Fractions are Hard!

4.1 Egyptian Fractions

Scholars of ancient Egypt (ca. 3000 BCE) were very practical in their approaches to mathematics and always sought answers to problems that would be of most convenience to the people involved. This led them to a curious approach to thinking about fractions.


Consider the problem: Share 7 pies among 12 boys.


Of course, given our model for fractions, each boy is to receive the quantity \(\dfrac{7}{12}\) of a pie. But this answer has little intuitive feel.


But suppose we took this task as a very practical problem. Here are the seven pies


Is it possible to give each of the boys a whole pie? No. How about the next best thing – each boy half a pie? Yes! There are certainly 12 half pies to dole out. There is also one pie left over yet to be shared among the 12 boys. Divide this into twelfths and hand each boy an extra piece.



Thus each boy receives \(\dfrac{1}{2}+\dfrac{1}{12}\) of a pie and it is indeed true that \(\dfrac{7}{12}=\dfrac{1}{2}+\dfrac{1}{12}\).



a) How do you think the Egyptian’s might have shared five pies among six girls?

b) How might they have shared twelve pies among seven students?


The Egyptians insisted on writing all their fractions as sums of fraction with numerators equal to \(1\). They did this because fractions of the form \(\dfrac{1}{N}\) have a good intuitive feel to them. For example


\(\dfrac{3}{10}\) was written as \(\dfrac{1}{4}+\dfrac{1}{20}\)


\(\dfrac{5}{7}\) was written as \(\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{70}\).


That is, to share three pies among ten students, the Egyptians said to give each student one quarter of a pie and one twentieth of a pie. To share five pies among seven students, the Egyptians suggested giving out half a pie, one fifth of a pie, and one seventieth of a pie to each student.


Question:  It is true that \(\dfrac{4}{13}=\dfrac{1}{4}+\dfrac{1}{18}+\dfrac{1}{468}\). What does this say about how the Egyptians may have advised sharing four pies among thirteen girls?


Comment: A fraction with numerator \(1\) is today called a unit fraction. The Egypyians denoted their unit fractions with a raised dot. For example, \(\dot{2}=\dfrac{1}{2}\) and \(\dot{103}=\dfrac{1}{103}\). Thus \(\dfrac{3}{10}\) was written as \(\dot{4}+\dot{20}\), and so on.

They expressed all fractions as sums of unit fractions – except for the fraction \(\dfrac{2}{3}\), which had its own special symbol. (Probably because this fraction arose so often in day-to-day work.)


Curiously, the Egyptians did not like to repeat fractions. Although it is obviously true that




the Egyptians really did think it better to give each person receiving pie piece as large as possible, and so preferred to work with




(even though it meant giving out a tiny piece of pie with that bigger piece).


Question: Consider the fraction \(\dfrac{2}{11}\) .

a)       Show that \(\dfrac{1}{5}\) is bigger than \(\dfrac{2}{11}\).

b)      Show that \(\dfrac{1}{6}\) is smaller than \(\dfrac{2}{11}\).

c)       Work out \(\dfrac{2}{11}-\dfrac{1}{6}\).

d) Use c) to write \(\dfrac{2}{11}\) the Egyptian way.


Question: Consider the fraction \(\dfrac{2}{7}\).

a)       What is the biggest fraction \(\dfrac{1}{N}\) that is still smaller than \(\dfrac{2}{7}\)?

b)      Write \(\dfrac{2}{7}\) the Egyptian way.



 Write \(\dfrac{17}{20}\) the Egyptian way.


Question: CHALLENGE  What is the largest value of \(n\) for which the fraction \(\dfrac{n-1}{n}\) can be written in the form \(\dfrac{1}{a}+\dfrac{1}{b}\) with \(a\) and \(b\) positive integers?


Question: CHALLENGE  Find a formula for the number of ways one can write \(\dfrac{1}{2^{n}}\)  in the form \(\dfrac{1}{a}+\dfrac{1}{b}\) with \(a\) and \(b\) positive integers, \(a<b\) ?


The Egyptians were adept at computing sums of unit fractions. As the exercise show, it is typically not easy to do.


EXAMPLE: Write \(\dfrac{4}{13}\) as a sum of distinct unit fractions.


Answer: The Egyptians preferred always “take out” the largest unit fraction possible from any given fraction at each stage. Note that \(\dfrac{4}{13}=\dfrac{1}{3\dfrac{1}{4}}\) which shows that \(\dfrac{1}{3}\) is larger than \(\dfrac{4}{13}\), but \(\dfrac{1}{4}\) isn’t. With some scratch-work on the side we see that \(\dfrac{4}{13}=\dfrac{1}{4}+\dfrac{3}{52}\).


Now \(\dfrac{3}{52}=\dfrac{1}{17\dfrac{1}{3}}\) which shows that \(\dfrac{1}{18}\) is the next largest unit fraction we can “take out.” We have \(\dfrac{3}{52}-\dfrac{1}{18}=\dfrac{1}{468}\) and so \(\dfrac{4}{13}=\dfrac{1}{4}+\dfrac{1}{18}+\dfrac{1}{468}\). We’re done!



EXERCISE: Write \(\dfrac{3}{7}\) as a sum of unit fractions by removing, at each stage, the largest unit fraction possible. Do the same for \(\dfrac{5}{11}\).


In 1202 Italian scholar Fibonacci questioned whether or not every fraction can be written as a sum of distinct unit fractions. Does the process of “taking out” the largest unit fraction always yield the desired outcome? He answered this question to the affirmative.


CHALLENGE: Fibonacci’s Proof


Suppose we are working with a fraction \(\dfrac{a}{b}\) with \(a<b\) .


a) Let \(N\) be the smallest positive integer with \(\dfrac{a}{b}>\dfrac{1}{N}\). (Thus \(\dfrac{a}{b}<\dfrac{1}{N-1}\).) Write \(\dfrac{a}{b}-\dfrac{1}{N}\) over a common denominator and show that this new fraction has numerator that is both positive and smaller than \(a\).


b) Explain why, if we repeat this process, we shall eventually obtain a fraction with numerator equal to \(1\).


c) Explain why \(\dfrac{a}{b}\) is then sure to equal a sum of (distinct) unit fractions.


Question:  Use Fibonacci’s method to write the fraction \(\dfrac{1}{1}\) as an infinite sum of distinct unit fractions. What appears? Can anything be said about the denominators that arise?
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