I was doing a bit of reading about Egyptian Fractions. For those not familiar with the concept, an Egyptian Fraction is a sum of distinct unit fractions, or reciprocals of positive integers.

The text that I read argued that since the number $1$ has a single egyptian fraction representation, it has infinitely many. This is because if $$1=\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}$$ where $a_i\lt a_{i+1}$, one can make the substitution $$1=\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\cdot 1$$ $$1=\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\bigg(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\bigg)$$ $$1=\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_na_1}+\frac{1}{a_na_2}+...+\frac{1}{a_n^2}$$ However, I was wondering if there exist infinitely many disjoint egyptian fractions for $1$ - that is, egyptian fractions that do not share any unit fractions.

Any ideas?



Note that every rational has an Egyptian decomposition (by the greedy algorithm if nothing else).

Suppose you have a collection of disjoint decompositions of $1$. Here's how to construct a new one, disjoint from the collection you have:

Let $N$ be larger than every denominator in your collection. Consider the sum $$H_{N,i}=\frac 1N+\frac 1{N+1}+\frac 1{N+i}$$ where $i$ is defined so that $$H_{N,i}<1≤H_{N,i+1}$$

Note that the divergence of the Harmonic series implies the existence of $i$.

If $H_{N,i+1}=1$ then use that as your decomposition. Otherwise we can assume it is $>1$. Then consider an Egyptian decomposition of $1-H_{N,i}$

It's clear that no fraction appearing in that can have a denominator less than or equal to $N+i$, and that is enough to prove what you want.

  • $\begingroup$ Ah, and this will always work b/c of the divergence of the harmonic series. Gotcha. $\endgroup$ – Frpzzd Oct 3 '17 at 23:58
  • $\begingroup$ Yes, that's the point. $\endgroup$ – lulu Oct 4 '17 at 0:00

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