Cryptography-Digest Digest #719, Volume #13 Tue, 20 Feb 01 00:13:00 EST
Contents:
Re: Is there an algorithm to sequentially enumerate all transcendental numbers?
("Dik T. Winter")
Re: Super strong crypto (Bryan Olson)
Re: Is there an algorithm to sequentially enumerate all transcendental numbers?
("Paul Lutus")
Re: A seriously different cipher concept (long) ("Scott Fluhrer")
Re: CipherText patent still pending (Benjamin Goldberg)
Re: Given any arbitrary numbers a and b. Can I ALWAYS find a transcendental number
between a and b? (Lee Rudolph)
Re: What's a KLB-7? (Steve Portly)
Re: __(?) MATRIX version of Fermat's Little Theorem (Benjamin Goldberg)
FAQ ("kwd_kwp0ee9j9")
Re: Is there an algorithm to sequentially enumerate all transcendental numbers?
(Virgil)
Re: Is there an algorithm to sequentially enumerate all transcendental numbers?
(Virgil)
Re: Is there an algorithm to sequentially enumerate all transcendental numbers?
(Virgil)
Re: "RSA vs. One-time-pad" or "the perfect enryption" ("Douglas A. Gwyn")
Re: Is there an algorithm to sequentially enumerate all transcendental numbers?
(Dave Seaman)
Re: Super strong crypto ("Douglas A. Gwyn")
Re: Is there an algorithm to sequentially enumerate all transcendental ("Douglas A.
Gwyn")
Re: Is there an algorithm to sequentially enumerate all transcendental numbers?
(Dave Seaman)
Re: What's a KLB-7? ("Douglas A. Gwyn")
Re: Is there an algorithm to sequentially enumerate all transcendental ("Douglas A.
Gwyn")
Re: Given any arbitrary numbers a and b. Can I ALWAYS find a transcendental number
between a and b? (Dave Seaman)
----------------------------------------------------------------------------
From: "Dik T. Winter" <[EMAIL PROTECTED]>
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
numbers?
Date: Tue, 20 Feb 2001 01:47:37 GMT
In article <8Oek6.8603$[EMAIL PROTECTED]> "Doom the Mostly Harmless"
<[EMAIL PROTECTED]> writes:
> It is impossible to create an ordinal set of trancendental numbers because
> between any two given, there is an infinite number.
I do not know what you mean by "ordinal set". But when you mean "enumeration"
you are correct, but for the wrong reason. For instance, the rational numbers
(that have the same property) *can* be enumerated. The same holds for the
algebraic numbers (i.e. those numbers that are roots of polynomials with
integer coefficients). As the transcendentals form the complement of the
algebraic numbers in the reals, the transcendentals can not be enumerated,
because the reals can not. Because suppose the transcendentals can be
enumerated. We know the algebraics can be enumerated (it is not so very
simple, but an enumeration can indeed be constructed). Now we make a
new sequence: A_1, T_1, A_2, T_2, ..., where A_i is the i-th algebraic
and T_i is the i-th transcendental. This would provide an enumeration
of the reals. But as the reals can not be enumerated this provides a
contradiction, so the premisse (the transcendentals can be enumerated)
is false.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
------------------------------
From: Bryan Olson <[EMAIL PROTECTED]>
Subject: Re: Super strong crypto
Date: Mon, 19 Feb 2001 17:48:00 -0800
"Douglas A. Gwyn" wrote:
>
> David Wagner wrote:
> > Ahh, but the proposal doesn't work. The proposed scheme is provably
> > *insecure*, in the information-theoretic threat model (where the
> > adversary has unbounded computational power).
>
> That is obvious, but you're talking about a theory in its
> unrealizable realm, not about real-world systems. The
> real-world criterion is that only a "negligible" fraction
> of the total traffic is expected to be recovered by the
> attack using as many resources as one can realistically
> afford.
I think you missed the point. The straw-man system
rests by unproven computational security.
Actually, the straw-man system loops out. Sending a
new key encrypted under the old key does not move away
from the unicity distance, so the system has to send
another immediately, then another, then another....
