Harumph. Suppose R = {1, 1/2, 1/3}, and the problem is to rerepresent 3/2? The starting rep is 1 + 1/2. We replace 1 with 1/2 + 1/3 + 1/6, giving 2*1/2 + 1/3 + 1/6. Replacing the 1/2 in turn with 1/4 + 1/6 + 1/12 isn't progress -- we'll never get rid of the duplicated children of the duplicate 2*1/2. Thm: Given any integer N, any rational q can be represented as the finite sum of unit fractions whose denominators are all distinct and >= N. The number of terms doesn't exceed <messy formula>. Proof sketch: Start with 1/N + 1/(N+1) + ..., debiting q until eventually q' < 1/N'. Then follow the usual greedy strategy of subtracting 1/ceiling[1/q']. The first step will terminate becuase sum(1/N) diverges. We can estimate the place of termination with the log estimate for sum(1/N). The numerator and denominator of q' are potentially huge, but there is a formulaic bound, some factorial. The ceiling step terminates because the numerator decreases at each stage. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Marc LeBrun Sent: Mon 9/12/2005 12:56 PM To: math-fun Cc: math-fun; metaweta@gmail.com Subject: Re: [math-fun] Egyptian fractions question Or: Should such an r appear, replace it with {r/2,r/3,r/6}. Lather, rinse, repeat. At 11:35 AM 9/12/2005, Richard Guy wrote:
Yes, since sum 1/n diverges, even though you omit a finitenumber of terms. R.
On Mon, 12 Sep 2005, Mike Stay wrote:
Given a finite set R of unit fractions, is it possible to represent all rationals as Egyptian fractions without using elements of R?
-- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
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