Nothing Over a Thousand

The Rhind papyrus opens with a table nobody asked for. It is the larger of the two great mathematical papyri of ancient Egypt, five metres of it, copied about 1550 BC from a document three centuries older, and before a single problem is posed it lays out forty-nine lines of arithmetic. Every one of those lines had a shorter answer available. The scribe wrote the short answer four times.

Egyptian arithmetic has one moving part. To multiply, you double.

You want twelve times seventeen. You write 1 beside 17, then keep doubling both columns: 2 beside 34, 4 beside 68, 8 beside 136. Then you tick the left-hand rows that add to 12, which is 4 and 8, and add the right-hand entries beside them. That is the whole of Egyptian multiplication, and it works for every pair of whole numbers because every whole number is a sum of distinct powers of two. More than three thousand years before anyone wrote the word binary, the arithmetic of the Nile ran on it.

Instrument one · the ladder

Now try it with a fraction. Egyptian notation has no numerator, with one exception we will come to: a fraction is a whole number with a mark over it, and it means one part in that many. To double a part, you halve the number underneath. Two lots of one eleventh is one and one half elevenths, which is not a thing the notation can say, so you write it as one sixth plus one sixty-sixth, and both of those are things it can say.

When the number underneath is even the halving is free. Two lots of one tenth is one fifth, one stroke, no thought. When it is odd, the whole system stops. There is exactly one hard step in Egyptian arithmetic and this is it: 2 × 1/n for odd n.

The Rhind papyrus deals with it by refusing to have it. Before problem one, the scribe lays out the answer for every odd n from 5 to 101. It is a lookup table for the inner loop of the whole system.

The table

Every line is a claim that some unit fractions add to exactly 2/n. Click one and it is checked in front of you, in exact whole-number arithmetic, in your browser. The greedy column is explained further down; for now, notice how much larger its numbers get.

Instrument two · the recto table, checked

2/nthe papyrustermslargestthe short answerlargest
Select a row to check it.

The short answer

There is a rule that solves every line of this table in one move, and it fits in a sentence: take the largest unit fraction that still fits, then write down whatever is left. For 2/n with n odd, the largest unit fraction that fits is 1/((n+1)/2), and the remainder is

2/n − 2/(n+1) = 2/(n(n+1)) = 1/(n(n+1)/2)

which is a unit fraction, because n+1 is even. So the answer is always two terms, always, for every odd n there has ever been. We checked all forty-nine: two terms every time.

The scribe used that answer for n = 5, 7, 11 and 23. The other forty-five times he wrote something longer.

The reason is visible the moment you look at the second denominator. For 2/13 the short answer is 1/7 + 1/91. For 2/29 it is 1/15 + 1/435. For 2/101, the last line of the table, it is

2/101 = 1/51 + 1/5151

and one fifty-one-hundred-and-fifty-first of a loaf is not a quantity anyone was ever going to hand to anyone. Across the forty-nine lines the short answer breaks a thousand twenty-nine times, starting at n = 45. The papyrus's own largest denominator, anywhere in the table, is 890.

A caution about what that does and does not show. The two-term identity is trivial to us and the papyrus uses it four times, so it was inside the scribe's reach at least in those cases. Whether he held it as a general rule and declined it, or simply never worked in that direction, is not something the document tells us. What is measurable is the shape of the result: a systematic preference for more terms and smaller numbers, over fewer terms and larger ones.

The story you have probably heard

There is a famous explanation for why Egypt wrote fractions this way, and it goes like this. Nine loaves, ten men. The modern answer gives everybody nine tenths, which means cutting loaves into tenths and handing the last man a heap of scraps. The Egyptian answer gives every man two thirds, a fifth and a thirtieth, so everybody gets the same three pieces, and justice is not only done but seen to be done. That last phrase is the story's own, and it travels with it everywhere; we could not confirm those exact words in Gillings attached to this problem.

It is a good story with a real pedigree. Otto Neugebauer put the practical version in a lecture at Kiel in 1926: "how is that supposed to be done in reality? Divide each loaf of bread into 10 pieces and then give each man 7 pieces? The Egyptian result is much more convenient." Richard Gillings gave it the fairness form in 1972, writing of this exact problem that "every man gets exactly the same number of pieces and exactly the same-sized pieces."

