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** The Math Forum Summer Institute Announcement **
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Date: Mon, 22 Apr 1996 18:37:49 -0400
From: Milo Gardner
To: Mathematics History
Subject: Egyptian Fractions, 31/311 in four terms ( 2 of 2)
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I thought reader's of MAA might enjoy Kevin Brown's recent review of the
31/311 4-term subject. It appears my search for a shorter or smaller 31/311
conversion to an Egyptian fraction is over. David Eppstein's answer appears
unbeatable.
Any comments?
> He (Eppstein) came up with a solution for 31/311 as others wrote in
> 10 very awkward Fibonacci like huge denominators.
> To cite Eppstein, he found:
> 31/311 = 14' 36' (9*311)' (28*311)' without using any of
> his Ten Egyptian Fractions Fractions. At first glance
> a fully mental procedure was used. A second glance found
> this problem solved mod 1244 by adding the 2-term solutions
> for 89/1244 and 35/1244 as calculated by Hultsh's 1895 rule...
That's interesting. So he used the facts that
89/1244 = 1/14 + 1/8708 and 35/1244 = 1/36 + 1/2799
to give the combined result
(89+35)/1244 = 31/311 = 1/14 + 1/36 + 1/2799 + 1/8708
Notice that since 1/14 + 1/36 = 1/12 + 1/63 we can also write this
in slightly modified form as
31/311 = 1/12 + 1/63 + 1/2799 + 1/8708
Interestingly, the same binomial product form that appeared in one of
your earlier tables seems to apply here as well, because we have
31 / 1 1 \ / 1 1 \ 1
---- = ( --- + --- ) ( --- + --- ) + ---
311 \ 4 311/ \ 9 28/ 16
(this issue was covered in the decoding of 4/17 and 8/17 cited
in the Akhmim Papyrus).
This gives a four-term solution because 1/(4*28) + 1/16 = 1/14.
The value of 1/4 in the left-hand parentheses comes empirically from
the fact that the formula
n / 1 1 \ / 1 1 \
---- = ( --- + --- ) ( --- + --- )
311 \ Q 311/ \ u v /
gives the greatest number of integer values of n when we set Q=4.
So, setting Q=4, we can scan through the values of u and v. There
is no combination of u and v that gives n=31, but instead we can
look for u,v that give n = 31 - 311/16, which equals 185/16. Then
we find that setting u=9 and v=28 gives n=185/16, so we have a
four-term solution.
> Oh yes, I am looking for a shorted or smaller last term 31/311
> to compete with Eppstein's Egyptian fractions. If any solutions
> jump into your mind please feel free to share them.
I ran a quick computer check on this, and it appears there is no
three-term expression for 31/311, nor a four-term expression with a
"smaller" last term than 1/8708. However, there are some interesting
facts about this particular fraction. For example, the nearest
three-term expansion differs from the true value by a unit fraction,
i.e.,
31 1 1 1 1
--- = --- + --- + ----- - ----------
311 11 115 13566 5337067890
where the final (negative) denominator is just the least common
multiple of all the other denominators.
Pretty, neat, right?
Milo Gardner
Sacramento, CA
Foundation.
Interested teachers should see the full description and on-line application
that are available on the Math Forum Web site:
http://forum.swarthmore.edu/workshops/advanced96.html
Thank you,
Stephen Weimar
The Math Forum
steve@forum.swarthmore.edu
Date: Mon, 22 Apr 1996 18:26:44 -0400
From: Milo Gardner
To: Mathematics History
Subject: Egyptian fractions, 31/311 in four terms ( 1 of 2)
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This post lists a Greek/Egyptian method to compute 31/311
that follows an interesting David Eppstein solution, as cited below.
The scope of this post is to probe the possibility that ancient Greeks
and ancients both looked for an found p/q exact conversions to short and
small unit fraction series. A formal paper is planned to be submitted to an
appropriate mathematics journal in the very near future.
Comments are therefore welcomed.
