Twelve is special
Erkki Kurenniemi
shared by any other natural number. What is it? (Of course you are
supposed not to refer to any explicit number, like saying Ôequals 3
times 4Õ or Ô...even prime multiplied by itself and the next prime...Õ
or any too exotic object etc. I understand that there are probably many
simply stated answers which go beyond my understanding. State in less
than ten words.)
Leong Weng Ng
It is probably the solution to Kepler's Sphere-Packing Problem, unless
someone unluckily manages to squeeze in a 13th sphere :)
--
____________________________________________________________________________
Ng Weng Leong E:--------|-------|---------|------|
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____________________________________________________________________________/
Kevin Brown
The only stable cycle of -1/sin(x)cos(x) has period 12.
Kevin Brown
EK> I am testing this little puzzle. The number 12 has a property not
EK> shared by any other natural number. What is it? State in less
EK> than ten words.)
The number and sum of its divisors are perfect.
1 2 3 4 5 6 7 8 9
Robert A. Meyer
Well, I am not going to be able to state my disagreement in 10 words or less,
but while the sum of the divisors of 12 is 28 which is a perfect number, the
number 12 itself is not perfect (in order for 12 to be perfect, the sum of its
divisors would have to be 24, not 28). The smallest perfect number is 6 and
28 is the second. Perhaps Kevin meant that half the number and the sum of its
divisors are both perfect. "Half of it and the sum of its divisors are
perfect" describes 12 but uses 11 words. "It has 6 letters and starts with
tw" works but is certainly not very interesting.
Robert A. Meyer Bellevue West Senior High
mey...@unomaha.edu Bellevue, Nebraska
Matthew D. Mains
Mike Rubenstein Phoenix Contract <rubenst%occs.nlm.nih.gov> wrote:
>
>--
>Mike Rubenstein
He didn't say 12 was perfect.
Write down the factors of twelve.
We get {1,2,3,4,6,12}
So the number of divisors is 6, and the sum of them is 28.
6=1+2+3 and 28=1+2+4+7+14, thus both 6 and 28 are perfect.
--
============================================================================
| |
| "Hold my head, we'll trampoline." |
| Pixies |
Mike Rubenstein Phoenix Contract
12 is perfect? Since when.
--
Mike Rubenstein
Jan Kok
Mike, as I read it it says: The number of its divisors is perfect,
the sum of its divisors is perfect.
That is more than ten words. Are brackets allowed?
--
--Jan Kok | =#===-=========##=
Mail: CWI (dpt. NW) | -- ,___@
P.O.Box 94079/NL-1090 GB Amsterdam | -- __/\
E-mail: Jan...@cwi.nl | ' /_
Harry J. Smith
Rubenstein Phoenix Contract) writes:
No, "The number and sum of its divisors are perfect." says that there
are 6 divisors of 12 (1, 2, 3, 4, 6, 12) whose sum is 28. It's 6 and 28
that are perfect, not 12.
--
| Harry J. Smith
| 19628 Via Monte Dr., Saratoga, CA 95070-4522, USA
| Home Phone: 408 741-0406, Work Phone: 408 235-5088 (Voice Mail)
| EMail: HJS...@ix.netcom.com on the Internet via Netcom NetCruiser
--
Bruce W Appleby
>> EK = Erkki Kurenniemi
>> EK> I am testing this little puzzle. The number 12 has a property not
>> EK> shared by any other natural number. What is it? State in less
>> EK> than ten words.)
>> The number and sum of its divisors are perfect.
>12 is perfect? Since when.
>Mike Rubenstein
The number of divisors of 12 is 6 (1,2,3,4,6,12) and the sum of its
divisors is 28 (1+2+3+4+6+12). Kevin asserted that 6 and 28 are perfect,
not 12. Is it obvious that there are no other such?
Bruce Appleby
Matthew P Wiener
Let's see. Let N be "the number", S "sum", D "its divisors". "And"
of course is addition, while "of", as with fractions, is multiplication.
So we have N+S*D. Now, while MDAS applies universally in algebra, it
does not in English. So you can parse the sentence as (N+S)*D=N*D+S*D.
