Subject: nonsense sequences
From: massimo ales
Date: Sun, 19 Jan 1997 10:08:25 -0800
19/01/97
This is a problem dealing with coding, information theory and
combinatorics.
I'm studing (binary) sequences without "nonsenses".
It means that in these sequences, every subtring (from a length
n),cannot appear reversed; or with some parts reversed ; or like a
combination of (few) other substrings; or is "codified" with small
codewords, and so on.. .
In few words without every transformation that applied i.e. to a text or
a flow of data received from a probe, or from a cd-rom or to DNA, would
give us a nonsense.
I think that these sequences are very interesting. From a certain point
of view every possible accepted sequences has to be a lot different from
each other like codewords.
And of course, they deal with combinatorics, and cryptography. But ,this
is the most important question, also with compression: in fact, deleting
the sequences with nonsenses we can use less bits for transmitting and
storing datas.
So can you suggest me if there is someone who studied these problems, or
is interesting in it? Can you help me to find the tools for studing
these sequences?
Can you give me any kind of help (url,etc) ?
Thanks everyone.:-).
********************************************************************
* massimo ales fax +39 432 950004 *
********************************************************************
Subject: Re: Elliptic y^2=x^3+x
From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Date: 19 Jan 1997 08:47:36 GMT
In article <5ataeb$jb@rzsun02.rrz.uni-hamburg.de>,
Hauke Reddmann wrote:
>Can I parametrize this equation?
>Be aware that I don't work in Q but
>in Q+i+sqrt(whatevermaycomeup), so
>a parametrization which contains
>additional sqrt signs is OK and
>EC theory doesn't apply directly.
Sure it does. "EC theory" is pretty vast (as opposed to some of us
who use it, who are only half-vast).
First of all, note that _every_ elliptic curve can be parameterized:
that's the point of the Weierstrass Pe-functions; these establish a
one-to-one correspondence between the cosets C/L of a lattice in the
complex plane on the one hand, and the points of an elliptic curve, on
the other. Both domain and range are complex manifolds; the parameterization
is analytic. Even better, this map is an isomorphism of groups. It's
even fairly easy to perform explicit computations with them; for
example, the PARI / GP package includes function calls for this
parameterization (and its inverse).
I'm guessing however that what you wanted to know is whether there is
a _rational_ parameterization (x(t), y(t)) of this curve (evidently
you are willing to allow the functions to have coefficients in some
algebraic number field). This you cannot do. If you had such a
function, it would inter alia be a meromorphic function C -> E, hence
(if not constant) necessarily surjective, thus a covering of a torus
by a sphere, which is impossible. (Indeed, this argument precludes
the parameterization of _any_ [nonsingular] elliptic curve.)
Perhaps you were thinking of situations in which an argument concludes,
"...this curve has no rational points, but it has some over the field ...".
The nearest analogy I can suggest might be for elliptic curves
over function fields, that is, curves given as
y^2 = x^3 + A x + B
where A and B are _rational functions_ (of T, say). If you consider
all the combinations of (x, y, T), you see that this equation defines
a surface (i.e., a 2-dimensional manifold, if nonsingular). Now it
is more challenging to ask if there are rational functions from C
to the surface, since these need not be onto. In fact, given such a
surface, one can actually ask several related questions:
1) Is there a rational curve on the surface? (i.e., can you
find rational functions (x(t), y(t), T(t)) whose image
is contained in the surface?)
2) Is there such a curve for which T(t) is invertible? (This
requires finding a curve taking each T value once and
only once.)
3) Viewing the surface as an elliptic curve over the field C(T)
of rational functions, does this curve have a rational
point? (This is actually equivalent to question 2)
4) Is it true that for (almost) every T0 there is a point
(x,y,T) on the surface with T=T0 ?
In every case, one can ask the same questions with the complex field C
replaced by a number field K; a "yes" for K means a "yes" for C, too.
The questions are often quite hard, and the relationships among them are
not easy to clarify. Surely (1)=>(2)<=>(3)=>(4), but the two end
implications are not reversible. If (3) holds over C, it holds over a
number field, but it's not immediately clear how to find that field.
In fact, computationally, these questions are rather a mess. (For example,
the (weak) Mordell-Weil theorem guaranteeing that the group of points on the
elliptic curve is finitely generated is true over function fields such
as C(T), just as it is over number fields; but the proof is harder
because there are more issues involving units and factorization.
Consequently, it's harder to convert the theory into algorithms.)
Question (1), for example, is expected to have an affirmative answer if
question (4) does, in general, but that's not proven and I certainly
have seen no algorithm for finding an embedded rational curve in an
elliptic surface, even if it's known to have many points (and not be split).
