Subject: ...004 Vietmath War: training bootcamp for p-adics
From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Date: 16 Nov 1996 20:43:56 GMT
Newsgroups: sci.math
Subject: Re: Primes in the 10-adics
Message-ID:
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 6 Oct 93 20:30:16 GMT
References: <1993Oct1.204009.23228@nevada.edu>
Organization: Beckman Instruments, Inc.
Lines: 17
In mmcconn@math.okstate.edu (Mark
McConnell) writes:
>Will somebody restate what people on this group mean by the 10-
>adics? Just the ring Z/10Z ? The ring generated by Z and 1/p, for
>all primes p different from 2 and 5? Some completion on the
>latter? Perhaps different correspondents mean different things :-) .
The completion of Z (well, N, actually) under the 10-adic metric;
d(n,m) = 10^-(# of zeros the decimal expansion of |n-m| ends in).
--
-------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: Convergence of positive Reals to 1. Transfinite Integers??
Date: 6 Oct 1993 08:30:06 GMT
Organization: Open Software Foundation
Lines: 25
Message-ID: <28tvme$34s@paperboy.osf.org>
References:
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
>The positive Reals after successive roots all converge to the
>number 1. My question is what do the Transfinite Integers--those
>infinite strings to the left. What do they converge to in successive
>roots? If a proof is easy please give.
(I'm not sure if you mean general nth roots. I'll restrict this
article to square roots.)
A positive Real number has two square roots. In the process you
describe, you always take the positive root because otherwise the
process would halt: you can't take the square root of a negative
number while staying within the Reals. (Although you could do it by
adjoining the element i, which gives you the Complex numbers.)
In the 10-adic numbers, unless you start with one of the fixed points,
the process always halts after a finite number of steps, because no
matter which square root you choose (and it's not always obvious: 31
and -31 are both square roots of 961, but there is no 10-adic square
root of 31, and there is a 10-adic square root of -31), you eventually
get to a point where you can't take another square root while staying
within the 10-adic numbers.
So the only way for it the sequence to "converge" is if it's constant.
In addition to 0 and 1, . . .92256259918212890625 and . .
.07743740081787109376 are the two other fixed points.
-------------------------------------------------------------
EMAIL
Date: Thu, 7 Oct 93 17:08:46 EDT
From: ÒTerry TaoÓ
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: Re: Convergence of positive Reals to 1. Transfinite Integers??
Newsgroups: sci.math
In-Reply-To:
References:
<28tvme$34s@paperboy.osf.org>
Organization: Princeton University
In article you write:
>In article <28tvme$34s@paperboy.osf.org> karl@dme3.osf.org (Karl >Heuer) writes:
>>In the 10-adic numbers, unless you start with one of the fixed
>>points, the process always halts after a finite number of steps,
>>because no matter which square root you choose (and it's not
>>always obvious: 31 and -31 are both square roots of 961, but there
>>is no 10-adic square root of 31, and there is a 10-adic square root
>>of -31), you eventually get to a point where you can't take another
>>square root while staying within the 10-adic numbers.
>
>If you restrict yourself to successive square roots (just to make it
>easier) and restrict to only positive roots, but include all-adic
>numbers, then do the successive square roots all converge to the
>number 1? Is a proof possible?
No. The set of squares comprises only half of all the 10-adic numbers
(or any -adic, actually); the set of fourth powers comprises only a
quarter, and so on. This means that virtually no number in the 10-adics
(or any other p-adic system) can be square-rooted indefinitely.
>In article <28tvme$34s@paperboy.osf.org> karl@dme3.osf.org (Karl >Heuer) writes:
>> . . .92256259918212890625 and . . .07743740081787109376
>>are the two other fixed points.
>
> Please, out of curiousity, does anyone know what the equivalent
>(analog?,podal?) of these two points for these two strings are in
>the Reals?
There are none. These are points such that x^2 = x, and x not equal to
0 or 1; there are no analogues of this in the reals.
The reals are not isomorphic to any p-adic. They may be isomorphic to a
product of the p-adics, the rationals, and some extra set which has yet
not been found. This I have told you about before, and it is an ongoing
research problem in algebraic number theory.
-------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: Convergence of positive Reals to 1. Transfinite Integers??
Date: 8 Oct 1993 06:31:11 GMT
Organization: Open Software Foundation
Lines: 35
Message-ID: <2931ff$mh9@paperboy.osf.org>
References:
<28tvme$34s@paperboy.osf.org>
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
>In article <28tvme$34s@paperboy.osf.org> karl@dme3.osf.org (Karl >Heuer) writes:
>>In the 10-adic numbers, unless you start with one of the fixed
>>points, the process always halts after a finite number of steps,
>>because no matter which square root you choose (and it's not
>>always obvious: 31 and -31 are both square roots of 961, but there
>>is no 10-adic square root of 31, and there is a 10-adic square root
>>of -31), you eventually get to a point where you can't take another
>>square root while staying within the 10-adic numbers.
I thought I had a proof of the above, but I found a flaw in it. The
conclusion might still be true, but at the moment I can't rule out the
possibility of some other 10-adic number from which you can take square
roots infinitely often. (Possibly looping back to the starting number
after a finite number of steps.)
>If you . . . restrict to only positive roots
The normal nonzero integers can be partitioned into "positive" and
"negative", but this breaks down in the p-adics. ...79853562951413 and
...20146437048587 are negatives of each other, but which one is the
"positive" of the pair?
(More generally, there's no way to compare two p-adic numbers and
decide which one is "smaller" in the usual sense. Instead, we talk
about which one is divisible by more powers of p.)
>> . . .92256259918212890625 and . . .07743740081787109376
>>are the two other fixed points.
>
> Please, out of curiousity, does anyone know what the equivalent >(analog?,podal?) of these two points for these two strings are in >the Reals?
One of them is "sort of like 0 and kind of like 1", while the other is
"sort of like 1 and kind of like 0". They aren't analogous to any
single Real.
-------------------------------------------------------------
EMAIL
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: p-adics
Date: Fri, 08 Oct 93 10:46:15 +0100
From: clauwe@argus.sci.kun.nl
By now you should have learnt that what you call `infinite integers'
are what everyone else calls `10-adic numbers'.
May I draw your attention to the following well known facts.
1. There is such a thing as `n-adic numbers' for EVERY number base n.
2. For a given number base these `n-adic numbers' form a `ring' i.e.
you can add them and multiply them like ordinary `Peano-style'
integers.
3. So there infinitely many kinds of your `infinite integers'! That's
why we specify n.
4. However if the number base n is not prime (for example for n=10 i.e.
for your infinite integers) then the equation `xy=0' has solutions for
which neither x nor y vanishes.
Similarly the equation `x*x=x' has solutions for which x is neither 0
nor 1.
Exercise: find these solutions.
5. For any of these number systems it is trivial to decide whether an
equation like `a^n+b^n=c^n' has solutions.
6. And that's why all those mathematicians (and Wiles in particular)
have tried to answer the question: can `a^n+b^n=c^n' be solved IN PEANO
INTEGERS.
-------------------------------------------------------------
Newsgroups: sci.math
From: william@fine.princeton.edu (William Schneeberger)
Subject: Re: Wiles's proof OK?
Message-ID: <1993Nov1.193645.26904@Princeton.EDU>
Organization: Princeton University
References:
<28sl60$22a@galaxy.ucr.edu>
Date: Fri, 8 Oct 1993 19:48:41 GMT
Lines: 13
In article <28sl60$22a@galaxy.ucr.edu> baez@guitar.ucr.edu (john baez)
writes:
[Stuff about Wolfskehl Prize Deleted]
>Since Fermat's last theorem is true your scenario is necessarily
>false. Since a false statement implies anything, the answer to your
>question is "yes."
Actually, this is not quite correct. This scenario would contradict
Peano's Axioms, but the Wolfskehl Prize exists in the physical universe
independently of any infinite axiom system.
--
Will Schneeberger DISCLAIMER: The above opinions are
not
william@math.Princeton.EDU necessarily those of Ludwig Plutonium
-------------------------------------------------------------
Newsgroups: sci.math
From: william@fine.princeton.edu (William Schneeberger)
Subject: Re: Convergence of positive Reals to 1. Transfinite Integers??
Message-ID: <1993Oct8.202611.26062@Princeton.EDU>
Organization: Princeton University
References:
<28tvme$34s@paperboy.osf.org>
Date: Fri, 8 Oct 1993 20:26:11 GMT
Lines: 31
In article <28tvme$34s@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
>In the 10-adic numbers, unless you start with one of the fixed
>points, the process always halts after a finite number of steps,
>because no matter which square root you choose (and it's not
>always obvious: 31 and -31 are both square roots of 961, but there
>is no 10-adic square root of 31, and there is a 10-adic square root
>of -31), you eventually get to a point where you can't take another
>square root while staying within the 10-adic numbers.
This is not true. This is true in the 2-adic case; if n is divisible by
2 it must be so infinitely often and thus be 0, and if n is congruent
to 2^i + 1 (mod 2^(i+1)) its square root must be congruent to 2^(i-1)
+- 1 (mod 2^i) , so that eventually you will be forced to take the
square root of a number congruent to 3 or 5 mod 8. So the only 2-adic
components that work are 0 and 1.
In the 5-adics, however, any number congruent to 1 mod 5 has a square
root congruent to 1 mod 5. So here the ones that can go infinitely
often are 0 and those congruent to 1 mod 5.
So the 10-adics from which you can take infinitely many square roots
are those whose 2-adic components are 0 or 1 and which end in 1 or 6,
and the two numbers 0 and . . .92256259918212890625.
But the infinite sequence of square roots will not converge very often.
--
Will Schneeberger DISCLAIMER: The above opinions are
not
william@math.Princeton.EDU necessarily those of Ludwig Plutonium.
-------------------------------------------------------------
Newsgroups: sci.math
From: jgk@versant.com (Joe Keane)
Subject: Re: Convergence of positive Reals to 1. Transfinite Integers??
Message-ID:
Summary: You can order them.
Organization: Versant Object Technology
References: <2931ff$mh9@paperboy.osf.org>
Date: Sat, 9 Oct 1993 03:05:48 GMT
Lines: 38
In article <28tvme$34s@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
>In the 10-adic numbers, unless you start with one of the fixed
>points, the process always halts after a finite number of steps,
>because no matter which square root you choose (and it's not
>always obvious: 31 and -31 are both square roots of 961, but there
>is no 10-adic square root of 31, and there is a 10-adic square root
>of -31), you eventually get to a point where you can't take another
>square root while staying within the 10-adic numbers.
I think the basic problem is that if X - 1 is divisible by 2^k, then
X^2 -1 is divisible by at least 2^(k+1).
In article mh9@paperboy.osf.org, karl@dme3.osf.org (Karl Heuer) writes:
>The normal nonzero integers can be partitioned into "positive" and
>"negative", but this breaks down in the p-adics. ...79853562951413
>and ...20146437048587 are negatives of each other, but which one
>is the "positive" of the pair?
IÕd say the first one is the positive one. The ordering is fairly
arbitrary, and doesn't have familiar properties, for example if you add
two positive numbers you're likely to get a negative number. But it is
useful to have a total ordering of numbers. For example, when computing
roots, different methods will give you different answers, but you can
normalize the roots by picking the most positive conjugate, and then
you will see if you get a consistent answer. In the ordering i use, the
most positive number is
. . .7534928015749953548048064091990916324448335217311978340148925781.
>One of them is "sort of like 0 and kind of like 1", while the other is
>"sort of like 1 and kind of like 0". They aren't analogous to any
>single Real.
But, as is common, you get a good model if you go to matrices. The
fixed points are much like [[1 0] [0 0]] and [[0 0] [01]]. Multiplying
by these numbers gives you a projection function that isolates either
the 2-adic or 5-adic component of numbers.
--
Joe Keane, amateur mathematician
jgk@versant.com (uunet!amdcad!osc!jgk)
-------------------------------------------------------------
EMAIL
From: "Terry Tao"
Subject: Re: IGNORE ANYTHING WRITTEN IN CAPITALS
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Sat, 9 Oct 93 16:10:04 EDT
In-Reply-To: <5818819@blitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Oct 8, 93 10:26 pm
>
>--- You wrote:
>HavenÕt you heard? Self-reference is always a good source of subtle
>humor.
>--- end of quoted material ---
>HI TERRY I AM NOT IGNORING YOU. Give me some help of what Dik is
>saying. Is he saying that yes an infinitude of primes of the form
>(2^n)-1 and (2^2^n)+1 are manufactorable in 10-adics OR NOT?
>
>> (1) Manufacture an infinitude of primes of the form (2^n)-1 in the
>> 10-adics.
>
>Depends on what you wish. As somebody else already remarked,
>there are only two finite primes in the 10-adics (2 and 5). Unless
>you allow for multiplication by units, in that case 6, 14, 15 etc. are
>also prime. Check some literature about prime ideals and stuff like
>that. So we look at the prime ideal generated by 5 and find the
>following elements that have the above form: 15, 255, 4095,
>65535, 1048575, etc. More general: (2^4n)-1 for n=1 . . . inf. Of
>these, those with n not a multiple of 5 are a "prime" according to
>the extended definition. (So, of those mentioned above only
>1048575 is not a "prime".) And these are all.
