Egyptian Mathematics

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EGYPTOLOGY

EGYPTIAN MATHEMATICS

MOKHTAR ELNOMROSSY

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Egyptology Egyptian Mathematics

Timeline of Ancient Egyptian Civilization

Egyptian Numerals

Egyptian Arithmetic

Egyptian Algebra

Egyptian Geometry

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Egyptian Mathematics

Timelineof

Ancient EgyptianCivilization

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Early Human

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Timeline ofAncient Egyptian Civilization

Prehistoric Era

Lower Paleolithic Age 200000 – 90000 B.C.

Middle Paleolithic Age 90000 – 30000 B.C.

Late Paleolithic Age 30000 - 7000 B.C.

Neolithic Age 7000 - 4800 B.C.

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EgyptEgyptian civilization begins more than 6000 years ago, with the largest pyramids built around 2600 B.C.

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Predynastic Period

Upper EgyptBadarian Culture 4800 – 4200 B.C.Amratian Culture (Al Amrah) 4200 – 3700 B.C.Gersean Cultures A & B (Al Girza) 3700 – 3150 B.C.

* 365 day Calendar by 4200 B.C.* From 3100 B.C exhibited numbers in millions

Lower EgyptFayum A Culture (Hawara) 4800 – 4250 B.C.Merimda Culture 4500 – 3500 B.C. (Merimda Bani Salamah)

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Egyptian Calendar

As early as 4241 B.C, the Egyptians had created a calendar made up of twelve months of 30 days, plus five extra days at the end of the year.

The Egyptian Calendar, dated 4241 B.C, is based on the solar year and daily revolution of the earth around the sun.

Evidently, to reach this feat in calculating the days the earth takes to move round the sun in one year, Egyptian by then must have possessed knowledge of astronomy and mathematics

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Dynastic Period

Early dynastic period 3150 – 2685 B.C. (Dynasties 1 & 2)Old Kingdom 2685 – 2160 B.C. (Dynasties 3 to 8) In about 2600 B.C, the Great Pyramid at Giza is constructed

First Intermediate Period 2160 – 2040 B.C. (Dynasties 9 to 11)Middle Kingdom 1991 – 1668 B.C. (Dynasties 12 & 13)

1850 BC “Moscow Papyrus” contains 25 mathematical problems

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Pyramid from Space

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Timeline ofAncient Egyptian CivilizationDynastic Period

Second Intermediate Period 1668 - 1570 B.C. (Dynasties 14 to 17)

1650 BC “Ahmes Papyrus’ contains 85 mathematical problems

New Kingdom 1570 - 1070 B.C. (Dynasties 18 to 20)Late Period 1070 - 712 B.C. (Dynasties 21 to 24)Dynasty 25 (Kushite domination) 712 – 671 B.C. Assyrian DominationSaite Period (Dynasty 26) 671 - 525 B.C. Dynasties 27 to 31 525 – 332 B.C Persian Period

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Ahmes Papyrus (Rhind)

Part of the Rhind papyrus written in hieratic script about 1650 B.C. It is currently in the British Museum. It started with a premise of “a thorough study of all things, insight into all that exists, knowledge of all obscure secrets.” It turns out that the script contains method of multiply and divide, including handling of fractions, together with 85 problems and their solutions.

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Egyptian Numerals

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Rosetta Stone & Egyptian Language

The stone of Rosette is a basalt slab (114x72x28cm) that was found in 1799 in the Egyptian village of Rosette (Rashid).

Today the stone is kept at the British Museum in London.It contains three inscriptions that represent a single text in three different variants of script, a decree of the priests of Memphis in honor of Ptolemalos V (196 BC).

The text appears in form of hieroglyphs (script of the official and religious texts), of Demotic (everyday Egyptian script), and in Greek.

The representation of a single text of the three script variants enabled the French scholar Jean Francois Champollion in 1822 to basically to decipher the hieroglyphs.

Furthermore, with the aid of the Coptic language, he succeeded to realize the phonetic value of the hieroglyphs. This proved the fact that hieroglyphs do not have only symbolic meaning, but that they also served as a “spoken language”.