By the way, you could try to "cheat" and view your original question in
this context (simply declare the coefficients of your curve to be
constant rational functions, rather than numbers). The resulting
surface would then be simply the product E x C. Such split elliptic
surfaces tend to behave differently from the general ones and give rise to
asterisks in the discussion of questions (1)-(4). So you get nowhere
by cheating.
Silverman's second book ("Advanced topics...") has a chapter on
elliptic surfaces which covers much of this.
dave
(I assume you know the curve has no _rational_ points apart from the
torsion subgroup { [0,0], O }. )
Subject: Re: Evidence for God's Existence - TRY Math
From: love@love.net (Anubis)
Date: Sun, 19 Jan 1997 04:21:54 -0500
> >>Ward Stewart wrote:
> >>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.
If we are going to play games with words, then please remember that in
German (which is the parent language of English), The word for "light" is
"hell"! Now if there was a Hell (and of couse there is not), what would
this tell you?
As for "order" in the Universe, Einstein once said "God not only plays
craps, but he throws the dice where we can't see them". Chaos is a major
factor in all of existence. Like the Yin and Yang, order cannot exist
without chaos, and without darkness we would have no light. God is only a
Human concept for things we are to feeble to grasp, and the Bible is just a
book, written by the Humans who's names appear in the verses.
It was not written by God, nor is there a reference to any words directly
from God, only words from Jesus, who was a Human, born on this earth, and
according to the records in the last dead sea scrolls found, not from a
virgin mother, and he lived the rest of his life in hiding in a monastery
in France (Rescued from the cave by his followers...he was the Holy
Grail...i.e. "Holy Blood"), and had a family, who's descendents are still
living in France. This does not take away any of the wonderful teachings
he gave us, but gets rid of the "hocus-pocus" that surrounds the whole
tale.
And as far as the "ten commandments" most theologians today feel that Moses
was not a real person, but one invented by the Hebrews, because they had no
heros. There is no records of his existence anywhere in the world.
So once more, no words from God in the Bible...just an UFO sighting of two.
And that has more to do with the whole phenomenon of our existence then
some grossly mistranslated, and edited (the church removed all references
to reincarnation!) book called the Bible!
Don't take my word for it of course, if you can dig your head out of the
Bible for a few minutes you can look all this stuff up. Religion is
supposed to be a "search" for truth, and as soon as we think we have the
answers we stop looking. At that point we are lost! We have no answers
only questions, and may never have any answers. The only sure thing in our
future is when the sun will die...and then so will the Human race! Of
course that will be a long time from now!
D~
BTW, What kind of batteries do you have in that flashlight? The bunny ones?
Subject: Re: Simple (I hope) math problem
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Date: 19 Jan 1997 03:17:17 -0500
In article <01bc05af$aa3ef680$374f22cf@default>,
Gene Koesling wrote:
>Hi all,
>I've got something thats driving me nuts. I have to determine whether a
>point is within a polygon (its for use in a computer program). My
>apologies in advance if I'm using inappropriate terminology or notation...
>its been about 18 years since my last math class.
>
>I have a polygon with n vertices and I know the x,y coordinates of the
>vertices. I'll be given a point with x,y coordinates. I need to determine
>whether the point is inside the polygon or not.
>
>e.g. a four sided polygon with vertices at A(1,1), B(2,5), C(3,3), D(7,3)
>and a series of points a(3,2), b(5,2), c(3,4). By drawing this on paper I
>can see that only 'a' is within the polygon ABCD but how would I determine
>that mathematically. (In the program I'd be testing one point at a time)
>
>Please note that for the purposes of my program, my polygons will have
>(potentially) between 3 and 256 vertices.
>
>Any help at all would be appreciated.
Here is a routine in MATLAB language - I presume it's easy to translate
to your preferred language. It uses Cauchy index formula, arctangent
function is assumed. The polygon need not be convex.
Index is 1 inside the polygon, 0 outside, and undefined on the boundary.
Hope it helps, ZVK (Slavek).
% runs.m calculates the index of a point P with respect
% to a closed polygonal path given by n points.
% ******> Input: XY , an n-by-2 matrix XY (the rows
% defining the points),
% P, a 2-component vector.
% ******> Output: r, the net number of runs of XY around P
% It is assumed that the last point is connected
% to the first point, and that the region os in the left side
% of the boundary.
% ******> Call: r=runs(XY,P)
% By Z.V. Kovarik, May 3, 1996
%
function r=runs(XY,P);
i=sqrt(-1);
P=P(1)+i*P(2);
Z=XY(:,1)+i*XY(:,2);
n=length(Z);
Z=[Z;Z(1)]-P; % closing up the path
Z=Z(1+(1:n))./Z(1:n); % complex rotations - unscaled
r=round(sum(atan(imag(Z)./(abs(Z)+real(Z))))/pi);
Subject: Re: Clinton pardons Ohair on condition she converts to buddhism!
