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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?

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Sure.

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.

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  • $\begingroup$ 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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