Egyptian Fractions

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The Universe of Discourse : Egyptian Fractions

The Universe of Discourse

Mark Dominus (陶敏修)

mjd@pobox.com

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Update: Here I am at the Sagrada Família<br>Egyptian fractions for 2/105<br>Did Ahmes find the best expansions for 2/n?<br>Programmers will document for Claude, but not for each other<br>How are John Waters movies like James Bond movies?<br>Documentation is a message in a bottle<br>Bo Diddley<br>Language models imply world models<br>John Haugeland on the failure of micro-worlds<br>Crooked politicians love crab cakes!<br>Almost-trivial theorems<br>An anecdote about backward compatibility

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Thu, 11 May 2006

Egyptian Fractions

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

A simple algorithm for calculating this so-called "Egyptian fraction<br>representation" is the greedy algorithm: To represent<br>!!\frac nd!!, find the largest unit fraction !!\frac 1a!! that is<br>less than !!\frac nd!!. Calculate a representation for<br>!!\frac nd -\frac 1a!!, and append !!\frac 1a!!. This<br>always works, but it doesn't always work well. For example, let's use<br>the greedy algorithm to find a representation for !!\frac 29!!. The largest<br>unit fraction less than !!\frac 29!! is !!\frac 15!!, and !!\frac 29 - \frac 15 = \frac{1}{45}!!, so we get<br>!!\frac 29 = \frac 15 + \frac{1}{45} = [5, 45]!!. But it also happens that !!\frac 29 = [6, 18]!!,<br>which is much more convenient to calculate with because the numbers<br>are smaller. Similarly, for !!\frac{19}{20}!! the greedy algorithm<br>produces

$$\begin{align}<br>\frac{19}{20} & = [2] + \frac{9}{20}\\\\<br>& = [2, 3] + \frac{7}{60} \\\\<br>& = [2, 3, 9, 180].<br>\end{align}$$

But even<br>!!\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<br>!!\frac 37 = \frac 13 + \frac{2}{21}!!, and that !!\frac{2}{21}!! can be expanded by the greedy method<br>as !![11, 231]!!, so !!\frac 37 = [3, 11, 231]!!.<br>But even !!\frac{2}{21}!! has better expansions: it's also !![12, 84], [14, 42],!!<br>and, best of all, !![15, 35],!! so !!\frac 37 = [3, 15, 35]!!. But better than all<br>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<br>question I had been wondering about for years, which is: why did Ahmes<br>come up with a table of representations of fractions of the form<br>!!\frac 2n!!, rather than the representations of all possible quotients?<br>Was there a table somewhere else, now lost, of representations of<br>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<br>table and find that !!\frac 27 = [4, 28]!!. So !!\frac 37 = [4, 7, 28]!!.<br>Done.

OK, suppose you want !!\frac 47!!. You look up !!\frac 27!! in the table and find that<br>!!\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]!!.<br>Then you double !!\frac 37!!, and get !!\frac 67 = \frac 12 + \frac 27 + \frac{1}{14}!!. Now you look<br>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.<br>It's longer than !![2, 3, 42]!!, but on the other hand the<br>denominators are smaller.

Anyway, the table of !!\frac 2n!! is all you need to produce...

frac jfmamj jasond fractions table egyptian

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