From: love@love.net (Anubis)
Date: Sun, 19 Jan 1997 04:37:27 -0500
In article <32DF1225.174C@mail.teleport.com>, salad@mail.teleport.com wrote:
> Your Name wrote:
> > But I guess God must approve of incest, as only a short time later -
> > Biblically speaking of course - He pares the population down to Noah,
> > Noah's wife and their children. So from this small, inter-related
> > coregroup, the world is repopulated.
>
> Well again you're missing the whole point:
Yeah and how about all the murder and sex and incest in Genesis?
> God makes everyone and everything. Do you demand Intel build
> processors by your method? Nah, they do it the way they want to.
Why bother! I use a PowerMac! Intel is as lame as born agin xians!
> > Do the math. And use math to prove the existence of God.
>
> OK, I have several calculators, what math?
New math? :-)
> > Then take the God the math's existence proves, and reconcile Him with
> > the tenets you would have us live by.
God is not a "him" or a "her"...thank you very much! I live by peace,
love, and understanding. Should I live by the Inquisitions rules? Or how
about the missionaries? Just beat God into the stupid savages? God makes
people hate other people who don't have the same God.
> Sorry to pop the thread bubble, but I don't use math to prove God,
> God proves himself daily and in the bible prophecies coming true.
Daily? You must live in Ohio! Not in New York it don't!
> > Somehow, the equation just doesn't balance.
>
> I have some old algebra books, what's the problem? It occurs to me
> you're probably discouraged from getting non-substantial answers
> to your math problems from the atheists and physicists frequenting
> this newsgroup. Sure I could solve your problem, but then you'd
> miss out on the knowledge you'd gain by doing it yourself.
> Good luck anyway (double check your answers).
For these people logic is not the answer! Oh by the way... I am a Buddhist!
D~
Subject: Re: Probability and Wheels: Connections and Closing the Gap
From: bm373592@muenchen.org (Uenal Mutlu)
Date: Sun, 19 Jan 1997 10:32:39 GMT
On 13 Jan 1997 02:25:48 GMT, Craig Franck wrote:
>>>consecutive drawings if you buy 1 ticket per draw. And even after 168
>>>draws, there still is 0.042398 probability of having lost all draws.
>>
>>That's saying 95.76 % chance of winning (>= 3) if playing the same 1 ticket
>>in 168 consecutive draws. IMHO an important conclusion from this would be:
>> Playing the same 1 ticket in x consecutive draws is better than playing
>> x different tickets (or a wheel) in 1 draw.
>>Isn't it?
>
>No. If the odds of winning a game are 1 in 54 million, then if you
>buy 54 tickets, your odds of winning are 1 in 1 million.
Sorry, I don't know what you exactly mean. What has this 54 million todo?
(I think you mean the "6/54" game, don't you?)
Then, why "1 in 1 million", shouldn't this be 54 in 54 million?
>If you buy 1 ticket for 54 drawings, your odds of winning never get above
>1 in 54 million.
You mean the single probability, but what about the probability over
this period (ie. after 54 drawings). There is a well known formula for
calculating of this, which I (and also Normand) have used to get the
above probability values.
I got also an interessting email from a statistician:
:Date: Mon, 13 Jan 1997 15:36:43 -0500 (EST)
:Message-ID: <970113150300_38246188@emout13.mail.aol.com>
:To: bm373592@muenchen.org
:Subject: Re: keno
:Content-Type: text
:
:In a message dated 97-01-11 10:08:54 EST, you write:
:
:<< In Lotto 6/49 the probability for at least 3 matching numbers is
:1/53.6551. My conclusion was: if 54 randomly chosen different tickets are
:played in 1 drawing then one should expect at least 3 matching numbers.
:Does this hold in probability terms? And, is this the same as playing 1 fixed
:ticket (ie. always the same numbers) in 54 consecutive (random) drawings? >>
:
:For the first question it depends on what you mean by "expect." If you buy 54
:tickets your number of tickets with 3 matches has an approximate Poisson
:distribution with a mean of 1. That means the probability of no winning
:tickets is 36.79%, some remaining probabilities are shown below:
:
:0 36.79%
:1 36.79%
:2 18.39%
:3 6.13%
:4 1.53%
:5 0.31%
:6 0.05%
:
:On average you will have one winning ticket but you can see you will often
:have 0, 2 or 3 (4 or more is unlikely).
:
:Yes, it is the same as playing a fixed ticket in many drawings.
:
Any comments?