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Egyptian Hieroglyphs

This is the hieroglyphic inscription above the Great pyramid’s entrance.

Egyptian written language evolved in three stages:

Hieroglyphs

Hieratic

Coptic (spoken only)

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Egyptian NumbersThe knob of King Narmer, 3000BC

The numerals occupy the center of the lower register. Four tadpoles below the ox, each meaning 100,000 record 400,000 oxen. The sky-lifting-god behind the goat was the hieroglyph for “one million”; together with the four tadpoles and the two “10,000” fingers below the goat, and the double “1,000” lotus-stalk below the god, this makes 1,422,000 goats.

To the right of these animal quantities, one tadpole and two fingers below the captive with his arms tied behind his back count 120,000 prisoners.

These quantities makes Narmer’s mace the earliest surviving document with numbers from Egypt, and the earliest surviving document with such large numbers from anywhere on the planet.

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Egyptian Numerals

Egyptian number system is additive.

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Egyptian Arithmetic

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Addition in Egyptian Numerals

365

+ 257

= 622

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Multiply 23 х 13

1 √

2

4 √

8 √

1 + 4 + 8 = 13

23 √

46

92 √

184 √

23+92+184 = 299 multiplier 13Result:

multiplicand

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Principles of Egyptian Multiplication Starting with a doubling of numbers from one,

1, 2, 4, 8, 16, 32, 64, 128, etc.

Any integer can be written uniquely as a sum of “doubling numbers”.

Appearing at most one time. 11 = 1 + 2 + 8

23 = 1 + 2 + 4 + 1644 = 4 + 8 + 32

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Binary Expansion

Any integer N can be written as a sum of powers of 2.

Start with the largest 2k ≤ N, subtract of it, and repeat the process.

147 = 128 + 19 ; 19 = 16 + 3 ; 3 = 2 +1So

147 = 128 + 16 + 2 + 1 with k = 7, 4, 1, 0

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Principles of Egyptian Multiplication

Apply distribution law:a x (b + c) = (a x b) + (a x c)

Example: 23 x 13 = 23 x (1 + 4 + 8)

= 23 + 92 + 184 = 299

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Division, 23 х ? = 299

1 √

2

4 √

8 √

23 √

46

92 √

184 √

Result: 23+92+184=299 Dividend: 1+4+8= 13

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Numbers that cannot divide evenly e.g.: 35 divide by 8

8 1

16 2

√ 32 4

4 1/2

√ 2 1/4

√ 1 1/8

35 4 + 1/4 + 1/8

doubling

half

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Unit Fractions

One part in 10, i.e., 1/10

One part in 123, i.e.,

1/123

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Egyptian Fractions

1/2 + 1/4 = 3/4

1/2 + 1/8 = 5/8

1/3 + 1/18 = 7/18

The Egyptians have no notations for general rational numbers like n/m, and insisted that fractions be written as a sum of non-repeating unit fractions (1/m). Instead of writing ¾ as ¼ three times, they will decompose it as sum of ½ and ¼.

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Practical Use of Egyptian Fraction

Divide 5 pies equally to 8 workers. Each get a half slice plus a 1/8 slice.

5/8 = 1/2 + 1/8

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Algorithm for Egyptian Fraction

Repeated use of

Example:

1 1 1

1 ( 1)n n n n

2 1 1 1 1 1

19 19 19 19 20 3803 1 1 1 1 1 1 1 1

5 5 5 5 5 6 30 6 30

1 1 1 1 1 1 1 1 1 1

5 3 15 5 6 30 7 42 31 930

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Egyptian Algebra

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Arithmetic ProgressionProblems 40 & 64 of RMP

.Now we follow the scribe’s directions word by word, but we substitute for the numbers he used those letters commonly used in modern algebraic treatment of arithmetic progression, thus:

a = first term (lowest)l = last term (highest)d = common differencen = number of termsS = sum of n terms

The scribe direct:Find the average value of the n terms = S/nThe number of differences is one less than the number of terms = (n-1)Find half of the common difference = d/2

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Arithmetic ProgressionProblems 40 & 64 of RMP