[...]
> My idea of "provably strong" allows for meeting a
> security threshold criterion, since for any practical
> system there always is such a threshold.
Can you state your criteria in a mathematically
respectable way?
> Now, my example was just a straw-man proposal, and perhaps
> one could demonstrate that *any* such embedded-key
> approach would be substantially weaker *in practice* than
> just using a single key beyond its natural lifetime, but
Is "natural lifetime" some property of a key?
> I don't see how that could be demonstrated.
So given systems for which computational security
cannot be determined, you can produce systems with the
same property.
--Bryan
------------------------------
From: "Paul Lutus" <[EMAIL PROTECTED]>
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
numbers?
Date: Tue, 20 Feb 2001 01:58:53 GMT
"Dik T. Winter" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]...
> Confusing isn't it? But that
> is mathematics.
I agree with your enlightening post, and I appreciate your taking the time
to write it. I came at the original post's terminology from more of a
computer science perspective, where "enumerate all" has a different shade of
meaning, and reasonable algorithms eventually stop.
Thanks again for taking the time.
--
Paul Lutus
www.arachnoid.com
"Dik T. Winter" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]...
> In article <%cck6.343$[EMAIL PROTECTED]> "Paul Lutus"
<[EMAIL PROTECTED]> writes:
> > But it does, by your own definition. If the set is infinite, then
> > enumerating "all possible transcendental numbers" is not an achievable
goal.
>
> Oh, it is not, but not for the reason you supply.
>
> > NO, the answer is "no." The problem is your use of the term "all
possible
> > prime numbers." The set is infinite, therefore there is no way to
assign a
> > meaning to "all."
>
> But there is. {p | p in N, p prime} is the set of all prime numbers.
> There is a definite meaning of the word all here, namely there are no
> prime numbers that are not in the given set.
> >
> > When you speak of an algorithm to enumerate "all possible" members of a
set,
> > the implication is that the algorithm will create the list and stop. If
the
> > set is infinite, this is not possible.
>
> Why do you think the algorithm should stop? Moreover, why do you think
> "enumerate" should implie "generate"? Given the infinite set
> {n | n in N, n square} you say there is no algorithm that "enumerates"
> all elements of that set, nevertheless, the algorithm (function)
> f: i (in N) -> i^2, is such an enumeration, it assigns an integer
> number to every element of the set.
>
> > I think the problem is in how you are describing the problem, not the
> > problem itself. You want to list an arbitrarily large number of
> > transcendental numbers, not "all possible."
>
> But that was indeed *not* what was asked. The question was about
> *all possible*, and there is no enumeration possible in the case of
> transcendental numbers. There is for rational numbers, algebraic
> numbers, and even for constructable numbers; but in the latter case
> the enumeration is not constructable. Confusing isn't it? But that
> is mathematics.
> --
> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
> home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
>
------------------------------
From: "Scott Fluhrer" <[EMAIL PROTECTED]>
Subject: Re: A seriously different cipher concept (long)
Date: Mon, 19 Feb 2001 18:06:59 -0800
Paul Pires <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]...
> The following is a description of a cipher concept I
> have been working on for quite awhile. It's reason
> for existence is to explore a different method that
> has the potential for being unusually fast. All rights
> are preserved. Some or all of this material may be
> covered by patents held by the author or unknown
> others. No presentment is made that the methodology
> contained herein is suitable, safe or unencumbered for
> commercial use.
An attacker can efficiently derive the entire cipher state with 512 bytes of
chosen plaintext.
Here's how it is done:
The attacker chooses two consecutive plaintext blocks of all zeros
(actually, any two blocks of constant 32 bit values would do -- zeros make
it slightly easier to explain). Then (using the notation you gave in a
later posting):
Step 1: Compute ciphertext block:
C[A[(i+32)]] = B[i]
Since the P[A[i]] term is always zero, we can remove it. Or, in other
words, the ciphertext block consists of the B array permuted as a function
of the A permutation.