We have to seperate the things:
- probability of 54 fixed tickets in 1 drawing
- probability of 1 fixed ticket in 54 drawings
- probability of 54 fixed tickets in 54 drawings
- probability of 54 random tickets in 1 drawing
- probability of 1 random ticket in 54 drawings
- probability of 54 random tickets in 54 drawings
(there are no duplicate tickets in each case)
Can someone calculate some or all of them?
Subject: Re: Probability and Wheels: Connections and Closing the Gap
From: bm373592@muenchen.org (Uenal Mutlu)
Date: Sun, 19 Jan 1997 10:32:28 GMT
On 14 Jan 1997 23:24:01 -0500, rhoads@sceloporus.rutgers.edu (Glenn Rhoads) wrote:
>bm373592@muenchen.org (Uenal Mutlu) writes:
>
>>LOTSIM - Simulation-Program for all pick-X type Lottery Games
>
>>[text deleted]
>
>> Draw numbers are generated by the standard RNG, ie. the rand()
>> function. Seed (srand(time)) is done once at pgmstart.
>
>Is this in C?
It's C++
>You should note that the pseudo-random number generator in most
>implementations of C is flawed. C returns an unsigned integer but
>the rightmost bits of the number returned are NOT RANDOM! Suppose
>C returns an integer in the range from 1 through 4,000,000,000 and
>you want to convert this to a number in the range from 1 through 49.
>Typically, programmers use the formula x = (n mod 49) + 1 to convert
>n, the number returned by pseudo-random number generator, to the
>number x of the desired range. This is a bad thing to do in C.
>This formula emphasizes the bits on the right end, namely those bits
>that are not random. To use the C's random number generator properly,
>you have to first get rid of the rightmost bits. (e.g. x >>= 8 will
>get rid of the 8 rightmost bits) If you want to do some serious
>simulations with lots of samples, you really shouldn't use any
>language's built-in generator and instead use a stronger generator.
>
>-- Glenn Rhoads
Thanks for the info, but I doubt this being still true with the
newer C++ compilers. Let me know if someone knows if it is also the
case with the Borland C++ compiler v4.52 in 32-bit target mode.
Nevertheless I'll check the workaround you gave, though I don't know
why even 8 (!) bits should be affected by this, whereas only 1 bit is
required for the sign...
Subject: Re: Why can't 1/0 be defined???
From: pausch@electra.saaf.se (Paul Schlyter)
Date: 19 Jan 1997 11:10:22 +0100
In article <32E0FE15.6D1C@mail.student.utwente.nl>,
Wilbert Dijkhof wrote:
> Paul Schlyter wrote:
>
>> In article ,
>> Travis Kidd wrote:
>>
>>> pausch@electra.saaf.se (Paul Schlyter) writes:
>>>>1 = oo * 0 = oo*(0 + 0) = oo*0 + oo*0 = 1 + 1 = 2
>>>
>>> What makes you think you can/must always substitute 1 for oo*0?
>>> oo*0=1/2 too, ya know.
>>
>> One fundamental algebraic rule is:
>>
>> A=B and A=C implies B=C
>>
>> Above you claim:
>>
>> oo*0 = 1 and oo*0 = 1/2
>>
>> This would imply: 1 = 1/2
>
> I don't say I agree with him,
> but what he says is: oo*0 = {1/2,1,and more}.
He also claims oo is a number.
Now for any two numbers x and y, x*y is single-valued. From this
follows that if oo is a number, then oo*0 must also be single-valued
since oo*0 is just a special case of the general x*y. Thus the only
way to make oo*0=1 and oo*0=1/2 is to also make 1=1/2.
As a matter of fact, the only way to make 1/0 even exist as a number
is to make any number equal to any other number. Such a number
system would not be particularly useful.
--
----------------------------------------------------------------
Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
Grev Turegatan 40, S-114 38 Stockholm, SWEDEN
e-mail: pausch@saaf.se psr@net.ausys.se paul@inorbit.com
WWW: http://www.raditex.se/~pausch/ http://spitfire.ausys.se/psr/
Subject: Re: Why can't 1/0 be defined???
From: pausch@electra.saaf.se (Paul Schlyter)
Date: 19 Jan 1997 11:11:36 +0100
In article ,
Travis Kidd wrote:
> Wilbert Dijkhof writes:
>>I don't say I agree with him,
>>but what he says is: oo*0 = {1/2,1,and more}.
>
> Exactly. "*0" is a relation. It maps oo to more than 1 number.
> So to say 0*oo=1 and 0*oo=2 does not imply that 1=2. Thus
> the two statements are not inconsistent.
This implies another thing: oo is not a number
For any two numbers x and y, x*y is single-valued.
For any number x, x*0 = 0
> Also, responding to Paul's e-mail to me, 0*oo is not a symbol.