>
>> (2) Manufacture an infinitude of prime numbers N of the form
>> (2^2^n) + 1 in the 10-adics.
>
>Similar.
>--
>dik t. winter, cwi, kruislaan 413, 1098 sj
>
Ludwig, do you know what a Principal Ideal Domain is? It will be much
easier to explain equivalence classes of primes if you do.
Terry
-------------------------------------------------------------
From: sichase@csa5.lbl.gov (SCOTT I CHASE)
Newsgroups: sci.physics,sci.math
Subject: p-adic numbers in physics
Date: 12 Oct 1993 15:47 PST
Organization: Lawrence Berkeley Laboratory - Berkeley, CA, USA
Lines: 21
Message-ID: <12OCT199315475102@csa3.lbl.gov>
With all this talk about 10-adics, etc., I thought you might all like
to know that the most recent Physics Reports article is entitled
"p-adic Numbers in Physics," and is somewhat amusing.
The authors treat string theory with p-adics, replacing the real line
which is the boundary of an open stringÕs world sheet with the p-adics,
thereby discretizing the manifold. To get there, they give some
background, and do some interesting games, coming up with p-adic
versions of the gamma function, solutions to FermatÕs Last Theorem,
etc.
I am surprised at how much of what has recently been discussed around
here is in this paper. The mathematics may be too steep for most
physicists, and the physics may be too intense for most mathematicians.
*I* certainly didnÕt follow either the math or the physics entirely.
But it was fun anyway.
-Scott
Subject: Vietmath War: ...003 bootcamp for p-adics
From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Date: 16 Nov 1996 20:31:40 GMT
Newsgroups: sci.math
From: william@zucchini.princeton.edu (William Schneeberger)
Subject: Re: Primes in the 10-adics
Message-ID: <1993Oct1.211636.11392@Princeton.EDU>
Organization: Princeton University
References: <1993Oct1.025325.13517@nevada.edu>
<1993Oct1.204009.23228@nevada.edu>
Date: Fri, 1 Oct 1993 21:16:36 GMT
Lines: 24
In article <1993Oct1.204009.23228@nevada.edu> jangel@nevada.edu (JEFF
ANGEL) writes:
>In article a_rubin@dsg4.dse.beckman.com (Arthur Rubin) writes:
AR>>Actually, I thought the only "primes" in the 10-adic numbers are
AR>>2 and 5; and every number ending in 1, 3, 7, or 9 is a unit.
(Or AR>>am I wrong?)
JA> [correction regarding associates]
Another point. As I remember these definitions, a prime is a number p
such that
p|ab implies p|a or p|b
whereas an irreducible is a number p with
p=ab implies one of a and b is a unit.
By these definitions, the only _irreducibles_ (up to association) are 2
and 5. The two idempotents . . .6259918212890625 and . .
.3760081787109376 are also prime, as are their associates. So this does
give an infinitude of 10-adic twin primes, multiplying the latter
number by a unit and adding or subtracting 2 to avoid ending in 0. In
fact this gives you an infinitude of triple primes.
--
Will Schneeberger DISCLAIMER: The above opinions are
not
william@math.Princeton.EDU necessarily those of Ludwig Plutonium.
-----------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: N CONJECTURE, O CONJECTURE
Date: 1 Oct 1993 21:48:28 GMT
Organization: Open Software Foundation
Lines: 26
Message-ID: <28i8jc$j7l@paperboy.osf.org>
References: <1993Sep27.133514.11964@Princeton.EDU>
<1993Sep29.164556.1890@Princeton.EDU>
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
>In article <1993Sep29.164556.1890@Princeton.EDU> >kinyan@fine.princeton.edu (Kin Chung) writes:
>>Observe that . . .99999 + 1 = 0
> Prove this Kin.
This is true, and in fact I think it's one of the beautiful things
about the set of p-adic numbers: it includes not only the positive
integers (in which all but a finite number of digits are 0), but also
the negative integers (in which all but a finite number of digits are
p-1), without having to use the minus sign.
Most computers already use this, with binary numbers encoded in two's
complement: the number -1 is actually . . .1111 (an infinite string of
one bits). Typically it's only maintained to one word of precision, but
there's usually some type of sign-extend function that will add another
word of 0 bits (for positive numbers) or 1 bits (for negative numbers).
Two proofs:
(1) Just add it up by hand. In the rightmost place you have 9+1 = 10,
put down 0 and carry the 1. In the next position you have 9 plus the
previous carry, again 9+1 = 10, put down 0 and carry the 1. Continue.
Clearly, you get a 0 in each position, so the result is . . .0000
exactly.
(2) . . .9999 = sum (k>=0) of 9*10^k; by the rule for geometric series,
this is 9/(1-10), which is -1.
-------------------------------------------------------
Newsgroups: rec.puzzles,sci.math
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Subject: Boundary between "finite" and "transfinite" integers (was Re:
Fermat Proof.
Message-ID:
Organization: Dartmouth College, Hanover, NH
References:
<28i0re$kj7@vtserf.cc.vt.edu>
Date: Sat, 2 Oct 1993 16:48:02 GMT
Lines: 8
In article <28i0re$kj7@vtserf.cc.vt.edu> hart@corona.math.vt.edu (Heath
David) writes:
> d. The finite integers have no maximal element. The element . .
> .99999 is a maximal element of the set of transfinite integers.
Can you prove it? And then please say what you think of the number
. . .99999.9999. . .
------------------------------------------------------
EMAIL
From: "Kin Y. Chung"
Date: Sat, 2 Oct 1993 22:27:01 -0400
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: Information about p-adics that could help you. (Seriously)
Newsgroups: rec.puzzles,sci.math
In-Reply-To:
References:
Organization: Princeton University
Cc:
In article you write: [stuff about
10-adics deleted]
(1) Let us assume that conversion between p-adics and 10-adics can be
done. Here's how it must work: suppose a number can be represented as
.....abc in the p-adics, then subtracting c from the number makes it a
multiple of p. Thus c must be the remainder upon division by p. So
given the 10-adic representation, just divide by p and see what
remainder you get and you have c. Now subtract c and divide by p and
the number must now have p-adic representation ....ab. You can now
repeat the process to get b, and then a, and so on. In fact, this is
how one converts between bases in normal arithmetic.
(2) given an appropriate metric defined on the integers (I won't
provide that now), the p-adics can be obtained as the metric completion
of the integers. Thus you can be justified in saying that the 10-adics
are the "completion" of the integers. However,
(3) the 10-adics do not form an integral domain, i.e. there are nonzero
10-adics a and b such that ab=0. When p is prime, there are no nonzero
p-adic numbers a and b such that ab=0. Therefore the 10-adics and the
p-adics (p prime) are NOT equivalent ("isomorphic"), that is, there is
no way to write every 10-adic uniquely as a p-adic.
(4) more about the topology of the p-adics: the metric from which they
arise is such that p^n approaches zero as n approaches infinity. This
is the reason why they "eventually return" to zero. This is also why
there cannot be a well-ordering on the p-adics that agrees with that
for the integers. For this reason also, you cannot just append the
noninteger real numbers with their usual ordering to the p-adics
because you are trying to mesh two incompatible topologies.
Here is some advice, although I can't see you accepting it: give up on
trying to formalise "infinite" arithmetic before you make a fool of
yourself on sci.math. The complex numbers are the algebraic closure of
the integers as they stand, and they are perfect in that regard. Many
great minds have already studied "infinite" arithmetic and this
resulted in the transfinite ordinals and cardinal. You have to abandon
just about every useful property of the integers in order to accomodate
"infinite" arithmetic, and all this for what? Just to redefine
countability so that the reals are countable? By doing so, you will
then find that the original set of integers is infinite but not
countable, and Cantor's result will appear before you again.
-------------------------------------------------------------
Newsgroups: rec.puzzles,sci.math
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Subject: Re: Fermat Proof.
Message-ID:
Organization: Dartmouth College, Hanover, NH
References:
Date: Sun, 3 Oct 1993 00:29:23 GMT
Lines: 48
In article dik@cwi.nl (Dik T. Winter) writes:
>Writing 10-adic numbers base 9 would be eh, problematical. When
>starting base 9 your extension leads to 9-adic numbers. A different
>beast at all.
> For instance, in the 10-adics the expression a^n + b^n = c^n is true
> for all n given the following values of a, b and c:
> a = . . .9977392256259918212890625
> b = . . .0022607743740081787109376
> c = 1
> by virtue of the fact that a, b and c are idempotents (e.g. axa = a)
> and a + b = c. And this is only possible because 10 is divisible by
> two distinctive primes. You will not find a similar example in the
>9-adics. If you do not accept that arithmetic you should leave out
>Karl Heuer from your list as he would agree (and did use similar
>arithmetic finding his examples).
Thanks for the concerns Dik. When a builder of a new house has
the foundation already cemented P-adics, but then needs to now
construct the house. My attention is focused on constructing the house.
The first priority of 10-adics is to realize that they complete
the positive numbers. Before these discussions about 10-adics on
sci.math the positive numbers were considered to go out on a straight
line from the Descartes coordinate system. That is a false notion. The
complete set of positive numbers is the 10-adics and positive
non10-adics Reals between the 10-adics. The old fossil definition of
10-adics. The 10-adics have the bizarre property of returning to 0,1
and 2. That is as it should be for the set of all positive numbers are
the arithmetic set description of Riemannian geometry. Before my
teachings the math community was using a partial, incomplete set of the
positive numbers. With 10-adics the positive numbers are complete for
they are homeomorphic with Riemannian geometry. Because the 10-adics
return to 0,1,2 is exactly as Riemannian geometry is. For Riemannian
geometry has podal and antipodal points. These points are
substitutable, switchable. Distinct but switchable points.
Now to answer the representation pseudoproblem. There are no
establish axioms for P-adics representation. When there will be the
number representation between 9-adic and 10-adic will be as pointless
as the claim that primality is affected between decimal representation
and binary representation. This is built into the Peano axioms. When
the extension of the Peano axioms which I am working on, which includes
10-adics is formulated then the representation issue will evaporate.
Dik your concerns about the adic representation are low priority in
the house building. I would place that priority at about where extra
caulking is done to seal the windows airtight.
-------------------------------------------------------------
EMAIL
From: "Terry Tao"
Subject: Re: R is Countable
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Mon, 4 Oct 93 20:41:22 EDT
In-Reply-To: <5705527@blitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Oct 4, 93 8:16 pm
>
>--- You wrote:
>Unfortunately no. The square root of 3, for example, exists in reals
>but not in 10-adics.
>--- end of quoted material ---
>thanks for this info. Terry I admit that you know way more about infinite integers than I do. I am just learning about them so I am going to send you many questions. Here is the first
> In article
>Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly) writes:
>
>>Is this really called for Ludwig? I could make any number of
>>insults about your appearance also. But instead of doing that, I
>>will prove what I said. By definition 1/3 is the multiplicative
>>inverse of 3. That is, in a ring, a number x is defined to be 1/3 if
>>3x is the multiplicative identity, which means that it is a number
>>such that for every y in the ring, (3x)y=y. But now if you multiply
>>...66666667 against ...00000003 (which we are calling 3) in your
>>infinite integers you get the infinite integer ...000001. If you
>>check you will see that ...00001 times y is y for all y in the
>>infinite integers. Therefore 1/3, in your ring of infinite integers,
>>is ...66667.
>
>Question 1) what Tilly has done here. are each and every infinite
>integer uniquely equal to some positive Real number? If you can
>prove please send also.
>
No - however, you can always divide one p-adic by another as long as
the denominator is co-prime to p.
Well, maybe it depends on what you mean by "equal". If you mean "has
the same properties as", then the answer is no; as I said before, there
is no square root of 3 in the 10-adics (just check the first digit).
However, there is a one-to-one correspondence.
>Question 2) When I construct the 10-adics I get unpleasant
>properties. When I construct the p-adics (with p a prime) I get a
>field (in analogy with the reals). I am asking you Terry what do I
>want to construct the infinite integers from 10-adics???? I am
>asking you what I want?? Are the 10-adics a natural extension of
>the decimal Natural numbers??
10-adics are somewhat unpleasant, true: they are the direct sum of the
2-adics and the 5-adics. As such they are a relatively bad number
system. Don't forget there are many, MANY number systems that are
bigger than the integers: the p-adics are just the start. You have the
Stone Cech compactification of the integers (a VERY ugly space),
various algebraic closures, .. many, many number fields. And of course
the rationals and the reals.
As to your question about a "natural extension", well it is an
extension in a sense - a closure. Let me digress and give a lecture on
metric spaces.
Suppose you have a set X with a metric on it, let's call it d(x,y):
d(x,y) is a real number denoting the distance between x and y. Thus for
example on the real line the most reasonable metric to take is d(x,y) =
|x-y|. (the || signs stand for absolute value). In Euclidean space you
have the Euclidean metric, involving square roots of sums of squares;
then you have spherical metrics, hyperbolic metrics, etc. There are
three properties that a metric should have: positivity, symmetry, and
the triangle inequality: I won't go through them here.