That last sentence is not true, and you can see it is not true without knowing anything about Egypt. Ten men each need a piece of two thirds. A loaf is one whole, and two thirds plus two thirds is more than one, so a single loaf can yield at most one two-thirds piece. There are nine loaves. Nine pieces, ten men. At least one man's two thirds has to arrive in more than one piece, so the shares cannot all be alike.

The standard 1987 British Museum edition of the papyrus says as much in one sentence, on these very problems. Robins and Shute: "It is difficult to imagine loaves being divided into thirtieths as a practical procedure, but there are, in fact, documentary records of fractions of loaves being issued to temple employees on the basis of so many loaves being available for a certain number of men." The bread is real. The knife is not. What those records show is an entitlement written down, and an entitlement does not have to be cuttable.

And the papyrus never claims otherwise, because the papyrus gives no reason at all. Peet noticed this in 1923: in the bread problems "the problem is never solved at all: the question is asked, the answer is at once stated, and a proof is given which consists in multiplying this by 10." The one place an editor does record a motive points somewhere else entirely. On problem 4, seven loaves among ten men, Chace notes that the answer "might have been written 1/2 1/5, but the Egyptian thinks of 2/3 as the largest fraction that there is and prefers to use it wherever he can." That is a preference about writing, not about bread.

So the loaf story is not internet invention, and it is not evidence either. It is a twentieth-century reading, argued by serious people, resting on no Egyptian statement, and its most-repeated example fails on the arithmetic. The rest of this page stays with what can be counted.

How the lines were built

There is a reconstruction of the method, found by Friedrich Hultsch in 1895 and found again by Evert Bruins in 1945, and it is the kind of thing you can only really believe by driving it. Pick a first denominator A between n/2 and n. Subtract:

2/n − 1/A = (2A − n) / nA

Now the trick. If you can write the numerator 2A − n as a sum of distinct divisors of A, then each divisor d you use contributes d/nA, which is 1/(n × A/d), and A/d is a whole number because d divides A. Every piece lands on a unit fraction. That is the whole method.

So the scribe's degrees of freedom are two: which A to open with, and which divisors to spend. Below, both are yours. The number in red is the one you have to build, and it is in red because that is the colour the scribe used: the auxiliary numbers in the papyrus are written in red ink, the titles too, and everything else in black.

Instrument three · the scribe's method, operable

Divisors of A. Tick a set that adds to the number in red.

divisors chosen add to
0
need
0
your line
·

assemble a line

Run it down the table and something becomes clear: the method reaches almost everything. It reproduces the papyrus's line exactly for forty-six of the forty-nine, and the three it misses are the three that scholarship has always singled out.

Two of the three come from a rule for products. If n = pq with both factors odd, then p + q is even and

2/pq = 1/(p·(p+q)/2) + 1/(q·(p+q)/2)

which gives 2/35 = 1/30 + 1/42 from 5 and 7, and 2/91 = 1/70 + 1/130 from 7 and 13. Those are exactly the papyrus's lines, and exactly the two the Hultsch and Bruins method cannot reach. The papyrus seems to know that 2/35 is the odd one out: it carries an additional line of explanation that appears nowhere else in the table, and it is on that line that Gillings hung his whole account of how the table was built.

The third is the last line, and it is the strangest thing in the table. The scribe writes

2/101 = 1/101 + 1/202 + 1/303 + 1/606

which looks like a shrug. It is not. It is the identity 1 + 1/2 + 1/3 + 1/6 = 2 divided through by n, and it works for absolutely any n at all. He kept it in his pocket for the whole table and spent it once.

He spent it once because at n = 101 he had run out of room. We enumerated every way to write 2/101 as at most four distinct unit fractions with no denominator above a thousand. There is exactly one. It is his.

The two families, checked against every member

One rule the papyrus does state out loud, and it is the only one. Problem 61B is about taking two thirds of a part: make its 2 times and its 6 times, it says, and then, unusually for this document, it generalises, adding that one does the same for every odd fraction that may occur. Two thirds of 1/n is 2/3n, so what that sentence says is 2/3n = 1/2n + 1/6n, which is the rule for every line of the table whose denominator is divisible by three. It holds for all sixteen of them without a single exception.