> It is 3 symbols. The 3 symbols should not be taken together to
> refer to a specific number. They should be taken at face value.
> They might refer to any number that you can get when you multiply
> oo by 0.
0*oo is indeed three symbols, representing the arithmetic operation
multiplication on the two values 0 and oo
I guess we agree that 0 is a number.....
You also claim oo is a number, while all mathematicians agree that
oo is not a number. Now if oo is a number then oo*0 must be 0,
since x*0=0 for any number x.
For any two numbers x and y, x*y is single-valued.
Thus if oo is a number, oo*0 must be single-valued and equal to zero.
---------------------------------------------------------------------
If you admit that oo is not a number, then you're free to invent
any arithmetic rules you want for oo, without breaking the
well-defined set of rules for numbers.
If you insist on considering oo a number, you're in big trouble.
The usual arithmetic rules:
1a: definition ( a+b and a*b are defined )
1b: commutative law ( a+b = b+a and a*b = b*a )
1c: associative law ( (a+b)+c = a+(b+c) and (a*b)*c = a*(b*c) )
2a: additive unit ( a+0 = a )
3a: distributive law ( a*(b+c) = (a*b)+(a*c) )
3b: additive inverse ( a+(-a) = 0 )
4a: multiplicative unit ( a*1 = a )
4b: multiplicative inverse except for 0 ( a*(a^(-1)) = 1 )
must then be re-defined to something like:
1a: definition ( a+b and a*b are defined )
1b: commutative law fails
1c: associative law fails
2a: additive unit ( a+0 = a ) fails, since a may be different from a
3a: distributive law fails
3b: additive inverse ( a+(-a) = 0 ) fails since oo + (-oo) may be nonzero
4a: multiplicative unit ( a*1 = a ) fails since a may be diff. from a
4b: multiplicative inverse, including 0 ( a*(a^(-1)) = 1 )
The only thing you'll gain is to have a multiplicative inverse also
for zero. But you pay a pretty high price for this, since 1b to 4a
will be broken just to have a definition of the inverse of 0.
--
----------------------------------------------------------------
Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
Grev Turegatan 40, S-114 38 Stockholm, SWEDEN
e-mail: pausch@saaf.se psr@net.ausys.se paul@inorbit.com
WWW: http://www.raditex.se/~pausch/ http://spitfire.ausys.se/psr/
Subject: Re: 1997 is a prime year
From: James Tuttle
Date: Sun, 19 Jan 1997 01:31:01 +0000
When I wrote the original post here, I made a couple of mistakes, the
most egregious of which was my inclusion of the number 1 as prime. Many
people have straightened me out on that. (1 is neither prime nor
composite, it is unity.)
The other item of interest is the maximum number of primes into which a
number can be split. As originally posted, a 4-digit number ABCD can be
split 10 ways: ABCD ABC BCD AB BC CD A B C D, and the year 1997 can be
split into 6 primes 1997 199 997 19 97 7 according to this rule.
Below, I look at all 4-digit numbers in the range 1000-9999 to see how
many exhibit characteristics similar to 1997. But first, some replies:
Henrik Christian Grove wrote:
>
> U Lange writes:
>
> Michael L. Siemon wrote:
> :
> : Henrik Christian Grove wrote:
> :
> : Will there ever be a year which can be split in 10 different
> : primes?
>
> : Yes, assuming you are just asking the arithmetic question, an
> : infinite number of such, the first being at roughly 6.5 billion
> : CE. But it is a pretty good bet that ...
>
> Apparently you have a different understanding of "splitting into
> primes" than the poster who started this thread. In the sense of
> the original poster, the question of Henrik is IMHO trivial: The
> year 113731, for example, can even be split into 12 different
> primes:
>
> 3 7 11 13 31 37 73 113 137 373 1373 113731
>
> In the same way, it is very easy to increase the number of primes
> just by adding a few digits to known small "prime rich" years, such
> as the year 1373 mentioned by the original poster.
>
> It seems I was a bit unclear when I asked my question - I meant a year
> with 4 digits in its decimal representation, such that all the
> "splits" were prime.
There are 57 4-digit numbers ABCD such that ABCD is prime, ABC is prime,
and BCD is prime. Each of those numbers also has zero, one, two, or
three 2-digit primes embedded. These numbers are:
With no 2-digit primes (8 numbers): 1277 1499 3491 3499 4877 7577
7691 7877
With one 2-digit prime (19 numbers): 1571 2239 2293 2773 3593 3677
4211 4919 5233 6733 6911 7433 8233 8293 8839 9199 9293 9677 9839
With two 2-digit primes (23 numbers): 1733 1997 2113 2719 3313 3373
3673 3733 3739 4337 4397 4673 5419 5479 6199 6311 6317 6619 7193 7331
7431 9419 9479
With three 2-digit primes (7 numbers): 1373 3137 3797 6131 6173 6197
9719
So 1997 is indeed rare. Of the 9000 4-digit numbers, exactly 30 (0.33%)
are 4-digit primes with two 3-digit and two 3-digit primes embedded. On
the average, this occurs once every 300 years, the last time being 1733
when George Washington (born 1732) was a baby and the next being 2113.
