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EGYPTOLOGY EGYPTIAN MATHEMATICS MOKHTAR ELNOMROSSY.

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1 EGYPTOLOGY EGYPTIAN MATHEMATICS MOKHTAR ELNOMROSSY

2 Egyptology Egyptian Mathematics
Timeline of Ancient Egyptian Civilization Egyptian Numerals Egyptian Arithmetic Egyptian Algebra Egyptian Geometry

3 Timeline of Ancient Egyptian Civilization
Egyptian Mathematics Timeline of Ancient Egyptian Civilization

4 Early Human Homo [houmou] : man, sapien [sæpiən]: wise. Homo sapiens sapiens stands for wise, wise man.

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. 1 million = 1,000,000; 1 billion = 1000 million.

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 – 4200 B.C. Amratian Culture (Al Amrah) – 3700 B.C. Gersean Cultures A & B (Al Girza) – 3150 B.C. * 365 day Calendar by 4200 B.C. * From 3100 B.C exhibited numbers in millions Lower Egypt Fayum A Culture (Hawara) – 4250 B.C. Merimda Culture – 3500 B.C. (Merimda Bani Salamah) 1 million = 1,000,000; 1 billion = 1000 million.

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 The great pyramid is located near Giza. It was built by the Egyptian pharaoh Khufu around 2560 BC over a period of 20 years. When it was built, the Great pyramid was 146m. Over the years, it lost for 10 m off the top. It is the tallest structure on Earth for 4300 years. The base line is 229 m in length. It is a square to within 0.1% accuracy.

9 Timeline of Ancient Egyptian Civilization
Dynastic Period Early dynastic period – 2685 B.C. (Dynasties 1 & 2) Old Kingdom – 2160 B.C. (Dynasties 3 to 8) In about 2600 B.C, the Great Pyramid at Giza is constructed First Intermediate Period – 2040 B.C. (Dynasties 9 to 11) Middle Kingdom – 1668 B.C. (Dynasties 12 & 13) 1850 BC “Moscow Papyrus” contains 25 mathematical problems 1 million = 1,000,000; 1 billion = 1000 million.

10 Pyramid from Space The great pyramid is located near Giza. It was built by the Egyptian pharaoh Khufu around 2560 BC over a period of 20 years. When it was built, the Great pyramid was 146m. Over the years, it lost for 10 m off the top. It is the tallest structure on Earth for 4300 years. The base line is 229 m in length. It is a square to within 0.1% accuracy.

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) – 671 B.C. Assyrian Domination Saite Period (Dynasty 26) B.C. Dynasties 27 to – 332 B.C Persian Period 1 million = 1,000,000; 1 billion = 1000 million.

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. Papyrus [pə’paiərəs]: paper made from the papyrus plant by cutting it in strips and pressing it flat; used by ancient Egyptians and Greeks and Romans. Tall sedge of the Nile valley yielding fiber that served many purposes in historic times. Rhind Papyrus perhaps is the oldest math text ever existed.

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”. decipherThe name Rosetta refers to the crucial breakthrough in the research regarding Egyptian hieroglyphs. It especially represents the "translation" of "silent" symbols into a living language, which is necessary in order to make the whole content of information of these symbols accessible. The name Rosetta is attached to the stone of Rosette. This is a compact basalt slab (114x72x28 cm) that was found in July 1799 in the small Egyptian village Rosette (Raschid), which is located in the western delta of the Nile. 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 honour of Ptolemaios V. (196 b.c.). 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 mentioned script variants enabled the French scholar Jean Francois Champollion ( ) in 1822 to basically decipher the hieroglyphs. Furthermore, with the aid of the Coptic language (language of the Christian descendants of the ancient Egyptians), 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 pyramid’s entrance. Egyptian written language evolved in three stages: Hieroglyphs Hieratic Coptic (spoken only) This is the hieroglyphic inscription above the Great Pyramid’s entrance. From Egyptian written language evolved in three stages, hieroglyphs, hieratic, and 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 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. The mace head recorded victory of the first King of Egypt. 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 Heh- 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.

17 Egyptian Numerals Egyptian number system is additive.
Additive means that the order of these symbols does not matter.

18 Egyptian Mathematics Egyptian Arithmetic

19 Addition in Egyptian Numerals
365 + 257 = 622 To this day, it is not entirely clear how the Egyptians performed addition and subtractions.

20 Multiply 23 х 13 multiplicand 23 √ 46 92 √ 184 √ 1 √ 2 4 √ 8 √
= 13 A check means that this number will be counted to add up the desired multiplier or results. If we rotate 90 degree of the above figure, and use 1 for the check, and 0 for the non-check, we get a binary number represent of the number 13. “Eureka”, the Egyptians could have discovered binary numbers. Result: = 299 multiplier 13

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 = 23 = 44 = This is nothing but representing any positive integer as a binary expansion.

22 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 = ; 19 = ; 3 = 2 +1 So 147 = with k = 7, 4, 1, 0 Power of 2 from k=0 to 8: 1, 2, 4, 8, 16, 32, 64, 128, 256.

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 Note that a + b = b + a is called commutative law, and a + ( b + c) = (a + b) + c is called associative law.

24 Division, 23 х ? = 299 23 √ 46 92 √ 184 √ 1 √ 2 4 √ 8 √ Division and multiplication use the same method, except that the role of multiplier and result are interchanged. Need guess work, or not? Result: =299 Dividend: 1+4+8= 13

25 Numbers that cannot divide evenly e.g.: 35 divide by 8
8 1 16 2 4 1/2 √ 2 1/4 √ 1 1/8 /4 + 1/8 doubling half Of course, the result is 4 + 3/8, or The Egyptians have not developed the concept of decimal fractions (0.375). They represent the result as 4 + ¼ + 1/8.

