7.2: Egyptian Fractions

The seven symbols

Ancient Egyptian hieroglyphic numerals use separate signs for successive powers of ten. That is, the Egyptians used a pictorial number system, with different symbols for every power of 10 to 1 000 000. Each sign is repeated as needed, and its value does not depend on position. The drawings below are simplified classroom illustrations of the traditional signs.

Example \(\PageIndex{1}\)

1 = |

4 = ||||

10 = ∩

23 = ∩ ∩|||

12 = ∩||

 =2343

Notice that this is a non-positional representation.

Addition:

Example \(\PageIndex{1}\)

Add \(28+17\).

Solution

There are three tens and fifteen ones before regrouping:
\(
2(10)+8+1(10)+7=3(10)+15=4(10)+5=45.
\)

Comparison with modern Hindu-Arabic notation:

Egyptian fractions

 A unit fraction has numerator 1 and denominator a positive integer greater than 1:\(\dfrac{1}{2},\quad\dfrac{1}{3},\quad\dfrac{1}{4},\quad\dfrac{1}{10},\ldots\)
For example, \(\boxed{\dfrac{3}{4}=\dfrac{1}{2}+\dfrac{1}{4}}\)

Egyptian scribes commonly represented fractional quantities using unit fractions, with a special representation for \(2/3\), but we will ignore this for now. The unit fraction is made by writing the number with a “mouth” symbol over the top:

Egyptians did not like repeating unit fractions; therefore, each unit fraction must be unique. As a result, any fraction with a numerator > 1 must be written as a sum of Egyptian unit fractions. As a result of this mathematical quirk, Egyptian fractions are a great way to test students' understanding of adding and combining fractions with different denominators (grade 5-6) and of the relationship between fractions with different denominators (grade 5). We can use manipulatives to explore Egyptian fractions. For a detailed activity, please see Activity.

Visual understanding

Party Time

Example \(\PageIndex{2}\): Party Time

You're having a pie party. Unfortunately, you forgot to cut the pies before the party started. How do you cut and split the pies evenly amongst your guests if there are 3 people and 2 pies \( \dfrac{2}{3} \)? How do you split the pies using Egyptian fractions?

Solution

Using our manipulative, we begin by laying out 2 full pies. The biggest Egyptian fraction that can split these pies is \(\dfrac{1}{2}\). This gives us \(\dfrac{1}{2}\) of a pie for each person and \(\dfrac{1}{2}\) of a pie leftover. We can split this last half piece into thirds again, giving us \(\dfrac{1}{6}\) of a pie for each person. That is \(\boxed{ \dfrac{2}{3}= \dfrac{1}{2} +\dfrac{1}{6}} \).

 

Example \(\PageIndex{3}\):

If we were asked to write \( \dfrac{3}{5} \) into a sum of Egyptian fractions, we could write \( \dfrac{1}{5} + \dfrac{1}{5}+ \dfrac{1}{5} \), since each fraction has a numerator of 1. The only problem with this is that they aren't unique fractions; therefore, it wouldn't work for the sum of Egyptian fractions.

\( \boxed{\dfrac{3}{5} = \dfrac{1}{2}+ \dfrac{1}{10}} \).

 

Example \(\PageIndex{4}\):

What if there were only 4 pizzas to be split amongst 8 friends?

\( \boxed{\dfrac{5}{8} = \dfrac{1}{2}+ \dfrac{1}{8}} .\)

Exercise \(\PageIndex{5}\):

You're having a pie party. Unfortunately, you forgot to cut the pies before the party started. How
do you cut and split the pies evenly amongst your guests?

1. There are 4 people and 3 pies \(\dfrac{3}{4} \). How do you split the pies using Egyptian fractions?
2. There are 6 people and 5 pies \( \dfrac{5}{6} \). How do you split the pies using Egyptian fractions?

Answer

1. Using our manipulative, we begin by laying out 3 full pies. The biggest Egyptian fraction that can split these pies is \(\dfrac{1}{2}\).
This gives us \(\dfrac{1}{2}\) a pie for each person and a whole pie leftover. We can split this pie into 4 equal pieces, giving us \(\dfrac{1}{4}\) of a pie for each person.

