Subject: Re: Time & space, still (was: Hermeneutics ...)
From: moggin@mindspring.com (moggin)
Date: 17 Nov 1996 11:33:10 GMT
moggin@mindspring.com (moggin):
>>>>> As
>>>>>I shouldn't have to point out, introducing the notion of gravitas in the
>>>>>_Principia_ would be enough, by itself, to commit him to action-at-a-
>>>>>distance, even in the absence of any other considerations, since it's a
>>>>>force that exerts itself across space without any mechanism to account
>>>>>for its workings. As his contemporaries didn't hesitate to object.
Mati:
>>>>You certainly shouldn't have to point it out, since it would've been a
>>>>total nonsense. Newton's law of gravity says that the planetary
>>>>motion is welll explained by a force proportional to the product of
>>>>the masses and inversely proportional to the square of the distance.
>>>>This is well supported by observations. And, that's important, it by
>>>>no means precludes the existance of an underlying, deeper mechanism,
>>>>just pleads ignorance of such (as Newton explicitly stated).
moggin:
>>> Yet Newton's theory presented a force, namely gravity, that had
>>>effects across immense distances, and no mechanism by which to
>>>apply itself. Any self-respecting mechanist would be horrified by
>>>that kind of nonsense, and many were.
Mati:
>>The rule of science is "if it works, use it". Moreover, if it works
>>in a way that's opposed to your common sense, then it is your common
>>sense thet needs to be modified. I don't see why this concepts are so
>>difficult to comprehend.
moggin:
> They're easily comprehensible. The question is, what makes you think
>they're relevant?
Mati:
:But of course they are relevant to science.
And yet irrelevant to the topic at hand, which is whether or not religious
mysticism has played a role in scientific thinking. You say not, but you're
arguing, "If it works, use it," which supports the conclusion that religious
mysticism is a useful part of science.
moggin:
>>Are you shaking your finger at Newton's colleagues for
>>not properly following your "rule of science"? Or do you believe
>> you're arguing with me?
Mati:
>Not really, because they are two different levels of argument. One is
>"does the proposed rule fit well the experimental data?" At this
>level it is a matter of comparing things. If it fits, it fits and the
>statement "but it doesn't make sense" doesn't make sense. Now, you
>may go to the next level and argue that while the rule works, it is
>but a reflection of a deeper, underlying, mechanism which will somehow
>make it "make sense". This was for example the position Einstein took
>towards Quantum Mechanics (the results are in, Einstein was wrong on
>this one). That's a legitimate argument but, first, it doesn't
>invalidate the rule and, second, the burden of proof is on its
>proponent.
O.k., so you're not arguing with Newton's colleagues. And I know
you're not arguing with me -- so who is this verbiage directed to,
and why should they care about it?
Mati:
>In any case a statement like "although it seems to work, we cannot use
>it because it doesn't make sense" is clearly unscientific.
And where do you find that statement? You're setting up another
strawman.
Mati:
>>>No more so then the remote control I talked about few days ago is an
>>>illustration of action at a distance. You use it and it works from a
>>>distance. This is a fact.
moggin:
>> That's an interpretation, although hardly worth the name. As I've
>>already pointed out, "action-at-a-distance" is an explanation of the
>>workings of the remote control (albeit an empty one). I ask, "Mati,
>>how does the remote work?" You tell me, "It exerts a force." I say,
>>"Mysterious forces -- yeah, right. O.k., how do _they_ work?" You
>>answer, "Action-at-a-distance," and I reply, "You sound more like
>>a psychic than a physicist!" If you really thought like Joe Friday,
>>you'd have said, "Well, it just works, that's all. Damned if I know
>>how. Just don't forget to change the batteries."
Mati:
>No, that's where you go wrong. I don't reply "action at a distance".
>I reply "it acts from a distance and I've no idea how it does it".
Glad I managed to convince you. Now just keep proclaiming your
ignorance, and you'll be o.k.
Mati
>>If it happens to contradict the writings of half a dozen of your favorite
>>philosophers, too bad, it is still a fact.
moggin:
> Are you talking about _my_ favorite philosophers? If so, why?