Clear?
--
-Matthew P Wiener (wee...@sagi.wistar.upenn.edu)
KPT Ben
>In <3khtek$f...@elaine4.Stanford.EDU> alg...@leland.Stanford.EDU >(Leong
Weng Ng) writes:
>
>>In article <3kgil4$f...@idefix.eunet.fi>,
>>Erkki Kurenniemi <kur...@kurtsi.pp.fi> wrote:
>>>I am testing this little puzzle. The number 12 has a property not
>>>shared by any other natural number. What is it? (Of course you are
>>>supposed not to refer to any explicit number, like saying equals 3
>>>times 4 or ...even prime multiplied by itself and the next prime...
>>>or any too exotic object etc. I understand that there are probably many
>>>simply stated answers which go beyond my understanding. State in less
>>>than ten words.)
>>It is probably the solution to Kepler's Sphere-Packing Problem, unless
>>someone unluckily manages to squeeze in a 13th sphere :)
>>--
>Or simply, that it is the only number (that I can find) for which the
>sum of the factors (excluding the number itself, of course) is greater
>than the number.
If memory serves, these types of numbers are commonly referred to as
"abundant" numbers, and there are infinitely many of them. For instance...
30 has factors 1,2,3,5,6,10,15 which sum to 42... (quick, someone call
Douglas Adams!)
180 has factors which sum to 366, making it (I believe) the first "doubly
abundant" number.
In fact, if X is an abundant number, then so is nX for any positive
integer n, since if factors of X [f1, f2, f3, .... fm] sum to >X, then nX
will have factors [nf1, nf2, nf3... nfm] which sum to n * [f1 + f2 + .. +
fm] which is greater than nX.
Along these lines, anyone want to offer a proof (or disproof) that for any
N, there can be found an integer K whose factors sum to > NK? I'm pretty
sure this is true, but if anyone has a proof, I'd like to see it.
But in any case, for puzzle purposes...
12 is the smallest abundant number ;)
--
Ben Weiss
HSC Software
Don N Page
writes:
> I am testing this little puzzle. The number 12 has a property not
> shared by any other natural number. What is it?
(Incidentally, it is thus appropriate that the largest square Fibonacci number
is a "gross.")
Dean Hickerson
Bruce W Appleby <bapp...@world.std.com> wrote:
>
>The number of divisors of 12 is 6 (1,2,3,4,6,12) and the sum of its
>divisors is 28 (1+2+3+4+6+12). Kevin asserted that 6 and 28 are perfect,
>not 12. Is it obvious that there are no other such?
No, in fact this is false. Let
126 3 5 7 19 31 61
n = 2 (2 -1) (2 -1) (2 -1) (2 -1) (2 -1) (2 -1) =
6086555670238378989670371734243169622657830773351885970528324860512791691264.
The 6 parenthesized factors above are all known to be prime, so the number of
divisors of n is 127 * 2 * 2 * 2 * 2 * 2 * 2 = 8128, which is perfect. The
sum of the divisors of n is
127 3 5 7 19 31 61 126 127
(2 -1) 2 2 2 2 2 2 = 2 (2 -1).
Since 2^127-1 is prime, this number is also perfect.
Dean Hickerson
drhic...@ucdavis.edu
Benjamin J. Tilly
kpt...@aol.com (KPT Ben) writes:
[...]
> But in any case, for puzzle purposes...
>
> 12 is the smallest abundant number ;)
These puzzles always seemed boring to me. We already have two
characterizations of 12 in less than 10 words. Here are some others.
"It is 12."
"The smallest integer larger than 11."
"The second smallest integer larger than 10"
.
.
.
"It is 11 + 1."
"It is 3*4"
.
.
.
"It is the first non-power divisible by 4."
Need I go on?
Ben Tilly
Aaron Bergman
(Robert A. Meyer) wrote:
The number of the divisors is 6 is what was meant.