But anyway - with a metric, you have two notions of convergence. We can
say that a sequence of points x_1, x_2, ... in X "converges to x",
where x is another point in X, if the metric d(x_n,x) goes to 0. This
is a very natural definition of convergence. But there is another
definition of convergence, we can say that a sequence x_1,x_2, ... in X
is a "Cauchy sequence" if d(x_n,x_m) goes to 0 as n,m go to infinity.
In other words, the x_n get closer together.
It is not hard to prove that a convergent sequence is also a Cauchy
sequence. But the reverse is not always true: for example, if X is the
rational real numbers, with the standard metric d(x,y) = |x-y|, then
the sequence
3, 3.1, 3.14, 3.141, 3.1415, 3.14159, etc
is a Cauchy sequence (i.e. i the sequence gets closer to itself), but
it has no limit in Q. It does however have a limit in R, namely pi. So
in a sense R is more "complete" than Q, in that it contains all the
limit points that Q should have: every Cauchy sequence in Q (or in fact
R), must converge in R. (In fact, this is practically part of the
definition of R). You can also think of R as the "closure" of Q - the
smallest complete space such that Q is dense in it.
If you look up an elementary topology book, you'll find the formal
notion of "completing" a metric space. But now I get to my point:
The 10-adics are the completion of the positive integers under the
metric d(x,y) = 1 / 10^n, where n is the highest degree of 10 dividing
|x-y|.
Put it another way. If you take the integers normally, you would think
that numbers would be far apart if the big digits differ, hence 100001
and 200001 are quite far apart. But now, we say that those numbers are
only .00001 apart, whereas 1 and 2 are distance 1 apart. You can think
of it as placing the integers in a strange new geometry, where the
first digits matter much more than the last. And now, you can see what
happens if you try to complete the space; if you take a sequence like
1, 11, 111, 1111, 11111, ...
then it is a Cauchy sequence (think about it.. the first few digits
remain constant, hence the distance between elements far advanced in
this sequence is very small), so there must be a limit in the
completion of Z, namely the 10-adic 111111........
If you use the normal metric for Z, then the integers are already
complete: there is no need for "extra" numbers. The integers can be
completed to the 10-adics by the 10-adic metric above; if you use the
9-adic metric, d(x,y) = 1/9^n where n is the highest degree of 9
dividing |x-y|, then you get a different space, the 9-adics.
So the 10-adics are ONE extension of the integers. There are many,
many, metrics you can put on Z, and different metrics usually give you
a completely different number system.
Cheers,
Terry
-------------------------------------------------------------
EMAIL
From: "Terry Tao"
Subject: Re: R is Countable
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Sun, 3 Oct 93 21:18:43 EDT
In-Reply-To: <5673804@blitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Oct 3, 93 7:50 pm
>
>--- You wrote:
>.88888888888888889000
>
>(it's analagous to subtracting a real decimal in [0,1] from 1.)
>--- end of quoted material ---
>Terry tell me what you know of this idea-- that every positive Real
>number has a 10-adic representation?
>
Unfortunately no. The square root of 3, for example, exists in reals
but not in 10-adics.
On the other hand, there is a 1-1 correspondence between the reals and
the 10-adics.
Terry
-------------------------------------------------------------
EMAIL
Subject: Re: 10-adics
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Sun, 3 Oct 1993 22:08:14 -0400 (EDT)
In-Reply-To: <5675106@blitzen.Dartmouth.EDU> from "Ludwig
Plutonium" at Oct 3, 93 08:59:23 pm
From: kinyan@math.princeton.edu
>Kin do all the 10-adics have inverses?
No; for example 5 does not have an inverse. Trivially, zero also
doesnÕt have an inverse, so I assume you meant whether all nonzero
10-adics have inverses.
By the way, why do you view 10-adics as such a useful thing? I am aware
that the p-adics (especially p prime) are of great importance in number
theory, but I want to know what you find great about them.
--
Kin Yan Chung (kinyan@math.princeton.edu)
kinyan@fine.princeton.edu (Kin Yan Chung) writes:
-------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Infinite in both directions? (was: PERHAPS A CONSTRUCTIVE
PROOF . . . )
Date: 3 Oct 1993 22:12:17 GMT
Organization: Open Software Foundation
Lines: 34
Message-ID: <28nio1$7lp@paperboy.osf.org>
References:
<1993Oct2.032035.11423@Princeton.EDU>
In article <1993Oct2.032035.11423@Princeton.EDU> tao@fine.princeton.edu
(Terry Tao) writes:
>Well, maybe he can salvage one. With his amazing new numbers, can
>[LP] find any digits at all to the square of this number:
>. . .1111111111111111.1010010001000010000010000001. . .
>where the string of zeroes increases
>by one each time?
>
>(Hint: do NOT try long multiplication.)
Actually, you made it too easy, since the left half is rational and
hence this can be collapsed to a traditional real number, which can
then be squared in the usual way.
Or, one could respond that you're not playing by the rules, since your
number is infinite in both directions. (P-adic numbers are
left-infinite and right-finite, while reals are left-finite and
right-infinite.)
However, this does lead to a question that I've been thinking about
lately. (For concreteness, I'll restrict myself to p=2 here, though the
question does generalize.) Is it possible to create a well defined
number system where each number is the sum of a 2-adic number and a
real number? Since the rationals are special cases of both, it would be
nice to have be 0 (and hence the representation of a number
as a doubly-infinite string of bits will not be unique).
Now, how about algebraic numbers that appear to be common to both
systems? Both the 2-adics and the reals can solve x^2=17; should one of
the 2-adic solutions (...11001101100100010111 or
...00110010011011101001) be considered equal to the positive real
solution 100.00011111100000111101...? If so, how can we choose which
one? (If not, then we have zero-divisors.)
Even after considering some 2-adic and real numbers to be "equal",
there will be numbers that are neither pure 2-adic nor pure real, e.g.
z=sqrt(-7) + sqrt (7) =
...10001100000010110111.101001010100111111110...). z^2/14 is sqrt (-1).
Can this be represented as the sum of a 2-adic and a real?
-------------------------------------------------------------
Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Subject: Re: 10-adics, manufacture primes of the form (2^n)-1
Message-ID:
Organization: Dartmouth College, Hanover, NH
Date: Mon, 4 Oct 1993 00:10:35 GMT
Lines: 5
(1) Manufacture an infinitude of primes of the form (2^n)-1 in the
10-adics.
(2) Manufacture an infinitude of prime numbers N of the form (2^2^n)+1
in the 10-adics.
-------------------------------------------------------------
Newsgroups: sci.math
From: dik@cwi.nl (Dik T. Winter)
Subject: Re: 10-adics, manufacture primes of the form (2^n)-1
Message-ID:
Organization: CWI, Amsterdam
References:
Date: Mon, 4 Oct 1993 02:01:14 GMT
Lines: 37
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
Ludwig, why are you concerned so much with the 10-adics? They do not
form a very nice ring at all! Remember what I posted a few days ago:
> For instance, in the 10-adics the expression a^n + b^n = c^n is true
> for all n given the following values of a, b and c:
> a = . . .9977392256259918212890625
> b = . . .0022607743740081787109376
> c = 1
> by virtue of the fact that a, b and c are idempotents (e.g. axa = a)
> and a + b = c. And this is only possible because 10 is divisible by
> two distinctive primes.
(I should have written: "because 10 as a (finite) integer is divisible
by two distinctive primes".)
To worry you a bit more: a*b = 0. {Easy: a*b = a * (1-a) = a - a*a =
a-a.} So we have here a ring with zero divisors.
> (1) Manufacture an infinitude of primes of the form (2^n)-1 in the
> 10-adics.
Depends on what you wish. As somebody else already remarked, there are
only two finite primes in the 10-adics (2 and 5). Unless you allow for
multiplication by units, in that case 6, 14, 15 etc. are also prime.
Check some literature about prime ideals and stuff like that. So we
look at the prime ideal generated by 5 and find the following elements
that have the above form: 15, 255, 4095, 65535, 1048575, etc. More
general: (2^4n)-1 for n=1 . . . inf. Of these, those with n not a
multiple of 5 are a "prime" according to the extended definition. (So,
of those mentioned above only 1048575 is not a "prime".) And these are
all.
> (2) Manufacture an infinitude of prime numbers N of the form
> (2^2^n) + 1 in the 10-adics.
Similar.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: dik@cwi.nl
-------------------------------------------------------------
EMAIL
Printed on Mon, Oct 4, 1993 at 10:33 PM
From: Ludwig Plutonium
To: tao@math.Princeton.EDU
Subject: Re: R is Countable
--- You wrote:
Eeek. I think the 10-adics satisfy vastly different axioms than the
natural numbers.
0 is not the successor of any number: this is now false. (....99999) If
two numbers have the same successor, then they are the same: this is
still true.
The successor of a, plus b, is equal to a plus the successor of b: this
is still true.
a plus 0 is 0: this is still true.
The principle of induction: this is certainly NOT true.
--- end of quoted material ---
This is beautiful Terry for where I want to end at is Riemannian Geom.
The gaussian positive curvature, just simply elliptic geometry. I want
to get podal and antipodal points but as you have informed me some of
the Reals have no equals in the infinite strings. I wonder if there is
a way to remedy that so that every Real (podal point) has a infinite
string (antipodal point). If I can see my way over that hurdle then
MOLTO VIVACE.
-------------------------------------------------------------
From: dreier@jaffna.berkeley.edu (Roland Dreier)
Newsgroups: sci.math
Subject: Re: Infinite in both directions? (was: PERHAPS A CONSTRUCTIVE
PROOF . . . )
Date: 4 Oct 93 21:11:04
Organization: U.C. Berkeley Math. Department.
Lines: 21
Message-ID:
References:
<1993Oct2.032035.11423@Princeton.EDU> <28nio1$7lp@paperboy.osf.org>
In article <28nio1$7lp@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
However, this does lead to a question that I've been thinking about
lately. (For concreteness, I'll restrict myself to p=2 here, though the
question does generalize.) Is it possible to create a well defined
number system where each number is the sum of a 2-adic number and a
real number? Since the rationals are special cases of both, it would be
nice to have be 0 (and hence the representation of a number
as a doubly-infinite string of bits will not be unique).
This is not quite an answer to the question you are posing here, but
(if you don't know about it already) you might consider learning about
the adele ring of a number field. It is a ring that contains
information about all completions of a number field, both infinite and
finite. So for the rationals, it is a certain subset of the direct
product of the reals with Z_p for every p. (It is really the
"restricted topological product"). References would be Cassels &
Frohlich's "Algebraic Number Theory", and Lang's book of the same name.
--
Roland "Mr. Excitement" Dreier dreier@math.berkeley.edu
-------------------------------------------------------------
EMAIL
Date: Mon, 4 Oct 93 21:21:07 EDT
From: "Terry Tao"
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: Re: Convergence of positive Reals to 1. Transfinite Integers??
Newsgroups: sci.math
In-Reply-To:
Organization: Princeton University
Cc:
In article you write:
> The positive Reals after successive roots all converge to the
>number 1. My question is what do the Transfinite Integers--those
>infinite strings to the left. What do they converge to in successive
>roots? If a proof is easy please give.
They don't converge, because the last digit (the most significant one,
remember), bounces around a lot. Even if the last digit is 1, then the
second last digit bounces around, unless it is 0, in which case the
third last digit bounces around, etc..
the upshot is that 1 is the only number whose roots (by the way, not
all the roots will exist, especially the k(p-1)^th roots if p is prime
(4kth roots for 10-adics)) converge.
Terry
-------------------------------------------------------------
EMAIL
From: "Kin Yan Chung"
Subject: Re: 10-adics
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Mon, 4 Oct 1993 23:17:54 -0400 (EDT)
In-Reply-To: <5706329@blitzen.Dartmouth.EDU> from "Ludwig
Plutonium" at Oct 4, 93 08:40:10 pm
>Kin I am very much behind the curve on adics. You know far more
>than I on these things. So I am going to keep asking you alot of
>questions. I do not know if 10-adics are what I want. I do know that
>I want an extension of the whole numbers--what I call infinite
>integers. Are they 10-adics for which noone has yet bothered to
>alter the Peano axioms in order to make the other adics 2-adics, 3-
>adics the same only a different base representation. The idea that
>numbers are not affected by representation.
OK, I now know what you want, but why do you want them? What good will
they do? Why do you think that mathematics will benefit from them?
>is every positive Real number equal to some 10-adic?
No. As I pointed out earlier, there is no 10-adic whose square is 3, so
sqrt(3) cannot be identified with any real number. I don't know why I
came up with 3 first, since 2 has the same properties.
I am quite convinced that 10-adics are not what you want, since 1 has
infinitely many divisors. I think this is undesirable for you because
then 1 will not be a perfect number. In fact, since 1 divides every
number, it follows that every 10-adic has infinitely many divisors.
There is also a problem with defining perfect (p-adic) numbers in terms
of positive divisors because there is no notion of positivity in the
p-adics.
I will also warn you that anything of the sort you are seeking will
necessarily lead to multiple infinite cardinalities. For instance, the
10-adics have an infinite subset that cannot be put into one-to-one
correspondence with the 10-adics.
