2 Egyptology Egyptian Mathematics Timeline of Ancient Egyptian Civilization Egyptian Numerals Egyptian Arithmetic Egyptian Algebra Egyptian Geometry
3 Egyptian Mathematics Timeline of Ancient Egyptian Civilization
4 Early Human
5 Timeline of Ancient Egyptian Civilization Prehistoric Era Lower Paleolithic Age – B.C. Middle Paleolithic Age – B.C. Late Paleolithic Age B.C. Neolithic Age B.C.
6 Egypt Egyptian civilization begins more than 6000 years ago, with the largest pyramids built around 2600 B.C.
7 Timeline of Ancient Egyptian Civilization Predynastic Period Upper Egypt Badarian 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 Egypt Fayum A Culture (Hawara) 4800 – 4250 B.C. Merimda Culture 4500 – 3500 B.C. (Merimda Bani Salamah)
8 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
9 Timeline of Ancient Egyptian Civilization 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
10 Pyramid from Space
11 Timeline of Ancient Egyptian Civilization Dynastic Period Second Intermediate Period B.C. (Dynasties 14 to 17) 1650 BC Ahmes Papyrus contains 85 mathematical problems New Kingdom B.C. (Dynasties 18 to 20) Late Period B.C. (Dynasties 21 to 24) Dynasty 25 (Kushite domination) 712 – 671 B.C. Assyrian Domination Saite Period (Dynasty 26) B.C. Dynasties 27 to – 332 B.C Persian Period
12 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.
13 Egyptian Mathematics Egyptian Numerals
14 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.
15 Egyptian Hieroglyphs This is the hieroglyphic inscription above the Great pyramids entrance. Egyptian written language evolved in three stages: Hieroglyphs Hieratic Coptic (spoken only)
16 Egyptian Numbers The 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 Narmers mace the earliest surviving document with numbers from Egypt, and the earliest surviving document with such large numbers from anywhere on the planet.
17 Egyptian Numerals Egyptian number system is additive.
18 Egyptian Mathematics Egyptian Arithmetic
19 Addition in Egyptian Numerals = 622
20 Multiply 23 х = = 299multiplier 13Result: multiplicand
21 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 = = =
22 Binary Expansion Any integer N can be written as a sum of powers of 2. Start with the largest 2 k N, subtract of it, and repeat the process. 147 = ; 19 = ; 3 = 2 +1 So 147 = with k = 7, 4, 1, 0
23 Principles of Egyptian Multiplication Apply distribution law: a x (b + c) = (a x b) + (a x c) Example: 23 x 13 = 23 x ( ) = = 299
25 Numbers that cannot divide evenly e.g.: 35 divide by /2 21/4 11/ /4 + 1/8 doubling half
26 Unit Fractions One part in 10, i.e., 1/10 One part in 123, i.e., 1/123
27 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.
28 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
29 Algorithm for Egyptian Fraction Repeated use of Example:
30 Egyptian Mathematics Egyptian Algebra
31 Arithmetic Progression Problems 40 & 64 of RMP. Now we follow the scribes 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 difference n = number of terms S = sum of n terms The scribe direct: Find the average value of the n terms = S/n The number of differences is one less than the number of terms = (n-1) Find half of the common difference = d/2
32 Arithmetic Progression Problems 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]
33 Geometric Progression Problems 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.
34 Geometric Progression Problems 76 & 79 of RMP 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, Thus, it is understandable that the Egyptians 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.
35 Geometric Progression Problems 76 & 79 of RMP 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+ 1 st term) = 6 The sum of the first 3 terms is 14 = 2x(1+1 st 2 terms) = 14 The sum of the first 4 terms is 30 = 2x(1+1 st 3 terms) = 30 The sum of the first 5 terms is 62 = 2x(1+1 st 4 terms) = 62 The sum of the first 6 terms is 126= 2x(1+1 st 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.
36 Equations of first degree Problems 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?
37 Equations of first degree Problems 24 to 34 of the RMP 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.
38 Equations of first degree Problems 24 to 34 of the RMP 1 8 \2\ /2 4 \1/4 \2 \1/8 \ 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/ Total 7 15 (1/2 1/2) (1/4 1/4) 1/ /2 1/8 The answer, then, is 16 1/2 1/8
39 Equations of first degree Problems 24 to 34 of the RMP 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
40 Equations of first degree Problems 24 to 34 of the RMP Third group: The third group consists of problems 30 – 34 Problem 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
41 Equations of second degree Simultaneous 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: x 2 + y 2 = 100 4x – 3y = 0 Set 2: x 2 + y 2 = 400 4x – 3y = 0
42 Egyptian Mathematics Egyptian Geometry
43 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.
44 Area of Rectangle The scribes found the areas of rectangles by multiplying length and breadth as we do today. Problem: 49 of RMP The 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.
45 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 1/3 4.Reckon with these 12, 1 1/3 times. Result 16 5.Calculate thou its angle (square root). Result 4 for the length. 6.1/2 1/4 is 3 for the breadth.
46 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.
47 Area of triangle For the area of a triangle, ancient Egyptian used the equivalent of the formula A = 1/2bh. Problem: 51 of RMP The 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 MMP The 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.
48 Area of Circle Computing 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 RMP A 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 = But the suggestion that the Egyptian used is π = 3 = 1/13 + 1/17 + 1/160 =
49 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