodak Company
Distribution: usa
Date: Fri, 10 Apr 1992 12:03:28 GMT


Thanks for the response.  It seems not to be factorable any further, since
all the variables are reals.



-- 
-Kevin Price		price@serum.kodak.com             
                                                          
 (please use the address above ^   The mailer may call me 
  nobody@kodak.com, but IMHO, I am not.)
From: ocengiz@math.okstate.edu (Ozgener Cengiz)
Subject: Ramsey Numbers
Organization: Oklahoma State University, Math Department
Date: Wed, 8 Apr 1992 15:50:07 GMT

Are the Ramsey numbers R(4,5), R(4,6), and R(5,5) known?
If so, what are they?  

Thanks in advance,

Altay Ozgener




Newsgroups: sci.math
From: cyeomans@ms.uky.edu (Charles Yeomans)
Subject: Re: A simple question about R2 - Q2 (irrational points of the plane)
Date: Wed, 8 Apr 1992 14:06:40 GMT
Organization: University Of Kentucky, Dept. of Math Sciences
X-Bytes: 

In article <1992Apr7.174510@masg1.epfl.ch> baumgart@masg1.epfl.ch (Marcel Baumgartner) writes:
>Keywords: In R2 - Q2, does there exist an oblique straight line ?
>

Certainly.  Let a = sqrt(2) and consider the line ax + y = 1.  Assume that
X, Y are rational numbers satisfying aX + Y = 1.  Then a = (1 - Y)/X and 
thus a is rational, contradicting the well-known fact that a = sqrt(2)
is not rational.  Thus there are no rational points on the line.

Charles Yeomans
cyeomans@ms.uky.edu
yeomans@austin.onu.edu

Newsgroups: sci.math
From: parson_r@cubldr.colorado.edu (Robert Parson)
Subject: Re: quaternians, tracking, 3 dimensions
Organization: University of Colorado, Boulder
Date: Tue, 7 Apr 1992 18:43:33 GMT

In article <1992Apr7.030814.19053@njitgw.njit.edu>, andrews@rehab.njit.edu (John Andrews) writes:
>      I am told the most
>      efficient method of computing the relative position of
>      the object is with units of measure called
>      "quaternians". Can anyone give me some initial insight
>      into what quaternians are, how they are used, why they
>      are efficient, and/or how cumbersome the calculations
>      of the above directional parameters can be using
>      quaternians math?

 The idea is to devise a singularity-free global coordinate system for
describing the orientation of a body in three-space. Euler angles, etc.
have singularities which slow down the numerical integration of the
equations of motion. In their place, one can introduce a set of four
parameters - known as Euler parameters or Cayley-Klein parameters (actually
these names refer to two choices which are not identical but are simply
related, the Euler parameters being the real and imag. parts of the complex
Cayley-Klein parameters) which are not all independent, but which give rise
to nonsingular equations of motion. One also gets rid of a lot of trig.
functions in the e. of. m. , so it's worthwhile having to carry an extra
equation along in the numerical integration. One may regard the euler or
Cayley-Klein parameters as mapping the orientation space of the body onto
the surface of a 4-dimensional sphere.

It turns out that the euler parameters transform as the components of a
quaternion. One can carry out many of the manipulations very efficiently by
using quaternion algebra.

For background material on Euler and Cayley-Klein parameters, look at a
graduate text on Classical Mechanics (such as Goldstein, _2nd Edition_,
or Whittaker's old book.) A nice review of their use in molecular dynamics
simulations is:
"On the representation of orientation space"
D. J. Evans, _Molecular Physics_  vol. ??,  p. 317 (1977)
(sorry - I have a copy of the paper but the volume # is missing. The year
might be 1978.)

Robert
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Newsgroups: sci.math
Subject: Re: polyomino classification
Keywords: polyomino
Date: 14 Apr 92 20:44:53 GMT
Organization: Beckman Instruments, Inc.

In <seghkINNd3c@agate.berkeley.edu> reid@durban.berkeley.edu (michael reid) writes:


>          quadrant  ==>  three-quarter plane  ==>  plane

>the first implication is immediate, but i need the axiom of choice
>to deduce the second.  (if anyone can prove the second without the
>axiom of choice, please let me and sci.math know.)  as a hint,
>try to prove that if a polyomino can cover an N-by-N square for all N
>(no overlap, and pieces may hang over the edges of the square), then
>it tiles the plane.

