Ancient Egyptian Mathematics

**By: Tao Steven Zheng (郑涛)**

Our knowledge of ancient Egyptian mathematics is limited to the small number of mathematical documents written on **papyrus**. Several mathematical papyri have been acquired by European archaeologists in the nineteenth and twentieth centuries; the most important being the *Rhind Papyrus*, the *Moscow Papyrus*, and the *Berlin Papyrus*. These papyri were mainly a compilation of solved problems in arithmetic, algebra, and geometry. The problems were intended for training civil servants in the art of calculation. Most of the problems are posed in a practical manner; however, there are certain problems that illuminate the theory behind Egyptian mathematics.

 The ***Berlin Papyrus*** is a mathematics text written around 1800 BC. Only two problems are readable from the surviving fragments. The first problem may suggest that the ancient Egyptians had some knowledge of the Pythagorean theorem.

## **Egyptian Numerals**

The ancient Egyptians created a base-ten **additive system**. This means the hieroglyphic symbols for the numbers were simply added together. For example, the number 23 is written using two symbols for 10 and three symbols for 1. It does not matter which order the numerals are written: you can write the tens first followed by the ones, or the ones first followed by the tens. In fact, Egyptian hieroglyphs can even be written vertically. The Egyptians used seven distinct symbols for writing numerals:

* 1 is a vertical stroke
* 10 is a hobble for cattle
* 100 is a coil of rope
* 1000 a lotus plant
* 10000 is a bent finger
* 100000 is a tadpole or frog
* 1000000 is a figure of the Egyptian god Hah with his arms raised above his head.

In Figure 2, the number 258458 is written using 32 symbols: two frogs for 200000, five bent fingers for 50000, eight lotuses for 8000, four coils of rope for 400, five hobbles for 50, and eight vertical strokes for 8.

## **Egyptian Fractions**

The ancient Egyptians expressed fractions as the sum of **unit fractions** (with the exception of 2/3). Unit fractions are fractions with 1 in the numerator. For example, the fraction 5/6 is expressed as the sum of 1/2 and 1/3.

1
2

1
3

=

5
6

{\displaystyle \frac{1}{2} + \frac{1}{3} = \frac{5}{6}}

A table of 2/n is given at the start of the *Rhind Papyrus* to aid such calculations.

How did the ancient Egyptians compile this table? The table suggests that Egyptian mathematicians discovered several formulas for generating the appropriate sums. Problem 61 of the *Rhind Papyrus* gives one formula for decomposing the fraction 2/n into the sum of two unit fractions:

2

3
(
2
k
1
)

=

1

2
(
2
k
1
)

1

6
(
2
k
1
)

{\displaystyle \frac{2}{3(2k+1)} = \frac{1}{2(2k+1)} +\frac{1}{6(2k+1)} }

where

2
k
1

{\displaystyle 2k+1}

 is an odd integer. Another formula can be observed from the decompositions of 2/101:

2
n

=

1
n

1

2
n

1

3
n

1

6
n

.

{\displaystyle \frac{2}{n} = \frac{1}{n} + \frac{1}{2n} + \frac{1}{3n} + \frac{1}{6n}. }

## **Egyptian Algebra**

### Linear Equations

There are many problems recorded on Egyptian papyri that involve linear equations. Among the earliest and most basic linear equations were called **Aha problems** (literrally, "stack" ). Problem 19 of the *Moscow Mathematical Papyrus* (c. 1850 BC), a good example of an Ahau problem, asks:

*A number taken* 1 *and* 1/2 *times and added to* 4 *to make* 10*. What is the number?*

The scribe gives the solution:

*Calculate the excess of this* 10 *over* 4*. The result is* 6*. You operate on* 1 and 1/2 *to find* 1*. The result is* 2/3*. You take* 2/3 *of this* 6*. The result is* 4*. Behold,* 4 *says it. You will find that this is correct.*

In modern notation, the equation is

x

1
2

x
# 4

{\displaystyle x + \frac{1}{2}x + 4 = 10. }

The ancient Egyptian scribe instructs the reader to subtract 4 from 10

x

1
2

# x

{\displaystyle x + \frac{1}{2}x = 6. }

After subtracting 4, the scribe noted that the reciprocal of

1

1
2

{\displaystyle 1 \frac{1}{2}}

is equal to

2
3

{\displaystyle \frac{2}{3}}

. The scribe then multiplies 6 by

2
3

{\displaystyle \frac{2}{3}}

to obtain the answer:

2
3

×
# 6

{\displaystyle \frac{2}{3} \times 6 = 4. }

The above solution is similar to the method taught in school mathematics today; however, the ancient Egyptians often utilized a different method when solving linear equations. This technique is called the **method of false position**, or *regula falsi* in Latin. The method of false position begins by assuming a convenient but probably incorrect answer. The incorrect answer is then adjusted by the correct proportion to arrive at the correct answer. A good example of this method can be found in Problem 26 of the *Rhind Papyrus* (c. 1650 BC):

*Find a quantity such that when it is added to* 1/4 *of itself the result is a* 15*.*

The scribe solves the problem as follows:

*Assume* [the answer is] 4*. Then* 1 *and* 1/4 *of* 4 *is* 5*. Multiply* [the number to] 5 *so as to get* 15*. The answer is* 3*. Multiply* 3 *by* 4*. The answer is* 12*.*

In modern notation, the equation is

x

1
4

# x

{\displaystyle x + \frac{1}{4}x = 15. }

The scribe makes an initial guess

# x
4

{\displaystyle x=4}

based on the denominator of the fraction 1/4, and then proceeds to calculate the left hand side of the equation:

4

1
4

×
# 4

{\displaystyle 4 + \frac{1}{4} \times 4 = 5. }

The scribe would notice that the initial guess gives a result that is one-third of 15. He then concludes that the correct answer must be triple the initial guess:

# x
3
×
# 4

{\displaystyle x = 3 \times 4 = 12. }

Therefore, the desired number is 12. The *Rhind Papyrus* has several problems of this type, all solved using the method of false-position. There is no discussion of how or when this method was discovered or any explanation of why it works. However, it is clear that the ancient Egyptian scribes understood the concept of proportionality of two quantities in a linear relationship of the form

# y
a
x

{\displaystyle y = ax}

, where

x

{\displaystyle x}

 is the variable to be solved,

y

{\displaystyle y}

is the given quantity, and

a

{\displaystyle a}

is the proportionality constant.

### Quadratic Equations

How did the ancient Egyptians solve quadratic equations? The near absence of quadratic problems from surviving Egyptian mathematics texts makes this question difficult to answer. Of all the the archaeological evidence available, the earliest extant quadratic equation problems from ancient Egypt can only be found on the surviving fragments of the *Berlin Papyrus* (c. 1800 BC). This document contains only two problems, of which Problem 1 states:

*You are told the area of a square of* 100 *square cubits is equal to that of two smaller squares. The side of one square is* 1/2 + 1/4 *of the other. What are the sides of the two unknown squares?*

**Solution**

Let

x

{\displaystyle x}

 be the side length of the small square and

y

{\displaystyle y}

 be the side length of the large square. According to the problem, we get a system of simultaneous equations:

{

x
2

y
2

=
100

# x

(

1
2

1
4

)

y

.

{\displaystyle \begin{align} \begin{cases} x^2 + y^2 = 100 \\ x = \left(\frac{1}{2} + \frac{1}{4} \right)y \end{cases} \end{align}. }

Note that the second equation is equivalent to

# x

3
4

y

{\displaystyle x = \frac{3}{4}y }

but is expressed as

# x

(

1
2

1
4

)

y

{\displaystyle x = \left(\frac{1}{2} + \frac{1}{4} \right) y }

 because the ancient Egyptians expressed their fractions as the sum of unit fractions. Interestingly, the ancient Egyptians applied the method of false-position on quadratic equations as well! The scribe who wrote the *Berlin Papyrus*, begins by assuming that the sides of the two squares are 1 and 3/4. He then calculated the sum of the areas of these two squares to be

1
2

(

3
4

)

2

=
1

9
16

.

{\displaystyle 1^2 + \left(\frac{3}{4} \right)^2 = 1 \frac{9}{16}.}

The scribe realized that he could not compare each area directly too 100 (the total area), so he must compare their sides. Subsequently, the scribe took the square root of

1

9
16

{\displaystyle 1\frac{9}{16}}

, which is

1

1
4

{\displaystyle 1\frac{1}{4}}

. The scribe than compared this to 10 (the square root of 100). Since 10 is 8 times larger than

1

1
4

{\displaystyle 1\frac{1}{4}}

, the scribe concluded that the sides of the two other squares must be 8 times the original guesses.

{

8
×
1

=
8

8
×

3
4

=
6

{\displaystyle \begin{cases}
8 \times 1 &= 8 \\
8 \times \frac{3}{4} &= 6
\end{cases}}

Thus, the side of the large square is 8 and the side of the small square is 6 cubits.