Mati:
>>>By acknowledging this fact you don't deny the possibility of an
>>>underlying mechanism (you don't affirm it, either). But you don't
>>>need to know the mechanism in order to acknowledge the fact.
moggin:
>> What "fact"?
Mati:
>In the specific case we're talking about, the fact that the planetery
>orbits were fully explained using the proposed law, which moreover was
>used to make successful predictions for things which couldn't be known
>in advance (Halley's comet was a pretty successful demonstration).
Oh, so now we're talking about a "specific case" again, huh? And there
I was, thinking you took all of science as your domain. Guess it depends.
Anyway, a law can't explain anything, as I've observed before -- it just
describes certain regularities. If you want to insist that you don't trade
in concepts or ideas, you've got to give up the particular, old-fashioned
notion that you can offer explanations. And of course none of this shows
that religious mysticism is unrelated to physics.
Mati:
>>>>>These are technicalities, though. What is important to understand is
>>>>>that the adoption of a physical law in this or other form in no way
>>>>>constitutes an acceptance of this or other philosophical principle.
>>>>>It just constitutes a recognition that said law fits well with
>>>>>available experimental evidence. In physics evidence is king, not
>>>>>philosophical ideas.
moggin:
>>>> That _is_ a philosophical idea, d00d -- not the brighest one in
>>>>the world, either.
Mati:
>>>That's your opinion. In my opinion it is brighter than anything that
>>>ever came out of philosophy. And, as I stated above, I value my
>>>opinions higher than yours.
moggin:
>> Again, your self-evaluation has no relevance here. And I see you
>>accept my point, since you're now claiming to have a philosophy
>>that's superior to any philosopher's, while before you argued that
>>physics was ruled by evidence, not "philosophical ideas."
Mati:
>I claim that empirical evidence carries more weight than the opinions
>of all philosophers that ever lived. If you want to call it "philosophy",
>be my guest.
It's sure not evidence -- it's an idea you're invested in. In specific,
it's a valuation. One that fits nicely into a philosophy of empiricism.
moggin:
>>>> But I've got no interest in arguing with you
>>>>about the philosophy of science (or the supposed lack of it). You
>>>>claimed that there was no relation between physics and religious
>>>>mysticism. That's false.
Mati:
>>>Statement from authority? And whose authority? You're too much in
>>>the habit of passing pronouncements, trying to act as a referee while
>>>taking a side in a debate, at the same time. But, I'm not impressed.
moggin:
>> I don't care _what_ your feelings are -- your assertion is false for
>>the reasons that I've offered.
Mati:
>Apparently you can't get over this habit. It may impress the folks at
>a.p. but not at sci.phys.
I don't want to generalize about the sci.physics population, since it's
easy to imagine that only the bottom of the barrel have contributed to
this discussion. But I agree that reasoning doesn't go a long ways with
the folks I've seen.
Mati:
>>>Now, if the above you mean to say that many physicists were (and some
>>>still are) motivated by mysticism (religious or otherwise) I'll
>>>certainly agree. This exchange started with me bringing an example
>>>for just such occurence.
moggin:
>> You're being duplicitous again. You cited the religious motivations of
>>Maupertois or Fermat (you weren't sure which) in order to deny that
>>mysticism had a meaningful relationship with physics. You did poorly.
>>Not only were you uncertain who you were actually talking about, you
>>had nothing to say about either their religious beliefs or their work as
>>physicists.
Mati:
>Frankly, If I'll elaborate on their work as physicists, I rather doubt
>whether you'll understand any of it. It takes some calculus, you
>know.
You claimed that religion was irrelevant to physics. As of yet, you
haven't either answered my objections or made an argument of your
own. Those examples were as close as you came, but you weren't sure
who you really meant to talk about, and in any case you had nothing to
say. Instead you've chanted your mantra: "Does it work or doesn't
it?" That's beside the point, as I've said, since it doesn't bear on the
question you raised, no matter how important it may be elsewhere.
-- moggin
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: ŇTerry 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?
IŐd 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:
>HavenŐt 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 stringŐs 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 FermatŐs 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 didnŐt 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
doesnŐt 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
Ňinside 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
Ňfar 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 canŐt be matched in any
useful manner (other than the canonical one-to-one mapping), and they
canŐt 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
