HDTenEightyP/NetNews huggingface dataset:
From: mab@wdl39.wdl.loral.com (Mark A Biggar)
Newsgroups: sci.math
Subject: Re: Rounding Rules
Date: 21 Dec 92 23:41:14 GMT
Organization: Loral Western Development Labs

In article <LOTTO.92Dec21062820@math.math.berkeley.edu> lotto@math.berkeley.edu (Ben Lotto) writes:
>>>>>> On Sun, 20 Dec 1992 00:30:18 GMT, shaw@toadflax.UCDavis.EDU (Rob Shaw) said:
>Rob> What is the rationale behind the following rounding rule?
>Rob> when dealing with 5's followed by all zero's, check the
>Rob> next digit to the left. If it's even, round down; odd,
>Rob> round up.
>I was told way back in my undergraduate days that one should generally
>"round to evens."  Someone recently explained the rationale for this
>as follows.  Suppose we start with 1.49 and round in the traditional
>manner twice.  First we would round to 1.5 and then to 2.  On the
>other hand, if we round just once we get 1.  Rounding to evens
>prevents us from getting different answers.

Round to even also has the nice property round(abs(x)) = abs(round(x)), which
given that most floating point representations on computers use  a 
signed-magnitude format means that the rounding hardware doesn't have to
look at the sign bit to do the rounding.

--
Mark Biggar
mab@wdl1.wdl.loral.com
Newsgroups: sci.math,sci.philosophy.tech
From: holmes@opal.idbsu.edu (Randall Holmes)
Subject: Re: Numbers and sets
Organization: Boise State University
Date: Wed, 23 Dec 1992 17:51:45 GMT

In article <Bzosz1.FMx@cantua.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>In article <1992Dec19.140927.18700@husc3.harvard.edu>, zeleny@husc10.harvard.edu (Michael Zeleny) writes: 
>
>|> The Axioms of Foundation and Choice are analytically
>|> true of sets;
>
>What does this sentence mean ? 
>Could someone please explain further ?
>Why can it be so blithely asserted ?
>
>--------------------------------------------
>Bill Taylor.       wft@math.canterbury.ac.nz 
>Bill Trylor. que  rwft@maih.casterkury.aa.n! 
>Tiel Tryloco quer rwst@maihuc sterkesy.ga.n!
>Thelworyd co quer rwsi@mvihus strikesy.gain!
>The world conqueror sig-virus strikes again!
>--------------------------------------------

What the sentence purports to mean is that the axioms of foundation
and choice are true of sets for the same kind of reason that bachelors
are necessarily unmarried; that they are included in the definition of
the notion of "set".  The sentence is false.

The power of blithe assertion is independent of the truth or
falsehood, plausibility or implausibility, of the sentences asserted.

The axiom of foundation asserts that each set is disjoint from at
least one of its elements; it ensures that sets are constructed in an
orderly fashion starting with the empty set or perhaps with atoms as
well, each set being a set of previously constructed sets.  This
prevents such oddities as sets which are elements of themselves
(which, I hasten to point out to the uninformed, are _not_ paradoxical
and do turn up in some set theories).

The axiom of choice asserts that, given any collection of disjoint
sets, there is some set which consists of exactly one element of each
element of the collection of disjoint sets; as Lord Russell put it, if
we have infinitely many pairs of socks, we may choose one sock from
each pair and form a set (in many different ways).

Both axioms have powerful common-sense arguments behind them, but it
is also the case that there are good arguments against both of them.
I don't think that either of them is part of the _definition_ of the
concept "set".  For the record, my official set theory includes choice
and denies foundation; I'm a professional set theorist, so you might
want to take this into account in evaluating Zeleny's claim.






-- 
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: ac999239@umbc8.umbc.edu (ac999239)
Subject: Re: Would Pi repeat, were it expressed in a base other than 10?
Organization: University of Maryland, Baltimore County Campus
Date: Tue, 22 Dec 1992 19:14:29 GMT

>>> That is say non-decimal.

>> Intuitively (not saying much, that is), I'd say yes, but that it would re-
>> quire a transcendental base....

> Try writing it in base 3.1415926--- :->

I don't know if one can truely say there is no base...I would wager to say
an integral base probably can be proved or has been proved not to repeat.
But as the guy above said, if you choose base pi then the number pi would
be 10.
From: geversti@oasys.dt.navy.mil (Gordon Everstine)
Newsgroups: sci.math
Subject: Re: Need: prpndcular dist from pt to line
Date: 30 Dec 92 19:36:38 GMT
Reply-To: geversti@oasys.dt.navy.mil (Gordon Everstine)
Organization: David Taylor Model Basin, Bethesda, Md.