Step 2: Update B:
B'[i] = (B[i]>>5) xor (B[i+1]<<27) xor C[i]
I'll use the B' notation to refer to the modified B array. And, again,
as the P[A[63-i]] term is always zero, we can remove it. The important
point here is the lower 27 bits of each element of B' consists of the upper
27 bits of the corresponding element of B xor'ed with a particular element
of C (which is known to the attacker)
Step 3 and 4: Update A
I'll idealize the tsel and updating A procedures into totally random
processes, so that the updated A' array is randomly chosen, and independent
of A or B.
Step 1 (second block):
C'[A'[(i+32]] = B'[i]
Again, the ciphertext block consists of the B' array permuted as a
function of the A' array.
At this point, the attacker has enough information to reconstruct the
contents of the A array. To be precise, for any x and y, he is able to test
the hypothesis that:
A[x] = y
If that hypothesis is true, then:
(taking i=x+32):
C[A[x]] = B[x+32]
or
C[y] = B[x+32]
and then,
B'[x+32] = B[x+32]>>5 xor B[x+33]<<27 xor C[x+32]
or
B'[x+32] = C[y]>>5 xor (5 random upper bits) xor C[x+32]
Or, in other words, the lower 27 bits of an element of B' consist of C[y]>>5
xor C[x+32].
And, since C' is a permutation of B', this implies there must be an element
of C' whose lower 27 bits consist of C[y]>>5 xor C[x+32].
If that hypothesis is false, then one can expect that to be true with
probability 2**-27 * 2**6 = 2**-21, which is sufficiently low that we can
ignore false hits.
And, since the attacker has the C and C' arrays, he can easily test this
hypothesis.
Then, once the attacker has reconstructed the A array, he can use the
relation:
C[A[(i+32)]] = B[i]
to reconstruct the B array, and that appears to be the entire cipher state.
> I would appreciate any comments or observations
> that any might have on this material.
> Thank you,
You're welcome!
--
poncho
------------------------------
From: Benjamin Goldberg <[EMAIL PROTECTED]>
Subject: Re: CipherText patent still pending
Date: Tue, 20 Feb 2001 02:34:01 GMT
John Myre wrote:
>
> Benjamin Goldberg wrote:
> <snip>
> > Any cipher which runs in polynomial time can be converted to a 3SAT
> > problem, where the number of terms is
> > O( (keysize + blocksize) * knownplaintexts ).
>
> The point is, merely knowing that P = NP isn't enough.
> If the way we know is an actual polynomial time algorithm
> for 3SAT, then you could do as you say, and solve (current)
> ciphers that way. More likely, the method for doing 3SAT
> would lead towards more specific cipher breaking methods.
Right. And since any form NP-complete problem can be converted into any
other form of NP-complete problem, if we have a proof by construction
that P=NP, then we will have an algorithm which does 3SAT in polynomial
time.
> But! It is entirely conceivable that proving P = NP would
> be non-constructive; for instance, by showing P != NP leads
> to a contradiction, in a very abstract way. Such a proof
> would give us nothing in the way of breaking ciphers.
Yes. However, I would be rather curious to see how we could prove that
P=NP, but still lack an algorithm for solving NP problems in P time.
> Actually, even a general "polynomial time" algorithm isn't
> necessarily the death knell of even 128-bit symmetric keys.
> Suppose the algorithm is O(N^100)? (That's even longer than
> 2^N, for N=128.)
Although there were a number of problems with it, there was a paper
(less than a year ago, IIRC) which claimed you could solve some type of
NPC graph problem in O(N^6) time.
Suppose that the algorithm could be fixed...
If the resultant fix takes O(N^100) time, then we have the comparison:
2^128 ? 128^100; 128 = 2^7; (2^7)^100 = 2^(7*100)
2^128 ? 2^(7*100)
2^128 < 2^700
So the NPC solver would take longer than brute force.
But suppose the NPC solver takes O(N^6) time.
2^128 ? 2^(7*6)
2^128 > 2^42
The NPC solver would take *much* less time than brute force.
What if the NPC solver takes O(N^18) time?
2^128 ? 2^(7*18)
2^128 > 2^126
The NPC solver lets us break the system in 1/4 the time of brute force.