--
Kin Yan Chung (kinyan@math.princeton.edu)
-------------------------------------------------------------
EMAIL
Date: Tue, 5 Oct 93 09:57:09 EDT
From: "Kin Yan Chung"
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: Re: Riemannian (elliptic,spherical) geom. are infinite
integers
Newsgroups: sci.math,sci.physics,sci.chem
In-Reply-To:
Organization: Princeton University
Cc:
In article you write:
> The wedding of Lobachevskian geometry to the set of negative
>numbers has not occurred yet. The big drawback is to reconfigure
>the complex numbers augmented onto the negative numbers. And
>also the negative numbers have infinite strings to the left, right.
>What a mess.
You said it: what a mess. I want to point out that you don't need the
negative numbers to motivate complex numbers since in order for 1 to
have 3 cube roots you need to introduce complex numbers anyway. In
fact, historically the acceptance of complex numbers did not arise from
the need to define square roots of negative numbers but from the desire
to use Cartan's formula for the solutions of a cubic equation. Cartan's
formula involved the taking of cube roots, and when there were three
real solutions the only way to make them materialize from the formula
was to use the complex cube roots.
Finally, I shall repeat one thing about 10-adics that I had said
earlier. The reason that the 10-adics "eventually return" to 0 is that
the topology associated with the 10-adics is such that 10^n gets closer
to zero as n get larger. This means that 100 is closer to 0 than it is
to 1, 1000 is even closer, and so on. Hardly consistent with the
familiar integers.
--
Kin Yan Chung (kinyan@math.princeton.edu)
-------------------------------------------------------------
EMAIL
Date: Tue, 5 Oct 93 14:51:01 EDT
From: "Terry Tao"
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: Re: Riemannian (elliptic,spherical) geom. are infinite
integers
Newsgroups: sci.math,sci.physics,sci.chem
In-Reply-To:
Organization: Princeton University
In article you write:
> Riemannian (elliptic, spherical) geom. are infinite integers
>unioned with the Reals. The infinite integers---infinite strings to
>the left usually have an equal Real. For example, ....6667. is 1/3.
>This is a most beautiful result for the meaning is that the positive
>numbers when they go out there really far, they come back onto
>themselves. This is
This is not true, because you are using two different metrics.
....66667 is the limit of 7, 67, 6667, ... under the 10-adics metric.
0.3333... is the limit of 0.3, 0.33, 0.333, ... under the classical
metric.
The closure of the integers and the terminating decimals in the 10-adic
and classical metrics respectively are similar in some respect (as you
mentioned, They both have a reciprocal of 3, for example), but there is
no duality between them.
A better analogy would be that, if you somehow turned the integers
Òinside outÓ so that they then resembled a Cantor set, then if you
follow a convergent sequence in this Cantor set you can get the
equivalent of Real numbers - sometimes. The number ....6666667 is not
Òfar outÓ from the finite integers if you use the 10-adic metric. And
that number is not somewhere between 0 and 1 either: the 10-adics
cannot be ordered.
To repeat: the 10-adics use different geometry, different ordering, and
have different properties from the reals. They canÕt be matched in any
useful manner (other than the canonical one-to-one mapping), and they
canÕt be welded into one geometry with any degree of union.
Terry
------------------------------------------------------------
Newsgroups: sci.math
From: william@zucchini.princeton.edu (William Schneeberger)
Subject: Re: Infinite in both directions? (was: PERHAPS A CONSTRUCTIVE
PROOF . . . )
Message-ID: <1993Oct5.175655.1721@Princeton.EDU>
Organization: Princeton University
References:
<1993Oct2.032035.11423@Princeton.EDU> <28nio1$7lp@paperboy.osf.org>
Date: Tue, 5 Oct 1993 17:56:55 GMT
Lines: 35
In article <28nio1$7lp@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
[random stuff about LP's numbers deleted]
>However, this does lead to a question that I've been thinking about
>lately. (For concreteness, I'll restrict myself to p=2 here, though
>the question does generalize.) Is it possible to create a well
>defined number system where each number is the sum of a 2-adic
>number and a real number? Since the rationals are special cases of
>both, it would be nice to have be 0 (and hence the
>representation of a number as a doubly-infinite string of bits will
>not be unique).
[similar questions deleted]
Not likely, depending on what you want. A continuos image of the
2-adics and the interval [0,1] identifying appropriate rationals must
have a really weak topology.
To prove this, note that a basic open set in the 2-adic metric of the
rationals looks like
{2^n a/b + u | a an integer, b an odd integer}
and that this set is dense in the usual metric on the rationals.
The inverse image of a closed set containing a nonempty open set must
be a closed set (in the real-metric sense) containing a basic open set
(in the 2-adic sense) of the rationals. Thus it must contain all of the
rationals, and since the rationals are dense in both metrics, it must
contain all of the reals and 2-adics.
If we assume that the map is surjective (we may, as above, look at the
image), this gives us that there are no proper closed sets containing
nonempty open sets.
--
Will Schneeberger DISCLAIMER: The above opinions are
not
william@math.Princeton.EDU necessarily those of Ludwig Plutonium
Subject: ...0005 Vietmath War: training bootcamp for p-adics
From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Date: 16 Nov 1996 21:09:54 GMT
Newsgroups: sci.math,sci.physics
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Subject: Re: . . .9999.999. . . is infinity; inf is a number &
property
Message-ID:
Organization: Dartmouth College, Hanover, NH
References:
<1993Oct9.203457.3936@Princeton.EDU>
<1993Oct10.150112.17440@Princeton.EDU>
Date: Thu, 14 Oct 1993 07:05:43 GMT
Lines: 88
In article sguare@ask.uio.no (Simen
Gaure) writes:
>If you look carefully, you may find that you're not in unexplored
>land. Your infinite integers have been explored before, under the
>name of 10-adics. Your notion of countability and of what's legal or
>not in mathematics is the topic of mathematical logic and
>set/model theory. You may very well create a universe without all
>power sets and prove theorems and do other mathematics in there.
>Or a universe without infinity if you wish. Or a universe in which
>everything is finitely constructible.
>
>However, this is not the universe most mathematicians work in. To
>claim the absolute truth or falsity of certain independent axioms is
>not done in mathematics. But you are free to drop e.g. the power set
>axiom.
>
>But if you claim the absolute truth or falsity of such an axiom, you
>are not doing anything new. You are repeating history, this was the
>state of mathematics for centuries. I.e. the belief that the models
>of mathematics had to correspond to certain observable phenomena.
>Your favorite phenomenon seems to be something connected with
>atoms. Older phenomena used to be line segments, ratios, various
>finite constructions, time etc.
>
>Keep up the good work, use your intuition; that's how mathematics
>is created. But be a little more careful with your statements. The
>history of mathematics is indeed very long, many brilliant thoughts
>have been thought. Don't assume you're the first one. This doesn't
>degrade your discoveries, they may still be the result of good
>mathematical reasoning. They many however not be so new as you
>think.
>
>Simen Gaure
>University of Oslo
I salute you Simen. I think I have given some people the impression
that I discovered P-adics which I called Infinite integers at first. I
had never heard of P-adics until here in sci.math of this year. I had
independently discovered them in order to show the Reals are countable.
After reading your post Simen I wen to look-up some of the P-adic
history.
I pay tribute to Kurt Hensel who created P-adic fields, with his
1908 work Theorie der Algebraischen Zahlen and his later expansion with
the work Zahlentheorie of 1913. KH defined the 4 basic operations with
these numbers. KH makes use of expansion of numbers into power series
of a prime number p. Other major contributors to p-adics were Paul
Seelhof, Francois Edouard, Anatole Lucas, Fortune Landry, AJC
Cunningham, FWP Lawrence and DN Lehmer.
I do not want to mislead anyone into thinking that I discovered
P-adic numbers. I have much to learn about them. But I do claim the
following 1) that the P-adic numbers are the true set of all Whole
Numbers for which the integers as per the Peano Axioms are just a crude
axiomatized subset thereof. 2) This number . . .9999.999. . . is
infinity itself, since when multiplied by 0 the product is 1. The
number . . .9999. is the last whole number and the number . .
.9999.999. . . is the max positive number. 3) That the set of all
positive numbers for which each one of those positive numbers is an
infinite string to the leftwards and rightwards of the decimal point is
the number system equivalent to Riemannian Geometry. 4) Hence I claim
the first unification of Riemannian Geometry with the positive number
system. Much of which has to be worked-out but the doors are now wide
open. (I had just read a posting by Scott Chase that P-adics were
printed in a physics journal. I am convinced that the authors were
aware of sic.math goings-on.) 5) If you accept the Peano axioms in a
sense you are saying that Riemannian Geometry is not spherical. And if
you accept Peano's axioms you are in a bind for the Peano integers
never end-hence they go off into those leftward strings. 6) It was here
in sci.math that I learned from others, namely Karl Heuer and Will
Schneeberger that P-adics are solutions to Fermat's Last Theorem, and
by deduction it is obvious to me that FLT is false. No proof of FLT
needs to be searched for, only the Peano Axiom of integers needs repair
to include all the Whole numbers, namely P-adics. 7) here in sci. math
I discovered that Wiles alleged proof most definitely is a fake, and I
am sure others will in time come to see the same. For if his fake proof
is accepted then those supporters must talk about the boundary at which
the so called "finite integers" do not work, and the "infinite
integers" do work. 8) But what is most important about the P-adics is
the unification of numbers with geometry and the open doors to finding
a better axiomatization for Whole numbers to include all the integers,
and also an assault into better definitions for dimension.
Simen thanks again. And I have talked to long now, but in closing I
think I had better post to sci.physics my thoughts on the "Meaning of
Time" and the "Meaning of Space and Geometry". And to post my thoughts
on the 3 Schools of Mathematics, for after 7 Nov I will no longer make
anymore posts.
-------------------------------------------------------------
Newsgroups: sci.math
From: shallit@jalapeno.cs.wisc.edu (Jeff Shallit)
Subject: Re: . . .9999.999. . . is infinity; inf is a number &
property
Message-ID: <1993Oct14.155441.18722@cs.wisc.edu>
Organization: University of Wisconsin, Madison -- Computer Sciences
Dept.
References:
Date: Thu, 14 Oct 1993 15:54:41 GMT
Lines: 14
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
> Other major contributors to p-adics were Paul
>Seelhof, Francois Edouard, Anatole Lucas, Fortune
>Landry, AJC Cunningham, FWP Lawrence and DN
>Lehmer.
These people are known for their work on the integer factoring problem.
All but one had little or nothing to do with p-adic number theory, the
exception being Lucas.
As for people who really *did* (or do) work in p-adic number theory,
the names Dwork, Katz, Koblitz, Amice, Mahler come to mind.
Jeff Shallit
-------------------------------------------------------------
From: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.physics,sci.math
Subject: Re: p-adic numbers in physics
Date: 14 Oct 1993 18:58:44 GMT
Organization: University of California, Riverside
Lines: 39
Message-ID: <29k7h4$oj6@galaxy.ucr.edu>
References: <12OCT199315475102@csa3.lbl.gov>
<29ig75$n2i@amhux3.amherst.edu>
In article <29ig75$n2i@amhux3.amherst.edu> mkrogers@unix.amherst.edu
(Michelle Rogers) writes:
> I was told Manin has suggested that string theory
>be done over the ring of adeles -- combination of
>the p-adic and real and complex fields. But what
>I know about the adeles does not make up for my
>ignorance of string theory.
What I know about everything else combined does not make up for my
ignorace of adeles OR string theory! :-) But. . . I think Witten
came before Manin in pondering "adelic string theory." Let me briefly
impart my minute understanding of his subject. Besides the usual notion
of absolute value on the rational numbers - let us call this |
|_{infinity} for some odd reason - there are a bunch of others called |
|_p, one for every prime number p. These also satisfy the triangle
inequality etc., so one can complete the rational numbers with respect
to these absolute values (i.e., make sure Cauchy sequences have limits)
and get a field, the p-adics, just as one can complete the rationals
with respect to the usual | |_{infinity} and get the reals. It is
actually nice to think of the reals as the p-adics where one uses the
prime p = infinity. One nice fact is that if one takes any rational n/m
and takes the product of |n/m|_p as p ranges over all primes, including
the prime at infinity, one gets 1. Or in other words, one can express
|n/m|_{infinity} in terms of all the |n/m|_p. This can be used to
reduce certain calculations in the real numbers to lots of calculations
in the p-adics. "Great," the physicists must be thinking, "instead of
doing one calculation in the real numbers I only have to do infinitely
many calculations in the p-adic numbers. That's really progress!" :-)
But the point is that if one is a sufficiently number-theoretic kind
of person this can actually make certain calculations doable. Witten
saw how to do this with certain calculations in string theory (I don't
know if he was the *first*). The way to systematically keep track of
such problems is with adeles, which are a beautiful big fat sort of
number simultaneously. So people got interested in "adelic string
theory." Manin, a mathematician who has done a lot in number theory,
gauge theory, and quantum groups (and has written a textbook in
mathematical logic, and is a very nice guy to boot), wrote some stuff
suggesting that maybe nature really *does* like p-adics just as much as
the reals.