Aaron
--
Aaron Bergman
matthew conroy
>N, there can be found an integer K whose factors sum to > NK? I'm pretty
>sure this is true, but if anyone has a proof, I'd like to see it.
>
Sure. Let n(k) be the product of the first k primes.
Let p(i) denote the i-th prime (p(1)=2,p(2)=3,...).
Let sigma(n) be the sum of the divisors of n, including n.
Then sigma(n(k))/n(k) tends to infinity as k does.
Why? sigma(n(k))/n(k) = product ( (p(i)+1)/p(i) , i=1..k) =
product(1+1/p(i), i=1..k) > sum(1/p(i),i=1..k).
As k tends to infinity, so does this last sum (i.e. the sum
of the reciprocals of the primes is divergent).
Hence we can make sigma(n(k))/n(k) as big as we want, just by choosing
k as large as necessary.
matthew
Kevin Brown
EK> I am testing this little puzzle. The number 12 has a property not
EK> shared by any other natural number. What is it? State in less
EK> than ten words.)
KB> The number and sum of its divisors are perfect.
BA = Bruce Appleby
BA> Is it obvious that there are no other such?
No, it isn't obvious. In fact, upon reflection, I don't even think it's
true. If we set N equal to
6086555670238378989670371734243169622657830773351885970528324860512791691264
then sigma(N) and tau(N) are both perfect.
For any integer N with prime factorization
N = p1^a1 p2^a2 ... pt^at
we have t / pj^(aj+1) - 1 \
sigma(N) = PROD ( --------------- )
j=1 \ pj - 1 /
Also, recall that every even perfect number is of the form
(2^s - 1) 2^(s-1) where 2^s - 1 is a prime. Thus an even perfect
number has exactly one odd prime factor.
Now suppose N is divisible by exactly k powers of 2. It follows that
sigma(N) is divisible by 2^(k+1) - 1, which is odd, so this must be a
prime (else it would factor into two odd primes). Also, all the other
factors of N must then contribute a combined factor of 2^k to sigma(N).
But each odd prime power p^d contributes a factor of
sigma(p^d) = 1 + p + p^2 + ... + p^d
to sigma(N), which can only be even if d is odd, in which case it
factors as
sigma(p^d) = (1+p) (1 + p^2 + p^4 + ... + p^(2t))
The right hand factor can be even only if t is odd, in which case it
factors as
sigma(p^d) = (1+p) (1+p^2) (1 + p^4 + p^8 + ... + p^(4s))
However, the factors (1+p) and (1+p^2) cannot both be pure powers of
2, because if p=2^r-1 then 1+p^2 = (2)(2^[2r-1] - 2^r + 1). Therefore,
if the number sigma((2^k)(G)) is perfect, then G must be the product of
distinct Mersenne primes pj = 2^(aj) - 1 where the exponents aj sum
to k, where k+1 is also a "Mersenne exponent".
Moreover, to make tau(N) perfect, it's necessary for k+1 itself to be
a Mersenne prime. Thus, to construct an example of this type, we need
to find a Mersenne exponent E that is also a Mersenne prime, and such
that E-1 is expressible as the sum of exactly log_2(E+1)-1 distinct
Mersenne primes.
One example is E=3, because 2 can be expressed as the sum of exactly
log_2(4)-1 = 1 Mersenne primes, i.e., 2=2. Thus, we have
N = (2^(3-1))(2^2-1) = 12. Other possible values of E are 7, 31,
127,..? Off hand I don't see how to express 6 as a sum of 2 Mersenne
primes, nor how to express 30 as a sum of four distinct Mersenne
primes. However, 126 can be expressed as a sum of six distinct
Mersenne primes, namely 61+31+19+7+5+3, so we have
N = (2^126)(2^61-1)(2^31-1)(2^19-1)(2^7-1)(2^5-1)(2^3-1)
which is the number cited above. Are this number and 12 the only two
integers with the property that tau(N) and sigma(N) are both perfect?
Andy Parkerson
>If memory serves, these types of numbers are commonly referred to as
>"abundant" numbers, and there are infinitely many of them. For instance...