If a P=NP proof by construction is found, approximatly what kind of
exponent might it be? Well, each increment of the exponent is
equivilant to one more nested loop. The N^6 algorithm has 6 nested
loops. Are we able to concieve of and understand algorithms which
*requires* that many or more nestings? Sure, anyone can write a trivial
algorithm which uses lots of nestings, but most of the time, it can be
simplified in some way to use fewer [levels of] loops. What is the
human limit of understanding here?
--
A solution in hand is worth two in the book.
------------------------------
From: [EMAIL PROTECTED] (Lee Rudolph)
Crossposted-To: sci.math
Subject: Re: Given any arbitrary numbers a and b. Can I ALWAYS find a transcendental
number between a and b?
Date: 19 Feb 2001 21:36:05 -0500
[EMAIL PROTECTED] (Dave Seaman) writes:
>Density has nothing to do with countability.
Oh, I don't know. Some of the people who have been posting idiocies
about countability lately seem pretty dense to me.
Lee Rudolph
------------------------------
From: Steve Portly <[EMAIL PROTECTED]>
Subject: Re: What's a KLB-7?
Date: Mon, 19 Feb 2001 21:49:54 -0500
Apparently this machine used a 5 bit cipher, does anybody have any
samples of PT and resulting CT?
David Hamer wrote:
> Richard Outerbridge wrote:
> >
> > -----BEGIN PGP SIGNED MESSAGE-----
> >
> > 2001-02-19 17:29:00 GMT
> > On board the H.M.S. Belfast today (a heavy-light cruiser
> > museum piece moored in the London Pool) in the Electronic
> > Warfare Shack, behind glass, I observed an obvious piece
> > of crypto equipment which proclaimed itself an instance
> > of NSA "KLB-7/T SEC" S#12405, best-delivery-before date
> > sometime in 1990 to somewhere in Whitehall.
> >
> > Can anyone say what this hardware was for? The Belfast
> > saw 'angry' service up to the end of the Korean War.
> >
> > outer
>
> Richard...
>
> It's a KL-7. There is a fair amount of information on
> this machine [with photographs] on Jerry Proc's page:
> <http://webhome.idirect.com/~jproc/crypto/kl7.html>
>
> KL-7 is also included, briefly, in 'Machine Cryptography
> and Modern Cryptanalysis by C.A. Deavours & Louis Kruh,
> Artech House, 1985'
>
> DHH
> --
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> David Hamer The Crypto Simulation Group
> [EMAIL PROTECTED] or [EMAIL PROTECTED]
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
------------------------------
From: Benjamin Goldberg <[EMAIL PROTECTED]>
Subject: Re: __(?) MATRIX version of Fermat's Little Theorem
Date: Tue, 20 Feb 2001 03:23:11 GMT
kctang wrote:
>
> Hi,
>
> We know that a^(p-1) = 1 mod p, where p is a prime, and p does not
> divide a.
>
> Is there a *Matrix* version of Fermat's little theorem? (e.g.
> mod p taking element-wisely, what is the index?)
Interesting question. Assuming there is such a thing, some things which
would have to be changed:
1) You would have to replace 1 in the original with something else --
probably I (the identity matrix).
2) You would have to replace "p does not divide a" with something else
-- probably "p does not divide the determinant of a" -- though I'm not
certain of this.
If anyone's done proofs for the problem you're interested in, I suspect
it will be a bit hard to find. Why don't you do a few tests? Pick a
value p, pick a 2x2 matrix a, make sure that p does not divide |a|, and
test to find out if a^(p-1) = I mod p. Do a few dozen trials... If
fermat's little theorem does not translate to matrices, and if you're
lucky, then you'll be able to disprove it by counterexample quite
quickly.
--
A solution in hand is worth two in the book.
------------------------------
From: "kwd_kwp0ee9j9" <[EMAIL PROTECTED]>
Subject: FAQ
Date: Mon, 19 Feb 2001 21:50:05 -0500
where can I find this newsgroup FAQ?