2. Using our manipulative, we begin by laying out \(5\) full pies. The biggest Egyptian fraction that can split these pies is \(\dfrac{1}{2}\).
This gives us \(\dfrac{1}{2}\) of a pie for each person and two whole pies left over. We can split these two pies into 6 equal pieces, giving us \(\dfrac{1}{3}\) of a pie for each person.

Bigger Parties, Bigger Problems

You are now having an even bigger party with a lot more pies. Can you find the sum of Egyptian fractions that will help split the pies evenly amongst your guests? Having manipulatives for these next questions would be unrealistic, as there would be too many pies. Instead, we can think in terms of factors of 100 and use them as the denominators of our Egyptian fractions.The factors of \(100\) are \(1, 2, 4, 5, 10, 20, 25, 50, \) and \(100.\)

Example \(\PageIndex{6}\):

There are \(100\) people and \(91\) pies \( \dfrac{91}{100} \). How do you split the pies using Egyptian fractions?

Solution

Looking at our factors of \(100\), we can begin by subtracting the biggest Egyptian fraction \(\dfrac{1}{2}\).

\( \dfrac{91}{100}-\dfrac{1}{2}= \dfrac{91}{100}-\dfrac{50}{100}=\dfrac{41}{100} \).

Next, we subtract \( \dfrac{1}{4}\).

\( \dfrac{41}{100}-\dfrac{1}{4}= \dfrac{41}{100}-\dfrac{25}{100}=\dfrac{16}{100} \).

Notice that we can't subtract Next, we subtract \( \dfrac{1}{5}= \dfrac{20}{100}\). Therefore, we subtract \( \dfrac{1}{10}\).

\( \dfrac{16}{100}-\dfrac{1}{10}= \dfrac{16}{100}-\dfrac{10}{100}=\dfrac{6}{100} \).

Next we subtract \(\dfrac{1}{20}\).

\( \dfrac{6}{100}-\dfrac{1}{20}= \dfrac{6}{100}-\dfrac{5}{100}=\dfrac{1}{100} \).

Thusm \(\boxed{ \dfrac{91}{100}=\dfrac{1}{2}+ \dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{100} }\).

Each person would get \(\dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{5}.\) and \(\dfrac{1}{100}\) of a pie.

Exercise \(\PageIndex{7}\):

There are \(100\) people and \(99\) pies \( \dfrac{99}{100} \). How do you split the pies using Egyptian fractions?

Answer

\( \dfrac{99}{100}=\dfrac{1}{2}+ \dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{25} \). Also note that it can be split into Egyptian fractions whose denominators are not factors of 100.

\( \boxed{\dfrac{99}{100}=\dfrac{1}{2}+ \dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{20}+ \dfrac{1}{150} }\).

 A reliable modern method:

The Greedy Algorithm

For a proper fraction \(0<\dfrac{a}{b}<1\), let \(n=\left\lceil \dfrac{b}{a}\right\rceil\), the smallest integer at least \(\dfrac{b}{a}\). Then:

1. Select the largest unit fraction that does not exceed the remainder, namely \(\dfrac{1}{n}\).

2. Subtract \(\dfrac{1}{n}\) from the remainder.

3. Repeat until the remainder equals zero.

The resulting denominators are strictly increasing. This is a modern algorithm, not a claim about the exact historical process used by Egyptian scribes.
 

Example \(\PageIndex{8}\)

Apply the method to \(\dfrac{3}{7}\):

\( n_1=\left\lceil\dfrac{7}{3}\right\rceil=3,
 \qquad \dfrac{3}{7-}\dfrac{1}{3}=\dfrac{2}{21}.
\)

\(
 n_2=\left\lceil\dfrac{21}{2}\right\rceil=11,
 \qquad \dfrac{2}{21}-\dfrac{1}{11}=\dfrac{1}{231}.
\)

Hence \(\boxed{\dfrac{3}{7}=\dfrac{1}{3}+\dfrac{1}{11}+\dfrac{1}{231}}\).