The matching pattern for five, 2/5k = 1/3k + 1/15k, does not. It holds at n = 5, 25, 65 and 85, and fails at 35, 55 and 95, which is four out of seven. Peet listed those same three exceptions in 1923, working by hand. It is worth noticing that the four lines where the five-rule holds are also four of the only five lines in the whole table that open on an odd denominator. The fifth is 2/101. Every other line in the table opens on an even number.

What was he choosing for?

That is the question the table has been asked since Chace published it in 1927, and the usual method of answering has been to name a criterion and show a few lines that obey it. Enumeration lets us do better: build every alternative the scribe could have written, then score each proposed rule against all forty-nine lines at once and report where it fails.

The envelope is the papyrus's own: at most four distinct unit fractions, no denominator above a thousand. Inside it there are 27,866 decompositions across the forty-nine lines. The papyrus's own line is inside the envelope every time, which is a check on the envelope rather than on the scribe.

That total is worth a footnote of its own, because the count has been argued about. Gillings ran the enumeration in the nineteen-sixties on an English Electric KDF-9 and reported 22,295. Bruckheimer and Salomon put it at about 28,000 in 1977, and Gillings replied in 1978. Neither paper was reachable from here, so we have the dispute at second hand, from Abdulaziz (2008) and from Robins and Shute, who confirm what was being counted: Gillings, "followed by Bruckheimer and Salomon, ran computer programs aiming to obtain a complete set of such expressions". Our own count is 27,866 over the forty-nine lines, or 27,932 if a row for 2/3 is included, and 27,852 under the strict reading of Gillings' own precept that no denominator be "as large as 1,000". All four numbers sit next to Bruckheimer and Salomon and a long way from 22,295. We cannot be certain the three envelopes are identical, so read this as a third count rather than a verdict, and note that ours is stated in full and reproducible in one command.

Two ways of scoring, and only the second is a prediction. A rule is consistent with a line if the papyrus's answer is among the rule's winners. A rule predicts a line only if it has exactly one winner and that winner is the papyrus's. A rule that says "fewest terms" is consistent with a great many lines and predicts almost none, because there are usually several two-term answers and it cannot tell you which.

Instrument four · fifteen rules, all forty-nine lines

rulepredictsconsistentfirst miss
Select a rule to see where it first parts company with the papyrus.

One of those fifteen is not our invention. Robins and Shute state the criteria plainly: "the smallest possible number of fractions should appear in the expression for 2/n, and fractional numbers that were odd or needlessly large should be avoided". Fewest terms, then small, then even. Scored against all forty-nine lines, the rules closest to that description sit at 24, 14 and 12 predictions, and the hedge in their own sentence, "usually, though not always", is doing exactly the work the numbers make visible.

Nothing gets past half. The best of the fifteen, "fewest terms, then the smallest largest denominator", predicts 24 of the 49. Adding a hard denominator ceiling and fitting it does not rescue it: sweeping the ceiling from 60 to 1000, the best score is 27 of 49, at a ceiling of 800, and the curve is flat and unconvincing on both sides. We also fitted the simplest possible generative model, "take the first opening denominator that works, subject to never writing a number above C", across the same range of C. It tops out at 19 of 49.

Gillings' canon, made executable

The most-cited attempt at the rule is a set of five precepts. They are worth reading in Gillings' own words, because their trouble is visible in them:

1. Of the possible equalities, those with the smaller numbers are preferred, but none as large as 1,000. 2. An equality of two terms is preferred to one of three terms, and one of three terms to one of four terms; but an equality of more than four terms is never to be used. 3. The unit fractions are set down in descending order of magnitude, that is, smaller numbers come first, but never the same number twice. 4. The smallness of the first number is the main consideration, but the scribe will accept a slightly larger first number if it will greatly reduce the last number. 5. Even numbers are preferred to odd numbers, even though they might be larger, and even though the number of terms might thereby be increased.

Precepts 1 and 3 are constraints, and this page has been working inside them all along. Precepts 2, 4 and 5 are objectives, and they do not agree: 5 explicitly overrides 2, 4 trades against both, and neither "slightly" nor "greatly" is a quantity. So the canon does not compose into a single rule, and to test it you must first decide what it says. Here are four readings of it, and, for contrast, the inversion of precept 4 proposed by Abdulaziz in 2008.

readingpredictsconsistent
precept 5, then 2, then 416/4916/49
precept 2, then 4, then the last denominator14/4914/49
precept 4, then 212/4912/49
precept 5 counted rather than absolute, then 2, then 43/493/49
Abdulaziz's inversion: the largest opening denominator, then precept 21/491/49

Sixteen of forty-nine is the best the canon does under any reading we could build from it. Imhausen's verdict on Gillings in 2016 was that "although these rules explain some of the choices, overall it is not possible to predict the answer that is found in the table based on these rules". That is now a number.