The outline of the paper takes into account Exact Sciences in Antiquity
(Neugebauer) and similar books like the Rhind Mathematical Papyrus
(Robins and Shute), Middle Kingdom Egyptians easily wrote composite
numbers near the form that Neugebauer stated, as I prefer to re-write as
a modern number theory composite by:
2/pq = (1/q + 1/pq)2/(p + 1), as used for all RMP 2/nth terms
except 2/35, 2/91 and 2/95.
It can be easily shown that 2/35 and 2/91 followed a unique partition
method as detailed in the 500 AD Hellenized Copt Akhmim Papyrus,
as I first read in Howard Eves undergraduate textbook, as:
n/pq = 1/pr + 1/qr, where r = (p + q)/n
The 2/95 case was simply computed by 2/19 times 1/5 where 2/19
was derived from the prime number algorithm suggested by Hultsch
in 1895. Bruins also ran across this form in 1952, as stated by:
2/p - 1/a = (2a -p)/ap where the divisors of a, the first partition
are used to compute 2a -p, the 2/nd - 4th partition value.
What is interesting to report is that the Egyptian Mathematical
Leather Roll (EMLR) 1/p and 1/pq forms confirms almost every major
point that is cited above.
So why has this easily managed set of ancient algorithms not been
agreed upon long ago as a method to convert p/q into exact unit
fraction series?
Well the problem seems to be certain p/q conversion that go beyond the
ancient texts like the RMP 2/3 - 2/101 table are difficult to compute.
Several p/q values have drawn a great deal of attention, being used as
counter examples, that attempted to infer that Egyptian really could not
convert all rational numbers into short and small unit fraction series.
The one I will discuss is 31/311, one that scholars have awkwardly
written as 9-terms, 10-terms and higher. Well it turns out that
David Eppstein found an Egyptian 4-term solution, as all other
p/q conversion to Egyptian fractions probably can be found, seen as:
31/311 = 14' 36' 2799' 8708'.
Eppstein found this by hand (with the machine only doing the arithmetic)
using the same "highly divisable number" method with 252 = 2^2 3^2 7 as
the highly divisable number as his discusses on his web paper "Ten
Algorithms For Egyptian Fractions". It also turns out that Eppstein, a
Mathematica expert (using a sophisticated computer program) could not find
" a good automatic method to generate this sort of thing", as ancient
Greeks and Egyptians may have understood very well.
Well, please note that there is an Egyptian/Greek method to compute
31/311, after all. Eppstein worked the problem mod 1244 by adding the
2-term solutions to 89/1244 and 35/1244 by using n/p - 1/a = (na - p)/ap
rule found in the Akhmim Papyrus, as calculated by:
89/1244 - 1/14 = 2/(1244*14)
= 1/(622*14)
= 1/(311*28) which means that
89/1244 = 14' (311*28)' as Greeks wrote was added to 35/1224 computed by:
35/1244 - 1/36 = 16/(1244*36)
= 4/(311*36)
= 1/(311*9) which means
35/1244 = 36' (311*9)' as Greeks may have also recognized by
31/311 = 14' 36' (311*9)' (311*28)'.
Eppstein's method is neat. However, Kevin Brown's confirmation
method is also neat, as the second post will detail.
For additional information on David Eppstein's work, he is found at UC
Irvine Dept. of Information & Computer Science
http://www.ics.uci.edu/~eppstein/numth/egypt/.
Respectfully,
Milo Gardner
Sacramento, CA
Date: Mon, 22 Apr 1996 18:37:49 -0400
From: Milo Gardner
To: Mathematics History
Subject: Egyptian Fractions, 31/311 in four terms ( 2 of 2)
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I thought reader's of MAA might enjoy Kevin Brown's recent review of the
31/311 4-term subject. It appears my search for a shorter or smaller 31/311
conversion to an Egyptian fraction is over. David Eppstein's answer appears
unbeatable.
Any comments?
> He (Eppstein) came up with a solution for 31/311 as others wrote in
> 10 very awkward Fibonacci like huge denominators.
> To cite Eppstein, he found:
> 31/311 = 14' 36' (9*311)' (28*311)' without using any of
> his Ten Egyptian Fractions Fractions. At first glance
> a fully mental procedure was used. A second glance found
> this problem solved mod 1244 by adding the 2-term solutions
> for 89/1244 and 35/1244 as calculated by Hultsh's 1895 rule...
