Newsgroup sci.math 157889

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Gary Hampson   wrote:
>In article <5ap3jf$1u53@b.stat.purdue.edu>, Herman Rubin
> writes
>>[...]
>>Unless there is a specific course, TV is not a means of learning a
>>subject.  At best it is a means of getting a low-level popularized
>>version of it.
It is possible to learn much from these versions if one develops the habit
of reading between the lines and understanding the context of the
presentation. Very few people seem to have that habit, though it is useful
for much more. 
>In the UK we have the Beeb (BBC) of course, in particular BBC2. Channel
>4 also does a load of educational stuff. Also in this centre of
>civilisation [ ;) ] we have the Open University which basically allows
>degree study at home with the help of tv programs and course books. The
>programs are often transmitted at the most obnoxious times, but if you
>can programme a video recorder you can still get a nights sleep (unless
>you're paranoid of course). It usually takes 5-7 years to study for a
>first degree using this method. I guess they may have a web page.
Where I grew up, the University of South Florida broadcast some
canned lectures on public television. I watched a course while skipping
school, and was fascinated enough to major in that area. Not only did I
learn the subject material, but before that random exposure I had no idea
what that discipline studied. 
Lecturers should ask themselves what they can do better than videos, and
exercise these possibilities or consign themselves to obsolescence.
Douglas Zare

Leonard Timmons  wrote:
>Is the duality between mind and matter equivalent
>to the duality between numbers and numerals?
The duality between mind and matter is isomorphous to the duality between 
fish and bicycles.
-- 
Alan "Uncle Al" Schwartz
UncleAl0@ix.netcom.com ("zero" before @)
http://www.ultra.net.au/~wisby/uncleal.htm
 (Toxic URL! Unsafe for children, Democrats, and most mammals)
"Quis custodiet ipsos custodes?"  The Net!

Stijn van Dongen  wrote:
>
>		Hi all,
>
> Does anyone know an answer to the question:
> Is it and if so how possible to enumerate all pythagorean triples (in N)?
>[...]
Define rational points to be those with both coordinates rational.
One extremely useful idea is to think of relatively prime pythagorean
triples as rational points on the unit circle, e.g., (3/5,4/5). Then note
that a line with a rational slope which intersects the circle in at least
one rational point intersects it in two rational points (possibly equal).
Finally, the slope between two distinct rational points is either rational
or infinite.
So, choose one rational point P on the unit circle. Then other rational
points correspond to slopes of lines through P, with P corresponding to
the tangent line.
This is often given as the first example of the fruitful combination of
number theory and geometry.
Douglas Zare
http://www.cco.caltech.edu/~zare

Leonard Timmons  wrote:
>Is the duality between mind and matter equivalent
>to the duality between numbers and numerals?
The duality between mind and matter is isomorphous to the duality between 
fish and bicycles.
-- 
Alan "Uncle Al" Schwartz
UncleAl0@ix.netcom.com ("zero" before @)
http://www.ultra.net.au/~wisby/uncleal.htm
 (Toxic URL! Unsafe for children, Democrats, and most mammals)
"Quis custodiet ipsos custodes?"  The Net!

In article <5bg1c8$ni1$2@gruvel.une.edu.au>,
ibokor  wrote:
:SAPDMCPC22@ wrote:
:: 
:: 
:: The different infinity cardinalities don't matter at all here.
:  'oo' is (here) the description of a "point" at infinity to get a 
:closured topological >: real space.
:
:
:But the set of all real numbers with its (usual) Euclidean
:topology is already a closed real space. No additional points
:are needed!
The underlying set of every topological space is closed per definitionem. 
In other words, being closed is not a big deal; you can make a lot of 
holes in R, declare the rest a new topological space, and it will be closed.
Let me guess:
-> you wanted to say that the set R of real numbers is closed in 
   every metric space into which it is isometrically embedded.
   True. This is a property of complete metric spaces.
-> or you wanted to say that R is closed in every metric space into which 
   it is homeomorphically embedded.
   False. There are embeddings of R, even into R (using arctan, for
   example) under which the image of R fails to be closed. The arctan 
   example leaves two "missing points" - boundary points of the 
   embedding which are not images of anything real.
   In metrizable spaces (and no doubt in other classes), being absolutely 
   closed under all homeomorphic embeddings means being compact.
-> Or: None of the above: if so, please specify.
Have fun, ZVK (Slavek).

