sums of unit fractions
Michael Barr
rational number is a sum of unit fractions. Can anyone supply a
reference for the fact that arbitrarily large numbers of unit
fractions are required to represent every rational number between 0
and 1? I am sure the fact is known, but I would like to know if the
elegant proof I have just discovered is new.
Arturo Magidin
If you mean, a sum of unit fractions using pairwise distinct
denominators, then the buzzword you should look for is "Egyptian
fractions."
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
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Arturo Magidin
mag...@math.berkeley.edu
Peter Webb
"Michael Barr" <ba...@barrs.org> wrote in message
news:19daab49.0403...@posting.google.com...
.111111111111111111111111111111111111
Bart Goddard
news:19daab49.0403...@posting.google.com:
This paper:
Martin, G. "Dense Egyptian Fractions." Trans. Amer. Math. Soc. 351,
3641-3657, 1999.
shows that you CAN have arbitrarily large numbers of unit fractions
in your representation, but I don't think it shows that this is
_required_. The website
http://mathworld.wolfram.com/EgyptianFraction.html
has several references.
Bart
Virgil
ba...@barrs.org (Michael Barr) wrote:
Every unit fraction is in the interval (0,1), so is required to
represent itself in that interval. Is there something about the question
that I am not understanding?
Gerry Myerson
Virgil <ITSnetNOTcom/vir...@COMCAST.com> wrote:
I took the question to mean, how can you show that for every k
there's a rational that requires more than k unit fractions?
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Gerry Myerson
"Peter Webb" <wabbfamily...@yahoo.com> wrote:
????
It's obvious that the number with 17 ones can be done with 17 unit
fractions - but is it so easy to see that it can't be done with 16?
You don't have to use the greedy method to choose the unit fractions.
Peter Webb
I thought that 0.111111 in base p = 1/p + 1/p^2 + ... 1/p^n obviously can't
be represented with fewer terms ... but even if its true, it isn't obvious.
Sorry
Michael Barr
Your take is correct. Here is the proof. Let X be the space
consisting of the 1/n together with 0. X is compact and so is X^k for
any integer k. There is a map X^k --> [0,k] that adds the coordinates
and the image is compact as is its intersection with [0,1]. The image
consists of all rationals that can be represented as the sum of k or
fewer unit fractions and obviously does not include all the rationals
in that interval.
Herman Rubin
How hard did you try? Consider 1/10 + 1/91 + 1/9990, this
is the sum of at most 7 more unit fractions. If we
consider the 17 series for p=10, the difference is
.00002200002200002088891088891..
Now the five 2's can be done with 5 terms, the repeating
888000888000... with 1 term, and the repeating 91000091...
also with 1 term.
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are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
The World Wide Wade
ba...@barrs.org (Michael Barr) wrote:
> Your take is correct. Here is the proof. Let X be the space
> consisting of the 1/n together with 0. X is compact and so is X^k for
> any integer k. There is a map X^k --> [0,k] that adds the coordinates
> and the image is compact as is its intersection with [0,1]. The image
> consists of all rationals that can be represented as the sum of k or
> fewer unit fractions and obviously does not include all the rationals
> in that interval.
Nice proof.
Gerry Myerson
ba...@barrs.org (Michael Barr) wrote:
Nice. There may be a constructive proof in Johannessen and Sohus,
On unit fractions, Nordisk Mat. Tidskr. 22 (1974) 103-107, 135,
MR 55 #252. The review in Math Reviews says that for all integers
p and k, p / (k p! + 1) requires p unit fractions. The paper is in
Norwegian.
It seems to me the following is true, and can be proved by your
method, but also "directly;" if A and B are bounded sets of reals
with countable closure then A + B has countable closure. Here A + B
means { a + b : a in A, b in B }. The unit fractions result follows.
David Eppstein
It's not as elegant as the compactness argument later in this thread, but
the method described briefly in the "EstimatedTerms" function of
http://www.ics.uci.edu/~eppstein/numth/egypt/Egypt.m can be used to show
that x/(x+1) requires Omega(log log x) terms.
--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer Science
Gerry Myerson
Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
> Nice. There may be a constructive proof in Johannessen and Sohus,
> On unit fractions, Nordisk Mat. Tidskr. 22 (1974) 103-107, 135,
> MR 55 #252. The review in Math Reviews says that for all integers
> p and k, p / (k p! + 1) requires p unit fractions. The paper is in
> Norwegian.
I must have misunderstood the review, as what I have written is
nonsense. E.g., 4/25 = 1/10 + 1/25 + 1/50 shows that it's false
for p = 4, k = 1. In fact, if the Erdos-Straus conjecture is
correct (it states that for every n there exist x, y, z such that
4/n = 1/x + 1/y + 1/z) then p / (k p! + 1) can't require p unit
fractions for any p > 3.