Ah Hah; I see it.  (Does that proof REALLY require the axiom of choice; I
would think that, since you are dealing with finite subsets of the plane
with integer sides, you should be able to order the choices, and take a
"first" choice at each induction step.)

--
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.
Ich bin ein Virus.  Mach' mit und kopiere mich in Deine .signature.
Newsgroups: sci.math
From: schwartz@symcom.math.uiuc.edu (Gary Keith Schwartz)
Subject: Re: A Proof about Groups
Organization: Math Dept., University of Illinois at Urbana/Champaign
Date: Wed, 15 Apr 1992 16:39:35 GMT

In article <X3F6iB1w164w@netlink.cts.com> teacher@netlink.cts.com (Daniel Pick) writes:
>In Joseph Landin's book, 'An Introduction to Algebraic Structures,'
>the author makes the following definitions:
>A group G which contains only a finite number of elements is a 
>finite group.
>An element x of a group G has finite order if and only if
>there is a positive integer n such that x^n = e, where e
>is the neutral element.
>Then he asks the reader to prove that if G is a finite group,
>then every element in G has finite order.
>This proof stumps me.  If you induct on the number of elements in
>the group, how do you show that each successive element has finite
>order?
>

Let x be an element of G and let S=the set of all powers of x (positive
only) since G is closed under multiplication, S is a subset of
G. Thus, there must be a least positive integer power of x which
which repeats ie there is an N such that for some M<N x^M=x^N.
Then x^(N-M) must be the neutral element.
Newsgroups: sci.math
From: sethb@fid.morgan.com (Seth Breidbart)
Subject: Re: On knowing who's who (was: Fermat's Last Theorem)
Organization: Morgan Stanley & Co., New York, NY
Date: Wed, 15 Apr 1992 21:58:33 GMT

In article <1992Apr13.080612.24177@etl.go.jp> andreas@etl.go.jp
(Andreas Knobel) writes:

>In fact Erdos is so famous that I am sure you are aware of the existence
>of the _Erdos distance_, defined on the set of mathematicians:
.
.
>I heard this function was first defined by Erdos.
>
>Andreas (Erdos distance 7)

It's generally referred to as Erdos number, not Erdos distance.

It was definitely _not_ first defined by Erdos (that would be rather
egotistical, wouldn't it?  Besides, if Erdos defined it, people like
me  wouldn't be able to calculate their Erdos numbers :-).  When I met
Erdos (in 1979, I think; it might have been 1978) he wasn't aware of
Erdos numbers.  (This is not folklore; I was there when he was told
about them.)

Seth		sethb@fid.morgan.com
Newsgroups: sci.math
From: obryankw@NeXTwork.Rose-Hulman.Edu (Kevin W. Obryant)
Subject: Re: Test wanted for "groupness"
Reply-To: obryankw@NeXTwork.Rose-Hulman.Edu (Kevin W. Obryant)
Organization: Rose-Hulman Institute of Technology
Date: Wed, 15 Apr 1992 21:13:44 GMT

In article <31661@sdcc12.ucsd.edu> ec120wee@sdcc4.ucsd.edu (Anthony  
Minkoff) writes:
> I want a way to determine efficiently whether a given set and
> operation constitute a group.  In particular:
> 
> Suppose I am given a set and an operation; the operation is
> described by specifying a*b for all ordered pairs (a,b) in SxS.
> 
> I can write down the definition of the operation as a matrix.  The
> element in the ith row and jth column is i*j.  Now I want to know,
> is it a group?
> 
> For example,
> 
>      1 2 3 4 5
>      2 1 5 3 4
>      3 4 1 5 2
>      4 5 2 1 3
>      5 3 4 2 1
> 
> (This particular example, as you may have realized, is not a group.)
> 
> Of course, it is simple to determine, at a glance, whether an
> identity exists, and whether each element has an inverse.  But
> associativity?  How am I to test associativity, without testing all
> ordered triples?  For a reasonably large group, there are
> approximately n^3 triples that need to be tested.
> 
> I have figured out some things to look for.  E.g., if any element is
> duplicated in any row or column, then it's not a group.  If the
> positions of the identities are not symmetric about the diagonal,
> then it's not a group.
> 
> I have also determined that certain triples certainly do not need to
> be tested, once the above conditions are satisfied.  For example, if
> e is the identity, then certainly (ae)b = a(eb), and there is no need
> to test such triples.  Furthermore, if a is the inverse of b, then
> there is no need to test aba.
> 
> Is there a way, short of testing all the triples, to determine
> whether it's a group?  A couple of professors I asked here at UCSD
> were unable to answer this question, and a couple of textbooks I
> consulted did not even address this issue.
> 
> I'm mostly looking for something I can use myself, to determine
> withing a reasonable amount of time whether a given operation,
> specified as above, is a group on a given set.  But, more
> rigorously,
> 
> Given an operation over the integers 1..n, specified as above (i.e.,
> a*b is given, for all a, b in 1..n), what is the most efficient
> algorithm for determining whether the operation and the integers
> 1..n constitute a group?  Is it possible to do better than o(n^3)?
> 
> My gut tells me that it must be possible to do better than o(n^3),
> but I don't know how.
> 
> Tony
> -- 
> Tony Minkoff
> aminkoff@sdcc13.ucsd.edu
> ec120wee@sdcc4.ucsd.edu