26 Unit Fractions One part in 10, i.e., 1/10 One part in 123, i.e., 1/123
A web page on Egyptian fraction:

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
5/8 = 1/2 + 1/8 Divide 5 pies equally to 8 workers. Each get a half slice plus a 1/8 slice.

29 Algorithm for Egyptian Fraction
Repeated use of Example: Other formulas are also available, e.g., 2/(3k) = 1/(2k) + 1/(6k), or 2/n = 1/n + 1/(2n) + 1/(3n) + 1/(6n).

30 Egyptian Mathematics Egyptian Algebra

31 Arithmetic Progression Problems 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 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 . Picture from Although they may count to two, they still have a very good sense of large or small. There are two aspects to numbers, cardinal (size, one, two, three, four, …) and ordinal (sequence, first, second, third, etc).

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] . Picture from Although they may count to two, they still have a very good sense of large or small. There are two aspects to numbers, cardinal (size, one, two, three, four, …) and ordinal (sequence, first, second, third, etc).

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. De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6.

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 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. De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6.

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+ 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. De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6.

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. Pr 26: A quantity and its ¼ added becomes 15. Pr 27: A quantity and its 1/5 added becomes 21. De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6.

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. De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6.

38 Equations of first degree Problems 24 to 34 of the RMP
\2 \16 1/ \1/4 \2 \1/8 \1 Total /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 (1/2 1/2) (1/4 1/4) 1/8 /2 1/8 The answer, then, is /2 1/8 De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6.

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. Both of these problems are discussed under “think of a number” De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6.

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? De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6. 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: x2 + y2 = 100 4x – 3y = 0 Set 2: x2 + y2 = 400 De-associate number from concrete objects is the first step in sharp the concept of numbers. See T. Dantzig, “Number, the Language of Science”, The Free Press, New York 1967, p.6.

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. This solid figure is also known as frustum. This problem was found in Moscow Papyrus. The Egyptians thought that the numbers and their mathematics are given by god; and they does not seem to have the need to justify their methods. Some of the formulas they devise may only be approximate. For example, in the Temple of Horus at Edfu delicatory inscription, area of the 4-sided quadrilateral was given the formula A = (a+c)/(b+d)/4, where a, b, c, d are the lengths of the consecutive sides, which is incorrect.

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. Method of calculating area of rectangle. If it is said to thee, a rectangle in 12 in the area is 1/2 1/4 of the length. For the breadth. Calculate 1/2 1/4 until you get 1. Result 1 1/3 Reckon with these 12, 1 1/3 times. Result 16 Calculate thou its angle (square root). Result 4 for the length. 1/2 1/4 is 3 for the breadth. This solid figure is also known as frustum. This problem was found in Moscow Papyrus. The Egyptians thought that the numbers and their mathematics are given by god; and they does not seem to have the need to justify their methods. Some of the formulas they devise may only be approximate. For example, in the Temple of Horus at Edfu delicatory inscription, area of the 4-sided quadrilateral was given the formula A = (a+c)/(b+d)/4, where a, b, c, d are the lengths of the consecutive sides, which is incorrect.

46 Area of Rectangle Problem: 6 of MMP (In modern form) A = L x b
L x b = 12 and b = (1/2 1/4)L Then, inverse of 1/2 1/4 is 1 1/3 L x L = 12 x 1 1/3 = 16 Therefore, L = 4 for the length And 1/2 1/4 of 4 is the breadth 3. This solid figure is also known as frustum. This problem was found in Moscow Papyrus. The Egyptians thought that the numbers and their mathematics are given by god; and they does not seem to have the need to justify their methods. Some of the formulas they devise may only be approximate. For example, in the Temple of Horus at Edfu delicatory inscription, area of the 4-sided quadrilateral was given the formula A = (a+c)/(b+d)/4, where a, b, c, d are the lengths of the consecutive sides, which is incorrect.

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. This solid figure is also known as frustum. This problem was found in Moscow Papyrus. The Egyptians thought that the numbers and their mathematics are given by god; and they does not seem to have the need to justify their methods. Some of the formulas they devise may only be approximate. For example, in the Temple of Horus at Edfu delicatory inscription, area of the 4-sided quadrilateral was given the formula A = (a+c)/(b+d)/4, where a, b, c, d are the lengths of the consecutive sides, which is incorrect.

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 = This solid figure is also known as frustum. This problem was found in Moscow Papyrus. The Egyptians thought that the numbers and their mathematics are given by god; and they does not seem to have the need to justify their methods. Some of the formulas they devise may only be approximate. For example, in the Temple of Horus at Edfu delicatory inscription, area of the 4-sided quadrilateral was given the formula A = (a+c)/(b+d)/4, where a, b, c, d are the lengths of the consecutive sides, which is incorrect.

49 Egyptian Geometry It is a whole science
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 This solid figure is also known as frustum. This problem was found in Moscow Papyrus. The Egyptians thought that the numbers and their mathematics are given by god; and they does not seem to have the need to justify their methods. Some of the formulas they devise may only be approximate. For example, in the Temple of Horus at Edfu delicatory inscription, area of the 4-sided quadrilateral was given the formula A = (a+c)/(b+d)/4, where a, b, c, d are the lengths of the consecutive sides, which is incorrect.

50 Egyptian Mathematics Thank You


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