>In fact, if X is an abundant number, then so is nX for any positive
>integer n, since if factors of X [f1, f2, f3, .... fm] sum to >X, then nX
>will have factors [nf1, nf2, nf3... nfm] which sum to n * [f1 + f2 + .. +
>fm] which is greater than nX.
No. nX will have factors [n, f1, f2, f3, ... fm]. (n^m)X *will* have said
factors, but one doesn't know whether n*[f1 + f2 + f3 + ... + fm] is greater
than (n^m)X.
Andy Parkerson
Texas A&M University
an...@tamu.edu
David B. Freiman, MD
(Don N Page) wrote:
> (Incidentally, it is thus appropriate that the largest square Fibonacci number
> is a "gross.")
Has this been proven? How? I have never heard this before.
(This thread is also of interest to me because twelve is far and away my
favorite number)
Timothy Y. Chow
David B. Freiman, MD <fre...@a1.mscf.upenn.edu> wrote:
>In article <3kq6c9$1j...@rover.ucs.ualberta.ca>, d...@page.phys.ualberta.ca
>(Don N Page) wrote:
>
>>(Incidentally, it is thus appropriate that the largest square Fibonacci number
>> is a "gross.")
>
>Has this been proven? How? I have never heard this before.
Here's some old sci.math stuff on the topic.
==========================================================================
From: ow...@andrew.cmu.edu (Oswald Wyler)
Message-ID: <oeAC8mq00...@andrew.cmu.edu>
Date: 6 Jun 92 16:05:06 GMT
This problem was solved in 1963/64 by three people, including myself.
Answer: F_{12} = 144 is the last square Fibonacci number.
My solution appeared in the Problems section of the Amer. Math. Monthly
in 1964. John H.E. Cohn published his (more complicated) solution in
Journal or Proceedings of the London Math. Soc. in 1964, and there
should also be a note by J.H.E.Cohn in the Fibonacci Quarterly, 1964.
After all this time, I do not have more precise references, nor do I
recall who the third solver of the problem was.
From: memb...@vax.oxford.ac.uk (Fausto H. Membrillo)
Message-ID: <1992Jun10....@vax.oxford.ac.uk>
Date: 10 Jun 92 12:16:16 GMT
Thanks to every one who answer my question on Fibonacci numbers. I
have found the right references and now I can use that on my own work.
I made a summary of all the answers.
Question: What are all the square Fibonacci numbers?
Answer: The square Fibonacci numbers are {0,1,144}. This was done by tree
people in 1964: JHE Cohn in the Journal of the London Math. Soc. in 1964, and
in a note in the Fibonacci Quarterly: J.H.E. Cohn, "Square
Fibonacci Numbers", Fibonacci Quart. 2, pp. 109-113, 1964.
Oswald Wyler in the Problems section of the Amer. Math. Monthly
in 1964. And a third person who I do not have a reference.
Moreover Ray Steiner proved in 1967 that the cube Fibonacci numbers are {0,1,8}
Article 54891 of sci.math:
Path: galois!bloom-beacon.mit.edu!grapevine.lcs.mit.edu!uhog.mit.edu!MathWorks.Com!news.kei.com!eff!usenet.ins.cwru.edu!howland.reston.ans.net!news.cac.psu.edu!news.pop.psu.edu!psuvax1!news.cc.swarthmore.edu!netnews.upenn.edu!news
From: lar...@math.upenn.edu (Michael Larsen)
Newsgroups: sci.math
Subject: Re: Is a gross the largest square Fibonacci number?
Date: 18 May 1994 00:29:34 GMT
Organization: University of Pennsylvania
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Message-ID: <2rbnhe$7...@netnews.upenn.edu>
References: <2raoim$f...@quartz.ucs.ualberta.ca>
NNTP-Posting-Host: ts7-55.upenn.edu
In article <2raoim$f...@quartz.ucs.ualberta.ca> d...@page.phys.ualberta.ca (Don N
Page) writes:
> I am now conjecturing that a gross (144)
> is appropriately named in that it is the largest Fibonacci number that is a
> perfect square. However, I couldn't find this in Guy. I bet a couple of
> graduate students $100 to $1 that my conjecture is true. Does anyone know a
> proof or a counterexample?
