TITLE: *Disjoint* Egyptian Fraction representations of $1$
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: 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 $$H_{N,i}=\frac 1N+\frac 1{N+1}+\cdots +\frac 1{N+i}$$  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.