I'm not sure of the results, but Gallian's "Contemporary Abstract Algebra"  
does address the issue.  As this book has an excellent elementary  
bibliography (chapter by chapter), there is likely a good reference as  
well as a brief description of the known results.

Kevin O'Bryant
obryankw@hydra.rose-hulman.edu
Newsgroups: sci.math
From: carabalo@phoenix.Princeton.EDU (David G. Caraballo)
Subject: Re: An elementary, "naturally occurring" problem:  oops
Originator: news@ernie.Princeton.EDU
Organization: Princeton University
Date: Tue, 7 Apr 1992 18:39:18 GMT

I have received nearly 20 replies about the first part of my question, but 
none about the correctly phrased part of the rest of it.  It is reproduced
below from a previous post of mine:

In article <1992Apr3.045519.9839@Princeton.EDU> carabalo@phoenix.Princeton.EDU
(David G. Caraballo) writes:

>Prove or disprove that there exists a bijection f:[0,1] -> [0,1] that is
>not continuous at any point and such that f has the following geometric
>property:  Any line on the plane that intersects the interior of [0,1]x[0,1]
>contains exactly one or two points of the graph of f.
>                     ^^^^^^
>David G. Caraballo



DGC














Newsgroups: sci.math
From: charlie@umnstat.stat.umn.edu (Charles Geyer)
Subject: Re: Zero, "Arabic" numerals, etc.
Organization: School of Statistics, University of Minnesota
Date: Fri, 10 Apr 1992 16:02:06 GMT

In article <1992Apr9.003728.7659@midway.uchicago.edu>
trivedi@curie.uchicago.edu (Anil Trivedi) writes:

> The PHILOSOPHICAL mystery that seems to intrigue you has shifted, as far
> as mathematics is concerned, [from the  number zero] to the concept of
> "null set". 

What's the difference?  I thought the number zero was the null set.

-- 
Charles Geyer
School of Statistics
University of Minnesota
charlie@umnstat.stat.umn.edu
Xref: sparky sci.math:5815 sci.crypt:1840
From: bs@gauss.mitre.org (Robert D. Silverman)
Newsgroups: sci.math,sci.crypt
Subject: Re: Tangent^2 re: Mersenne
Date: 9 Apr 92 14:01:21 GMT
Article-I.D.: linus.1992Apr9.140121.25691
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA

In article <1992Apr6.231148.10786@husc3.harvard.edu> elkies@ramanujan.harvard.edu (Noam Elkies) writes:
:In article <1992Apr1.185322.22929@linus.mitre.org>
:bs@gauss.mitre.org (Robert D. Silverman) writes:
:
:> Now, if [p is prime and] 2p+1 is prime, then it will divide 2^p-1
:>with probability 1/2. (Theorem! This arises from quadratic reciprocity
:>and the fact that 2 is a quadratic residue of 1/2 the primes).
:
:Whoa --- one also needs that half the primes p *such that 2p+1 is prime*
:are congruent to 1 mod 4; it is still a big open problem to prove that
:there are infinitely many such p to begin with!  I certainly have no
:reasonable doubt that there are infinitely many, and that they are
:correctly distributed mod 4, but a "Theorem!" it isn't.
:
:--Noam D. Elkies (elkies@zariski.harvard.edu)
:  Department of Mathematics, Harvard University


I did not claim that there were an infinite number of p such that 
p and 2p+1 were both prime. Merely that IF p and 2p+1 are both prime
then it divides 2^p-1 with probability 1/2 and 2^p+1 with probability
1/2. It must divide one or the other (Lagrange's theorem) and divides
on the minus side when 2 is a residue and on the plus side when it is
a non-residue.
 
