Egyptian Fractions
The Ahmes papyrus is one of the very oldest extant mathematical documents. It was written around 3800 years ago. As I mentioned recently, a large part of it is a table of the values of fractions of the form !!\frac 2n!! for odd integers !!n!!. The Egyptians, at least at that time, did not have a generalized fraction notation. They would write fractions of the form !!\frac 1n!!, and they could write sums of these. But convention dictated that they could not use the same unit fraction more than once. So to express !!\frac 35!! they would have needed to write something like !!\frac 12 + \frac{1}{10}!!, which from now on I will abbreviate as !![2, 10]!!. (They also had a special notation for !!\frac 23!!, but I will ignore that for a while.) Expressing arbitrary fractions in this form can be done, but it is non-trivial.

A simple algorithm for calculating this so-called "Egyptian fraction representation" is the greedy algorithm: To represent !!\frac nd!!, find the largest unit fraction !!\frac 1a!! that is less than !!\frac nd!!. Calculate a representation for !!\frac nd -\frac 1a!!, and append !!\frac 1a!!. This always works, but it doesn't always work well. For example, let's use the greedy algorithm to find a representation for !!\frac 29!!. The largest unit fraction less than !!\frac 29!! is !!\frac 15!!, and !!\frac 29 - \frac 15 = \frac{1}{45}!!, so we get !!\frac 29 = \frac 15 + \frac{1}{45} = [5, 45]!!. But it also happens that !!\frac 29 = [6, 18]!!, which is much more convenient to calculate with because the numbers are smaller. Similarly, for !!\frac{19}{20}!! the greedy algorithm produces $$\begin{align} \frac{19}{20} & = [2] + \frac{9}{20}\\\\ & = [2, 3] + \frac{7}{60} \\\\ & = [2, 3, 9, 180]. \end{align}$$ But even !!\frac{7}{60}!! can be more simply written than as !![9, 180]!!; it's also !![10, 60], [12, 30]!!, and, best of all, !![15, 20]!!.

So similarly, for !!\frac 37!! this time, the greedy methods gives us !!\frac 37 = \frac 13 + \frac{2}{21}!!, and that !!\frac{2}{21}!! can be expanded by the greedy method as !![11, 231]!!, so !!\frac 37 = [3, 11, 231]!!. But even !!\frac{2}{21}!! has better expansions: it's also !![12, 84], [14, 42],!! and, best of all, !![15, 35],!! so !!\frac 37 = [3, 15, 35]!!. But better than all of these is !!\frac 37 = [4, 7, 28]!!, which is optimal.

Anyway, while I was tinkering with all this, I got an answer to a question I had been wondering about for years, which is: why did Ahmes come up with a table of representations of fractions of the form !!\frac 2n!!, rather than the representations of all possible quotients? Was there a table somewhere else, now lost, of representations of fractions of the form !!\frac 3n!!?

The answer, I think, is "probably not"; here's why I think so.

Suppose you want !!\frac 37!!. But !!\frac 37 = \frac 27 + \frac 17!!. You look up !!\frac 27!! in the table and find that !!\frac 27 = [4, 28]!!. So !!\frac 37 = [4, 7, 28]!!. Done.

OK, suppose you want !!\frac 47!!. You look up !!\frac 27!! in the table and find that !!\frac 27 = [4, 28]!!. So !!\frac 47 = [4, 4, 28, 28] = [2, 14]!!. Done.

Similarly, !!\frac 57 = [2, 7, 14]!!. Done.

To calculate !!\frac 67!!, you first calculate !!\frac 37!!, which is !![4, 7, 28]!!. Then you double !!\frac 37!!, and get !!\frac 67 = \frac 12 + \frac 27 + \frac{1}{14}!!. Now you look up !!\frac 27!! in the table and get !!\frac 27 = [4, 28]!!, so !!\frac 67 = [2, 4, 14, 28]!!. Whether this is optimal or not is open to argument. It's longer than !![2, 3, 42]!!, but on the other hand the denominators are smaller.