-------------------------------------------------------------
From: brock@ccr-p.ida.org (Bradley Brock)
Newsgroups: sci.physics,sci.math
Subject: Re: p-adic numbers in physics
Date: 15 Oct 1993 11:39:09 -0400
Organization: IDA - Center for Communications Research, Princeton
Lines: 32
Message-ID: <29mg6t$ft@tang.ccr-p.ida.org>
References: <12OCT199315475102@csa3.lbl.gov>
<29ig75$n2i@amhux3.amherst.edu> <29k7h4$oj6@galaxy.ucr.edu>
In article <29k7h4$oj6@galaxy.ucr.edu>, john baez
wrote:
> In article <29ig75$n2i@amhux3.amherst.edu>
> mkrogers@unix.amherst.edu (Michelle Rogers) writes:
> One nice fact is that if one takes any rational n/m and takes the
> product of |n/m|_p as p ranges over all primes, including the prime
> at infinity, one gets 1. Or in other words, one can express
> |n/m|_{infinity} in terms of all the |n/m|_p. This can be used to
> reduce certain calculations in the real numbers to lots of
> calculations in the p-adics.
One must be a little careful here to normalize things properly. In fact
define |p|_p=1/p and |a|_p=1 if gcd(a,p)=1 and extend the definition to
all rationals by the multiplicative property |ab|_p=|a|_p|b|_p. With
this definition the product over all "absolute values" is one. Hence,
two numbers are close p-adically if their difference is divisible by a
large power of p.
One interesting thing about p-adics is that it takes more steps to get
to a complete algebraically close field. For the usual absolute value
the process takes two steps, namely complete the rationals to get the
reals and then algebraically close the reals to get the complexes.
However, in the p-adics this process takes four steps (if I remember
correctly), namely one needs to complete the rationals to get the
p-adics Q_p then algebraically close the p-adics to get \bar{Q_p} which
is not complete and then repeat both steps again. See KoblitzÕs book on
p-adics for details.
Some calculations in the rationals cannot be reduced to calculations in
the p-adics. For example a rational curve, i.e. a curve of genus zero,
has a rational point iff it has a p-adic point for all p. However, if
the genus>0 this is no longer true. For example, the Fermat curve has
p-adic points for all p but no rational point.
--
Bradley W. Brock | ÒAll they asked was that we should
brock@ccr-p.ida.org | continue to remember the poor, the very
IDA/CCR Princeton, NJ | thing I was eager to do.Ó - a Tarsian
tentmaker
--------------------------------------------------------------------
Newsgroup: sci.math
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Subject: PROOF THAT PEANO AXIOMS GO INTO THE P-ADICS
Message-ID:
Organization: Dartmouth College, Hanover, NH
Date: Mon, 18 Oct 1993 15:36:11 GMT
Lines: 5
Such a proof in favor or disfavor will settle the issue of
counterexamples for Fermat's Last Theorem. It would settle the proof of
FLT once and for always. Prove that given 1 and being able to always
add 1 yields not only the infinite string leftwards of 0's but all the
P-adics. I am soliciting help for this proof.
------------------------------------------------------------
Newsgroups: sci.math
From: Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)
Subject: Re: PROOF THAT PEANO AXIOMS GO INTO THE P-ADICS
Message-ID:
Organization: Dartmouth College, Hanover, NH
References:
Date: Mon, 18 Oct 1993 18:40:09 GMT
Lines: 15
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
> Such a proof in favor or disfavor will settle the issue of
>counterexamples for Fermat's Last Theorem. It would settle the
>proof of FLT once and for always. Prove that given 1 and being able
>to always add 1 yields not only the infinite string leftwards of 0's
>but all the P-adics. I am soliciting help for this proof.
Actually the result that you want is false. As a matter of fact it is
easy to show with induction that every integer has only a finite number
of nonzero digits. But I am glad that you appear to recognize that
there is a difference between the p-adics and the usual integers.
Ben Tilly
-------------------------------------------------------------
Newsgroups: sci.math
From: hahn@newshost.lds.loral.com (Karl Hahn)
Subject: Re: PROOF THAT PEANO AXIOMS GO INTO THE P-ADICS
Message-ID: <931018174013@are107.lds.loral.com>
Lines: 46
Organization: Loral Data Systems
References:
Date: Mon, 18 Oct 93 22:40:13 GMT
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
> Such a proof in favor or disfavor will settle the issue of
>counterexamples for Fermat's Last Theorem. It would settle the
>proof of FLT once and for always. Prove that given 1 and being able
>to always add 1 yields not only the infinite string leftwards of 0's
>but all the P-adics. I am soliciting help for this proof.
Someone already posted the outline of a proof to the contrary, but it
appears that LP requires this spelled out in detail (if even that will
persuade him):
Peano postulates the existence of a nonempty set N (the natural
numbers) and a function s(n).
Axiom 1: for all n in N, s(n) is also in N.
Axiom 2: if m and n are in N and s(m) = s(n) then m = n.
Axiom 3: there exists a unique element of n (called 1) such that 1 !=
s(n) for any n in N.
Axiom 4: if X is a subset of N, and X contains 1, and for every x in X,
s(x) is also in X, then X = N.
Here, s(n) is intended to be the familiar function of adding 1 to a
number.
Now for the proof:
Let X be the set of all x in N such that its decimal representation
terminates leftward in all 0's. Clearly 1 has this property, thererfore
1 is in X. Let x be an arbitrary element of X. This means it terminates
leftward in all 0's. Clearly s(x) also teminates leftward in all 0's.
Hence, by axiom 4, X = N. This means that X, the set of all leftward
terminating natural numbers, completely exhausts N, the set of all
natural numbers. There is no room left for the p-adics or anything
else.
--
------------------------------------------------------------
EMAIL
From: "William Schneeberger"
Date: Mon, 18 Oct 93 16:46:53 EDT
To: Ludwig.Plutonium@Dartmouth.edu
Subject: Re: PROOF THAT PEANO AXIOMS GO INTO THE P-ADICS
Newsgroups: sci.math
In-Reply-To:
Organization: Princeton University
In article you write:
> Such a proof in favor or disfavor will settle the issue of
>counterexamples for Fermat's Last Theorem. It would settle the
>proof of FLT once and for always. Prove that given 1 and being able
>to always add 1 yields not only the infinite string leftwards of 0's
>but all the P-adics. I am soliciting help for this proof.
You will likely get no help from anyone. If you succeed in this,
though, you will truly have accomplished your goal of breaking the
established mathematics.
--
Will Schneeberger Terry told me that I should
william@math.Princeton.EDU change my .signature .
-------------------------------------------------------------
Newsgroups: sci.math
From: tao@lentil.princeton.edu (Terry Tao)
Subject: Re: PROOF THAT PEANO AXIOMS GO INTO THE P-ADICS
Message-ID: <1993Oct22.074804.18852@Princeton.EDU>
Summary: Read and learn, Ludwig.
Organization: Princeton University
References:
<1993Oct21.021204.29615@Princeton.EDU>
Date: Fri, 22 Oct 1993 07:48:04 GMT
Lines: 163
This thing about N! = 0 is very simple to resolve. The salient point is
that the number system you get by completing the naturals depends
heavily on the metric you want to use. Put it another way - it depends
on what you mean by "closer".
Consider this question:
which number is closer to 0, 10! or 1000! ?
There have been two views about this (both, incidentally, expoused by
LP):
1. 1000! > 10!, i.e. 1000! is larger than 10!, so 10! should be the
closer to 0.
Here, we are using the standard metric (or order) for the naturals: we
say that x is closer to z than y if |x-z| < |y-z|. Perfectly good
metric, and the one that most people are used to. For example, 1001 is
fairly close to 1000, whereas 1000000 is not. But in this metric, N!
does not tend to 0, and we do not get any 10-adics. The natural numbers
are complete under this metric. This is the metric that LP intuitively
uses, while trying to grapple with the other metric (see below). It is
also the order induced by the Peano axioms. (x > y if x is an eventual
successor of y).
2. 1000! is closer to 0 than 10!, because it has more 0s at the end.
This is the definition of "closer" that LP uses to derive that N! goes
to 0: x is closer to ) than y if x has more trailing 0s. More
generally, we can say that x is closer to z than y if |x-z| has more
trailing 0s than |y-z|. Again, this is a perfectly good metric. No
problem at all. and yes, the 10-adics come out as the completion of the
naturals under this metric. This is in fact one standard definition of
the 10-adics. (The other being the direct limit of Z/10^n, as n tends
to infinity.)
The problem of course is, that with this metric you have to chuck out
the notion of "order". 1000 is now close to 1000000 but is really far
away from 1001 (in fact no number can be further away from 1000 than
1001). Knowing that xy and z>w no longer guarantee that x+z > y+w. Take y=w=0, z = 1, and
x = . . . . . .99999.
Similarly, x>y>0 and z>w>0 do not guarantee that xz>yw. Example: x = .
. . . .88889, y = . . . . . .666666. z = 9, w = 3.
Every number is > 0. In fact, we have -1 > 0. (-1 = . . . .9999).
Put it another way: for every x, both x and -x are positive.
There exists two numbers, neither of which is larger than the other.
Which one is bigger, . . . .01010101010101 or
. . . .10101010101010?
Finally, . . . . .9999 is not the best "bound" for the finite integers.
Every finite integer is "less" than . . . . . .111111, for instance.
Or even . . . . .101010101, etc, etc. Indeed, there is no "lowest"
infinite integer.
In short, there is no order that can be imposed on the 10-adics that
makes any sense: certainly, it doesn't satisfy many of the laws that we
expect of it. At best, we have transitivity and anti-symmetry, and
that's it.
Summing up again: using this metric gives you a p-adic system, which is
complete and very beautiful and useful in its own right, but you have
to give up (among other things, like induction) the notion of order.
You cannot have your cake and eat it too.. you cannot have the
old-style order of view 1 while still trying to create p-adics using
view 2. In particular, there isn't anything very interesting about the
10-adic . . . . .9999. It's just -1. It may "look" large, but so
what? 99999999999 looks larger than 111111111111111, but the second
number is larger (in the old-style order, of course.)
Finally: the p-adics do not have "logarithmic spirals" and "Riemmanian
curvature". They have a geometry equivalent to that of a Cantor set.
For example, the 2-adics are topologically equivalent to Cantor's
"middle thirds" set: the set obtained by considering the interval [0,1]
and taking out its "middle third" (1/3,2/3), and then considering the
two remaining intervals and taking out their middle thirds, etc. In
other words, the set of all numbers in [0,1] whose digital
representation in base 3 consists only of 0s and 2s: no 1s. The
natural numbers get mapped into the terminating decimals of the Cantor
set, and they look sort of like this:
[ . . . . . . . . . . . . . . . .the interval [0,1] . . . . . . . . . .
. . . . . . . . . . . . . . . . . . ]
0 -2 1
-1
2
3
4 6 5
7
8 12 10 14 9
13 11 15
. . . . . .
or, flattening them out,
0 8 4 12 2 10 6 14 -2 1 9 5 13 3 11
7 15 -1
(essentially, by flipping the 2-ary digit string, converting it to base
3, and then multiplying by 2.)
(Incidentally, this Cantor set induces an order, which is slightly less
useless for the p-adics than the naive order. This order is superior,
seeing as addition and multiplication are actually continuous with
respect to the topology generated by this order - though order still
does not preserve addition or multiplication.)
And one last thing that should be repeated: the natural numbers are of
course a different set from say, the 10-adics. . . . . . .111111 is
not a natural number: its digit string does not terminate in a sequence
of 0s. They are two markedly different number systems. One happens to
be imbedded in the other, but that's about it. The 10-adics of course
are a much larger space than the naturals, cordinality wise: if there
was a mapping f from the naturals N to the 10-adics Z_10, then the
element A of Z_10, defined to be
9- (the 1st digit of f(1)) + 10*(9-(the 2nd digit of f(2)) +
100*(9-(the 3rd digit of f(3)) + . . .
is an element which is not in the range of f. Hence there is no
surjective mapping from N to Z_10.
Both number systems exist in their own right, but are different. They
are equally consistent. Certainly the naturals, WHEN ENDOWED WITH THE
10-ADIC METRIC (this is important), are dense in the 10-adics. If they
are endowed with their normal, old-style metric, then they can still be
homeomorphically imbedded in the 10-adics (for example using the map n
-> 10^n), but any such imbedding will no longer be dense. (N having
only one possible accumulation point).
Integers with the standard metric, integers with the 10-adic metric,
10-adics with the 10-adic metric - all three are perfectly good
topological rings. None of them is the "best". (This stuff about
"true math" is bogus. Any logical system is true math. Not to be
confused with mathematical models of the physical world of course.
That is applied math. :-) Certain statements can be true in some of
those systems and false in others. (e.g. FLT true in 1 and 2, false in
3; N! -> 0 false in 1, true in 2 and 3; Peano axioms true in 1 and 2,
false in 3; Topological completeness true in 1 and 3, false in 2.).
So of course Ludwig is wasting his time trying to deny that one of
these systems exists, or attempting to prove that they blend into each
other (the second system is dense in the third, admittedly), or that
two of them are the same (I've given a proof above that the 10-adics
cannot be put in 1-1 correspondence with the naturals, hence with the
integers). This in part is due to Ludwig's confusions, particularly
having two different metrics on the same space. (It's very easy to
prove anything you want if you give a word (like "close") two different
definitions at the same time.) In keeping his old definitions of
closeness, he has an image of a real line; using the 10-adic
definition, he can see loops. The point is that the first image is
incompatible with 10-adics, though he tries to fit it in with thoughts
of "logarithmic spirals" and so forth. Reals and 10-adics are
incompatible, as can be seen by the impossibility of multiplying . . .