Indeed, it is a well known oustanding conjecture that there ARE an
infinite number of such pairs, p and 2p+1 both prime.


An expert in sieve methods tells me that if one assumes
that there are an infinite number of p such that p and 2p+1 are both
prime, then it is possible to show that they must be correctly 
distributed mod 4. That it would violate some known results in sieve
methods if they were not. Since I am a novice to sieve methods
[I am just starting to learn what they are about] I don't know this for
myself. Apparently one looks at some Dirchlet character sums
Sum(x < N) chi(x) where chi(x) is the appropriate Dirchlet character mod 4.
One then follows through with Dirichlet's proof for primes in an arith.
prog. subject to the side condition that p and 2p+1 are both prime.
One applies a weighted sieve to the character sum and shows that it has
the right order of magnitude as N --> infty. 

I can follow the general gist of this argument [I've worked through
Dirichlet's proof], but don't know enough math to follow the details.

However, I'm learning!! [at least I hope I am]. It's fascinating stuff.
 
Apparently some of Heath-Brown's recent techniques that he developed
when he proved Artin's conjecture are relevent.
--
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: scott@ferrari.LABS.TEK.COM (Scott Huddleston)
Newsgroups: sci.math
Subject: Re: factoid about new prime
Date: 10 Apr 92 19:02:14 GMT
Reply-To: scott@crl.labs.tek.com
Organization: Computer Research Laboratory, Tektronix, Inc., Beaverton OR

In article <MARTIN.92Apr10092553@lyra.cis.umassd.edu> martin@lyra.cis.umassd.edu (Gary Martin) writes:
+In article <1992Apr9.111809.19221@linus.mitre.org> bs@faron.mitre.org (Robert D. Silverman) writes:
+   :>In article <3987@crl.LABS.TEK.COM> scott@ferrari.LABS.TEK.COM (Scott Huddleston) writes:
+   :>:It was mentioned on CNN that the new prime number discovered
+   :>:recently is four times bigger then the previous record.
+
+   4 times as big as P does NOT equal 4P. Size is measured by number of
+   bits. 4 times as big means 4 times as many bits. CNN got it RIGHT.
+   Most people here are getting it wrong.
+   --
+What are the chances that CNN understood why what they said was right?
+(Particularly considering that there are people here who don't.)
+If one takes a skeptical view of the popular media's understanding
+of math, it does make a good joke.

And the joke works under either interpretation of size.  When the CNN joke
came out, I cancelled my background job searching for primes of the form
4*n.  After Bob clarified size for us, I had to cancel another background
job searching for primes of the form n^4.
--
Scott Huddleston
scott@crl.labs.tek.com
From: rmallier@mafalda (Roland  Mallier)
Newsgroups: sci.math
Subject: theta functions
Date: 14 Apr 92 21:32:19 GMT
Organization: McGill University Computing Centre

Can anyone answer the following questions about theta functions?
 (they're quasi-doubly periodic functions such that the
Jacobian elliptic functions cn, sn, dn, etc are ratios of two
theta functions)

i) what is theta_{s}(x + i y) * theta_{s}(x-iy) in terms of functions of
x and functions of y only ?

ii) what is the integral of (theta_{s} (z))^{-2}) ?

I've tried Abramowitz & Stegun and Gradstheyn and Ryzhik without success

-Roland
From: wsadjw@rw7.urc.tue.nl (Jan Willem Nienhuys)
Newsgroups: sci.math
Subject: Re: Fermat's Last Theorem
Date: 7 Apr 92 14:38:23 GMT
Reply-To: wsadjw@urc.tue.nl
Organization: Eindhoven University of Technology, The Netherlands

In article <STROMME.92Apr6130830@jeeves.math.utah.edu> stromme@math.utah.edu (Stein Stromme) writes:
>Why not just answer the question? After all, the poster says "Pardon
>my ignorance", and then he does not deserve to be run over like that.
>
>Stein 

Well, the poster is the [expletive deleted] Derek Abbott who has 
developed a habit of asking silly questions in a tone as if he is
a kind of know-all, and that about a wide range of subjects, among
which quantum mechanicss, free will, evolution, entropy, and 
various instances in sci.math.  
Let's say, some people are like that.
The respondee, Bob Silverman is known for the somewhat caustic way
of expressing himself (with a dash of jusitified superciliousness),
but in this case his response showed remarkable restraint. Of course,
ignoring D.A. altogether would have been even better.

JWN
Xref: sparky sci.math:5758 sci.logic:861