That's interesting. So he used the facts that
89/1244 = 1/14 + 1/8708 and 35/1244 = 1/36 + 1/2799
to give the combined result
(89+35)/1244 = 31/311 = 1/14 + 1/36 + 1/2799 + 1/8708
Notice that since 1/14 + 1/36 = 1/12 + 1/63 we can also write this
in slightly modified form as
31/311 = 1/12 + 1/63 + 1/2799 + 1/8708
Interestingly, the same binomial product form that appeared in one of
your earlier tables seems to apply here as well, because we have
31 / 1 1 \ / 1 1 \ 1
---- = ( --- + --- ) ( --- + --- ) + ---
311 \ 4 311/ \ 9 28/ 16
(this issue was covered in the decoding of 4/17 and 8/17 cited
in the Akhmim Papyrus).
This gives a four-term solution because 1/(4*28) + 1/16 = 1/14.
The value of 1/4 in the left-hand parentheses comes empirically from
the fact that the formula
n / 1 1 \ / 1 1 \
---- = ( --- + --- ) ( --- + --- )
311 \ Q 311/ \ u v /
gives the greatest number of integer values of n when we set Q=4.
So, setting Q=4, we can scan through the values of u and v. There
is no combination of u and v that gives n=31, but instead we can
look for u,v that give n = 31 - 311/16, which equals 185/16. Then
we find that setting u=9 and v=28 gives n=185/16, so we have a
four-term solution.
> Oh yes, I am looking for a shorted or smaller last term 31/311
> to compete with Eppstein's Egyptian fractions. If any solutions
> jump into your mind please feel free to share them.
I ran a quick computer check on this, and it appears there is no
three-term expression for 31/311, nor a four-term expression with a
"smaller" last term than 1/8708. However, there are some interesting
facts about this particular fraction. For example, the nearest
three-term expansion differs from the true value by a unit fraction,
i.e.,
31 1 1 1 1
--- = --- + --- + ----- - ----------
311 11 115 13566 5337067890
where the final (negative) denominator is just the least common
multiple of all the other denominators.
Pretty, neat, right?
Milo Gardner
Sacramento, CA
Date: Wed, 24 Apr 1996 07:33:06 -0400
From: Donald Campbell
To: nctm-l@swarthmore.edu
cc: Math-History-List
Subject: Cirriculum History
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I have heard that NCTM has an archive of publications concerning math,
can someone tell me how to acess it? I am particularly interested in the
history of math education and how we came to choose the cirriculum of
today for 9-12. For instance, if algebra provides a symbolic
representation of geometry, then why do we teach algebra first? I am
particularly interested in the Texas TEKS, -how did we get here?
Thanks,
Don Campbell
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Date: Wed, 24 Apr 1996 09:54:31 -0400
To: math-history-list@maa.org
From: michael quinn <91137277@groucho.mit.csu.edu.au>
Subject: complex numbers
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can anyone recommend any good solid sources on the history of complex
numbers. im looking at covering their development from the work done on
cubics by the ancient greeks, thru cardano until the current notation and
methods we use today. anyone done any work in this field or know of a good
starting point?
peace...
michael quinn }{ 61+63327394 }{
http://athene.mit.csu.edu.au/~91137277/index.html
site of the week : http://www.wpi.edu/ftp/starwars/
song of the week : NPG-"Count the Days"
...are u still saving our planet ???
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Date: Wed, 24 Apr 1996 11:32:40 -0400
To: math-history-list@maa.org
From: s-kutler@sjca.edu (Samuel S. Kutler)
Subject: Re: complex numbers
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Friends:
I too would like to see a good source for the history of complex numbers.
Here is one reason why. I asked Barry Mazur whether or not he was going to
give his students in a complex variable course that he was teaching the
discovery by Gauss of which regular polygons can be constructed by
euclidean means. Instead of answering my question, he expressed his wonder
about why Euler didn't discover it first. Could it be as simple as that
Euler didn't represent the complex numbers geometrically & hence didn't
investigate the geometric meaning of the cyclotomic equation?