------------------------------
From: Virgil <[EMAIL PROTECTED]>
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
numbers?
Date: Mon, 19 Feb 2001 20:56:25 -0700
In article <[EMAIL PROTECTED]>, Steve
Leibel <[EMAIL PROTECTED]> wrote:
> > : Let me clarify what I mean by sequentially enumerate.
> > : By sequentially enumerate I mean successively enumerate
> > : all possible transcendental numbers starting from zero
> > : to infinity.
There are uncountably many transendentals. Any well-ordering of them
cannot be of the same order type as the order type aleph-null, of the
natural numbers, so they cannot, in your sense, be successively
enumerated.
------------------------------
From: Virgil <[EMAIL PROTECTED]>
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
numbers?
Date: Mon, 19 Feb 2001 21:01:47 -0700
In article <[EMAIL PROTECTED]>, jtnews
<[EMAIL PROTECTED]> wrote:
> Jan Kristian Haugland wrote:
> >
> > jtnews wrote:
> >
> > > Is there an algorithm to sequentially enumerate
> > > all possible transcendental numbers?
>
> Thanks for the quick response!
>
> Is there some reference anyone can give
> where I can find mathematical proof of this?
The first proofs of the uncountability of the reals are due to Georg
Cantor. Then you need a proof of the countability of the algebraic
numbers to prove the uncountability of the transcendentals.
(transcendental = real but not algebraic)
------------------------------
From: Virgil <[EMAIL PROTECTED]>
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
numbers?
Date: Mon, 19 Feb 2001 21:03:03 -0700
In article <[EMAIL PROTECTED]>, stanislav shalunov
<[EMAIL PROTECTED]> wrote:
> No. There are 2^{\aleph_0} transcendental numbers.
Only if you assume the continuum hypothesis.
------------------------------
From: "Douglas A. Gwyn" <[EMAIL PROTECTED]>
Subject: Re: "RSA vs. One-time-pad" or "the perfect enryption"
Date: Tue, 20 Feb 2001 04:27:34 GMT
"John A. Malley" wrote:
> Are there other books on the subject you'd recommend - especially any
> covering cryptanalysis with hidden markov models?
There is now a vast literature on technology and applications of
HMMs, but there are only a couple of articles I know of in the open
literature specifically addressing application to cryptanalysis, and
unfortunately I can't recommend them, because they don't show any
real advantage over other well-known approaches. That's not to say
that there *aren't* good applications for the technology, just that
it is not evident from what has been published.
A couple of years ago I ran across an interesting application of HMMs
to the problem of "hands" in the Voynich manuscript, and managed to
get it declassified (except for one example and a few names); there
is a brief summary of its results in Jim Reed's Voynich bibliography
at URL http://www.research.att.com/~reeds/voynich/bib.html
Maybe some day I'll make a copy available on a Web site.
------------------------------
From: [EMAIL PROTECTED] (Dave Seaman)
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
numbers?
Date: 19 Feb 2001 23:38:43 -0500
In article <[EMAIL PROTECTED]>,
Virgil <[EMAIL PROTECTED]> wrote:
>In article <[EMAIL PROTECTED]>, stanislav shalunov
><[EMAIL PROTECTED]> wrote:
>> No. There are 2^{\aleph_0} transcendental numbers.
>Only if you assume the continuum hypothesis.
No. CH is irrelevant to this assertion.
--
Dave Seaman [EMAIL PROTECTED]
Amnesty International calls for new trial for Mumia Abu-Jamal
<http://www.amnestyusa.org/abolish/reports/mumia/>
------------------------------
From: "Douglas A. Gwyn" <[EMAIL PROTECTED]>
Subject: Re: Super strong crypto
Date: Tue, 20 Feb 2001 04:40:48 GMT
Bryan Olson wrote:
> Is "natural lifetime" some property of a key?
No, it's a property of the encryption method.
Surely you know? that real symmetric-key systems
require the key to be changed at intervals calculated
to resist cryptanalysis. That is a given; what I
am particularly interested in is, *assuming a correct
assessment of that property*, is there appreciable
actual weakness that could be exploited in practice,
caused by piggybacking the distribution of additional
key material on the existing channel? This is not an
academic exercise about unrealizable infinities; it's
an important cryptoengineering issue that is relevant
for certain kinds of communication systems. It would
be useful to definitely know the answer.