Christopher Hillman (hillman@math.washington.edu) wrote:
[...]
> Nonetheless, a particle falling into the BH (or the matter of the star
> itself as the hole is being formed) experiences nothing strange as it passes
> through the event horizon.  The event horizon is an artificial mental construction
Let me ask a follow up question. In Schwarchild's coordinates the 
particle passes through the event horizon at time t = infinity. 
However, Hawking has shown that a black hole cannot last until 
t = infinity, it will evaporate first. If by the time the particle 
enters the black hole it does not exist any more, how can it do it? 
Someone told me that the "paradox" comes from an improper mixture of 
classical and quantum physics, but I would appreciate any more detailed 
explanation about how matter can fall inside an evaporating (non rotating 
and non charged) black hole.
Miguel A. Lerma

In article <32DB507E.41C67EA6@clipper.ens.fr>
David Madore  writes:
> Oh, by the way, you might like the following definition: a sequence
> consisting
> of a 2-adic number, a 3-adic number, a 5-adic number,... and a real
> number,
> such that all its components except at most a finite number are
> integers, is
> called an "adele". Adeles form a ring under pointwise addition and
> multiplication.
> They were introduced by Andr\'e Weil. The invertible elements of the
> ring of
> adeles, that is those sequences all of whose components are non zero,
> and
> all but a finite number are units (a p-adic integer is said to be a unit
> iff
> it is not a multiple of p) are called ideles, they form a group and they
> were
> introduced by Claude Chevalley.
  Thanks but no thanks. In math as in life, it is best to stick to main
essentials and not be sidetracked with tourist curios knick knacks.

Subject:      Re: Infinitude of Primes in P-adics
From:         dik@cwi.nl (Dik T. Winter)
Date:         1997/01/13
Message-Id:   
Sender:       news@cwi.nl (The Daily Dross)
References:   <5b8tan$7ao$1@dartvax.dartmouth.edu> 
<5bbsh4$s10$1@dartvax.dartmouth.edu>
Organization: CWI, Amsterdam
Newsgroups:   sci.math,sci.physics,sci.logic
 > 
 >   Different primes p, q, then p-adics not isomorphic to q-adics, but
 > p-adics have a multiplicative inverse in the q-adics
I do not understand what you are meaning here at all.  What is the
multiplicative inverse of the 2-adic 11 in the 3-adics?
****
 Sorry Dik, my mistake, that was all wrong. I was confused with Q and Z
of adics. Today I was looking for the 3-adic multiplicative inverse of
"12",  110, which of course is 1/3 X 1/4 and finding these inverses is
not an easy math problem. I suppose I need some expansion series on
1/4, oh well, that is not important for me, but please tell if you
know, what is the proof trick that one can know that all 3-adic
Integers have a mult. inverse except the 3. There must be some quick
logic that one can see that all of these integers have an inverse
except the 3. 
  Thanks for teaching me some more about p-adics, Dik. I am trying to
see if this imaginary 3', the multiplicative inverse of 3 in 3-adic
Integers possesses any of the characteristics that i possesses for
Reals? Could there be some e^i pi = -1, for 3' ?
   Could there be one and only one imaginary integer to cover all Z_p
that would make all of the Z_p-adics a field. Or do I need to append a
3' to 3-adics, a 5' to 5-adics. Thereby requiring an infinitude of
imaginary integers.

In article , Nick Sexton
 writes
>In article <5as1hf$7jm@news.fsu.edu>, Jim Carr 
>writes
>>Rebecca Harris  writes
>>}STARGRINDER  writes
>>}>
>>}>get a life!
>>} 
>>} Hear Hear!
>>
>>Goddess  writes:
>>>
>>>Yeah! I don't see why they bother with these posts on here. Why don't they 
>post
>>>it on some maths chat group?
>>
>> To those of us reading the crosspost in sci.physics or sci.math, the 
>> concern is that completely misleading junk such as 
>>
>>: Comment: Note that atoms (atoms) = atoms
>>:
>>:       It seems that squaring an item (not a unit of
>>: measurement) equals the item.  What do you think?
>>
>> is being posted in k12 groups where it could confuse impressionable 
>> children.  If Kaufman was only talking to teachers, who should have 
>> the sense to ignore him, it would not be quite so bad. 
>>
>
>I'd just like to make a point. To whoever posts the educational stuff.
>Listen up.
>
>k12.chat.junior is a chat group. People talk and stuff. What really
>annoys people here is the educational stuff that gets posted. I don't
>think many people read it, anyway. (And those who do are probably on
>k12.ed.math anyway) So if you want to make us happy, then _please_ don't
>send stuff to this group. s'just a thought.
Don't write 'it's just a thought'! Sweetie, nobody's goin' to listen to ya if ya
say that! You gotta tell em out right. Just like that! It's not just a thought,
cause everybodys thinkin' it, so SPEAK OUT!
-- 
Goddess
The girl who cried "MONSTER!" and got her brother....
E-mail : goddess@segl.demon.co.uk
Homepage: http:/www.segl.demon.co.uk/frances