Yes, the statement is true. Siegel's theorem implies that there are
only finitely many square Fibonacci numbers but doesn't say exactly
how many. Here's a sketch of a proof.
The n'th Fibonacci number, F_n, can be written (a^n - b^n) / sqrt(5),
where a and b are the roots of the polynomial x^2 - x - 1.
From this, the following claims are easy exercises:
1. F_m divides F_{mn} for all m, n.
2. If a prime p divides F_m and p does not divide n, then F_{mn}/F_m is
prime to p.
3. For all k, F_{4k-1} and F_{4k+1} are congruent to -1 modulo
L = F_{4k} / F_{2k}. In particular, the period of the sequence
F_n (mod L) divides 8k but not 4k.
4. For all m > 2, F_{2^m} / F_{2^{m-1}} is congruent to 2 (mod 5).
5. If p is prime and congruent to 2 or 3 (mod 5), and n is the smallest
positive integer such that p divides F_n, then n divides p+1.
If p is congruent to 1 or 4, then n divides p-1.
Numbers 1 through 4 are entirely elementary. Number 5 is an easy
application of the law of quadratic reciprocity and the theory of
finite fields.
Suppose F_n is a perfect square. Let p be the largest prime divisor
of n. Assume p > 5. By #5, every prime divisor q of F_p is greater
than p. By #2, the highest power of q dividing F_n is the same
as the highest power of q dividing F_p. As F_n is a square,
F_p must be square as well. So (modulo the assumption p > 5)
we have reduced to the problem of showing that F_p cannot be square
when p > 2 is prime.
Let L_m = F_{2^m} / F_{2^{m-1}}. By #3, for fixed m > 1, the residue of
F_n (mod L_m) depends only on n (mod 2^{m+1}). For example, when m = 2,
L_m = 3, and the residue of F_n (mod 3) depends only on n (mod 8).
We readily check that F_3 and F_5 are congruent to 2 (a quadratic
non-residue) (mod 3), so if F_p is a perfect square, p is congruent to
plus or minus 1 (mod 8).
Let m denote the highest power of 2 dividing p+1 or p-1, which ever gives
the larger value. We've seen m > 2. By #4, this means L_m is congruent
to 2 (mod 5). So L_m has some prime factor which is congruent to 2 or 3
(mod 5). Let q be that factor. By #5, 2^m divides q+1, so in particular,
q is congruent to 3 (mod 4). On the other hand, by #3, F_p is congruent
to -1 (mod q). Therefore, F_p is a quadratic non-residue of q,
which implies F_p is not a square.
Thus we only have to consider the case that n = 2^a 3^b 5^c.
By considering the highest power of p dividing F_n, as p ranges over
{5, 7, 11, 3001}, we see a < 3, b < 2, c < 1. Checking the remaining
cases, we obtain the theorem.
-Michael Larsen
--
Tim Chow tyc...@math.mit.edu
Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs
30 tons, computers in the future may have only 1,000 vacuum tubes and weigh
only 1 1/2 tons. ---Popular Mechanics, March 1949
KPT Ben
>>In fact, if X is an abundant number, then so is nX for any positive
>>integer n, since if factors of X [f1, f2, f3, .... fm] sum to >X, then
nX
>>will have factors [nf1, nf2, nf3... nfm] which sum to n * [f1 + f2 + ..
+
>>fm] which is greater than nX.
>No. nX will have factors [n, f1, f2, f3, ... fm]. (n^m)X *will* have
said
>factors, but one doesn't know whether n*[f1 + f2 + f3 + ... + fm] is
greater
>than (n^m)X.
I don't follow that logic... we're not talking about just prime factors
here.
12 is an abundant number, having factors 1, 2, 3, 4, 6, which sum to 16.
Therefore 12n has (non-prime) factors n, 2n, 3n, 4n, 6n (among others),
which sum to at least 16n, for any n. Hence 12n is abundant for any n.