------------------------------
From: "Douglas A. Gwyn" <[EMAIL PROTECTED]>
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
Date: Tue, 20 Feb 2001 04:47:42 GMT
jtnews wrote:
> Is there an algorithm to sequentially enumerate
> all possible transcendental numbers?
No, not even in an infinite time.
An enumeration would have measure 0 but the
set of transfinites has measure 1.
The idea of using compact "programs" as key generators
doesn't actually work, since the interpreter would be
part of the "general system" and so all the secrecy
would reside in the compact program (parameters),
which is undoubtecly harder to memorize than a typical
password containing the same amount of entropy.
------------------------------
From: [EMAIL PROTECTED] (Dave Seaman)
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
numbers?
Date: 19 Feb 2001 23:49:49 -0500
In article <xbkk6.1257$[EMAIL PROTECTED]>,
Paul Lutus <[EMAIL PROTECTED]> wrote:
>"Dik T. Winter" <[EMAIL PROTECTED]> wrote in message
>news:[EMAIL PROTECTED]...
>> Confusing isn't it? But that
>> is mathematics.
>I agree with your enlightening post, and I appreciate your taking the time
>to write it. I came at the original post's terminology from more of a
>computer science perspective, where "enumerate all" has a different shade of
>meaning, and reasonable algorithms eventually stop.
Do you know what it means for a set to be recursively enumerable? The
terminology comes from computability theory, which might be considered to
be at the intersection of mathematics and computer science.
It is possible for an infinite set to be recursively enumerable. It
simply means there is a Turing machine that generates members of the set
in such a way that each member is generated at some finite time. Notice
that there is not necessarily a way to find out that a given number is
*not* in the set, since you could never be sure whether it might be the
next number to turn up.
--
Dave Seaman [EMAIL PROTECTED]
Amnesty International calls for new trial for Mumia Abu-Jamal
<http://www.amnestyusa.org/abolish/reports/mumia/>
------------------------------
From: "Douglas A. Gwyn" <[EMAIL PROTECTED]>
Subject: Re: What's a KLB-7?
Date: Tue, 20 Feb 2001 04:52:20 GMT
Steve Portly wrote:
> Apparently this machine used a 5 bit cipher, does anybody have any
> samples of PT and resulting CT?
The PT would be a message in Baudot code, such as a position
and status report. (Baudot code is explained in many places.)
The CT would be so close to random noise that there would be no
point in a sample of it.
------------------------------
From: "Douglas A. Gwyn" <[EMAIL PROTECTED]>
Crossposted-To: sci.math
Subject: Re: Is there an algorithm to sequentially enumerate all transcendental
Date: Tue, 20 Feb 2001 04:58:35 GMT
"Douglas A. Gwyn" wrote:
> set of transfinites has measure 1.
I of course meant transcendentals (which is what was
asked about). Sorry!
------------------------------
From: [EMAIL PROTECTED] (Dave Seaman)
Crossposted-To: sci.math
Subject: Re: Given any arbitrary numbers a and b. Can I ALWAYS find a transcendental
number between a and b?
Date: 19 Feb 2001 23:56:09 -0500
In article <[EMAIL PROTECTED]>,
John Savard <[EMAIL PROTECTED]> wrote:
>On Mon, 19 Feb 2001 15:01:18 -0500, jtnews <[EMAIL PROTECTED]>
>wrote, in part:
>>Can I ALWAYS find a transcendental number
>>between a and b?
>Yes.
Agreed.
>The algebraic numbers are nowhere dense.
No, the algebraic numbers are everywhere dense. So are the
transcendentals.
The Cantor set is nowhere dense.
--
Dave Seaman [EMAIL PROTECTED]
Amnesty International calls for new trial for Mumia Abu-Jamal
<http://www.amnestyusa.org/abolish/reports/mumia/>
------------------------------
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