Considering the full set of 57 numbers, the longest gap between
successive numbers is 652 (from 5479 until 6131). On the other hand, a
person born around 3670 could experience four of these years (3673,
3677, 3733 and 3739) as could a person born in the 62nd century (6131,
6173, 6197 and 6199).
The seven numbers in this list with three 2-digit primes have a total of
28 digits, and 21 of them are 1, 3 or 7, with four 9s and three 6s
sprinkled in for good measure. Only four of these numbers (1373, 3137,
3797 and 6173) have two single-digit primes.
The year 1373 continues to be fascinating in world history because the
medieval era was breaking up and there were signs here and there of the
early modern era. Most historians point to circa 1500 as the beginning
of the modern era, but it's instructive to look back a century earlier.
The world was in turmoil in 1373. Ashikaga Japan was in civil war,
France and England were in the midst of the Hundred Years War which
starred Joan of Arc, England was at the beginning of the period covered
by Shakespeare's histories (Richard II), Europe was one generation
removed from the Black Death, the Ming replaced the Mongol Yuan in
China, Tamerlane's incipient invasion of India would spawn the Mughal
Empire, the Ottomans were conquering the Balkans, Islam was spreading
along trade routes to Southeast Asia and West Africa, the increasing use
of paper (introduced from China by way of the Moors) in the West paved
the way for Gutenberg, the Roman Catholic Church was in schism, Wycliffe
and Petrarch were preparing the way for the Protestant Reformation and
the early Italian Renaissance, the 500-year-old Koryo was giving way to
the 500-year Yi dynasty in Korea, the Duchy of Muscovy (Russian Empire)
began to expand, and in another generation Portuguese navigators
commenced the Age of Discovery heralding a half-millennium of European
global domination.
Subject: Re: Why can't 1/0 be defined???
From: Wilbert Dijkhof
Date: Sun, 19 Jan 1997 13:28:26 +0000
James Hamblin wrote:
>
> All right, let me get my proverbial feet wet in this one.
>
> The field axoims (from an analysis book):
IF it forms a field, but ok, suppose it does ...
> i. + is commutative and associative
> ii. there is an element 0, such that for all x, x + 0 = x
> iii. every element x has an additive inverse, -x, s.t. x + (-x) = 0
> iv. * is commutative and associative
> v. there is an element 1, s.t. x * 1 = x
> vi. every element except 0 has a reciprocal 1/x s.t. x * 1/x = 1
> vii. x*(y+z) = x*y + x*z
>
> Clearly you intend to remove the restriction "except 0" in (vi) and
> define the symbol "oo" to be 1/0.
Ok.
> Let's make it a separate axiom:
>
> viii. there is some element oo such that 0 * oo = 1.
>
> Let's call this object a fauxld (just to have a name for it).
>
> Unfortuately, I can prove that for all x, x*0 = 0 without even using
> axiom (vi).
>
> Lemma 1: a+b = a+c iff b = c.
>
> b = b + 0 (ii)
> = b + (a + (-a)) (iii)
> = (b + a) + (-a) (i)
> = (a + b) + (-a) (i)
> = (a + c) + (-a) (given)
> = (c + a) + (-a) (i)
> = c + (a + (-a)) (i)
> = c + 0 (iii)
> = c.
>
> Claim: x*0 = 0.
> x*0 + x*x = x*(x+0) (vii)
> = x*x (ii)
> = 0 + x*x (ii)
>
> So, by the lemma, x*0 = 0.
Ok.
> Thus, oo * 0 = 0, contradicting (viii).
Why, what he said is that 00*0 isn't single-valued.
> So the "fauxld" is inconsistent, and worthy of its name.
>
> * and + are _functions_, and must be single valued. More importantly,
> you seem to want to introduce a definition for 1/0, even if it means
> breaking something as fundamental as transtivity of =. IMHO, I don't
> think it's worth it.
I don't see your problem, there are more functions that are multi-valued.
For example, e^z or z*(-1)^1/2.
> James
> --
>
> James Hamblin jeh13@cornell.edu
> ------------------------------------------------------------------------
> "Only one human captain has ever survived battle with the Minbari fleet.