Best wishes,
Sam
>can anyone recommend any good solid sources on the history of complex
>numbers. im looking at covering their development from the work done on
>cubics by the ancient greeks, thru cardano until the current notation and
>methods we use today. anyone done any work in this field or know of a good
>starting point?
>
>peace...
>michael quinn }{ 61+63327394 }{
>http://athene.mit.csu.edu.au/~91137277/index.html
>site of the week : http://www.wpi.edu/ftp/starwars/
>song of the week : NPG-"Count the Days"
> ...are u still saving our planet ???
Date: Wed, 24 Apr 1996 12:55:21 -0400
From: JUDITH GRABINER
Subject: Quipus
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Dear Colleagues,
We've all learned a lot about Inca quipus from the work in the
'80s of Marcia and Robert Ascher.
Is there anything new and important since then that I should
be aware of?
Thanks
--Judy Grabiner
From: "Lingard, David"
To: math-history-list
Subject: The Goldbach conjecture
Date: Tue, 23 Apr 1996 22:00:00 -0400
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Following Wiles' proof of Fermat's last theorem, some of my students were
asking what's next ?, ie. what remains to be proved ?!
My initial response was ' Well I don't think anyone has proved Goldbach's
conjecture yet ' ( Every even number greater than two is the sum of two
primes. )
However, thinking about this afterwards I realised that I wasn't sure, and
although the few references that I've checked since ( eg. Keith Devlin's '
The New Golden Age', page 7 ) confirm my original assertion, I wasn't sure
where to turn for an authoritative and uptodate position.
Perhaps someone has shown it is unproveable ( ugh ); perhaps another Wiles
is beavering away on it in secrecy ? Have there been recent ( failed)
attempts to find a proof ? ( not by computer / exhaustion please ! )
Can anyone clarify the picture for me and others here ?
Thanks in advance if so.
David Lingard
( D.Lingard@shu.ac.uk )
Date: Wed, 24 Apr 1996 19:13:34 -0400
Illegal-Object: Syntax error in From: address found on math.maa.org:
From: William G.Dubuque
^ ^-illegal period in phrase
\-phrases containing '.' must be quoted
From:
To: D.Lingard@shu.ac.uk
Cc: math-history-list@maa.org, math-fun@cs.arizona.edu,
wgd@martigny.ai.mit.edu
In-Reply-To: "Lingard, David"'s message of Tue, 23 Apr 1996 22:00:00 -0400 <317E7039@cerberus.shu.ac.uk>
Subject: The Goldbach conjecture
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>From: "Lingard, David"
>Date: Tue, 23 Apr 1996 22:00:00 -0400
>To: math-history-list
>
>Following Wiles' proof of Fermat's last theorem, some of my students were
>asking what's next ?, ie. what remains to be proved ?!
>
>My initial response was ' Well I don't think anyone has proved Goldbach's
>conjecture yet ' ( Every even number greater than two is the sum of two
>primes. )
>
>However, thinking about this afterwards I realised that I wasn't sure, and
>although the few references that I've checked since ( eg. Keith Devlin's '
>The New Golden Age', page 7 ) confirm my original assertion, I wasn't sure
>where to turn for an authoritative and uptodate position.
> ...
>Can anyone clarify the picture for me and others here ?
Goldbach's Conjecture is still open. The standard reference for
Unsolved Problems in Number Theory is Guy's book of this title,
now in its 2nd (1994) edition. Also very stimulating is Shanks'
"Solved and Unsolved Problems in Number Theory".
>Perhaps someone has shown it is unproveable ( ugh );
Keep in mind that any universal statement undecidable in Peano
Arithmetic is necessarily true. Thus a proof of undecidablity
of Goldbach's conjecture implies its truth -- in contrast to
non-universal statements, which may be undecidable and either
true or false (e.g. the Twin Prime Conjecture). Why? Recall that
a universal statement S is a formula all of whose free variables
are in the scope of universal quantifiers, i.e. whose prenex
normal form is S = (for all) x1,...,xn P(x1,...,xn). If such
a statement were false, then its negation -S would be a true
existential statement. But true existential statements are always
provable (this is known as Sigma-1 completeness). Thus
S false => -S provable => S decidable
or, contrapositively,
S undecidable => S true.