For any number X which is abundant, using similar logic, it can easily be
shown that nX is abundant for any positive integer n.
Andy Parkerson
You are correct. I apologize for my mistake. I realized it shortly after I
posted. Too bad you cannot unpost :-).
Andy Parkerson
Erkki Kurenniemi
The rather trivial answer I had in mind, in five words: (12 is the)
"smallest non-power divisible by power." In 18 words "Twelve is the
smallest positive integer which is properly divisible by a power
without itself being a power."
Many solutions to the problem refer implicitly to specific numbers.
Some people have referred to the Kepler sphere packing problem, but
they are of course smuggling in the dimension 3. "Smallest Fibonacci
square" is fine although one has to give two seeds (eg. 0 and 1).
Peano's axioms refer only to 1 and a successor function and
12 is (((((((((((1)+)+)+)+)+)+)+)+)+)+)+
where (n)+ denotes the successor of n. The expression contains 3 words
if you consider the 11-fold application of the successor funtion to 1
to be one word. Otherwise, it exceeds the 10-word limit. And is not too
elegant.
Questions:
1) Have I smuggled?
2) Has it been proved that 144 is the only Fibonacci square?
3) Is it possible to (axiomatically or otherwise) characterize natural
numbers without (explicitly or implicitly) referring to any particular
number?
And, if you care, my next problem: Is 8640 special?
I don't know but would like to. The funny thing is that its divisors
give quite a long stretch of the musical diatonic scale (with a certain
catch, a reference is: Thomas D. Rossing, The Science of Sound,
Addison-Wesley, 1982, p. 155). In Mathematica, evaluate:
d = Divisors[8640] ; Table[d[[i+1]]/d[[i]],{i,15,41}]
KPT Ben
(snip)
And, if you care, my next problem: Is 8640 special?
Yes, 8640 is special because...
it is the only number contained in this post. ;)
--
Ben Weiss "Say the secret word, and Wakko will
HSC Software "hit you on the head with a mallet!" -Yakko
Matthew P Wiener
>The rather trivial answer I had in mind, in five words: (12 is the)
>"smallest non-power divisible by power." In 18 words "Twelve is the
>smallest positive integer which is properly divisible by a power
>without itself being a power."
>Many solutions to the problem refer implicitly to specific numbers.
>Some people have referred to the Kepler sphere packing problem, but
>they are of course smuggling in the dimension 3. "Smallest Fibonacci
>square" is fine although one has to give two seeds (eg. 0 and 1).
Why are two seeds for a recurrence worse than one seed (powers)?
Andre Mutovic
equations.. preferable for windows.. DOS ok...
-Andre
JRA...@utcvm.utc.edu
Proof by Contradiction:
Assume that there exists a set of non-special positive integers.
Since this is a subset of the positive integers, this set must be
well-ordered. By the well-ordering property, there exists a number
N which is the least integer in the set. But the any integer with
the property "least non-special integer" is itself special by virtue
of having this unique property. Since this is a contradiction, the
set of non-special positive integers must be the null set.
Actually, I've always wondered if this was kosher or not...seems
all-to-similiar to the exam paradox to me...
Aaron Bergman
:In article <3l344v$h...@idefix.eunet.fi>
What is "the smallest number that cannot be expressed in under twelve words?"
Hint: count the number of words in quotes.
(BTW--I've never seen a satisfactory resolution for this paradox....)
Benjamin J. Tilly
JRA...@utcvm.utc.edu writes:
Well...if you really try to do it, in order to avoid self-reference in
the definition of "special" you find that the "counter-example" is
"meta-special", so it does break down.
It also says a lot about the mindset of people who find these numbers
special that they do find them special. :-)
Ben Tilly
Martin Cohen
Its value is twelve :)
1 2 3 4 4.5
--
Marty Cohen (mco...@nrtc.northrop.com) - Not the guy in Philly
This is my opinion and is probably not Northrop Grumman's!