. .10001010110 and .01101010001. . . . Abandon the real line idea,
and everything becomes perfectly consistent. The 10-adics are
geometrically a Cantor set, nothing more.
Now. I hope this stops the meaningless debate about p-adics. LP has a
very fuzzy picture of them at the moment (still clinging to the real
line, and notions of order), but they are a very useful and nice space,
and their elementary properties have been studied to a depth far far
greater than anything Ludwig has ever contemplated. But they are not a
replacement for the integers, which have their own fascinating
properties.
Terry
.sig donated to fight the federal deficit
-------------------------------------------------------------
EMAIL
From: "Kin Yan Chung"
Subject: Re: primes
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Fri, 22 Oct 1993 10:57:44 -0400 (EDT)
In-Reply-To: <6146902@blitzen.Dartmouth.EDU> from "Ludwig Plutonium" at
Oct 22, 93 02:56:44 am
>Thanks Kin. Perhaps the my tone is very aggressive but if you really
>knew me I am shy passive type of person.
Well it certainly doesn't seem that way from your articles.
>Anyway I need the following help--- primes
>of the form (2^n)-1 & (2^2^n)+1 in the 10-adics.
I can't really help you too much on this because the notion of prime
needs to be formulated for the 10-adics first. I remember Will having
said that the only primes in the 10-adics are 2, 5 and their associates
(my memory could be faulty here); the problem with primes in 10-adics
is that there is a 10-adics x such that 3x = 1, so 3 is a unit and not
a prime.
Personally, I think that talking about primes in the 10-adics is a bit
of a waste since most numbers have inverses (and are therefore not
primes). If what I recalled above is correct, then the only finite
primes are numbers such as 15, 6, 18, etc which are of the form 2A or
5A where A is a finite number not divisible by 2 or 5.
>You could help me alot if you listed say TEN 10-adics out to say 20
>digits showing my a pattern of MANUFACTURING an infinitude of
>primes of the form (2^n)-1
All the examples that you can manufacture using 2^n-1 would be (finite)
integers since exponentiation isn't well defined for anything else
(except infinite cardinals/ordinals which you don't accept). Basically,
to make sure that 2^n - 1 is prime (as a 10-adic), you only have to
ensure that it is divisible by 5 but not by 25. Simple number theory
shows that 2^n - 1 is divisible by 25 if and only if n is divisible by
4. Also, 2^n - 1 is divisible by 25 if and only if n is divisible by
20. Therefore 2^n - 1 is prime (assuming everything said earlier is
correct) if and only if n is a multiple of 4 but not a multiple of 20.
>Then please do the same for (2^2^n)+1
This is trickier. By 2^2^n I assume you mean 2^(2^n) in agreement with
the standard convention. For n>1, 2^n is a multiple of 4. Therefore
2^2^n - 1 is divisible by. This means that 2^2^n + 1 is not divisible
by 5, and it is also clearly not even. Hence 2^2^n + 1 is cannot be
prime if n>1. When n=1, we get 5 which is prime.
--
Kin Yan Chung (kinyan@math.princeton.edu)
-------------------------------------------------------------
Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Subject: Re: PROOF THAT PEANO AXIOMS GO INTO THE P-ADICS
Message-ID:
Organization: Dartmouth College, Hanover, NH
References:
<931021124251@are107.lds.loral.com>
Date: Fri, 22 Oct 1993 02:08:00 GMT
Lines: 32
In article <931021124251@are107.lds.loral.com>
hahn@newshost.lds.loral.com (Karl Hahn) writes:
>By what metric does N! approach ..000? Certainly not by the metric
>|x-y|. Not by the difference in measure of how many digits you have
>to traverse before you get to the region of all zeros leftward (in N!,
>this measure grows without limit, in ...000. it's always 0). You can't
>say x approaches y without some definition of what that means.
Thanks for the analysis Karl. If you buy that ...9999. +1 is proved
to equal to 0. Then likewise, N! is ...000. It is inexorable.
So if you do not like it, well, go into art. Must I go into some
type of discussion for ...9999.+1 =0 about x approaches y decorated
with definitions? 0 carry the 1, 0 carry the 1, . .
In article <931021124251@are107.lds.loral.com>
hahn@newshost.lds.loral.com (Karl Hahn) writes:
>In order to show that the Peano Naturals go into the 10-adics (or
>P-adics as you call them), you must find a finite integer, n, such that
>n+1 is a 10-adic that does not terminate leftward in all zeros. I
>would even accept it if you could find a finite n such that n! was
>nonterminating.
No I do not. For this line implies a boundary. A break in the Whole
numbers. The trouble here Karl is that you are keeping in the old.
The standard proof to show one set equal to another is proper subset
method. That I have already done.
But in writing this reply to you Karl I just intuited something
important about the P-adics. P-adics do not just circle back in one
circle through the negative numbers ending at -1=...9999. But the
P-adics logarithmically spiral back to -1 through many turnings.
-------------------------------------------------------------
Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Subject: Re: PROOF THAT PEANO AXIOMS GO INTO THE P-ADICS
Message-ID:
Organization: Dartmouth College, Hanover, NH
References:
<1993Oct21.021204.29615@Princeton.EDU>
Date: Fri, 22 Oct 1993 02:19:52 GMT
Lines: 26
In article <1993Oct21.021204.29615@Princeton.EDU>
kinyan@fine.princeton.edu (Kin Yan Chung) writes:
>Which is why the Peano axioms are what they are. Whoever heard of . . .999
>apples? Also, can you tell me the . . .444th digit of the 10-adic number . .
>.987654321098765432109876543210? (The 10-adics are defined to be
>sequences of digits, so given a "whole" number n, you should be able to tell
>me the nth digit of any 10-adic.)
These are your sentiments Kin. They are void of math content.
Whoever heard of the decimal number system during Archimedes?
There is a separate definition of Whole number and P-adic. Whole
number is a broader definition for it is the total possible arrangement
of all ten decimal digits.
In article <1993Oct21.021204.29615@Princeton.EDU>
kinyan@fine.princeton.edu (Kin Yan Chung) writes:
>First of all, N! does not approach zero as N increases in the usual topology of >the natural numbers. Certainly, nobody will disagree with you that under a
>different metric, for instance that of the p-adics, N! does approach zero as N >increases.
Yes it does for as you yourself pointed out that ...9999. +1 is
proved equal to 0 because carry the 1 leaving 0, carry the 1 leaving 0.
Likewise N! as it increases equals ...0000. ATOM
---------------------------------------------------------
EMAIL
From: "Terence C. Tao"
Date: Sun, 24 Oct 1993 17:48:33-0400
To: Ludwig. Plutonium @ Dartmouth.EDU
Subject: Re: FLT counterexamples neither a,b,c are = 1??
Newsgroups: sci.math
In- Reply-To:
Organization: Princeton University
In article you write:
>ln article
>Ludwig.Plutonium @ dartmouth.edu (Ludwig Plutonium) writes:
>
È I please need help in finding P-adic counterexamples of FLT
È a^n+b^n=c^n, where none of the a,b,c are equal to 1.
If you consider the solution (a,b,c) as the same as (ka,kb,kc) for
nonzero k, then there are no other solutions in the n-adics, as long as
n has at most 2 prime factors (for example, the lO-adics).
This is because one of a,b,c is not divisible by 2 or 5, hence is
invertible, and hence can be scaled so that it is 1. If a,b,c are all
divisible by one of 2 or 5, then two of them must be divisible by 2
(say), which means they all are divisible by 2, and hence you can
divide everything by 2 and repeat.
Terry
----------------------------------------------------------
EMAIL
From: "Kin Y. Chung"
Date: Sun, 24 Oct 1993 23:14:40-0400
To: Ludwig. Plutonium @ Dartmouth. EDU
Subject: Re: There exists a P-adic = to i?
Newsgroups: sci.math
In-Reply-To:
Organization: Princeton University
In article you write:
>I am looking to replace the imaginary number i with a P-adic, or hybrid
>P-adic with infinite string to the right of decimal point. Any help in
>advance is appreciated.
What is so great about finding p-adics that play the role of i? The
thing is (and I've already said this a few times) the p-adics are
different for different p, and they do not even form a field! When p is
not prime, the p-adics don't even form an integral domain. Notice the
name "integral domain". This means that when p is not prime, the
p-adics do not have arguably the most important property of the
integers, namely cancellation. All the p-adics are different for
different p, so why should one be preferred over another? Also, it is
routine to show that there is no p-adic x such that x^2 = -1 for
various
p. There may exist such x for other p, but not for all p (eg p=3
doesn't work).
Trust me, you cannot replace the real numbers by the p-adics.
--
Kin Yan Chung (kinyan@math.princeton.edu) 0 0 0 Sydney
------------------------------------------------------------
EMAIL
Date: Mon, 25 Oct 1993 01:56:28-0400
From: somos@kleinrock.cba.csuohio.edu (Michael Somos)
To: ludwig.plutonium@Dartmouth.EDU
Subject: 5-adic sqrt(-1)
Ludwig, in case you are interested, it is possible to have roots of
unity in p-adic integers. For example, the 5-adic numbers have a
square root of -1. An approximation to it is 7 since 7*7 = 49 = -1+
2*5*5. You can get an arbitrary degree of approximation in several
ways. It would be nice to get a lO-adic approximation, but that is not
possible. Note that this does not really "repIace" i, but that is
probably too hard for you to understand. Shalom, Michael Somos
--------------------------------------------------------------
Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Subject: There exists a P-adic = to i?
Message-ID:
Organization: Dartmouth College, Hanover, NH
Date: Mon, 25 Oct 1993 02:17:01 GMT
Lines: 3
I am looking to replace the imaginary number i with a P-adic, or hybrid
P-adic with infinite string to the right of decimal point. Any help in
advance is appreciated.
----------------------------------------------------------
EMAIL
From: "Terry Tao"
Subject: Re: FLT counterexamples neither a,b,c are = 1??
To: Ludwig.Plutonium @ Dartmouth. EDU (Ludwig Plutonium)
Date: Mon, 25 Oct 93 19:01:12 EDT
In-Reply-To: <6220937@bIitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Oct 25, 93 6:05 pm
Sure. Take the idempotent a such that the last digit of a (base 30) is
15, and the idempotent b such that the last digit of b is 10. Then a+b
is also an idempotent, so a^n + b^n = (a+b)^n for all n.
(An idempotent is a number such that a*a=a. In the 30-adics there are
eight idempotents, whose last digits are 0,1,6,10,15,16,21, and 25.
From the last digit of an idempotent you can determine the others
successively.)
Terry
-------------------------------------------------------------
EMAIL
From: "Terry Tao"
Subject: Re: FLT counterexamples neither a,b,c are = 1??
To: Ludwig. Plutonium @ Dartmouth. EDU (Ludwig Plutonium)
Date: Mon, 25 Oct 93 19:04:41 EDT
In-Reply-To: <6220937@blitzen.Dartmouth.EDU>; from "Ludwig PIutonium"
at Oct 25, 93 6:05 pm
The reason why this works is because Z_30 (the 30-adics) is essentially
the direct sum of Z_2, Z_3, and Z_5. What this means is that for every
number in Z_30, there corresponds a triplet of numbers, the first one
in Z_2 the next in Z_3 and the last in Z_5, such that addition and
multiplication are preserved. The idempotents ending in
0,1,6,10,15,16,21,25 then correspond to the triples
(0,0,0), (1,1,1), (0,0,1), (0,1,0), (1,0,0), (0,1,1), (1,0,1),
(1,1,0)
respectively. Incidentally, Z_10 is the direct sum of Z_2 and Z_5: if
you let a be the idempotent ending in 6 and b the idempotent ending in
5, then the map is
(x,y) <=> ax + by.
for all x in the 2-adics, and y in the 5-adics.
Terry
-----------------------------------------------------------
EMAIL
From: "Terry Tao"
Subject: Re: FLT counterexamples neither a,b,c are = 1??
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Mon, 25 Oct 93 19:26:48 EDT
In-Reply-To: <6220937@blitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Oct 25, 93 6:05 pm
In any event, the p-adics are very similar to the reals for p prime.
(The lO-adics, being the direct sum of two p-adics, is more like the
number
system R^2. This is the system of "numbers" which are ordered pairs of
real numbers, with addition and multiplication defined componentwise.
Hence, for example
(2,3) + (6,-3) = (8,0)
(11,2) * (3, 9) = (33, 18).
The idempotent counterexamples to FLT are analagous then to the fact
that
(1,0)^n + (0,1)^n = (1,1)^n.)
Counterexamples to FLT in say the 5-adics are about as worthwhile as
counterexamples to FLT in the reals. Most numbers have square or cube
roots or nth roots in the p-adics for prime p (in fact, if n is coprime
to p-1, then all numbers in the p-adics have nth roots, except those
which are divisible by p.) Similarly, most numbers have square or cube
roots in the reals (and if n is odd, then all numbers have nth roots.)
This isn't anything too earthshattering.