Note: Different decompositions may represent the same value:
\(\dfrac{2}{3}=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}.\)

Verify the second: \(\dfrac{4}{12}+\dfrac{3}{12}+\dfrac{1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\).

Do not confuse a modern decomposition of \(\dfrac{2}{3}\) with the special sign commonly used for \(\dfrac{2}{3}\) in ancient Egyptian writing.

The value of a rational number is unique, but its Egyptian unit fraction representation need not be unique. The splitting identity gives a systematic way to construct infinitely many representations.

Example \(\PageIndex{9}\)

Consider the fraction \(\dfrac{3}{4}\).

We have
\(
\boxed{\dfrac34=\dfrac12+\dfrac14.}
\)

Another decomposition is
\(
\boxed{
\dfrac34=\dfrac12+\dfrac15+\dfrac1{20}.
}
\)

A third decomposition is
\(
\boxed{
\dfrac34=
\dfrac12+\dfrac15+\dfrac1{21}+\dfrac1{420}.
}
\)

Theorem \(\PageIndex{1}\): Splitting identity

For every positive integer \(n\),
\[
\boxed{
\dfrac1n=
\dfrac1{n+1}+\dfrac1{n(n+1)}.
}
\]

Proof

By adding the fractions,
\begin{align*}
\dfrac1{n+1}+\dfrac1{n(n+1)}
&=\dfrac{n}{n(n+1)}
+\dfrac1{n(n+1)}\\
&=\dfrac{n+1}{n(n+1)}\\
&=\dfrac1n.
\end{align*}

This identity allows us to replace one unit fraction by two smaller unit fractions. For example,
\(
\dfrac14=\dfrac15+\dfrac1{20}.
\)

We can apply the identity repeatedly to produce additional decompositions.

Theorem \(\PageIndex{2}\)

Every positive rational number has infinitely many distinct Egyptian fraction decompositions.

Example \(\PageIndex{10}\)

Repeated Splitting:

Start with
\(
\dfrac35=\dfrac12+\dfrac1{10}.
\)

Since
\(
\dfrac1{10}=\dfrac1{11}+\dfrac1{110},
\)
we obtain
\(
\boxed{
\dfrac35=\dfrac12+\dfrac1{11}+\dfrac1{110}.
}
\)

Next,
\(
\dfrac1{110}=\dfrac1{111}+\dfrac1{12210}.
\)

Therefore,
\(
\boxed{
\dfrac35=
\dfrac12+\dfrac1{11}+\dfrac1{111}
+\dfrac1{12210}.
}
\)

All denominators are distinct.

 The following theorem gives a way to express \( \dfrac{2}{n}\), where \(n \) is an odd integer greater than \(1\)

Theorem \(\PageIndex{3}\)

Any fraction \( \dfrac{2}{n}\), where \(n \) is an odd integer greater than \(1\), can be expressed as a sum of two Egyptian unit fractions.

Proof

Let \(n\) be an odd integer greater than \(1\). Since \(n\) is odd, there exists an integer \(k\geq 1\) such that \(n=2k+1\).

Observe that
\begin{align*}
\dfrac{1}{k+1}+\dfrac{1}{n(k+1)}
&=\dfrac{n+1}{n(k+1)}\\
&=\dfrac{2k+2}{n(k+1)}\\
&=\dfrac{2(k+1)}{n(k+1)}\\
&=\dfrac{2}{n}.
\end{align*}

Hence,
\(
\boxed{
\dfrac{2}{n}
=
\dfrac{1}{(n+1)/2}
+
\dfrac{1}{n(n+1)/2}.
}
\)

Both denominators are distinct positive integers. Therefore, \(\dfrac{2}{n}\) is the sum of two distinct Egyptian unit fractions.

• Thanks to Jillian Periliat for all the diagrams.

   

SOURCE for some images

• https://www.zenius.net/blog/rumus-pecahan/