. Multiply that (n-1) by d/2 = (n-1)d/2

Then either add this to the averaged = S/n + (n-1)d/2, this is the highest term l, then, l = S/n + (n-1)d/2 or it can be written as S/n = l – (n-1)d/2, hence,

S = n/2[2l – (n-1)d]

Or subtract this from the averaged value = S/n – (n-1)d/2, this is the lowest term a, then, a = S/n – (n-1)d/2 or it can be written as S/n = a + (n-1)d/2, hence,

S = n/2[2a + (n-1)d]

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Geometric ProgressionProblems 76 & 79 of RMP

Because of the Egyptian method of performing all multiplications by continued doubling, it was natural enough that they should be interested in numbers arranged serially, and especially the series

1, 2, 4, 8, 16, …. This series is called a geometric progression whose first

term is 1 and whose common multiplier is 2 It has a special property, which the Egyptians were aware of

and which is today made use of in the design of modern digital computers

This property is that every integer can be uniquely expressed as the sum of certain terms of the series. Thus, an integral multiplier, when partitioned in this form, can be used in Egyptian multiplication.

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The importance of this property to the Egyptian scribes lies in the uniqueness of the partitioning of any multiplier.

For example: The multiplier 26 can be expressed as the sum of terms of this in one way only, namely, 2+8+16

Thus, it is understandable that the Egyptian’s attention would quite naturally be directed to the sum of certain terms of this and other series, and that those properties of progressions that could be used in subsequent calculations would interest them deeply.

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Therefore, let us look at the property of GP described in RMP 76, namely,

2, 4, 8, 16, 32, 64, …. The sum of the first 2 terms is 6 = 2x(1+ 1st term) = 6 The sum of the first 3 terms is 14 = 2x(1+1st 2 terms) = 14 The sum of the first 4 terms is 30 = 2x(1+1st 3 terms) = 30 The sum of the first 5 terms is 62 = 2x(1+1st 4 terms) = 62 The sum of the first 6 terms is 126= 2x(1+1st 5 terms) = 126

Then, by inductive reasoning the Egyptian scribe have concluded that, in any GP, whose common multiplier is the same as the first term, the sum of any number of its terms is equal to the common ratio times one more than the sum of the preceding terms.

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Equations of first degreeProblems 24 to 34 of the RMP

The eleven problems deal with the methods of solving equations in one unknown of the first degree.

Based upon the order of difficulty and method of solution, these problems fall into three groups.

The first group: Pr 24: A quantity and its 1/7 added becomes 19. What is the quantity? Pr 25: A quantity and its ½ added becomes 16. What is the quantity? Pr 26: A quantity and its ¼ added becomes 15. What is the quantity? Pr 27: A quantity and its 1/5 added becomes 21. What is the quantity?

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Each of these problems is solved by the method known as “false position’, and each deals with abstract numbers unrelated to loaves bread, hekats of grains, or the area of fields.

The scribe is showing with four similar problems, but different numbers, a general method of solution for this type of problems.

The number “falsely assumed” in each case is the simplest that could be chosen, namely, 7, 2, 4, 5 respectively.

Problem 24; Assume the false answer 7. then, 1 1/7 of 7 is 8. Then as many times as 8 must be multiplied to give 19, just so many times 7 be multiplied to give the correct number.

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1 8\2 \16

---------------------------------------- 1/2 4\1/4 \2\1/8 \1

----------------------------------------Total 2 1/4 1/8 19

Now, multiply 2 1/4 1/8 by 7\1 \2 1/4 1/8\2 \4 1/2 1/4

\4 \9 1/2 ------------------------------------------------

Total 7 15 (1/2 1/2) (1/4 1/4) 1/8 7 16 1/2 1/8

The answer, then, is 16 1/2 1/8

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Second group:Two problems constitute the second group. They are:Problem 28: A quantity and its 1/3 added together, and from the sum a third of the sum is subtracted and 10 remains. What is the quantity?

Problem 29: A quantity and its 1/3 are added together, 1/3 of this added, then 1/3 of this sum is taken, and the result is 10. What is the quantity?

