The Problem Pit: The Pit of Egyptian Fractions

 

The Pit of Egyptian Fractions: Sploding Rational Numbers Into Unit Fractions

There are problems that hide their teeth. They grin at you like puzzles for children, then drag you into quicksand. Egyptian fractions are one of those problems.

The rule is simple:

Problem. Write any rational number $\tfrac{p}{q}$ as a sum of distinct unit fractions:

$$\frac{p}{q} = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_k}, \quad a_i \in \mathbb{Z}^+ \text{ distinct}.$$

The Egyptians did this thousands of years ago. They refused to write $\tfrac{3}{4}$ as “three-fourths.” Instead, they would say:

$$\frac{3}{4} = \frac{1}{2} + \frac{1}{4}.$$

At first, I thought: how charming, how elementary. Surely this would be the lightest kind of math to play with.

Instead, it became a pit.

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Step 1: The Honeymoon Phase

I began with easy cases:

• $\tfrac{2}{3} = \tfrac{1}{2} + \tfrac{1}{6}.$

• $\tfrac{3}{4} = \tfrac{1}{2} + \tfrac{1}{4}.$

• $\tfrac{5}{6} = \tfrac{1}{2} + \tfrac{1}{3}.$

So clean! I imagined there must be a simple formula for everything.

Then $\tfrac{4}{5}$:

$$\frac{4}{5} = \frac{1}{2} + \frac{1}{4} + \frac{1}{20}.$$

Not bad. Slightly more work, but still tidy.

I was lulled into confidence.

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Step 2: The Greedy Algorithm

The classic method is the greedy algorithm:

1. Given $\tfrac{p}{q}$, pick the largest unit fraction $\tfrac{1}{n}$ such that $\tfrac{1}{n} \leq \tfrac{p}{q}$.

2. Subtract it.

3. Repeat with the remainder.

It always works. That sounds comforting. But in practice, it’s a slippery rope down the pit.

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Example A: $\tfrac{5}{121}$

• Largest unit fraction ≤ $\tfrac{5}{121} \approx 0.0413$ is $\tfrac{1}{25} = 0.04$.

Subtract:

$$\frac{5}{121} - \frac{1}{25} = \frac{5\cdot25 - 121}{3025} = \frac{125 - 121}{3025} = \frac{4}{3025}.$$

• Now largest ≤ $\tfrac{4}{3025}$ is $\tfrac{1}{757}$.

Subtract:

$$\frac{4}{3025} - \frac{1}{757} = \frac{4 \cdot 757 - 3025}{3025 \cdot 757} = \frac{3028 - 3025}{2292925} = \frac{3}{2292925}.$$

• Next: largest ≤ $\tfrac{3}{2292925}$ is $\tfrac{1}{764309}$.

Subtract:

$$\frac{3}{2292925} - \frac{1}{764309} = \frac{3\cdot764309 - 2292925}{2292925 \cdot 764309} = \frac{2292927 - 2292925}{1753637943825} = \frac{2}{1753637943825}.$$

• Next: largest ≤ $\tfrac{2}{1753637943825}$ is $\tfrac{1}{876818971912.5}$.

And so on…

What started as $\tfrac{5}{121}$ quickly exploded into denominators in the billions. The fraction itself wasn’t exotic. But the greedy expansion became grotesque.

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Example B: $\tfrac{3}{7}$

Greedy steps:

• Largest ≤ $\tfrac{3}{7} \approx 0.428$ is $\tfrac{1}{3} \approx 0.333$.

Subtract:

$$\frac{3}{7} - \frac{1}{3} = \frac{9 - 7}{21} = \frac{2}{21}.$$

• Next: largest ≤ $\tfrac{2}{21}$ is $\tfrac{1}{11}$.

Subtract:

$$\frac{2}{21} - \frac{1}{11} = \frac{22 - 21}{231} = \frac{1}{231}.$$

So:

$$\frac{3}{7} = \frac{1}{3} + \frac{1}{11} + \frac{1}{231}.$$

Neat! This one ends quickly. The greedy method feels like magic here.

But you never know which way it will go: tidy or monstrous. That unpredictability is part of the pit.

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Step 3: False Hope in Patterns

I started wondering: maybe there are formulas?

Yes, sometimes:

• $\tfrac{2}{n} = \frac{1}{\lceil n/2 \rceil} + \frac{1}{n \cdot \lceil n/2 \rceil}.$

• $\tfrac{3}{n}$ has known formulas too.

But patterns fall apart fast. For instance:

$$\frac{4}{17} = \frac{1}{5} + \frac{1}{30} + \frac{1}{510}.$$

Looks systematic. Then try $\tfrac{4}{19}$… and you’re in chaos.

I scribbled pages of attempts, trying to find clean rules. Nothing stuck.

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Step 4: The Live Wires — Open Problems

That’s when I realized this wasn’t just a playground. Egyptian fractions touch open research.

1. The 2/n Problem (still unsolved):
   For every integer $n$, can $\tfrac{2}{n}$ be written as a sum of three unit fractions?
   Verified for vast ranges, but no proof.

2. Erdős–Graham Conjecture (proved in 2003 by Erdős–Straus–Spencer team):
   If you partition the integers into finitely many classes, one class contains a finite subset whose reciprocals sum to 1. Egyptian fractions everywhere, hiding in Ramsey theory.

3. Denominator Growth:
   How fast can denominators blow up in the worst case? The greedy algorithm guarantees a solution, but at the cost of numbers so huge they look absurd.

4. Efficiency:
   What’s the minimal number of terms needed for a given $\tfrac{p}{q}$? Even simple fractions defy quick answers.

These are not solved by clever tricks. They’re still open territory.

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Step 5: Climbing Out (Sort Of)

By the end of my descent, I had learned:

• The greedy algorithm always finds a decomposition — but often drags you into enormous denominators.

• Some fractions collapse beautifully, others expand hideously.

• Egyptian fractions aren’t just historical curiosities; they’re tied to deep modern number theory.

The problem looks like a toy. In reality, it’s a pit that connects elementary arithmetic with unsolved conjectures.

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Takeaway

What drew me into this pit was the illusion of simplicity. It looked like a problem about splitting candy bars. Instead, I met the Hydra of denominators, the chaos of greedy expansions, and open problems that mathematicians are still wrestling with.

That’s what The Problem Pit is all about: stepping into traps you didn’t know were traps.

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