> He is behind me. You are in front of me. If you value your lives, be
> somewhere else." -- Ambassador Delenn, "Babylon 5: Severed Dreams"
Wilbert
Subject: Re: Why can't 1/0 be defined???
From: Wilbert Dijkhof
Date: Sun, 19 Jan 1997 13:45:55 +0000
Paul Schlyter wrote:
>
> In article <32E0FE15.6D1C@mail.student.utwente.nl>,
> Wilbert Dijkhof wrote:
>
> > Paul Schlyter wrote:
> >
> >> In article ,
> >> Travis Kidd wrote:
> >>
> >>> pausch@electra.saaf.se (Paul Schlyter) writes:
> >>>>1 = oo * 0 = oo*(0 + 0) = oo*0 + oo*0 = 1 + 1 = 2
> >>>
> >>> What makes you think you can/must always substitute 1 for oo*0?
> >>> oo*0=1/2 too, ya know.
> >>
> >> One fundamental algebraic rule is:
> >>
> >> A=B and A=C implies B=C
> >>
> >> Above you claim:
> >>
> >> oo*0 = 1 and oo*0 = 1/2
> >>
> >> This would imply: 1 = 1/2
> >
> > I don't say I agree with him,
> > but what he says is: oo*0 = {1/2,1,and more}.
>
> He also claims oo is a number.
>
> Now for any two numbers x and y, x*y is single-valued. From this
> follows that if oo is a number, then oo*0 must also be single-valued
> since oo*0 is just a special case of the general x*y. Thus the only
> way to make oo*0=1 and oo*0=1/2 is to also make 1=1/2.
>
> As a matter of fact, the only way to make 1/0 even exist as a number
> is to make any number equal to any other number. Such a number
> system would not be particularly useful.
>
> --
> ----------------------------------------------------------------
> Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
> Grev Turegatan 40, S-114 38 Stockholm, SWEDEN
> e-mail: pausch@saaf.se psr@net.ausys.se paul@inorbit.com
> WWW: http://www.raditex.se/~pausch/ http://spitfire.ausys.se/psr/
See responds to another post.
Wilbert
Subject: Re: Why can't 1/0 be defined???
From: Wilbert Dijkhof
Date: Sun, 19 Jan 1997 13:44:39 +0000
Paul Schlyter wrote:
>
> In article ,
> Travis Kidd wrote:
>
> > Wilbert Dijkhof writes:
> >>I don't say I agree with him,
> >>but what he says is: oo*0 = {1/2,1,and more}.
> >
> > Exactly. "*0" is a relation. It maps oo to more than 1 number.
> > So to say 0*oo=1 and 0*oo=2 does not imply that 1=2. Thus
> > the two statements are not inconsistent.
>
> This implies another thing: oo is not a number
>
> For any two numbers x and y, x*y is single-valued.
>
> For any number x, x*0 = 0
>
>
> > Also, responding to Paul's e-mail to me, 0*oo is not a symbol.
> > It is 3 symbols. The 3 symbols should not be taken together to
> > refer to a specific number. They should be taken at face value.
> > They might refer to any number that you can get when you multiply
> > oo by 0.
>
> 0*oo is indeed three symbols, representing the arithmetic operation
> multiplication on the two values 0 and oo
>
> I guess we agree that 0 is a number.....
>
> You also claim oo is a number, while all mathematicians agree that
> oo is not a number. Now if oo is a number then oo*0 must be 0,
> since x*0=0 for any number x.
That's not true, oo is sometimes treaten as a number.
For example in complex-analysis or measure-theory.
> For any two numbers x and y, x*y is single-valued.
>
> Thus if oo is a number, oo*0 must be single-valued and equal to zero.
If you define that x*y (with x and y numbers) must be single-valued, that
yes oo isn't a number. But why do you insist that x*y must be single-valued.
The function x^y (for example (-1)^(1/2)) isn't single-valued, although
apperently nobody has problems with this.
> ---------------------------------------------------------------------
>
> If you admit that oo is not a number, then you're free to invent
> any arithmetic rules you want for oo, without breaking the
> well-defined set of rules for numbers.
>
> If you insist on considering oo a number, you're in big trouble.