Since the available information included the coordinates of the point 
P and the end points A and B of the line, another algorithm would be 
to use the vector cross product to compute twice the area of the triangle 
PAB and divide by the length of the vector AB.
From: max@west.darkside.com (Erik Max Francis)
Newsgroups: sci.math
Subject: Re: Self-Avoiding Random Walks
Date: 23 Dec 92 18:56:23 GMT
Organization: The Dark Side of the Moon +1 408 245 SPAM

pkeller@sam.math.ethz.ch (Philipp Keller) writes:

> It seems to be that the average length of a path is
> about 116 units.

That's pretty neat.

> I was wondering if there is any proof of this or if
> this a yet unsolved problem and if there is some good
> literature about this.

It would depend on a few things.  For instance, how big is the screen (in 
pixels)?  Does it wrap around?  Where does the walker initially start?  
(The last question doesn't matter if the screen wraps toroidally.)

More important, perhaps, is:  How is movement actually done?  Does it 
pick a random direction and walk one space, then pick another and walk 
that space, etc.?  What happens if it tries to step on a pixel it already 
has?  Nothing, or does it automatically have to pick another direction, 
counting as the same step?


  ____  Erik Max Francis -- ..!apple!uuwest!max -- max@west.darkside.com   __
  \  /  1070 Oakmont Dr. #1  San Jose, CA  95117  37 20 N 121 53 W  ACGT  /  \
   \/   Like strategic interstellar conquest games?  Ask about UNIVERSE!  \__/
        Omnia quia sunt, lumina sunt.   All things that are, are lights.
Newsgroups: sci.math,rec.puzzles
From: asymptot@sbe (Mike Mathews)
Newsgroups: sci.math
Subject: Re: What is the number
Date: 31 Dec 92 19:03:01 GMT
Organization: School of Business, California State University, Hayward

In article <BzE0v9.C1F@unccsun.uncc.edu> sbardhan@uncc.edu writes:
>Whose sum of the factorials of the digits is the number itself
>( no programming please !).

Well, 1 and 2 are trivial answers to this...

mmathews@sbe.csuhayward.edu
Newsgroups: sci.math,rec.puzzles
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Subject: Probability of volunteered information is not known  Was: Re: Marilyn Vos Savant's error?
Organization: Purdue University Statistics Department
Distribution: na
Date: Fri, 1 Jan 1993 12:41:36 GMT

In article <1992Dec31.203934.1@stsci.edu> zellner@stsci.edu writes:
>>   1) You meet a man on the street and ask him how many children he has.
>>      He replies "two, and one is a boy."  What is the probability that
>>      his other child is also a boy?

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

>>The answer to problem 1 is 1/3, while the answer to problem 2 is 1/2.

>> The answer to both is 1/2.  One child whose gender is not known, and 
>> two choices for that gender.  It's a 50-50 proposition.

The first result for this is correct, BUT for the situation in which 
The scenario is as follows:

Q1:  How many children do you have?

A2:  Two.

Q2a:  Is one a boy?

A2a:  Yes.

In this case the probability that the other child is a boy is 1/3.

However, if

Q2b:  What is the sex of the older child?

A2b:  Male.

In this case the probability (excluding twins) that the other is a boy is 1/2.

However, if the result is volunteered, we need to know the probability,
given the sexes of the children, that the information will be volunteered.
That the information is volunteered, rather than given as the result of a
question, makes quite a difference.  This is the main reason why randomized
experiments are performed.

-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@snap.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!snap.stat!hrubin(UUCP)
From: vounckx@goya.esat.kuleuven.ac.be (Johan Vounckx)
Newsgroups: sci.math
Subject: algorithm : strictly ordered relation
Date: 22 Dec 92 14:40:44 GMT
Organization: Catholic University of Leuven, Belgium

Hi,

I'm looking for algorithms that solve the following problem :
	given :	a set and a relation defined between the elements.
	question : determine a strictly ordered subrelation,
		   that connects all elements, otherwise
                   stated : order the elements, according to
		   some subrelation of the given relation.

Can anyone give me some algorithms and their complexity, or
some references where I can find them.

Thank you very much !