The fourth precept is the one that has been fought over, and it can be measured directly. Gillings says the scribe wants the smallest opening denominator he can get away with. Bruins said the precept is violated repeatedly, and Abdulaziz went further, writing that the scribe "chose, as always, the one with the largest a". For each line we listed every opening denominator the Hultsch and Bruins method could have used under the ceiling, and asked where the scribe's actual choice sits in that list. It is the smallest workable start in 16 lines, the largest in none, and on average it sits 13 per cent of the way up. Read as a claim about the full list of workable starts, Gillings has the direction right and Abdulaziz's inversion does not survive: the scribe stays near the bottom. Abdulaziz makes his remark while comparing a handful of candidates he has already narrowed by other criteria, so this tests a stronger reading of it than he may have meant, and it should be read that way.

So the honest result of the whole search is negative, and it is worth saying plainly: no rule we tested reproduces the table. The lines are reachable, all forty-nine of them, by three named constructions with nothing left over. But which line to write, given the choice, is not the output of an objective function.

You can watch that operate on a single line. Take 2/13. The short answer is 1/7 + 1/91: two terms, largest denominator 91. There is a three-term answer with a smaller largest denominator, 1/10 + 1/26 + 1/65. The scribe wrote neither. He wrote 1/8 + 1/52 + 1/104, which is longer than the first and coarser than the second, and whose only visible virtue is that every number in it is even. Then, four lines later, at 2/25, he opened with an odd number.

Instrument five · the whole field, for one line

decompositions in the envelope
·
the papyrus's line
·
smallest largest denominator available
·
did he take it?
·
Choose a line and press the button. Everything below is computed here, now, in your browser.

And if you would rather not take the aggregate on trust, run it. The button below repeats the entire enumeration, all forty-nine lines and all 27,866 decompositions, in this page, and then compares its own total against the number printed above. It takes about half a minute and it will make your laptop warm.

Instrument six · re-run the whole search here

lines done
0 / 49
decompositions found here
0
against the published total
·

Still open

The question the table poses did not close. Every fraction can be written as a sum of unit fractions, and the greedy rule proves it constructively: the numerator of the remainder strictly decreases at every step, so the process must stop. Fibonacci gave a procedure with the greedy rule at its heart in 1202, and Sylvester published the argument in 1880.

But ask for a bound and the ground goes soft. In 1948 Erdős and Straus conjectured that for every n greater than 1,

4/n = 1/x + 1/y + 1/z

has a solution in positive whole numbers. Three terms, always. It has been verified up to 1017 by Salez in 2014, nobody has proved it, and it is the same question the scribe was answering by hand: given a numerator you cannot write, how few pieces will do? He needed at most four for a numerator of two. Whether three always suffice for a numerator of four is open.

Instrument seven · find the three pieces

Enter any n above 1. The search opens on a first denominator and then solves the remaining two-term problem exactly by factoring, so it is finding the answer, not looking it up.

The apparatus

What the source says, and where it disagrees with itself

The table transcribed here is the recto of the Rhind Mathematical Papyrus as printed in Chace, Bull and Manning, The Rhind Mathematical Papyrus, Volume I (1927), pages 37 and 38, under the heading "Division of 2 by Odd Numbers". That edition is in the public domain and its page scans were read directly. Every one of the forty-nine lines was then cross-checked against four independent secondary tables, and each was verified to sum to 2/n in exact rational arithmetic.

Two of those secondary tables contain arithmetic errors, which is a small demonstration of why the check is worth running. The Clark University table prints 2/45 = 1/30 + 1/60 (which equals 1/20) and 2/67 = 1/40 + 1/355 + 1/536 (which is not 2/67). The University at Buffalo table prints 2/35 = 1/25 + 1/30 + 1/42 (which is nearly double 2/35). None of the three survives contact with a common denominator, and all three are against Chace, so all three were rejected as page typos rather than variant readings.