> The usual arithmetic rules:
Suppose oo is a number
> 1a: definition ( a+b and a*b are defined )
> 1b: commutative law ( a+b = b+a and a*b = b*a )
> 1c: associative law ( (a+b)+c = a+(b+c) and (a*b)*c = a*(b*c) )
> 2a: additive unit ( a+0 = a )
> 3a: distributive law ( a*(b+c) = (a*b)+(a*c) )
> 3b: additive inverse ( a+(-a) = 0 )
> 4a: multiplicative unit ( a*1 = a )
> 4b: multiplicative inverse except for 0 ( a*(a^(-1)) = 1 )
>
> must then be re-defined to something like:
>
> 1a: definition ( a+b and a*b are defined )
> 1b: commutative law fails
> 1c: associative law fails
> 2a: additive unit ( a+0 = a ) fails, since a may be different from a
> 3a: distributive law fails
> 3b: additive inverse ( a+(-a) = 0 ) fails since oo + (-oo) may be nonzero
> 4a: multiplicative unit ( a*1 = a ) fails since a may be diff. from a
> 4b: multiplicative inverse, including 0 ( a*(a^(-1)) = 1 )
Suppose why define it different:
Definitions concerning oo:
1) k*oo = oo (for k real except k=0)
2) 0*oo = oo*0 = 0
3) oo-oo = 0
4) If a=oo then the product a*b is multivalued.
Must you still redefine things?
> The only thing you'll gain is to have a multiplicative inverse also
> for zero. But you pay a pretty high price for this, since 1b to 4a
> will be broken just to have a definition of the inverse of 0.
What you gain with this is a different case.
> --
> ----------------------------------------------------------------
> Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
> Grev Turegatan 40, S-114 38 Stockholm, SWEDEN
> e-mail: pausch@saaf.se psr@net.ausys.se paul@inorbit.com
> WWW: http://www.raditex.se/~pausch/ http://spitfire.ausys.se/psr/
Wilbert
Subject: Re: Why do Black Holes Form at all? (A slight deviation)
From: "John DeHaven"
Date: 19 Jan 1997 11:46:22 GMT
Christopher Hillman wrote in article
<5bpurr$2u3@nntp1.u.washington.edu>...
> On the other hand, according to the same model, if we stopped
> hovering and allowed ourselves to fall radially into the hole,
> we would pass through the horizon without noticing anything
> particular and quickly find the gravitational forces (radially
> expansive and tangentially compressive) increasing without limit,
> and in finite time what was left of our corporeal remains would
> impact the singularity. If we tried to avoid this fate after
> passing through the horizon, we could only DECREASE the time
> until our deaths (since by not resisting the fall, we follow
> a geodesic, the path of longest "length", whereas by accellerating
> away from this geodesic we are following a SHORTER world line
> which cannot avoid eventually striking the singularity.
--
What an interesting thread! Makes me want to ask some related questions.
1) The point of view of almost everyone in almost every non-suicidal
situation is for sure "outside" the S-radius of a black hole, and hovering
isn't necessary. Looking thru a telescope would be good enough to observe
it. If I understand this thread right, it is hard to see how a black hole
would seem ever to "grow" from this viewpoint. Time would "stop" (from our
viewpoint) for the thing the instant the developing black hole condensed
enough to "close up" and form an S-radius. This would seem to imply that
every black hole is some kind of fossil of the instant of its creation.
(Being really simplistic about it and excluding Hawking radiation, etc.)
Infalling matter (from our point of view) wouldn't _reach_ the radius. And
if time (from our point of view) has stopped at the radius, it necessarily
would be (from our point of view) changeless. That means it could not
expand (from our point of view). But it does (from our point of view), or
so they say. How?
2) And for that matter I don't understand Hawking radiation. Stop me if I'm
wrong, but it boils down to some occasional quantum virtual particle pairs
forming near the surface _just so_ that they don't recombine again, but one
of the pair is captured by the hole and the other escapes. But if, in such
an event, one particle _enters_ the hole, it would seem that it should
_increase_ the mass and the hole would _grow_ instead of shrink as a result
of Hawking radiation! Net gain for the universe: a more massive hole and
more radiation outside the hole! An interesting effect, if true, but I'm
sure this is wrong somehow. I even think I knew why once, but right now
I've forgotten.
3) Everyone talks about the singularity at the middle, of zero size, and
the very much non-zero S-radius that defines the "surface." What's in
between? If we managed to dive through, we surely could not cover that
distance instantly, FTL-wise. What _would_ our (or some godlike-being's)
experience be as we (or it) fell further towards that singularity? What is
the spacetime picture inside the hole? _Is_ there even any "distance?" Why
_isn't_ the volume between the singularity and the S-radius populated with
falling stuff? Or would it be?
4) I am skeptical (at the moment) about singularities, at least as they are
usually characterized. A singularity is a purely relativistic idea. But as
a blob of stuff collapses towards a limiting size of zero, it seems to me
that it must (eventually) collapse through epochs where _quantum_ spacetime
effects completely overwhelm relativistic spacetime effects. I don't think
you can even speak coherently about a "point in space" in quantum spacetime
at very small scales. Is that right? If so, then there would be some
super-intense, and very "singular," quantum-foam thingamajig at the center
of a black hole, but it would not be a dimensionless point.
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