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Let $\frac{a}{b}$ and $\frac{p}{q}$ be rational numbers in the interval $\left(0,1\right)$ such that $\frac{a}{b}+\frac{p}{q}<1$, and such that: $$\frac{a}{b} = \frac{1}{u_{1}}+\cdots+\frac{1}{u_{M}}$$ $$\frac{p}{q} = \frac{1}{v_{1}}+\cdots+\frac{1}{v_{N}}$$ are Egyptian fraction representations of minimal length, where, for any distinct $j,k$, we have that $u_{j}\neq u_{k}$, $v_{j}\neq v_{k}$, and that none of the $u_{j}$s is a $v_{k}$, and vice-versa.

I have two questions:

I. Is: $$\frac{1}{u_{1}}+\cdots+\frac{1}{u_{M}}+\frac{1}{v_{1}}+\cdots+\frac{1}{v_{N}}$$ necessarily a minimal-length Egyptian fraction for $\frac{a}{b}+\frac{p}{q}$?

II. In case (I) is false, will my conclusion hold if I suppose further that all the $u_{m}$s and $v_{n}$s are positive integer powers of a fixed integer $\lambda\geq2$?

MCS
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1 Answers1

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I

No. In fact, for any integer $n\ge 2$, $$\frac1{n(n-1)}+\frac1n=\frac1{n-1}$$

ajotatxe
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  • @hardmath Denominators must be different. – ajotatxe Sep 07 '17 at 23:20
  • Yeah. All of the denominators are different. – MCS Sep 07 '17 at 23:55
  • If all the denominators are different powers of $\lambda \ge 2$, then the "Egyptian fractions" considered are radix $\lambda$ expansions that terminate, and these are unique (so the minimality of any of these is automatic). – hardmath Sep 07 '17 at 23:59
  • Huzzah! So, the radix expansions do actually have the same length as the minimal egyptian fraction length? – MCS Sep 08 '17 at 00:09
  • @MCS: Yes, your intuition did not fail you there. Of course by "radix expansions ... length" here is meant the number of ones (equiv. nonzero place values) appearing in the expansion. – hardmath Sep 08 '17 at 00:30