Alison Brooks  wrote:
> Interestingly enough, while the US was busy getting bogged down in
> Vietnam, the UK was engaged in fighting in Borneo, in remarkably similar
> political situations. The UK military position wasn't as good as that of
> the US; the Borneo border was massively longer than that which the
> Americans had to deal with, and the terrain very much harder.
> 
> Nonetheless, the UK was successful.
> 
> One can debate why this should be; however, there was no great "anti-
> war" debate in the UK. I suspect that this was in part because of
> different attitudes.
The main difference was that there was no land border with a Communist
state a few miles away. The Communist guerrillas in Malaya were forced
to operate entirely on their own resources. The Viet Cong were supported
by hundreds of thousands of North Vietnamese troops and lavish quantities
of arms.
The Borneo border was exposed; but the Indonesian side of it is far
more remote from any civilized base area than was the inland border of
South Vietnam. Sarawak was not the key part of Malaysia, and the
infiltrators had relatively little local support. (My understanding
is thyat he Dyaks took to collecting the heads of the infiltrators.)
Also, after 1966, the Indonesian government no longer supported
insurrection in Malaysia, abandoning Suharto's "konfrontasi" and
expelling or killing his Communist friends.
-- 
Rich Rostrom  | You could have hit him over the head with it and he
              | wouldn't have minded. He never did mind being hit
R-Rostrom@    | with small things like guns and axe handles.
bgu.edu       | - Ellis Parker Butler, "That Pup of Murchison's"

In article <853089576.987@dejanews.com>, sarahjane@kwic.com says...
>
>A boat is being pulled into a dock with a rope.  The rope passes through a 
>pulleyon the dock that is 1.5 metres above the bow of the boat.  If the 
>boat is approaching at a rate of 50 cm/s, how fast is the rope being 
>pulled in when the boat is 5 m from the dock?
>
Let the length of rope between the pulley and boat be y, and the distance 
between the dock and boat be x.  What you want to find out is the rate of 
change of the rope length dy/dt.  By Pythagoras, if the distance to the dock is 
x + dx and the rope is y + dy then ((x + dx)^2 + 1.5^2) = (y + dy)^2.  
Simplifying (x + dx)^2 gives x^2 + 2xdx + dx^2 and you can omit the dx^2, the 
same for (y + dy)^2 = y^2 + 2ydy.  Since x^2 + 1.5^2 = y^2, you can simplify to 
2xdx = 2ydy, or dy/dx = x/y = x/sqrt(x^2 + 1.5^2), and dy/dt = dy/dx * dx/dt.  
Since dx/dt = 0.50 and x = 5, dy/dt = (5 * 0.5)/sqrt(5^2 + 1.5^2) = 0.479 m/s.  
If you are pulling the rope from the boat of course it will be twice that.
Allen
-- 
Allen Windhorn
Kato Engineering (A Rockwell Automation Company)
P.O. Box 8447
N. Mankato, MN  56002-8447

Hi, I was having a discussion with a friend and was wondering:
On a theoretical roullette wheel with even number of squares, half red, half
black, is it a safe bet to say that if 5 reds had been rolled, that the next
roll isn't likely to be red again?  I know the previous results don't affect
the next one (ie 50%), but aren't the chances of rolling 6 reds in a row
slim (i.e 1 in 64) ?
email me response please, thanks.
Peter

In article <01bc026d$054f38a0$267e6bcf@rauhala.tyenet.com> "Daryl Rauhala"  writes:
>
>I am in need of a proof by contradiction that the set of prime numbers is
>infinite.
>
>Starts by assuming that the set of primes is finite and the largest prime
>is P.
>Let x = P! and  let y = x + 1. 
So far so good.
The ! above means factorial,
so x = 1 times 2 times 3 times ... times P.
Now consider y, there are 2 cases:
Case 1: y is prime.  y is greater than P.  Contradiction.  We are done.
Case 2: y is not prime.
        Therefore, y has at least one prime factor.
        But, y = P! + 1 so this prime factor must be greater than P,
        because y/z has remainder 1 if 2 <= z <= P.
        Contradiction.
        We are done.
-- 
dennis@netcom.com (Dennis Yelle)
"You must do the thing you think you cannot do." -- Eleanor Roosevelt

