# Does there exist an operation which partitions any fraction into the sum of the minimum number of unit fractions?

Motivation : I've been interested in finding an operation which partitions a fraction into unit fractions. The following is one of the operations which I've found.

Let's start a rational number $q_0$ which satisfies $0\lt q_0\lt 1$.

Step 1 : Let $i=0$.

Step 2 : Finding an integer $k_i$ which satisfies $\frac{1}{k_i}\le q_i\lt \frac{1}{k_i-1}$, let $q_{i+1}=q_i-\frac{1}{k_i}.$

Step 3 : If $q_{i+1}\not=0$, then go back to step 2 with changing $i$ to $i+1$. If $q_{i+1}=0$, then $\frac{1}{k_0}+\frac1{k_1}+\cdots+\frac1{k_i}$ is what you want.

Though we can show that this operation does end, I'll skip the details.

By the way, my operation does not give the sum of the minimum number of unit fractions.

For example, in my operation, we get $$\frac{19}{88}=\frac15+\frac1{63}+\frac1{27720}.$$ On the other hand, the sum of the minimum number of unit fractions is $$\frac{19}{88}=\frac18+\frac1{11}.$$

I've been finding an operation which partitions any fraction into the sum of the minimum number of unit fractions, but I'm facing difficulty.

Here is my question.

Question : Does there exist an operation which partitions any fraction into the sum of the minimum number of unit fractions?

• When I compute $\frac18+\frac1{11}$ I get $\frac{19}{88}$, not $\frac{11}{88}$. – Henning Makholm Sep 19 '13 at 14:46
• Special cases are known, so for e.g. $p, q$ odd primes, we have $\dfrac{2}{p} = \dfrac{1}{(p+1)/2} + \dfrac{1}{p(p+1)/2}$ and $\dfrac{2}{pq} = \dfrac{2}{p+q}(\dfrac{1}{p} + \dfrac{1}{q})$ etc. – Macavity Sep 19 '13 at 15:07