Adventures in Egyptian Mathematics

Dr N K Srinivasan

Introduction

This exploration is mainly to find how they

developed mathematics--- to find areas and

volumes of solid objects, how they

approximated for pi and how they worked out

slopes of pyramids and how they did many

calculations using multiplications.

They did not use angles as such but only

slopes or gradients.

The way they handled fractions is also

ingenious based on the simple fraction 2/n

where n is an odd number , 3,5, 7 upto 101.

They had a nice table for these in terms of

fractions only!

The Egyptians were very practical people:

they developed and used math for

agriculture, to measure lands and to store

grains. ---and of course, to build pyramids

for their pharoahs.

Historical Introduction

This article is based on the material found

in the famous "Rhind Papyrus".

Alexander Henry Rhind , a scottish man,

bought a piece of papyrus or Egyptian

manuscript [a scroll of 18feet and width 13

in] in the year 1858 in Luxor, Egypt.

Later this document was bequeathed to the

British Museum from his estates.

The scribe who wrote this papyrus is one

Ahmes, around 1650 BC. He stated that he

derived the material available about 200

years earlier. So the origin of this

mathematics is around 1850 BC. This period

corresponds to Middle Kingdom of Egypt.

This book, RP of math was first published

in Germany in 1873. There were hot debates

on this between German mathematicians and

British historians.!

There is a similar Egyptian papyrus in

Moscow and we shall refer only to one

problem from that manuscript in this

article.

What we learn from Rhind Payrus?

This manuscript contains 64 problems of

arithmetic and geometric ones---much of

practical value to Egyptians at that time.

This also reveals the status of mathematics

at that period in history. One can study

its relation to Greek mathematics which

developed almost a thousand years later.

I was first intrigued to learn how they

computed or approximated 'pi' -- a value we

take it to be 3.14156. Modern computers

approximate pi to million digits! Greeks

used the simple ratio of 22/7 or similar

fractions.The Hindus and the Chinese also

used also some fractions . What did the

Egyptians use for pi?

I was also interested in knowing the

various formulae or equations they had

known.It was surprising to learn that they

did know the formula for volume of a

truncated pyramid .

The value of Pi

In Book II of Rhind Papyrus [RP, for

short], we find this problem of

"mensuration" --- meaning 'measurement

related'. The Egyptians were interested in

the volume of cylindrical and rectangular

granaries.

For a cylinder they wrote:

volume V = [ (1-1/9)d ]2 h

where d is the diameter and h the height.

A school girl would immediately write this

as: V = pi x radius2 x height

So, they approximated pi = 4 (8x8)/(9x9)=

256/81

Note that 256/81 can be written as

(4x4x4x4)/(3x3x3x3)

This approximation is convenient for many

computations --since Egyptians reduced many

measurements to factors of 3 and 4.

Pi approximated to 256/ 91 = 3. 1605

Taking the more accurate value as 3.14156,

we note that the error in their

approximation is about 0.6% or less than

1%.! This is indeed remarkable for their

work.

Area of a circle

The Egyptian mathematician knew the formula

for the area of a triangle:

A = (base x height)/2

The procedure used for the area of a circle

and thereby approximate for 'pi" is as

follows: we would call this "numerical

method" or numerical approximation, that

computer scientists use often.

1. Take a square of side 9 units; its area

a= 9x9 = 81 units

2 Fit a regular octagon with side 3 units

inside this square.

3 Cut and remove the four triangles

[equilateral triangles of side 3 units] at

the four corners.

4 The area of the octagon is then:

A = a - 4(3x3)/2= 81 - 18 = 63

5 Then the mathematician took A= 64 instead

of 63: area of the octagon = 64 units = 8x8

units. [This is an error of [1/63] 100 =

1.59% ]

6Fit a circle of radius 4.5 units inside

and then area of the circle can be

"approximated" to the area of octagon:

Area = 64 = pi (9/2)2

pi = (64 x 4)/81 = 256/81 = (4x4x4x4)/(3x3x3x3)

As we noted , pi = 256/81 = 3.1605

I am mentioning this from another angle. The Greek mathematicians nearly

1000 years later

followed the same method of approximation

with regular polygons of not eight sides,

but 96 sides and 360 sides,using perimeters

of these polygons.

Archimedes worked out the value of pi using

a regular polygon of 96 sided or 96-gon. He

found a value of pi approximated to 3

10/71.

If we reduce this to 3 + 1/7, we get the

common value of 22/7 or 3.14285.

Archimedes helped Ptolemy , a

mathematician and an astronomer, who used a

regular 360-gon and got the value of pi as

follows:

pi = 3.1416

[ As a matter of history, the Chinese

mathematician Tsu Chung-chih used the ratio

pi=355/113 = 3.1415929 [480 AD] and the

Hindu mathematician Bhaskara used the ratio

pi=3927/1250 = 3.1416 around 1150AD.]

So it is obvious that the Greek

mathematicians essentially followed the

method originally developed by the Egyptian

mathematicians.

Problem 50 in RP reads thus: A round field

has a diameter of 9 khet. What is its area?

Area = (256/81) (9x9)/(2 x2)

= (4x4x4x4)(9x9)/(3x3x3x3x 2x2)

= 4x4x4 =64 square khet.

To return to Pi: the Egyptians would use

the following fractional additions:

pi= 3 +1/9 + 1/27 + 1/81 = 3.1605

The 2/n tables

The Egyptians developed a remarkable method

to do multiplication or divison, using

fractions of the kind 1/n or 2/n.

They used 2/n tables where n is an odd

number from 3 to 101.