NCC-1701-E  wrote:
>[...]
>Prob : prove that C(2n, n) is divisible by (n+1). 
>
>Note : C(a,b) refers to the Binomial coefficient of x^b in (1+x)^a.
>
>Proof: C(2n,n)/(n+1) = 2C(2n, n) - C(2n+1, n+1) q.e.d
I think that this statement assumes too much. The quotients are the
Catalan numbers, which count, among other things, the number of lattice
paths from (0,0) to (n,n) which do not go above the main diagonal. The
following is in Graham, Knuth, and Patashnik's _Concrete Mathematics_ and
Richard Stanley told me it was known to Cayley, though I have seen it
published as new in a combinatorics journal in 1993:
There are C(2n,n) lattice paths from (0,0) to (n,n). Let us construct
equivalence classes of size n+1 such that each equivalence class contains
precisely one path which does not go above the main diagonal. For each
lattice path, write the sequence of horizontal and vertical steps, and add
a horizontal step at the end. Then repeat this sequence of length 2n+1.
n+1 lattice paths will give this one (one for each horizontal step which
could have been added). There is a unique highest point per period where a
line of slope n/n+1 can rest on the path. It is followed by a horizontal
step, and this is the added step of the unique member of the equivalence
class which does not go above the diagonal.
I would like to have a similarly combinatorial proof that 
(3a! * 2)/(a! a+1! a+2!) is an integer.
Douglas Zare
http://www.cco.caltech.edu/~zare

	Thankyou for picking up this message!
	I'm looking for a topic to write a short paper on for my 300 level
Diff.Eqn. class(undergrad).  I'm really interested in plants, and would
love to know if there are any SIMPLE applications of Diff.Eqn. in botany.
If anyone out there has any ideas please drop me a line.
		-Jessica  Fried
		 Grand Valley State University

Peter Bui wrote:
> 
> Hi, I was having a discussion with a friend and was wondering:
> On a theoretical roullette wheel with even number of squares, half red, half
> black, is it a safe bet to say that if 5 reds had been rolled, that the next
> roll isn't likely to be red again?  I know the previous results don't affect
> the next one (ie 50%), but aren't the chances of rolling 6 reds in a row
> slim (i.e 1 in 64) ?
The probability of the next one being red is still 50%.  The chances of
rolling the first five reds in a row was 1/32, so when you multiply by
the 50% on the 6th roll, you get the 1/64 that you mention--the key
there is that by rolling five reds, you've already satisfied most of
that 1/64.
-Paul

How induction works:
1.  Assume that it must be true for case (n) { this is clearly the easy
part }
2.  Prove that if it IS true for case (n) it MUST be true for case (n+1)
3.  Prove that it IS true for case (n=1)
[Why: It was true for 1.  We proved if it is true for 1, then it must be
true for n+1 = 2.  Since it is true for n+1, it also must be true for 3,
for 4, for 5 ...]
I won't insult you by doing the problem for you.
stan  francuz  wrote in article
<32dbcf78.1098335@tar-news.tpgi.com.au>...
> would  appreciate  help  with  this  MI  question .
> 
> Consider  the  Fibonnaci  sequence  1,1,2,3,5,8,....
> 
> which  may  be  defined  as  t(1) = t(2) = 1 , 
> 
>  and    t(n+2) = t(n+1) + t(n)
> 
> Prove   by  MI 
> 
> t(2n)  =  t(n){t(n+1)  +t(n-1)} 

In article <5avac8$1rs6@news.doit.wisc.edu>,
Brent Hetherwick  wrote:
>               I thought that she was quite the sharp cookie until I read 
>her little paperback on Fermat.
    I was pretty skeptical after her "tree falls in a forest" answer
    (which I don't recall any longer, I just remember seeing my
    fluff-o-meter needle jumping off the scale).
>If you haven't read her book on Fermat's Last Theorem (_The World's Most
>Famous Math Problem_, or something like that), do so immediately.
    Yeesh.  I wouldn't advocate reading it immediately.  It just isn't
    worth the time.  Most of the book seemed to stem from mangling
    Godel's Incompleteness Theorem into "we can't ever be sure" and
    then extending that to "so what's the point?"
>Thus, it seems more than likely that Martin Gardner has quite a few 
>"friends" on Marilyn's bookshelf, which would explain her cryptic answer 
>to the pyramid problem.
    I admit that Martin Gardner has quite a few "friends" on my
    bookshelf, too [actually, he did until I lent all of him to
    a friend].  But, I certainly wouldn't have written her FLT book.
    8^).
    alter,
    pat
-- 
I live in another Dimension, But I Have a Summer Home in Reality