Anyway, the table of !!\frac 2n!! is all you need to produce Egyptian representations of arbitrary rational numbers. The algorithm in general is:

• To sum up two Egyptian fractions, just concatenate their representations. There may now be unit fractions that appear twice, which is illegal. If a pair of such fractions have an even denominator, they can be eliminated using the rule that !!\frac 1{2n} + \frac 1{2n} = \frac 1n!!. Otherwise, the denominator is odd, and you can look the numbers up in the !!\frac 2n!! table and replace the matched pair with the result from the table lookup. Repeat until no pairs remain.
• To double an Egyptian fraction, add it to itself as per the previous.
• To calculate !!\frac ab!!, when !!a = 2k!!, first calculate !!\frac kb!! and then double it as per the previous.
• To calculate !!\frac ab!! when !!a!! is odd, first calculate !!\frac{a-1}{b}!! as per the previous; then add !!\frac 1b!!.

So let's calculate the Egyptian fraction representation of !!\frac{19}{20}!! by this method:

• !!\frac{19}{20} = \frac{18}{20} + \frac{1}{20}!!
• !!\frac{19}{20} = \frac{9}{10} + \frac{1}{20}!!
  • !!\frac{9}{10} = \frac{8}{10} + \frac{1}{10}!!
  • !!\frac{9}{10} = \frac 45 + \frac{1}{10}!!
    • !!\frac 45 = \frac 25 + \frac 25!!
      • !!\frac 25 = [3, 15]!! (from the table)
    • !!\frac 45 = [3, 3, 15, 15]!!
    • !!\frac 45 = \frac 23 + \frac{2}{15}!!
      • !!\frac 23 = [2, 6]!! (from the table)
      • !!\frac{2}{15} = [12, 20]!! (from the table)
    • !!\frac 45 = [2, 6, 12, 20]!!
  • !!\frac{9}{10} = [2, 6, 10, 12, 20] !!
• !!\frac{19}{20} = [2, 6, 10, 12, 20, 20] !!
• !!\frac{19}{20} = [2, 6, 10, 10, 12] !!
• !!\frac{19}{20} = [2, 5, 6, 12] !!

(The Egyptians would have been happy with !!\frac 23!! in the middle step there, and would have ended up with !!\frac{19}{20} = \frac 23 + [5, 12]!!.) Our final result is suboptimal; to fix it, we need to notice that !![6, 12] = [4]!! and get !!\frac{19}{20} = [2, 4, 5]!!. But even without this, the final result is pretty good, and required no understanding or tricky reasoning; just a lot of grinding.

An alternative algorithm is to expand the numerator as a sum of powers of !!2!!, which the Egyptians certainly knew how to do. For !!\frac{19}{20}!! this gives us $$\frac{19}{20} = \frac{16}{20} + \frac{2}{20} + \frac{1}{20} = \frac 45 + [10, 20].$$ Now we need to figure out !!\frac 45!!, which we do as above, getting !!\frac 45 = [2, 6, 12, 20]!!, or !!\frac 45 = \frac 23 + [12, 20]!! if we are Egyptian, or !!\frac 45 = [2, 4, 20]!! if we are clever. Supposing we are neither, we have $$\begin{align} \frac{19}{20} & = [2, 6, 12, 20, 10, 20] \\\\ & = [2, 6, 12, 10, 10] \\\\ &= [2, 6, 12, 5] \\\\ \end{align}$$ as before.

(It is not clear to me, by the way, that either of these algorithms is guaranteed to terminate. I need to think about it some more.)

Getting the table of good-quality representations of !!\frac 2n!! is not trivial, and requires searching, number theory, and some trial and error. It's not at all clear that !!\frac{2}{105} = [90, 126]!!.

Once you have the table of !!\frac 2n!!, however, you can grind out the answer to any division problem. This might be time-consuming, but it's nevertheless trivial. So Ahmes needed a table of !!\frac 2n!!, but once he had it, he didn't need any other tables.

20260317

More about this.

20260617

I didn't realize it at the time, but the algorithm I hypothesized above really is very much like how the Egyptians did multiplication! I suggested, essentially, that for something like !!\frac{19}{20}!! they would expand the !!19!! in binary and then use doubling and adding !!1!! to build up the result from even multiples of !!\frac1{20}!!. And actually, the Egyptians normally multiplied using binary! To do !!21\times23!! they would expand !!19=16+4 + 1!! and then compute !!16\times 23 + 4\times 23 + 23!!.

I wrote this up in some detail in a new article.

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