Terry
-------------------------------------------------------------
From: gerry@macadam.mpce.mq.edu.au (Gerry Myerson)
Newsgroups: sci.math
Subject: Re: There exists a P-adic = to i?
Date: 26 Oct 1993 06:15:11 GMT
Organization: CeNTRe for Number Theory Research
Lines: 40
Message-ID:
References:
In article ,
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) wrote:
>I am looking to replace the imaginary number i with a P-adic, or >hybrid P-adic with infinite string to the right of decimal point. Any >help in advance is appreciated.
LP has asked an interesting question here, which I think deserves
better than the sarcastic replies I have seen (the newsfeed is
unreliable here; there may be some serious answers I haven't seen).
If p is a prime which leaves the remainder one when you divide by 4,
then there is a p-adic integer X which corresponds to i in the sense
that its square is minus one. You go like this:
First, there's an (ordinary) integer a whose square is one less than a
multiple of p (it's a theorem that there is such a thing if p is 1 mod
4, and not if p is 3 mod 4). Then it's easy to see that there is an
integer b such that the square of a + pb is one less than a multiple of
the square of p. In fact, it's not just easy to see the existence, it's
easy to compute b. Then, it's easy to see and compute c such that the
square of a + pb + ppc is one less than a multiple of the cube of p.
And so on; the p-adic integer X = a + pb + ppc + pppd + . . . will
satisfy xx = -1.
For example, take p = 5. We can take a = 2. Then (2 + 5b)^2 = -1 (mod
25) simplifies to 1 + 4b = 0 (mod 5), whence b = 1. Then (7 +25c)^2 =
-1 (mod 125) simplifies to 2 + 4c = 0 (mod 5), whence c = 2. Then (57
+125d)^2 = -1 (mod 625) simplifies to 1 + 4 d = 0 (mod 5), whence d =
1. Continue forever to get X = 2 + 1x5 + 2x25 + 1x125 + . . . . The
pattern 2, 1, 2, 1 does ÒnotÓ continue.
Now, thereÕs a problem with ÒreplacingÓ i with this X. Go back to where
I said, ÒWe can take a = 2.Ó We can also take a = 3, and get a
different X. Actually, of course, we get -X. But which of these should
replace i, and which should replace -i? ThereÕs no good reason to
prefer either of the alternatives to the other. So any system which
purports to replace i with some p-adic (and what happens for 5, happens
for all p = 1 mod 4) will have at its very foundation an arbitrary
decision.
Gerry Myerson
-------------------------------------------------------------
From: gsmith@uoft02.utoledo.edu
Newsgroups: sci.math
Subject: Re: There exists a P-adic = to i?
Message-ID: <1993Oct26.234745.6904@uoft02.utoledo.edu>
Date: 26 Oct 93 23:47:45 EST
References:
Organization: University of Toledo, Computer Services
Lines: 13
In article ,
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
>I am looking to replace the imaginary number i with a P-adic, or
>hybrid P-adic with infinite string to the right of decimal point. Any
>help in advance is appreciated.
2.121342. . . and 3.323102. . . are the two 5-adic square roots of -1,
written (as I prefer) with the infinite string to the right.
I am expecting you to prove something profound with this.
--
Gene Ward Smith/Brahms Gang/University of Toledo
gsmith@uoft02.utoledo.edu
-------------------------------------------------------------
From: edgar@math. ohio-state. edu (Gerald Edgar)
Newsgroups: sci math
Subject: Re: There exists a P-adic = to i?
Date: 27 Oct 1993 08:09:29 -0400
Organization: The Ohio State University, Dept. of Math.
Lines: 15
Message-ID: <2alodp$i9t@math.mps.ohio-state. edu>
References:
<19930ct26.234745.6904@uoft02 .utoledo.edu>
In <19930ct26.234745.6904@uoft02 .utoledo.edu> gsmith@uoft02
.utoledo.edu wrote:
>2.121342... and 3.323102... are the two 5-adic square roots of -1,
>written (as I prefer) with the infinite string to the right.
>
These are non-periodic expansions... So we conclude that i is
"irrational".
Gerald A. Edgar Internet: edgar@math.ohio-state.edu
Department of Mathematics Bitnet: EDGAR@OHSTPY
The Ohio State University telephone: 614-292-0395 (Office)
Columbus, OH 43210 -292-4975 (Math. Dept.) -292-1479 (Dept. Fax)
-------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: There exists a P-adic = to i?
Date: 1 Nov 1993 06:22:07 GMT
Organization: Open Software Foundation
Lines: 43
Message-ID: <2b29uf$52u@paperboy.osf.org>
References:
<1993Oct26.234745.6904@uoft02.utoledo.edu>
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
>I am suddenly asking if this 5-adic [square root of -1] is >substitutable in DeMoivreÕs theorem e^(pi(i))=-1?
We can define i as the 5-adic square root of -1, and exp() using the
power series, but what 5-adic number corresponds to pi?
Interestingly enough, this is the sort of thing that got me started
looking at p-adic numbers again a couple of years ago.
ThereÕs a recurring thread in sci.math and sci.physics which asks ÒWhat
would be the value of pi under different physical assumptions?Ó --
usually this means curved space, such as in an intense gravitational
field. As the mathematicians are quick to point out, the question as if
space is curved; the question which was usually meant is Òwhat is the
circumference/diameter ratio of a circle in this other spaceÓ or Òwhat
is the area of a unit circleÓ. Usually, the Òpseudo-piÓ value depends
on which definition youÕre using, and it often depends on the radius of
the circle or sphere that youÕre measuring it with.
Anyway, this discussion led me to wonder whether it makes sense to ask
ÒWhat would be the value of pi in a p-adic space?Ó -- or, more
precisely, ÒIs there a p-adic number which plays the same role that pi
does for the reals?Ó Again, it depends on the definition of pi. We
could try the sum pi = 4/1 - 4/3 + 4/5 - 4/7 + 4/9 + . . ., but this
doesnÕt converge in the p-adics, and besides, itÕs not clear why this
one should be any more fundamental than any of the other series (or
products) that yield pi.
LetÕs jump straight to the equation exp(2xpixi) = 1, which some sources
use as the definition of pi. This makes sense if exp() is a periodic
function in the p-adics the way it is in the complex numbers. So we
need to find a nonzero number z such that exp(z) = 1; in other words,
solve (exp(z)-1)/z = 0.
Unfortunately, it appears that there is no such value. Even if we make
an algebraic extension to the p-adics (just as we had to adjoin
sqrt(-1) to the reals in order to solve same the equation there), it
seems that the power series for (exp(z)-1)/z, when it converges, always
has a limit of 1 + something divisible by a positive power of p; hence
it is never 0.
So, my tentative conclusion is that there is no pseudo-pi in the
p-adics (for any p).
-------------------------------------------------------------
Newsgroups: sci.math
From: william@fine.princeton.edu (William Schneeberger)
Subject: Re: There exists a P-adic = to i?
Message-ID: <1993Nov1.193645.26904@Princeton.EDU>
Sender: news@Princeton.EDU (USENET News System)
Organization: Princeton University
References: <1993Oct26.234745.6904@uoft02.utoledo.edu>
<2b29uf$52u@paperboy.osf.org>
Date: Mon, 1 Nov 1993 19:36:45 GMT
Lines: 17
In article <2b29uf$52u@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
>We can define i as the 5-adic square root of -1, and exp() using the
>power series, but what 5-adic number corresponds to pi?
[snip]
Note there is no natural automorphism of the 5-adics taking i to -i, so
one ought to be specific about which square root of -1 one wants.
On a related point, is there any nice way to extend the definition of
exp where the power series doesnÕt converge so that, e.g.,
exp(z) : =exp(pz)^(1/p) ?
--
Will Schneeberger Terry told me that
I should
william@math.Princeton.EDU change my .signature .
-----------------------------------------------------------
Newsgroups: sci.math
From: kuangj@cda.mrs.umn.edu (Jinghua Kuang)
Subject: Re: There exists a P-adic = to i?
Message-ID:
Organization: University of Minnesota - Morris
References: <1993Oct26.234745.6904@uoft02.utoledo.edu>
<2b29uf$52u@paperboy.osf.org>
Date: Thu, 4 Nov 1993 17:20:56 GMT
Lines: 10
How would you define exp(x) in p-adic field? You said it is done by the
power series of 1+x/1!+x^2/2!+. . . . But this series is not
convergent on the p-adic field. I guess you have to adopt IwasawaÕs
definition of exp(x) in his Ôp-adic L-functionÕ book (Princeton Study
series, #=?, sorry, I forget). Good luck! After all, the problem of
pseudo-pi is interesting. But I think your claim of non-existence may
not be so true.
JHK.
-----------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: P-adics
Date: 5 Nov 1993 04:39:07 GMT
Organization: Open Software Foundation
Lines: 10
Message-ID: <2bcldb$dbf@paperboy.osf.org>
In article iachetta@bcrvmpc2.vnet.ibm.com
writes:
>No, post to the net. I am dying to know what a p-adic is myself.
>Never taught us engineers about them.
HereÕs the (very) informal version. A p-adic number is a string of
digits written in base p (normally a prime), similar to a real number,
except that youÕre allowed to have infinitely many digits to the ÒleftÓ
of the radix point and only finitely many to the ÒrightÓ. A p-adic
number is ÒsmallÓ if it ends with a lot of zeroes, i.e. if itÕs an
integer multiple of a large power of p. (This allows you to define
limits.)
-----------------------------------------------------------
Newsgroups: sci.math
From: william@fine.princeton.edu (William Schneeberger)
Subject: Re: p-adic exp (was Re: There exists a P-adic = to i?)
Message-ID: <1993Nov5.163732.24908@Princeton.EDU>
Sender: news@Princeton.EDU (USENET News System)
Organization: Princeton University
References: <2b29uf$52u@paperboy.osf.org>
<1993Nov1.193645.26904@Princeton.EDU> <2bcop7$dl5@paperboy.osf.org>
Date: Fri, 5 Nov 1993 16:37:32 GMT
Lines: 40
In article <2bcop7$dl5@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
>In article <1993Nov1.193645.26904@Princeton.EDU> william@fine.princeton.edu (William Schneeberger) writes:
[snip]
WS>>On a related point, is there any nice way to extend the
WS>>definition of exp where the power series doesn't converge so
WS>>that, e.g.,
WS>> exp(z) :=exp(pz)^(1/p) ?
KH>I suspect that what happens here is that exp(pz) yields a number
KH>that has no p'th root.
This is true, in the p-adic rationals. We have exp(z) converges iff
|z| However, you could extend the field by adjoining such a p'th root.
Yes you could. The question is, is there a 'nice' way of choosing the
proper pth root?
KH>It's not clear to me whether the definition of convergence can be
KH>tweaked so that the original divergent series can be said to
KH>pseudo-converge to this value in the extended field.
Not without altering norms. A non-Cauchy sequence does not converge in
a metric space.
KH>If so, then maybe there's still hope for finding a pseudo-pi.
I don't think so. If there were a nonzero exponent z with exp(z)=1, we
would have exp(p^k z)=1 in the power series for some sufficiently large
k.
--
Will Schneeberger Terry told me that I should
william@math.Princeton.EDU change my .signature .
-------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: PROOF OF INFINITUDE OF CONSTRUCTIBLE REGULAR POLYGONS
Date: 7 Nov 1993 04:21:03 GMT
Organization: Open Software Foundation
Lines: 15
Message-ID: <2bht3f$1rl@paperboy.osf.org>
References:
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
> PROOF OF THE INFINITUDE OF CONSTRUCTIBLE REGULAR-N-SIDED
> POLYGONS. PROOF: We can manufacture an infinitude of primes of
> the form (2^2^n)+1 in p-adics.
> QED.
We can? Why/how?
[And considering the dual problem with -1 instead of +1:]
> In P-adics it is straightforward to manufacture an infinitude of
> primes of the form (2^2^n)-1.
Again, why do you believe this? And how can it possibly be true, since
(after n>0) these numbers are all divisible by 3? (This is the case in
both the p-adic and the Natural numbers.)
-------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: PROOF OF INFINITUDE OF CONSTRUCTIBLE REGULAR POLYGONS
Date: 7 Nov 1993 04:26:46 GMT
Organization: Open Software Foundation
Lines: 51
Message-ID: <2bhte6$1rv@paperboy.osf.org>
References:
(I already posted one reply to this, but I decided to add this
independent thread as well.)
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
>We can manufacture an infinitude of [ . . . ] in p-adics.
Extending the Natural numbers to the p-adic numbers doesn't
automatically solve a problem of this sort. (In fact, the question
often doesn't make any sense in the p-adics, especially when we're
talking about primes.)
But you may be interested to hear that this sort of approach is
meaningful in a different "infinite integer" system, namely the
Hypernatural numbers N*. Induction doesn't work on the p-adic numbers
-- for example, you can prove by induction that every Natural number
has a first (leftmost) digit, but this isn't true for the 10-adic
number x=. . .1111 (the number which satisfies 10x+1 = x). In the
Hypernaturals, this isn't a counterexample: although you can construct
a number containing an infinite number of 1's and no other digit, it
will still have a leftmost digit: y=1. . .111; and 10y+1 is not y but
another Hypernatural number containing one more digit. (The number of
digits is "infinite", but is actually a (smaller) Hypernatural number
itself.)