Alan "Uncle Al" Schwartz wrote:
> 
> Leonard Timmons  wrote:
> >Is the duality between mind and matter equivalent
> >to the duality between numbers and numerals?
> 
> The duality between mind and matter is isomorphous to the duality between
> fish and bicycles.
Hey, I think you are making fun of me.  Someday, when I start taking myself
seriously, I'm going to be upset. ;-)
In the mean time, though ...
Does anyone out there believe that numbers (not numerals) actually 
exist (what ever that means) and on what basis are you making that 
claim?
My second question:  Does anyone out there believe that numerals
actually exist and on what basis are you making that claim?
Go ahead, make fun of me.  I can take it.
-leonard

Ward Stewart wrote:
> 
> fireweaver@insync.net (erikc) wrote:
> 
> >>|
> >>|The "proof" is gibberish. Any mathematician knows there can be no proof of
> >>|God. In fact, there cannot even be a definition of God...for purposes
> >>|of mathematical proof... for no one knows His character in fullness.
> 
> >Then why even bother to debate the issue at all.  As far as I am concerned, the
> >existance of god is undecideable, and any such discussions are noting more than
> >an exercise in mental masturbation.  You either believe or do not believe.
> >Contrary to all the bullshit I was raised with as a kid, nothing in my personal
> >life has ever given me any reason to suspect that there might be a god as
> >theists believe in it or that it would have any of the characteristics ascribed
> >to it by theists even if it did.
> 
> >Erikc -- firewevr@insync.net
> 
> We debate it because these odd folks, supposing that their
> ommnipotent deity is too lame to manage his own vinyard have
> decided that THEY must do the weeding.  THEY have the power
> to determine who has been anughty and who has been nice.
> 
> THEY are on thin ice, theologically and morally and had best
> watch their step.
> 
> ward
> 
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>            Ward and George
>                 40 years,
>            yet strangers before
>                  the law.
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
God made light.  By it we see.  Without it we cannot focus are attention
on externals.  God made the world.  We see it in light.  
An infinite multitude of world exists in darkness and we see only what
the flashlight of our preception can capture momentarily.  Math is a
vain attempt to place order on that which is already ordered.  God is
said to be within.  God cannot be seen in the light because he is not in
the darkness.  God is known through faith, and faith is grown in love.
Love is the only "light" that can reveal the existance of God.

andrew harford (cs4-03@cslan.ud.ie) wrote:
: Does anyone know if the following is true?
: 
: If A is a connected subset of a real vector
: space then for any elements x,y in A, there
: exists a continuous function f:[0,1]->A so
: that f(x)=0 and f(1)=y (using the usual
: topology). 
: 
False.
Take X:= { (0, v) | -1 < v < 1 } U {  (u, sin(1/u))| 0< u < 10}
which is a connected subset of R^2.
The is no such function as you seek when x = (0,0) and y=(1, sin(1)).
d.A.

does anyone have or no of precalculus software. i am looking for it to
help a friends daughter do well in her high school course and any help
would be appreciated. major retailers did not carry any pre-cal software
only full blown calculus.
thank you
santa@ix.netcom.com