Both of these problems are discussed under “think of a number”

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Third group:The third group consists of problems 30 – 34Problem 30: If the scribe says, “What is the quantity of which (1/3 1/10) will make 10”.Problem 31: A quantity, its 1/3; its 1/2 and its 1/7 added becomes 33. What is the quantity?Problem 32: A quantity, its 1/3 and its ¼ added to become 2. What is the quantity?Problem 33: A quantity, its 1/3; its 1/2 and its 1/7 added becomes 37. What is the quantity?Problem 34: A quantity, its 1/2 and its 1/4 added becomes 10. What is the quantity?

The scribe solved these problem by a different method,That of division

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Equations of second degreeSimultaneous equations

Two problems in the Berlin Papyrus appear to deal clearly with the solution of simultaneous equations, one being of the second degree.

The scribe proposed to solve the following two sets of equations:

Set 1: x2 + y2 = 100 4x – 3y = 0

Set 2: x2 + y2 = 400 4x – 3y = 0

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Egyptian Geometry

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Egyptian Triangle

Surveyors in ancient Egypt has a simple tool for making near-perfect right triangle: a loop rope divided by knots into twelve sections. When they stretched the rope to make a triangle whose sides were in the ratio 3:4:5, they knew that the largest angle was a right angle.

The upright may be linked to the male, the base to the female and the hypotheses to the child of both. So Ausar (Osiris) may be regarded as the origin, Auset (Isis) as the recipient, and Heru (Horus) as perfected result.

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Area of Rectangle

The scribes found the areas of rectangles by multiplying length and breadth as we do today.

Problem: 49 of RMPThe area of a rectangle of length 10 khet (1000 cubits) and breadth 1 khet (100 cubits) is to be found 1000x100= 100,000 square cubits.

The area was given by the scribe as 1000 cubits strips, which are rectangles of land, 1 khet by 1 cubit.

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Area of Rectangle

Problem: 6 of MMP

Calculation of the area of a rectangle is used in a problem of simultaneous equations.

The following text accompanied the drawn rectangle.

1.Method of calculating area of rectangle.2.If it is said to thee, a rectangle in 12 in the area is 1/2 1/4 of the length.3.For the breadth. Calculate 1/2 1/4 until you get 1. Result 1 1/34.Reckon with these 12, 1 1/3 times. Result 165.Calculate thou its angle (square root). Result 4 for the length.6.1/2 1/4 is 3 for the breadth.

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Area of Rectangle

Problem: 6 of MMP(In modern form)

1)A = L x b

2)L x b = 12 and b = (1/2 1/4)L

3) Then, inverse of 1/2 1/4 is 1 1/3

4)L x L = 12 x 1 1/3 = 16

5)Therefore, L = 4 for the length

6)And 1/2 1/4 of 4 is the breadth 3.

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Area of triangle

For the area of a triangle, ancient Egyptian used the equivalent of the formula A = 1/2bh.

Problem: 51 of RMPThe scribe shows how to find the area of a triangle of land of side 10 khet and of base 4 khet.

The scribe took the half of 4, then multiplied 10 by 2 obtaining the area as 20 setats of land.

Problem: 4 of MMPThe same problem was stated as finding the area of a triangle of height (meret) 10 and base (teper) 4.

No units such as khets or setats were mentioned.

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Area of CircleComputing π

Archimedes of Syracuse (250BC) was known as the first person to calculate π to some accuracy; however, the Egyptians already knew Archimedes value of

π = 256/81 = 3 + 1/9 + 1/27 + 1/81

Problem: 50 of RMPA circular field has diameter 9 khet. What is its area?The written solution says, subtract 1/9 of the diameter which leaves 8 khet. The area is 8 multiplied by 8 or 64 khet. This will lead us to the value of

π = 256/81 = 3 + 1/9 + 1/27 + 1/81 = 3.1605

But the suggestion that the Egyptian used is π = 3 = 1/13 + 1/17 + 1/160 = 3.1415

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Egyptian Geometry

Ancient Egyptian had a very large knowledge about Volumes.

The knowledge of the Egyptians about the geometry of the Pyramids and Frustums was very elaborated

It is a whole science

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Egyptian Mathematics