In fact, you get the Hypernaturals by adding an infinite integer to the
Peano axioms. So induction does work in this system, because it's one
of the axioms on which they're constructed; although you do have to be
careful to *not* draw a barrier between the finite and infinite. (In
the Hypernatural realm, N is not a valid subset of N*, because it's not
a definable set at all.)
Now, in the Hypernaturals, if you can identify *one* instance of an
infinite value satisfying some relationship (such as 2^2^n+1 prime, for
n a non-finite Hypernatural), then this would imply that it must be
true infinitely often in the finite Natural numbers. (Proof: suppose
there are only finitely many instances in N. Then you could prove% the
statement "k is the largest value of this type" in N. But the same
proof would carry over into N*, since it has all of the same axioms,
and hence "k is the largest value of this type" must be true in N*. But
in that case we couldn't have found an infinite value satisfying it,
since any infinite value would be larger than the finite k we
constructed in N.)
In practice, this doesn't do you much good, because the easiest way to
prove that a particular infinite example exists is to show that the
property is unbounded in N.
For more information, look for a book on Non-Standard Analysis. (The
Hyperreals R* have gotten more attention than N*, but the principle is
the same.)
--------------------------------------------------------
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Newsgroups: alt.sci.physics.plutonium,sci.math
Subject: Re: THE CLEANING OF THE MATH COMMUNITY
Date: 25 Sep 1994 22:39:41 GMT
Organization: Dartmouth College, Hanover, NH
Lines: 52
Message-ID: <364u7d$ps1@dartvax.dartmouth.edu>
References: <35t9ts$ahj@dartvax.dartmouth.edu>
<361m65$2hi@dartvax.dartmouth.edu>
In article
jgreene@frodo.d.umn.edu (John Greene) writes:
> I think you are right here that if naturals = adics, then you must rethink
> the definition of "primeness." Isn't it reasonable that you do this before
> even saying that you have a valid proof that there are infinitely many primes?
> At this stage, it may not even be obvious that there is such a thing as a prime!
>
> Most people would say that Unique Factorization is a more important result
> than the infinitude of the primes. My understanding is that Gauss was the
> first to realize how fundamental this idea was, and the first to give a
> proof
> of it. As you point out, his proof uses mathematical induction. I think
> people would be very interested if you could give a correct, one paragraph
> proof of unique factorization which did not rely on induction. Should you
> decide to do so, I would ask that you keep in mind my example above of a
> factorization of 2. Also, since unique factorization is again a statement
> concerning primes, no proof can be given untill primes are carefully
> defined.
Here is a one sentence proof of UPFAT and which is true for both
prime numbers themselves and not prime numbers, i.e.--- composite
numbers. PROOF BY CONTRADICTION: Suppose not true; implies there exists
a number which is simultaneously different yet equal, contradiction, .
. QED
We can make the very best type of definition possible for REALS via
a constructive definition. Dedekind cuts or others.
Adics are base dependent. Let us get rid of base dependency for
ADICS = NATURALS = INFINITE INTEGERS. Let us call ADICS the INFINITE
INTEGERS via this construction. Let us use the Reals as the decimal
Reals with finite string to the left of the decimal point and infinite
string to the right of the decimal point. Let us manufacture the
INFINITE INTEGERS as infinite strings to the left of the radix and
finite string to the right. Let us define the operations on the
INFINITE INTEGERS by converting them into the REALS, perform the
operations and then reconvert back.
What is primeness for INFINITE INTEGERS? Good question. The only
thing I can think of is a special class of transcendental numbers of
the Reals and hence of the Infinite Integers. Not pi since it is evenly
divisible by 2 (Proof: semicircle). But a number like e. The number e
for the Reals is transcendental and prime (as far as I know).
How to convert the 5-adic sqrt -1 to INFINITE INTEGERS? Then we may
be able to solve e^(ixpi) = -1.
Here is a counterexample to Goldbach's Conjecture of which an
infinite number of counterexamples are generated. The Infinite Integer
produced by e. The Infinite Integer e is even. Note: .......17.2. By
Zeus,thanks for the definition of evenness is still intact. This even
integer which is e flipped around cannot be the addition of two primes
(Proof: if so, then e in the Reals would be algebraic). And the
construction of an infinite cases of counterexamples are e raised to
the power of another transcendental number.
Archimedes Plutonium A@P
------------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: There exists a P-adic = to i?
Date: 1 Nov 1993 06:22:07 GMT
Organization: Open Software Foundation
Lines: 43
Message-ID: <2b29uf$52u@paperboy.osf.org>
References:
<1993Oct26.234745.6904@uoft02.utoledo.edu>
In article
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
>I am suddenly asking if this 5-adic [square root of -1] is
>substitutable in DeMoivreÕs theorem e^(pi(i))=-1?
We can define i as the 5-adic square root of -1, and exp() using the
power series, but what 5-adic number corresponds to pi?
Interestingly enough, this is the sort of thing that got me started
looking at p-adic numbers again a couple of years ago.
ThereÕs a recurring thread in sci.math and sci.physics which asks ÒWhat
would be the value of pi under different physical assumptions?Ó --
usually this means curved space, such as in an intense gravitational
field. As the mathematicians are quick to point out, the question as if
space is curved; the question which was usually meant is Òwhat is the
circumference/diameter ratio of a circle in this other spaceÓ or Òwhat
is the area of a unit circleÓ. Usually, the Òpseudo-piÓ value depends
on which definition youÕre using, and it often depends on the radius of
the circle or sphere that youÕre measuring it with.
Anyway, this discussion led me to wonder whether it makes sense to ask
ÒWhat would be the value of pi in a p-adic space?Ó -- or, more
precisely, ÒIs there a p-adic number which plays the same role that pi
does for the reals?Ó Again, it depends on the definition of pi. We
could try the sum pi = 4/1 - 4/3 + 4/5 - 4/7 + 4/9 + . . ., but this
doesnÕt converge in the p-adics, and besides, itÕs not clear why this
one should be any more fundamental than any of the other series (or
products) that yield pi.
LetÕs jump straight to the equation exp(2xpixi) = 1, which some sources
use as the definition of pi. This makes sense if exp() is a periodic
function in the p-adics the way it is in the complex numbers. So we
need to find a nonzero number z such that exp(z) = 1; in other words,
solve (exp(z)-1)/z = 0.
Unfortunately, it appears that there is no such value. Even if we make
an algebraic extension to the p-adics (just as we had to adjoin
sqrt(-1) to the reals in order to solve same the equation there), it
seems that the power series for (exp(z)-1)/z, when it converges, always
has a limit of 1 + something divisible by a positive power of p; hence
it is never 0.
So, my tentative conclusion is that there is no pseudo-pi in the
p-adics (for any p).
-------------------------------------------------------------
Newsgroups: sci.math
From: william@fine.princeton.edu (William Schneeberger)
Subject: Re: There exists a P-adic = to i?
Message-ID: <1993Nov1.193645.26904@Princeton.EDU>
Sender: news@Princeton.EDU (USENET News System)
Organization: Princeton University
References: <1993Oct26.234745.6904@uoft02.utoledo.edu>
<2b29uf$52u@paperboy.osf.org>
Date: Mon, 1 Nov 1993 19:36:45 GMT
Lines: 17
In article <2b29uf$52u@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
>We can define i as the 5-adic square root of -1, and exp() using the
>power series, but what 5-adic number corresponds to pi?
[snip]
Note there is no natural automorphism of the 5-adics taking i to -i, so
one ought to be specific about which square root of -1 one wants.
On a related point, is there any nice way to extend the definition of
exp where the power series doesnÕt converge so that, e.g.,
exp(z) : =exp(pz)^(1/p) ?
--
Will Schneeberger Terry told me that
I should
william@math.Princeton.EDU change my .signature .
--------------------------------------------------------------------
Newsgroups: sci.math
From: kuangj@cda.mrs.umn.edu (Jinghua Kuang)
Subject: Re: There exists a P-adic = to i?
Message-ID:
Organization: University of Minnesota - Morris
References: <1993Oct26.234745.6904@uoft02.utoledo.edu>
<2b29uf$52u@paperboy.osf.org>
Date: Thu, 4 Nov 1993 17:20:56 GMT
Lines: 10
How would you define exp(x) in p-adic field? You said it is done by the
power series of 1+x/1!+x^2/2!+. . . . But this series is not
convergent on the p-adic field. I guess you have to adopt IwasawaÕs
definition of exp(x) in his Ôp-adic L-functionÕ book (Princeton Study
series, #=?, sorry, I forget). Good luck! After all, the problem of
pseudo-pi is interesting. But I think your claim of non-existence may
not be so true.
JHK.
------------------------------------------------------------------
From: bernardi@mathp7.jussieu.fr (Dominique Bernardi)
Newsgroups: sci.math
Subject: Re: The two 5-adic sqrt -1
Followup-To: sci.math,sci.math.num-analysis,sci.math.symbolic Date:
Thu, 07 Apr 1994 15:50:34 +0100
Organization: Theorie des Nombres, Universite Pierre & Marie Curie
Lines: 62
Distribution: inet
Message-ID:
In article <2o02al$b9i@dartvax.dartmouth.edu>,
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) wrote:
> I please need help in extending these two strings, both out to
> 1000 places leftwards from the radix point. Any computer help?
> . . .243121r2 and . . .201323r3 are the two 5-adic square roots of
> -1
> The purpose is that I am looking for a correlation, and if I can
> stare at them it may help.
Here they are (from left to right, sorry), you can stare as long as you
dare...
33231021412240312040104032030331303024331122040204
13204114144413133414410311104224243403300234244140
00424243131240230101323111334132240141323322314123
21413141441403332212002214433021104311210431204321
11434140213402312410420004234221031320231214002203
21333413042344124233211100442012420011310044412431
22031433201410124424000213333432410434233221123404
42143230100410420334203424100032234444224314211134
30043114414204130142242310240033430142334143134044
34124000314134442112203220440401423331244432340112
30222012001411114001311402311204222201440332220204
03441013402041421431114122232231314404431040203313
42342124033202411203314310333210434302444430231004
41341230110400344411141422114221231410120410221024
30333101301443120201230101434024414211021132214302
04423204010424021022443220422101332002200320033414
21211203143222020224422142231041301241413141242020
04212233022222432322013342021301231201224210143322
22140133204244142433200340043122431430141034100312
31024311342003003302402213102032343130032423312100
21213423032204132404340412414113141420113322404240
31240330300031311030034133340220201041144210200304
44020201313204214343121333110312204303121122130321
23031303003041112232442230011423340133234013240123
33010304231042132034024440210223413124213230442241
23111031402100320211233344002432024433134400032013
22413011243034320020444231111012034010211223321040
02301214344034024110241020344412210000220130233310
14401330030240314302202134204411014302110301310400
10320444130310002332241224004043021113200012104332
14222432443033330443133042133240222243004112224240
41003431042403023013330322212213130040013404241131
02102320411242033241130134111234010142000014213440
03103214334044100033303022330223213034324034223420
14111343143001324243214343010420030233423312230142
40021240434020423422001224022343112442244124411030
23233241301222424220022302213403143203031303202424
40232211422222012122431102423143213243220234301122
22304311240200302011244104401322013014303410344132
13420133102441441142042231342412101314412021132344
-------------------------------------------------------------
From: Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
Newsgroups: alt.sci.physics.plutonium,sci.math,sci.physics
Subject: MATH#22: ASSAULT ON e^(ixpi) = -1
Date: 28 Sep 1994 10:58:19 GMT
Organization: Dartmouth College, Hanover, NH
Lines: 20
Message-ID: <36bi8b$4o7@dartvax.dartmouth.edu>
In article <36bhcf$3qg@dartvax.dartmouth.edu>
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
> In article <36bdbd$i8@dartvax.dartmouth.edu>
> Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
>
> > In article <36bciv$t4a@dartvax.dartmouth.edu>
> > Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
> >
> > > It was around a year ago that I saw and read this thread in
> > > sci.math. Let me follow-up in various posts on my thinking a year
> > > later.
Taking e as a growing logarithmic spiral of Riemannian Geometry.
Taking (ixpi) as a growing Lobachevskian Geometry. Taking -1 as the -1
of Loba.
Taking e as a growing logarithmic spiral of Riemannian Geometry.
Taking (ixpi) as a growing rectilinear Euclidean Geometry. Taking -1
as the -1 of Eucl.
-------------------------------------------------------------
Newsgroups: alt.sci.physics.plutonium,sci.math
From: dik@cwi.nl (Dik T. Winter)
Subject: Re: p-adics question of digit representation say 17-adics
Message-ID:
Organization: CWI, Amsterdam
References: <3rt9co$9pl@dartvax.dartmouth.edu>
Date: Sat, 17 Jun 1995 02:04:54 GMT
Lines: 9
In article <3rt9co$9pl@dartvax.dartmouth.edu>
Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) writes:
> How does one represent the digits 10,11,12,13,14,15,16 in
17-adics?
As 10,11,12,13,14,15 and 16. Or, if you wish to go to 17-base notation
you can chose every 7 symbols you most particularly like. You can do
17-adics in base 10 notation.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924098
home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: dik@cwi.nl