Johan

--------------------------------------------------------------------
Johan Vounckx
K.U.Leuven - ESAT Laboratory      Tel. +32-16-22 09 31
Kard. Mercierlaan 94              Fax. +32-16-22 18 55
B-3001 Heverlee, Belgium          email: vounckx@esat.kuleuven.ac.be
--------------------------------------------------------------------
Newsgroups: sci.math
From: ronl@tessi.com (Ron Lunde)
Subject: I imagine this comes up all the time...
Summary: What's the smallest N with 15 unique factors other than 1 and N?
Keywords: unique factors
Organization: Test Systems Strategies, Inc., Beaverton, Oregon
Date: Sun, 3 Jan 1993 00:35:21 GMT

I'm not a mathematician, so I'd appreciate it if anyone could point me to
any references on this topic (that I might be able to understand :-)).

The question I was toying with last night was: Given a number N,
what is the smallest number M such that aside from 1 and M, M has exactly
N unique factors (I guess I'm probably using "factor" in a funny way here,
since I'm not referring to primes, necessarily).  Obviously there *is* such
a number, since we can construct at least one by multiplying the first N
primes.  It seems odd that the first ones are all fairly small, but I can't
find the 15th (at least nothing smaller than 614889782588491410):

N Divisors Smallest M  Factors
2          6           2, 3
3          16          2, 4, 8
4          12          2, 3, 4, 6
5          64          2, 4, 8, 16, 32
6          24          2, 3, 4, 6, 8, 12
7          36          2, 3, 4, 6, 9, 12, 18
8          48          2, 3, 4, 6, 8, 12, 16, 24
9          1024        2, 4, 8, 16, 32, 64, 128, 256, 512
10         60          2, 3, 4, 5, 6, 10, 12, 15, 20, 30
11         4096        2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048
12         192         2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96
13         144         2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72
14         120         2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60

Anyone know about this?
Thanks!

Ron Lunde
email: ronl@tessi.com
Newsgroups: sci.math
Subject: Re: Rounding Rules
From: news@massey.ac.nz (USENET News System)
Date: Mon, 21 Dec 92 20:47:19 GMT
Organization: Massey University

In article <LOTTO.92Dec21062820@math.math.berkeley.edu>, lotto@math.berkeley.edu (Ben Lotto) writes:
> 
> I was told way back in my undergraduate days that one should generally
> "round to evens."  Someone recently explained the rationale for this
> as follows.  Suppose we start with 1.49 and round in the traditional
> manner twice.  First we would round to 1.5 and then to 2.  On the
> other hand, if we round just once we get 1.  Rounding to evens
> prevents us from getting different answers.

Thanks, I always wondered.

I just have one little problem, what about double rounding of 2.51?
From: kosowsky@minerva.harvard.edu (Jeffrey J. Kosowsky)
Newsgroups: sci.math
Subject: Re: Philosophy of Pi
Date: 21 Dec 92 21:05:15 GMT
Article-I.D.: minerva.KOSOWSKY.92Dec21160515
	<1992Dec18.114450.22206@lth.se> <1992Dec20.212707.9225@massey.ac.nz>
Organization: Harvard Robotics Lab, Harvard University

In article <1992Dec20.212707.9225@massey.ac.nz> news@massey.ac.nz (USENET News System) writes:
> A number is said to be 'normal' in base 10 if every finite string of digits
> occurs eventually and with the same limiting relative frequency as any
> other sequence of the same length. E.g. 25369 should appear with frequency
> --> 10^(-5) in the first n digits as n --> oo. Similarly for other bases.
> A normal number is one which is normal in all bases.

>   Normal numbers behave like random numbers. In fact a random sequence of
>   digits will form a normal number. 


Not any random sequence. First, you must assume that the digits are
uniformly distributed on {0,1,2,3,4,5,6,7,8,9}, second you must assume
some type of independence. Furthermore, the statement is still then
only true with probability one. e.g. 0000000000... is a random
sequence and certainly is not normal.

Jeff Kosowsky
Newsgroups: sci.math
From: jbaez@riesz.mit.edu (John C. Baez)
Subject: Re: Tensor derivatives
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Mon, 21 Dec 92 18:13:34 GMT

In article <9235215.10767@mulga.cs.mu.OZ.AU> psm@burra.ee.mu.OZ.AU (Phil Malin) writes:
>Hi all.
>
>This question may not make sense, but anyway...
>
>I've read about covariant and contravariant derivatives.  Does it
>make sense to have mixed tensor derivatives?  i.e. taking the derivative
>of a function w.r.t. a type (1,1) tensor, etc... if the function is
>a function of this tensor?  Does this question make sense?

>If, on the other hand, this idea shows a total lack of understanding
>of the subject please inform me gently (no flames :-).

