How this instrument works
Ancient Egyptian mathematics had no notation for a general fraction like 2⁄3 or 5⁄6 — every fraction besides 2⁄3 itself had to be expressed as a sum of DISTINCT unit fractions (fractions with a numerator of 1), each denominator used only once. The greedy algorithm reconstructs one such decomposition: at each step, subtract the largest unit fraction that still fits without overshooting, then repeat on whatever remains.
The largest unit fraction that fits a proper fraction b's-worth is found by rounding b divided by a UP to the next whole number — that rounded value becomes the first term's own denominator. Whatever's left over after subtracting it becomes a new, smaller fraction, and the identical largest-fit search repeats on that remainder to find the second term.
This page reports only the FIRST TWO terms of that expansion. Some fractions resolve cleanly in exactly two terms with nothing left over; others, like 3⁄7, still have a smaller remaining piece after two terms that would need a third (or further) unit fraction to fully account for.
- Enter the fraction's numerator into the Numerator field (it must be smaller than the denominator).
- Enter the denominator into the Denominator field.
- Read First unit fraction's denominator: the largest single unit fraction that fits.
- Read Second unit fraction's denominator: the next largest unit fraction, once the first has been subtracted out.
Worked example — decomposing 2⁄3
For 2⁄3: the largest unit fraction that fits is found by rounding 3÷2=1.5 up to 2, so the first term is 1⁄2. Subtracting it leaves 2⁄3−1⁄2=1⁄6 exactly, so the second term is 1⁄6 — and nothing remains after that. In full: 2⁄3 = 1⁄2 + 1⁄6, a clean two-term decomposition confirmed by adding the two unit fractions back together.
Not every fraction resolves so neatly in two terms. 3⁄7's first term is 1⁄3 (rounding 7÷3=2.33 up to 3), and its second term is 1⁄11 — but 1⁄3+1⁄11 alone doesn't yet reach the full 3⁄7, meaning a smaller third unit fraction, not computed by this page, would still be needed to finish the decomposition exactly.
Questions
What is an Egyptian fraction?
A sum of distinct unit fractions (fractions with numerator 1, each denominator used only once) that together equal a given fraction — the only form ancient Egyptian mathematics had for expressing fractions besides 2⁄3, which had its own dedicated symbol.
How does the greedy algorithm find each term?
At every step, it picks the LARGEST unit fraction that still fits without exceeding what's left, subtracts it, and repeats the same search on the remainder — a simple, repeatable rule, though not always the shortest possible decomposition available.
Why does this page only show two terms?
Some fractions, like 2⁄3 or 5⁄6, resolve completely in exactly two greedy terms; others need three or more to fully account for the original value, so this page's two reported terms are sometimes the complete answer and sometimes just the start of one.
Does the greedy method always give the shortest possible decomposition?
No — it's simple and guaranteed to terminate, but it can sometimes produce more terms, or larger denominators, than a cleverer hand-picked decomposition of the same fraction would.
Why does the fraction have to be proper (numerator smaller than denominator)?
The greedy method as implemented here assumes a value between 0 and 1; a numerator equal to or larger than the denominator would need the whole-number part handled separately first, before this term-by-term unit-fraction search applies.