Use this material of your own free will
Matthew D. Mains
Robert A. Meyer <mey...@cwis.unomaha.edu> wrote:
>kevi...@delphi.com writes:
>> EK = Erkki Kurenniemi
>> EK> I am testing this little puzzle. The number 12 has a property not
>> EK> shared by any other natural number. What is it? State in less
>> EK> than ten words.)
>>
>> The number and sum of its divisors are perfect.
>> 1 2 3 4 5 6 7 8 9
>
>
>but while the sum of the divisors of 12 is 28 which is a perfect number, the
>number 12 itself is not perfect (in order for 12 to be perfect, the sum of its
>divisors would have to be 24, not 28). The smallest perfect number is 6 and
>divisors are both perfect. "Half of it and the sum of its divisors are
>perfect" describes 12 but uses 11 words. "It has 6 letters and starts with
>tw" works but is certainly not very interesting.
>
>mey...@unomaha.edu Bellevue, Nebraska
>
>
What Kevin said is right. The number and sum of its divsors means the
number of divisors and the sum of those divisors are both perfect.
--
============================================================================
| |
| "Hold my head, we'll trampoline." |
| Pixies |
Henning Makholm
>What is "the smallest number that cannot be expressed in under twelve words?"
>Hint: count the number of words in quotes.
>(BTW--I've never seen a satisfactory resolution for this paradox....)
Can one actually express such a paradox in any useful first-order logic
system (that is, replacing "words" with "symbols" and "twelve" with some
number)? It seems to require a formal notion of a string of symbols
"expressing" particular number, and thus to be invalidated by Tarski's
result that truth cannot be defined.
--
_ Henning Makholm - Math and CS student - University of Copenhagen
|_|_\/| Email: mak...@diku.dk - Fido: 2:236/151.12 - Phone +45 42845582
| |_ | Snail-mail: Aamosevej 44, DK-2610 Rodovre - HAM callsign: OZ2HEM
Kouhia Juhana
Finnish magazine Archimedes about some numbers, not
only about the number 12.
It is actually a list of numbers and what is associated
to them.
I try find the article and give reference.
Juhana Kouhia
Willy - Bernhard Strothmann
Willy
Bruce W Appleby
>So, are there nonspecial reals?
>Willy
Not if they are well-ordered :-)
Bruce
RENE GROTHMANN
>Hint: count the number of words in quotes.
>(BTW--I've never seen a satisfactory resolution for this paradox....)
You will most probably start an endless thread of articles by this single
remark.
To me, the resolution of this paradox is, that the word "EXPRESSED" is much
to vague in the quoted sentence. If you were to describe precisely, what the
phrase "can or cannot be expressed" means, you would notice, how vague the
statement is.
Now, if you define any set of rules, which describe, how numbers are to be
expressed, than you immediately have a number, which cannot be expressed by
the set of rules (using the quoted sentence). This just means, that no set
of rules is fully covering the natural meaning of the statement "can be
expressed". I guess, we now arrived at the heart of Goedels work.
Rene.
Thomas Andrews
RENE GROTHMANN <mga...@eo-nwfs-1.ku-eichstaett.de> wrote:
>>What is "the smallest number that cannot be expressed in under twelve words?"
>>Hint: count the number of words in quotes.
>>(BTW--I've never seen a satisfactory resolution for this paradox....)
>
>remark.
>
>To me, the resolution of this paradox is, that the word "EXPRESSED" is much
>to vague in the quoted sentence. If you were to describe precisely, what the
>phrase "can or cannot be expressed" means, you would notice, how vague the
>statement is.
>
One interesting side-affect of this paradox is that if you try to define
"expressed" in a rigorous way, you find that the paradox is reached *if*
you can solve the halting problem. Therefore, the paradox leads one
to conclude that the halting problem is unsolvable.
--
Thomas Andrews
CenterLine Software
Aaron Bergman
mga...@eo-nwfs-1.ku-eichstaett.de (RENE GROTHMANN) wrote:
:>What is "the smallest number that cannot be expressed in under twelve words?"
:>Hint: count the number of words in quotes.
:>(BTW--I've never seen a satisfactory resolution for this paradox....)