I have heard about covariant and contravariant tensors, and covariant
derivatives, but not contravariant derivatives.  So this post 
shows a total lack of understanding of the subject -- either on Malin's
part or my own, but I'm not sure which.  :-)  What would a contravariant
derivative be, anyway?  Ordinarily one takes a derivative with respect
to a tangent vector, so would this amount to taking the derivative with
respect to a cotangent vector?  Most odd.  Of course if you have a
Riemann metric around you can convert tangent vectors into cotangent
vectors and vice versa ad libitum, so that *could* give "contravariant
derivative" a meaning.
From: arodgers@dcs.qmw.ac.uk (Angus H Rodgers)
Newsgroups: sci.math
Subject: Re: Negative Zero
Date: 24 Dec 92 23:52:58 GMT
Organization: Computer Science Dept, QMW, University of London

[This is getting silly, but what the heck, because what am I doing
here on Christmas Eve anyway?]

In <1992Dec23.211510.21083@massey.ac.nz>
news@massey.ac.nz (USENET News System) writes:

>I'd better explain the joke - obviously some people need it :-)

>After a while the counter reached 32767. On the next iteration it
>became -32768.

>Of course he might have used LONGINT, in which case the program
>would run a little longer before cycling.

>To be serious for a moment :-), a computer is a finite state machine.

Yes, and some programming languages allow you to create "counters"
which you can repetitively increment for a VERY long time --  at
least as long as it takes for the machine to rust away.

You forgot the :-), which really was needed in this case.

As my (true) story indicates, some people really are just as crazy
as your fictional student.

And indeed, on two separate occasions recently, ideas which I've
proposed on Usenet as obviously crazy jokes have turned out to have 
already been put into practice, somewhere. (misc.kids readers may
remember.)

Maybe that's how religions get started. ... Indeed, isn't that
*exactly* how Scientology got started?

;->
--
Gus Rodgers,  Dept. of Computer Science, |  [Profound thought for 1993]
Queen Mary & Westfield College, Mile End |     "Philishave is about
Road, London, England.   +44 71 975 5241 |      personal choice" 
E-mail (JANET):   arodgers@dcs.qmw.ac.uk |             -- TV advert
Xref: sparky sci.math:17521 sci.physics:21914
From: columbus@strident.think.com (Michael Weiss)
Newsgroups: sci.math,sci.physics
Subject: Banach-Tarski paradox (was: Bayes' theorem and QM)
Date: 30 Dec 92 17:03:54
Organization: Thinking Machines Corporation, Cambridge MA, USA
	<1ht1arINNf8a@chnews.intel.com>
NNTP-Posting-Host: strident.think.com
In-reply-to: bhoughto@sedona.intel.com's message of 30 Dec 1992 20:37:47 GMT


Description of the Banach-Tarski paradox by Daryl McCullough omitted.

Blair P. Houghton replies:

   I'll buy the explanation (conditioned on the conjecture
   that such a situation might exist, which as yet isn't
   anything more than conjecture), but not the example.

   The two spheres thus formed do indeed have the same volume
   sum as the original sphere, but each has less volume than
   the original, or else they weren't composed of a finite
   number of pieces[*], or else when constructed they contained
   gaps, internally.

   That, or their "volume" is fractal, being the effective
   volume of a convoluted surface, and therefore isn't
   actually of order 3 but of some order less than 3 and
   greater than 2. [...]

I know it sounds unbelievable.  The proof relies on (and in
fact requires) the axiom of choice, so that's one way around it.

But if you do accept the "usual axioms of mathematics" (i.e., ZFC,
Zermelo-Frankel set theory with the axiom of choice), then you're stuck
with the BT paradox, which is indeed a theorem of ZFC.

You can find a more extensive discussion of the BT theorem in the FAQ for
sci.math.  Here is a statement of one form of the result.

    Let S, S1, and S2 be three spheres of radius 1.  Let B be the set of
    all points on or inside S; ditto for B1, B2.  Assume S, S1, and S2 are
    placed so that B, B1, and B2 are pairwise disjoint.

    The sets B, B1, and B2 can each be partitioned into a finite number of
    subsets:  (here, "+" stands for disjoint union)

    	B  = P_1 + ... + P_m + P_{m+1} + ... + P_{m+n}
    	B1 = Q_1 + ... + Q_m
    	B2 =                   Q_{m+1} + ... + Q_{m+n}
    
    such that P_i is congruent to Q_i for i=1 to m+n.  "Congruent" means
    that there is a rigid motion taking P_i onto Q_i.

