Ancient Egyptian Fractions
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Last time, we learned how the ancient Egyptians divided using as an example 351 divided by 13. This was a somewhat convenient choice because 13 divides neatly into 351. However, what if we had chosen numbers that were not nice, 358 divided by 13, for instance?
Knowing that \(351/13 = 27\), it isn’t difficult to see that \(358/13 = 27\frac{7}{13}\), but at this point an ancient Egyptian scribe would become confused.
You see, the ancient Egyptians viewed fractions as each having their own indivisible form or identity. For example, there was the half, the third, the fourth, etc. There was not, however, \(\frac{2}{3}\), because that would be two of the third, and how could the third be in the same place twice?
Instead of grappling with the existential terror of the parts suddenly gaining doppelgangers, the ancient Egyptians only represented fractions as the sum of unequal unit fractions (i.e. fractions with 1 in the numerator). For example, $$\frac{3}{4} = \frac{1}{2} + \frac{1}{4}$$ $$\frac{2}{3} = \frac{1}{2} + \frac{1}{6}$$ $$\frac{7}{11} = \frac{1}{2} + \frac{1}{11} + \frac{1}{22}$$
Note that it is easy to show that this representation is not unique for any fraction (even for a fraction that is already unit!). Another note is that what we’re doing is somewhat cheating, because we’re just taking modern notation and translating it into a form that an ancient Egyptian scribe might understand. Let’s not think about that, too much, okay? Instead, let’s look at some general algorithms of finding unit fraction representations!
Method 1: The “Greedy” Algorithm
In this method, we find the largest unit fraction less than our given fraction and subtract it from the given fraction. If the result is a unit fraction, stop. Otherwise, that result becomes the next unit fraction.
Example 1.1:Â \(\frac{3}{4}\).
The largest unit fraction less that \(\frac{3}{4}\) is \(\frac{1}{2}\), so we write $$\frac{3}{4} = \frac{1}{2} + \left(\frac{3}{4} - \frac{1}{2}\right) = \frac{1}{2} + \frac{1}{4}.$$
Example 1.2:Â \(\frac{3}{7}\)
The largest unit fraction less than \(\frac{3}{7}\) is \(\frac{1}{3}\), so we write $$\frac{3}{7} = \frac{1}{3} + \left(\frac{3}{7} - \frac{1}{3}\right) = \frac{1}{3} + \frac{2}{21}.$$Â The largest unit fraction less than \(\frac{2}{21}\) is \(\frac{1}{11}\), so we write $$\frac{2}{21} = \frac{1}{11} + \left(\frac{2}{21} - \frac{1}{11}\right) = \frac{1}{11} + \frac{1}{231}.$$Â Then, $$\frac{3}{7} = \frac{1}{3} + \frac{1}{11} + \frac{1}{231}.$$
This algorithm seems to work relatively well, but as we saw in Example 1.2, the denominator can become somewhat unwieldy for relatively simple non-unit fractions. Another potential issue would occur if there was a fraction such that the “Greedy” Algorithm didn’t terminate (i.e. returned an infinite series of unit fractions).
Exercise 1:
Does the “Greedy” Algorithm always terminate? If not, give an example. If so, prove it. (Answer under the cut)
Method 2: The “Splitting” Algorithm
Wouldn’t it be nice to not need to worry about whether our algorithm terminated or not? Let me introduce you to the “Splitting” Algorithm, which makes use of the fact that for any positive integer \(n\), $$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)}.$$ Let’s illustrate this method with an example!
Example 2.1:Â \(\frac{2}{3}\)
$$\frac{2}{3} = \frac{1}{3} + \frac{1}{3} = \frac{1}{3} + \frac{1}{3+1} + \frac{1}{3(3+1)} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12}.$$
Wasn’t that fast an easy? It’s simple to see that this algorithm will always terminate, and we didn’t need to go to all the trouble of finding largest unit fractions or subtracting fractions or any of the other headaches from the “Greedy” Algorithm. Don’t get too excited, though. Let’s try one more example first.
Example 2.2:Â \(\frac{4}{7}\)
We write out $$\frac{4}{7} = \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{1}{7} + \frac{1}{8} + \frac{1}{56} + \frac{1}{8} + \frac{1}{56} + \frac{1}{8} + \frac{1}{56},$$
and Yikes. If you would like to go through the process of finding this expansion using only the splitting algorithm, be my guest. If you’re sane, the answer is $$\frac{1}{7} + \frac{1}{8} + \frac{1}{56} + \frac{1}{9} + \frac{1}{72} + \frac{1}{57} + \frac{1}{3192} + \frac{1}{10} + \frac{1}{90} + \frac{1}{73} + \frac{1}{5256} + \frac{1}{58} + \frac{1}{3306} + \frac{1}{3193} + \frac{1}{10192056}.$$ Meanwhile, using the “Greedy” Algorithm, $$\frac{4}{7} = \frac{1}{2} + \frac{1}{14}.$$
In fact, if you don’t take advantage of obvious reductions, it isn’t difficult to show that given a fraction with numerator \(m\), the “Splitting” Algorithm would return the sum of \(2^{m}-1\) distinct unit fractions.
So, which method is better? Depends on the situation. In fact, in most cases the best method is probably some combination involving simplification.
A better question: What does this have to do with modern mathematics? I have no idea--it’s just interesting! It leads to a lot of wild questions like “Given any fraction, how can it be represented by the sum of the fewest number of unit fractions?” or “How many ways are there to write a given fraction as the sum of \(n\) distinct unit fractions?”
Anyway, I hope you enjoyed. This will probably be the last post in my Ancient Egyptian Mathematics series, and I want to thank everyone who read and gave feedback along the way. Until next time!