For instance, an Egyptian boy would write:

2/3 = 1/2 + 1/6

2/5 = 1/3 + 1/15

2/7 = 1/4 + 1/28

2/9 = 1/6 + 1/18

---------------------

2/15 = 1/10 + 1/30

----------------------

2/101 = 1/101 + 1/202 + 1/303 + 1/606

Milo Gardner had broken the code of this

2/n table around 2002.

Finding the slope of a pyramid

The Egyptians did not use the angle

measurements, but instead measured the

slope or gradient with run/rise ratios.

Take a pyramid with base 2b and height h.

Take the ratio of b/h as the "seked" or

slope. Further, for height h they used 1

cubit , a standard unit ;

1 cubit = 7 palms

1 palm = 4 digits.

To find the 'angle of slope , take the

height as 1 cubit or 7 palms. Take the run

or half-base b in palms. Then b/h = x palms

/7 palms or simply x palms since the

denominator is always kept at 1 cubit.

In modern terms their 'seked' measures the

cotangent of the sloping angle.[cotangent =

adjacent side/opposite side of a right

triangle]

For calculating the ratio of b/h, they

cleverly used the 2/n tables, as the next

example from RP shows:

If a pyramid is 250 cubits high, and of side 360

cubits, what is its seked?

In this case, b= 180, h= 250

slope = s = 180/250

In this math: 180/250 = 1/2 + 1/5 + 1/50

[180/250 = 125/250 + 50/250 + 5/250]

We convert this into palms:

1 cubit 7 palms

1/2 3 + 1/2

1/5 1 + 1/3 + 1/15

1/50 1/10 + 1/25

So S = 180/250 = 3 + 1/2 + 1 + 1/3 + 1/15 +

1/10 + 1/25 = 5 + 1/25 palms

[We can check our result: (5+1/25)/7 =

5.04/7 = 0.72 180/250 = 0.72]

We have used their way of computing slope

and also dividing by the use of 2/n tables.

This method is indeed very slow for modern

computer-geeks but alright for slow moving

Egyptians of Middle kingdom! The Egyptian

boys had never worried over multiplication

tables.

Solving equations

The Egyptians were fond of posing word

problems, as in Algebra 1 in the USA.. I am

sure that the young school children did not

like that very much. But then, they had to

get the math skills needed for their way of

life.

1 Find x if 2/3 + 1/15 + x = 1

[What you should add to complete 2/3 and

1/15 to get 1?]

The answer is easily found:

2/3 = 10/15 1= 15/15 so x= 4

Volume of a frustum

This problem is found in the papyrus kept

in Moscow museum ,called Russian Papyrus.

we wish to find the volume of a truncated

pyramid or frustum with square base of side

b and a square top of side a.

Then the clever Egyptian mathematician told

:1 find the area of base (bxb)

2 find the area of top (axa)

3 find the product of the two widths axb)

4 add them up, multiply with height h

divided by 3.

So you are using the formula :

V = (1/3) h ( a.a + a.b + b.b)

There is one such problem in RP:

Find the volume of the pyramid with base 4

units, top 2 units and height 6 units:

V = (1/3) 6 (16 + 8 + 4)=56 units.

Comment: Did the Egyptians use a formula as

we have written here or just an algorithm,

with this example.

This is a moot point. They must have used

this method several times to calculate, as

they built hundreds of pyramids.

But William Dunham , in his remarkable book

"Journey through Genius" [Penguin Books

1990] surmises that most probably they did

not abstract the computation with a

"formula" for the equation as we would for

the volume. They gave only prescriptive

procedure or algorithm. He even goes on to

add that the mindset in those days was to

just follow a method instead of deriving

it!.This was a 'dogmatic approach 'in

keeping with 'the authoritarian society

that was pharaonic Egypt.' As Dunham

himself states , this may be a 'sweeping

conclusion' based on one example. My own

view is that it is dangerous to make such

conclusions because manuscripts such as RP

which could have been only a class room

text book for students learning math and

not a treatise.May be RP was used as a

training manual for artisans.

Designing a Pyramid

Now we know their understanding of "pi", we

can learn how they designed pyramids.One

common method was to make the height of

pyramid 4 units while half the base would

be pi units.

Then the ratio of base to height =b/h

b/h = pi/4 = (256/81)/4 = 64/81 = 0.79

tan (slope angle) = 1/.79 = 1.266

The slope angle becomes 52 deg--a figure

often found in pyramid slopes.

This also removes the mystery behind the

selection of this angle as 52 deg.

Another word problem from RP:

There were seven houses;each had seven

cats,each had caught seven mice, each mouse

had seven spelt and each spelt seven

hekat.How many total were there?

Answer: Sum = 1 + 7 + 7.7 + 7.7.7 + 7.7.7.7

+ 7.7.7.7.7 = 28403

Another word problem:

700 loaves of bread are to be divided among

recipients in the proportion of

2/3:1/2:1/3:1/4.

Add 2/3 + 1/2 + 1/3 + 1/4 = 7/4

Divide 700 by 7/4 ,called the base number:

base number = 700/ (7/4) = 400

Now multiply 400 with the ratios:

a = 400 (2/3) b= 400(1/2) c=400(1/3) d=

400(1/4)

The numbers are: 266,200,133,100

{There will be a loss of one loaf in the

calculation.]

Some historical notes

Some writers claim that the greek

mathematicians like Pythagoras, Thales and

Eudoxus went to Egypt to learn much about

these math methods.

Did the Egyptians know Pythagorian theorem>

Most probably they knew the triple --3,4,5

---of a right triangle sides, as they did

much work with right traingles. But they

may not have used this as such.

References:

1 William Dunham Journey through genius

2 Internet Websites

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