One says that B and B1+B2 are "equivalent by finite decomposition".  There
are no gaps in B1 or B2--- they each consist of *all* points inside or on
their respective spheres.  There are indeed a finite number of pieces in
each partition (i.e., m and n are finite--- in fact I think m and n can be
chosen rather small, something like 20 or so.)

Note that the statement of the BT theorem does not mention the word
"volume".  B, B1, and B2 are really quite ordinary point sets, and I don't
believe there is any reasonable sense in which one could say they have
"fractal volume".  Of course, the P's and Q's are a different story...
From: keithw@ccorp.uucp (Keith Weintraub - dpr2)
Subject: test
Organization: Citicorp, New York City
Date: Mon, 21 Dec 92 16:41:02 GMT

test.
-- 
Keith Weintraub (KW) -- Citicorp |    I told you a trillion times:
keithw@Citicorp.COM              |	"Don't exaggerate."
uunet!ccorp!keithw               |
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Two (I think) interesting Problems
Date: 28 Dec 92 01:38:45 GMT
Organization: U of Cambridge, England

1. To fold a sheet of paper into a regular pentagon: If it's an A4 sheet
of paper (or any other of the A... sizes; what matters is that the ratio
of the sides should be root(2)) there is a fairly elegant way to do this
with perfect accuracy. It goes like this:
 - Find the centre of the piece of paper by folding both pairs of opposite
   sides together.
 - Fold all four corners to the centre, producing a sort of hexagonal
   shape.
 - Fold this in half along what was the shorter of the two "orthogonal"
   folds in the first step.
 - You now have a sort of five-sided greenhouse shape. Fold the two walls
   of the greenhouse in so that the top of each wall is on the line of
   symmetry of the shape, and so that the line joining that to the corner
   newly created by this fold is perpendicular to that line of symmetry.
 - You now have a regular pentagon.
The last step sounds intimidating, but if you've got the earlier stages
right it's easy because you can see it's producing a regular pentagon.
This produces a pentagon with a couple of "flaps" folded back; if you fold
the pentagon in the right way (in the second stage, fold two opposite
corners and then the other two opposite corners) you can then stick 12 of
these things together and make a quite rigid dodecahedron!

2. You cannot draw an equilateral triangle all of whose corners are at points
of a square grid. (Proof: Set up coordinates so that the "grid points" are
exactly the points with integer coordinates. Now, on the one hand the area
of the triangle is root(3)/4 times square of side length = root(3)/4 times
an integer; on the other, if the vertices are (x1,y1) (x2,y2) and (x3,y3)
you can check that the area is (plus or minus) the determinant of the matrix
(1 1 1   /   x1 x2 x3   /   y1 y2 y3) over 2, which is half an integer. Hence
root(3) is rational. But it isn't: contradiction.)
You can get pretty good approximations, of course: for instance, put two
vertices 8 squares apart along a grid line, and the other one half-way between
but displaced by 7 squares. I'm sure you can do better than this without too
big an increase in size...

-- 
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@cus.cam.ac.uk  Cambridge University, England.    [Research student]
Newsgroups: sci.math
Subject: Re: () -> [()] -> {[()]} -> ???
From: chisnall@cosc.canterbury.ac.nz (The Technicolour Throw-up)
Date: Mon, 21 Dec 1992 10:18:37 GMT
Organization: Computer Science,University of Canterbury,New Zealand

From article <1g1h1kINN45s@DOLPHIN.ZOO.CS.YALE.EDU>, by yan-dicky@cs.yale.edu (Dicky Yan):
> In article <92336.221030CCB104@psuvm.psu.edu> <CCB104@psuvm.psu.edu> wrote:
> 
>>Parentheses get enclosed inside brackets, which get enclosed by
>>braces. What next? Large parentheses, perhaps? What, please, is the
>>convention on this? Thanks....
> 
> When I was at primary school, I was taught that you put a line above
> the part of the expression you want to be performed before anything
> else is done. This usually goes inside the "()" brackets. I have
> never used it since then. Don't know whether it is a universal
> notation though. It is something like:
> 
>                                   ___
>               {2+3-[4 x 6 + 2 x(1+3-8+6)]-1}
> 

The overline is known as a vinculum and is never used these days.  Presumably
the overline that you find in radical signs (e.g. in sqrt(b + 4ac)) is some
vestige of the vinculum.

--
Just my two rubber ningis worth.
Name: Michael Chisnall  (chisnall@cosc.canterbury.ac.nz)
I'm not a .signature virus and nor do I play one on tv.
Xref: sparky sci.math:17405 sci.logic:2478