There is one real disagreement, and it is about the first line. Several widely read modern pages give the table's opening entry as 2/3 = 1/2 + 1/6. Chace does not. His Answer column for divisor 3 reads simply 2/3, because two thirds was itself a primitive of Egyptian arithmetic, the one non-unit fraction the papyrus uses, written with a sign of its own, and so needed no decomposition. MacTutor's account agrees, describing the doubling table as running from 5 to 101. We follow the edition we read. That is why every count on this page is out of forty-nine and not fifty, and why the first line here is 2/5.

What is attested and what is reconstruction

The table is attested: it is on the papyrus. The method is not. The papyrus shows auxiliary numbers for some lines but never states a rule, and the divisor procedure driven in instrument three is a reconstruction, published by Hultsch in 1895 and independently by Bruins in 1945. It was contested from the start. Neugebauer's objection in 1926 was that the resulting system of rules "certainly cannot be memorized by any modern European, and therefore also probably not by an ancient Egyptian". Peet, in 1923, had already put the deeper warning better than we can:

"Even could we show that all the results corresponded to a formula this would not prove that the Egyptian worked by this formula… The Egyptian, far from employing a formula, probably had no conception that the resolution could be accomplished by a single method in all cases. His method was undoubtedly that of trial."

That is the right frame for everything above. The standing of the divisor method on this page is exactly what the page demonstrates and no more: it reaches 46 of the 49 lines, and two further identities reach the other three. That is a statement about arithmetic, not about a man's intentions. It is also, read the other way, consistent with Peet: a table built by trial can still land inside a formula's reach without the formula ever having existed.

The same caution applies harder to the negative result. "No rule we tested reproduces the table" is a claim about fifteen rules plus five readings of Gillings' canon, all listed above and coded in the open. It is not a proof that no rule does. Someone may yet find one, and if they do, the enumeration here is the thing to test it against.

What we did not check

We did not read Bruckheimer and Salomon (1977) or Gillings' reply (1978); both were behind paywalls, and the dispute over the count is reported here at second hand. We have not seen the papyrus, or a photograph of the hieratic; the reading of record here is Chace's printed transcription, and the one place it matters, the first line, is flagged above. Gillings' precepts are quoted from his own restatement of them in the Dictionary of Scientific Biography rather than from the 1972 book, which we could not verify page by page.

Robins and Shute (1987) needs its own note, because we have quoted it and we have not read it. The book is print-only and the Internet Archive scan of it is lending-restricted. What is public is the Archive's full-text index over that scan, which returns the matched line of OCR for a phrase query. Every Robins and Shute sentence on this page was recovered that way, and can be recovered again by anyone, one phrase at a time, at openlibrary.org/search/inside.json with q=identifier:rhindmathematica0000robi_h8l4 AND "<phrase>". Two limits come with that. The OCR loses the overbars that mark Egyptian unit fractions, so their printed 2/n arrives as "2n", and we have restored it silently inside the quotation marks. And an absence of hits is weaker evidence than a presence: that index finds no occurrence of "justice", "cutting" or "equitable" anywhere in the book, which is suggestive about what they do not argue, but a search index is not a reading.

The numbers on this page

Reproduce it

Three scripts, no dependencies, exact integer arithmetic throughout, about four minutes in total. The first builds the whole field and scores every rule; the second fits the one-parameter ceiling model; the third counts the field four ways and puts Gillings' canon and precept 4 on the scales.

node research/egyptian-fractions/experiment.mjs node research/egyptian-fractions/fit-ceiling.mjs node research/egyptian-fractions/gillings.mjs

The page itself is checked end to end by verify-nothing-over-a-thousand.mjs, which drives it in a real browser, operates the instruments, and confirms that what is rendered here is what the notebook computes.

Sources

The scribe of the Rhind papyrus names himself in its colophon: copied in regnal year 33 under Awserre, the Hyksos king Apophis, from an ancient copy made in the time of Nimaatre, Amenemhat III; the scribe Ahmose writes this copy. The Lahun fragment UC 32159 carries the same decompositions for the odd numbers up to 21, which suggests the table was standard rather than personal. So it is at least twice as old as its copy, and the taste this page measures may not be his at all. We have called him "the scribe" throughout for that reason.