Can we partition the reciprocals of $\mathbb{N}$ so that each sum equals 1

Question

Let $S = \{1, 1/2,1/3,\dots\}$

Can we find a partition $P$ of $S$ so that each part sums to 1, e.g. $$P_1 = {1}$$

$$P_2 = { 1/2,1/5,1/7,1/10,1/14,1/70}$$

$$P_3 = {1/3,1/4,1/6,1/9,1/12,1/18}$$

$$P_4 = \{... \}$$

I think I could go on, technically, but it got increasingly complex.

Answer 1

The answer is yes!

Given already chosen disjoint subsets $P_1,\dots,P_k$ of $S$, here’s how to choose the next one $P_{k+1}$:

• Let $m$ be the smallest positive integer whose reciprocal is not already in $P_1\cup\cdots\cup P_k$.
• Let $\dfrac1N$ be the minimum element of $P_1\cup\cdots\cup P_k$ (the one with the largest denominator), and let $n$ be the largest integer such that $\displaystyle \sum_{k=N+1}^{N+n} \frac1k < 1 - \frac1m$.
• The standard greedy algorithm for Egyptial fractions will produce a representation of the form $\displaystyle 1 - \frac1m - \sum_{k=N+1}^{N+n} \frac1k$ as $\dfrac1{d_1} + \cdots + \dfrac1{d_\ell}$ where the integers $d_1,\dots,d_\ell$ are distinct and (necessarily) larger than $N+n$.
• Then set $P_{k+1} = \biggl\{\dfrac1m, \dfrac1{N+1},\dfrac1{N+2}, \dots, \dfrac1{N+n}, \dfrac1{d_1}, \dfrac1{d_2}, \dots, \dfrac1{d_\ell}\biggr\}$.

Always including the smallest integer $m$ available ensures that the union of the $P_j$ will equal all of $S$; choosing the other denominators of $P_{k+1}$ to be larger than all the denominators in $P_1\cup\cdots\cup P_k$ ensures that the $P_j$ are pairwise disjoint, so that this is a true partition.