Bruce C. Fielder (bfielder@quadrant.net) wrote:
: I'm shamelessly moving into this (hopefully well explained) thread to
: ask a question which has recently bothered me:
: If the gravitation of a black hole is such that anything falling into a
: black hole will have its "time" slowed the closer it comes to the event
: horizon, how does the thing form in the first place?  Surely as the
: original mass contracts, it should slow (from our point of view) until 
: the original mass remains "waiting" (sorry about all the quotation
: marks) at the event horizon?
: As far as I can see, the same should hold true with the mass inside the
: (soon to be) event horizon; the acceleration and gravity increases and
: slows the time to infinity.  So how does the thing ever form in our
: universe?
From the point of view of a distant observer, it doesn't exactly complete 
forming.  What happens is that it becomes closer and closer to the final 
state, with the image of this process being spread out over an 
arbitrarily long period of time as measured by a distant observer.  (For 
an observer collapsing in with the surface of the collapsing body, of 
course, this diesn't prevent the black hole from forming from their point 
of view because time dilation for them and the surface is identicle.)
Applying a little QM explains why this never-quite-formed black hole gets 
to look black anyway:  The surface has some temperature from its own 
point of view, and so is emiting photons of any given energy at some rate 
from its own point of view.  When you apply the redshifting and time 
dilation effects to the energy and emmision frequency of this emited 
radiation, however, it turns out that the emission becomes arbitrarily 
sparse and arbitrarily reddened as time passes.  For any finitely large 
detection threshhold, the emmision frequency and emmision energy of the 
radiation that escapes the collapsing body always falls below it within 
finite time.  Thus, the collapsing body can become arbitrarily like the 
ideal black hole final state.  (In models that attempt to incorporate 
quantum effects, it turns out that after finite time, Hawking radiation 
begins to outshine the radiation from the time-dilated surface, so it 
effectively becomes lost in the emmision noise.)
--
---------------------------+--------------------------------------------------
Ring around the neutron,   |  "OK, so he's not terribly fearsome.
A pocket full of positrons,|   But he certainly took us by surprise!"
A fission, a fusion,       +--------------------------------------------------
We all fall down!          |  "Was anybody in the Maquis working for me?"
---------------------------+--------------------------------------------------
"I'd cut down ever Law in England to get at the Devil!"
"And what man could stand up in the wind that would blow once you'd cut 
down all the laws?"
------------------------------------------------------------------------------
This message may not be carried on any server which places restrictions 
on content.
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e-mail will be posted as I see fit.
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Recently I have been thinking a lot about how to find a curve that best
matches a given set of points.  Today I got an idea.  Why not try
to find the differential equation that best describes a set of points?
It's fairly trivial to find the best simple harmonic oscillator diff. eq. 
With a little more wok you can find the best damped oscillator.  What
I was wondering is if any work has been done in this area before? 
-- 
Etherman

In article <32D68CB9.694D@lancnews.infi.net>,
John   wrote:
)Jean-Christophe Janodet wrote:
)> 
)> It's easy to explain why the sum of two negative
)> numbers is negative, using the example of a barometer.
)> Does anybody have a similar example to justify that
)> the product of two negative integers is positive ?
)> Or else, what are the theoretical reasons ?
)> 
)> Sincerely, Jean-Christophe.
)Hi Jean-Christophe,
)This difficulty is similar to the "1/0" flurry now on this site.
)
)The explanation I found is:
)"... To include the new symbols -1, -2, -3, ... in an enlarged
)arithmetic which embraces both positive and negative integers, we must,
)of course, define operations with them in such a way  that the original
)rules of arithmetical operations are preserved. For example (-1)*(-1) =
)1, which we set up to govern the multiplication of negative integers is
)a consequence of our desire to preserve the distributive law...."
)
)This was not an easily found rule and once found it was not easily
)explained. Euler had trouble with it so why not you and I?
)It appears that (-1)*(-1) = 1 is sort of an axiomatic afterthought.
No, it is not.
Now we -definitely- want (-1)(1) = (-1) and (-1)(0) = 0.
And he distributive law is:
	a(b+c) = ab+ac
If this is to hold, then so must the following
	(-1)((-1)+1) = (-1)(0) = 0
		     = (-1)(-1)+(-1)(1)
So
	(-1)(-1) = - (-1)(1) = -(-1) = 1
Mike
-- 
----
char *p="char *p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}
This message made from 100% recycled bits.
I don't speak for DSC.         <- They make me say that.

In article <32DB0B90.3A6F@quadrant.net>, Bruce C. Fielder  wrote:
)I'm shamelessly moving into this (hopefully well explained) thread to
)ask a question which has recently bothered me:
)
)If the gravitation of a black hole is such that anything falling into a
)black hole will have its "time" slowed the closer it comes to the event
)horizon, how does the thing form in the first place?  Surely as the
)original mass contracts, it should slow (from our point of view) until 
)the original mass remains "waiting" (sorry about all the quotation
)marks) at the event horizon?
)
)As far as I can see, the same should hold true with the mass inside the
)(soon to be) event horizon; the acceleration and gravity increases and
)slows the time to infinity.  So how does the thing ever form in our
)universe?
You are confusing proper time and time as measured by an observer. The
collapse proceeds very speedily in proper time (i.e. time as actually
seen by the one falling into the black hole). You might investigate the
Kruskall coordinates for a black hole.
Mike
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I don't speak for DSC.         <- They make me say that.