:
:You will most probably start an endless thread of articles by this single
:remark.
:
:To me, the resolution of this paradox is, that the word "EXPRESSED" is much
:to vague in the quoted sentence. If you were to describe precisely, what the
:phrase "can or cannot be expressed" means, you would notice, how vague the
:statement is.
Actually, two people explained it to me in e-mail. It made sense.
Aaron
--
Aaron Bergman -- The Number Two Poster on RASFWR-J.
Check out my homepage:
<http://www.cis.yale.edu/~abergman/abergman.html>
Could you do me a favor and fill out a survey? E-mail me if you
want it. :)
Kouhia Juhana
written by Finnish mathematician; here is the reference
Robert Piche (pi...@butler.cc.tut.fi) mailed to me. Thank you.
Archimedes 1993 vol 45 No. 2, p. 97-115: Lukujen mystiikkaa, Osmo Pekonen
The title is "Mysticism of the numbers". There might be a photograph
of him in some issue of The Mathematical Intelligencer journal within
a mathematical ghost story or such.
Juhana Kouhia
Henning Makholm
>So, are there nonspecial reals?
Intuitively yes. I wouldn't accept any mathematical object as 'special' if
it had a unique, decripable, attribute (so it was, so to say, possible to
point to it). Since any description of interesting attribute must consist
of finitely many symbols from a finite language, there can at most be
countably many special reals. The set of reals is uncountable, hence many of
them are nonspecial. Q.E.D.
(Exercise for the interested student: We now know that nonspecial reals
exist. Find one.)
Hershel Walters
>In article <mga010.31...@eo-nwfs-1.ku-eichstaett.de>,
>RENE GROTHMANN <mga...@eo-nwfs-1.ku-eichstaett.de> wrote:
>>>Hint: count the number of words in quotes.
>>>(BTW--I've never seen a satisfactory resolution for this paradox....)
>>remark.
>>To me, the resolution of this paradox is, that the word "EXPRESSED" is much
>>to vague in the quoted sentence. If you were to describe precisely, what the
>>phrase "can or cannot be expressed" means, you would notice, how vague the
>>statement is.
>"expressed" in a rigorous way, you find that the paradox is reached *if*
>you can solve the halting problem. Therefore, the paradox leads one
>to conclude that the halting problem is unsolvable.
>--
>Thomas Andrews
>CenterLine Software
Oh come on, this can be solved easily, or rather, "resolved" easily.
The phrase "the smallest number that cannot be expressed in under twelve words"
does not specify a language. We can safely assume the English language.
However, the same paradox could apply in general to a number of languages.
Thus, let's "invent" a language. Let's call it "subEnglish1".
The language "subEnglish1" is merely a subset of the English language. The
only things which are legal to express in "subEnglish1" are the numbers one,
two, three, four, twelve, twenty eight vingintillion nine hundred ninety nine
duodevingintillion nine hundred ninety nine octillion nine hundred ninety nine
trillion nine hundred ninety nine thousand nine hundred ninety nine (ie, just
some dumb arbitrary number which takes more than 12 words to express),
the phrase "the smallest number that cannot be expressed in under twelve
words", and let's just say one specific really huge number larger than that
other one and greater than 12 words.
Now, the phrase "the smallest number that cannot be expressed in under twelve
words" refers to that explicitely defined number... except... that, wait, it
refers to that number in under twelve words! So, thus, it's not the smallest
number... we have to look for something larger. So it must refer to that
other huge number... but... wait... if it does, then it expresses that
larger number in less than twelve words, and... hold on... twenty eight
vingintillion nine hundred... takes more than twelve words to express.
As you can see, the phrase only contradicts itself, by indirectly saying that
if it is true for any x, then it is false for that x, else if it is false for
that x, it is true for that x. In other words, it's like inductively applying
the "this statement is not true" paradox over numbers. This does not change
the meaning of numbers, nor does it allow subEnglish1 to express an infinite
number of numbers using a finite number of words.
This isn't in the FAQ is it? :)
--
"Let the fools have their tartar sauce." Montgomery Burns
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