This is a simple result that you will find in any good first
course in logic. It is also discussed in many popular accounts,
e.g. Hofstadter's "Godel, Escher, and Bach", although you would
do better to consult Hofstadter's logical source, namely
DeLong's "A Profile of Mathematical Logic", or even further,
Kemeny's paper "Undecidable Problems of Elementary Number
Theory", Math. Annalen, 135 (1958) 160-169, which discusses how
Henkin's completeness theorem lends further intuition in these
areas when one employs nonstandard models.
In fact one can go much further in applying logic to study such
problems and this has been one of the main focuses of
contemporary model theory and proof theory. One need only consult
the long series of papers by George Kreisel that touch on many
related issues. For example, one celebrated conjecture of Kreisel
states that if there is a uniform bound to the lengths of shortest
proofs of instances of S(n), then its universal generalization,
(for all) n S(n), is necessarily provable (in Peano Arithmetic).
For an elementary discussion see J. Dawson, "The Godel
Incompleteness Theorem from a Length of Proof Perspective",
Amer. Math. Monthly, Nov. 1979, 740-747. Kreisel's conjecture
was settled in the affirmative circa 1988 by M. Baaz, thus
resolving one of the major open problems in proof theory.
>perhaps another Wiles is beavering away on it in secrecy ?
>Have there been recent ( failed) attempts to find a proof ?
>( not by computer / exhaustion please ! )
There is a mathematician named H. A. Pogorzelski who claims to have
proved Goldbach's Conjecture, see the message below. Since I sent
that message I searched Math Reviews for further work of Pogorzelski
and found that he has published three volumes of a series entitled
"Foundations of Semiological Theory of Numbers" (Univ. of Maine at
Orono) 1982, 1985, 1988, MR 84g:03019, 86k:03003, 89k:03001. I've
found no opionions in print by any respected number theorist as to
the validity of his work, so perhaps Shanks' statement below still
holds true, i.e. that most number theorists will not accept
Pogorzelski's claimed 'proof' of Goldbach's Conjecture.
-Bill
From: Bill Dubuque
To: math-fun@cs.arizona.edu
Subject: Goldbach Conjecture: proved by Pogorzelski?
H. A. Pogorzelski, in the prestigious Crelle's Journal, 292, 1977, 1-12,
claims to have proved the Goldbach Conjecture based upon:
The Consistency Hypothesis
The Extended Wittgenstein Thesis
Church's Thesis
This is mentioned on p. 222 of Shanks' "Solved and Unsolved Problems
in Number Theory", where Shanks remarks "It seems unlikely that (most)
number theorists will accept this as proof of the Goldbach Conjecture
but perhaps we should wait for the dust to settle before we attempt
a final assessment".
Surely after twenty years the dust must have settled.
Does anyone know if this proof is accepted by any number theorists?
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Date: Thu, 25 Apr 1996 00:03:51 -0400
To: math-history-list@maa.org
From: michael quinn <91137277@groucho.mit.csu.edu.au>
Subject: complex numbers II
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dear friends..
following my earlier posting, i felt i should explain my situation a little
bit. i am preparing a major work for my course "the history and survey of
mathematics". i have chosen as my major topic the history and development
of complex numbers. i chose this area because i had a little work on it
last year in a curriculum subject.
i thought i might put up the work i did last year - it was a brief time line
of complex numbers. you can find it off my home page [below]. it should be
there by the time u get this posting. if anyone has any comments please
forward them to me.
thanks.
peace...
michael quinn }{ 61+63327394 }{
http://athene.mit.csu.edu.au/~91137277/index.html
site of the week : http://www.wpi.edu/ftp/starwars/
song of the week : NPG-"Count the Days"
...are u still saving our planet ???
Last modified: Thu Jan 8 15:48:06 MET 1998