Egypt and Mesopotamia Flashcards

Question

IMPORTANT

Answer 1

7 marks question on Egypt need to be able to write a paragraph on the rhind papyrus: Henry Rhind felt ill, doctor told him to take holidays further and further away. Winter in egypt. Henry Rhind bought the Papyrus at Luxor in 1858. Rhind papyrus is in two parts and listed as two numbers in the british museum, found half . Fragments from the break are found by the NY Hisorical society in 1922, now in Brooklyn Museum. This is a training manual for scribes, copies by Ahmes in 1650 BC comprises 84 worked problems and promises insight into all that exists. Translated by Eisenlohr, Peet and CHace Robins and Shute's Rhinf Mathematical Papyrus has 24 coloured plates of the Papyrus. red ink for problem and black ink for answer

Answer 2

* 24 to 27 aha problems false position, algebra * 48,50 area of a circle (implicit pi) * 41,42,43 vlumes of cylindrical granaries * 56-60 seqts, slopes, pyramids (trigonometry) *79 geometric progression (recreation) NO PROBLEM ON THE RMP or MMP concerns finding the circumference of a circle

Answer 3

round field diameter 9. Find its area subtract 1/9 of diameter, 1. Remainder is 8 X 8 to make 64. This is the area. ``` Do it thus: 1 9 1/9 1 is taken away leaves 8. 1 8 2 16 4 32 \8 64 ``` Area is 64 * passing from the diameter 9 of the circle to the side 8 of the square is affected by subtracting 1/9 of 9 from itself to leave 8. The Egyptians had no concept of the fractions such as 8/9, operating only with unit fractions, like 1/9 here. The diagram shows a circle enclosing a number 9

Answer 4

a cubit is 7 hands pyramid builders measured vertical distances in cubits and horizontal distances in hands, slopes are measured in seqts. (the seqt s of a face of a pyramid P is half its base in hands / height in cubits) the sect of a pyramid is the ratio of the horizontal displacement of a face (half base) enhance to its vertical displacement (height) in qubits, and is an inverse measure of steepness with gentler slopes having higher seqts. Seqt of a pyramid with height h cubits and base b cubits is 7(0.5b)/h hands per cubit The angle of slope theta of P is the angle between one of its faces P and its horizontal base, tan theta=2h/b = 7/s and s =7cot theta. RMP problems 56/59 concerning the seqts of pyramids, lie one below each other at the end of the recto of BM10057. Each is accompanied by a sketch of the pyramid, the numerical values of its base and height written alongside. RMP problem 57 the seqt of a pyramid is 5 hands and one finger per cubit, the side of its space is 140 cubits. What is it height? A finger is a quarter of a hand. Solution in modern Let pyramid have height hqbutts. Then the sect is 7(140/2)/h = 21/4 whence h=(7x140x4)/(2 x 21)=93 1/3

Answer 5

48:ONLY PROBLEM WITH NO RED INK, NO STATEMENT OF PROBLEM compare the area of a circle and that of its circumscribing Square ``` circle diameter 9 1 8 2 16 4 32 \8 64 ``` Square side 9 ``` \1 9 2 18 4 36 \8 72 sum 81 ``` Area circle ~octagon = 63~64 taking the area of the circle to be 64 * only RMP problem with no statement - no red ink Diagram is square circumscribing irregular polygon , octagon? with a nine drawn inside. Beneath, 2 multiplications are set out : 9x9 =81 and 8x8=64 by doubling and edition with check marks

Answer 6

RMP problem 79:concerns the geometric series. It is unclear what the problem is (as there is no red ink??? only 48??) describe sets out in a column: 7 houses, 49 cats, 343 mice, 2401 ears of Grain, 16807 hekats of grain, total 19607. another column shows 19607 was found as 7(1+7+49+343+2401), so the Egyptians knew something of geometric progressions. the problem may just be a diversion, perhaps a precursor of the nursery rhyme: as I was going to St Ives, I met a man with seven wives; each wife had seven sacks, each sack had seven cats, each cat had 7 kits; kits, cats, sacks and wives, how many we're going to St Ives?

Answer 7

coined the greatest Egyptian pyramid by ET Bell. This problem calculates the exact volume V = 56 of the frustum of a square pyramid of height h = 6, side of an upper face a = 2, and side of lower face b = 4, using the equivalent of the correct formula: V= (h/3)(a^2+ab+b^2) a diagram is given along with the full details of all the sub calculations performed, written both in and around it. On two occasions, a pair of legs walking from left to right indicates operation of squaring. *the frustum can be seen to have been cut from a pyramid of height 12 and base 4. This had seqt (7x2)/12 = 7/6 hands per cubit.

Answer 8

concern pyramids as shown by triangles illustrating them. Pyramid builders used seqts to maintain constant slopes in their constructions. let pyramid have height h cubits and angle of slope theta. Then seqt = 5 1/4 = (7x70)/h and h= (280/3) = 93 1/3 cubits. Now tan theta= height/ 0.5base = (280/3)/70 = 4/3 and theta = tan -1 (4/3) = 53 to the nearest degree

Answer 9

problem 15 of the rhind papyrus, implicit values of Pi, finding the area of a circle diagram of problem 48 on the Rhind papyrus finding areas of the circle: only one that does not have the red ink, all it has is a diagram of the square circumscribing an irregular polygon which is possibly an octagon with a number 9 script inside of it. underneath are the multiplications of 9 x 9 and 8 by 8. Approximates area. fraction notation in egyptian * *7 marks is on Egyptian mathematics * *16 marks on Mesopotamia?

Answer 10

The land between the rivers 3000 BC Sumerian civilization flourished the fertile crescent between the Tigris and Euphrates rivers, now in Iraq. Babylonian refers to any people occupying the region up to 539 BC, then Babylon fell to Persia. Babylonian maths continued through persia and greek rule up to Christianity. ``` 3000 BC SUmerians 2500BC cuneiform script sexagesimal, place-value number system 1800-1600BC old babylonian 539 BC Persia defeats Babylon 330 BC Alxander the Greats conquest 312 BC+ Seleucid period:0 symbol 1849 AD Rawlinson deciphers cuneiform 1935 AD+ Neugebauers pioneering work ``` *history of Mesopotamia is a story of Invaders who attracted by the richness of the Land conquered their predecessors only to be overcome by new Intruders

Answer 11

the Babylonians inscribe their numbers in clay using a triangular stylus, producing a wedge-shaped script known as cuneiform. ``` Babylonic cuneiform (wedge-shaped) script made by impressing triangular stylus into moist clay, then baking. ``` inscribed with the clay is still moist and baked hard in an oven or by The Heat Of The Sun

Answer 12

our knowledge of that mathematics rests on 400 also clay tablets either TABLE TEXTS OR PROBLEM TEXTS from the old Babylonian period (1800 -1600) BC *published in 20th-century by assyriologists such as: Thureau-Dangin, Sachs, Neugebauer (last century and a half) *Exhibits of Babylonian Cuneiform Tablets National Museums: Berlin, London, Paris US Universities: Columbia, Pennsylvania, Yale inscribed on bottom, sides etc

Answer 13

what number means related to where placed unit symbols were placed immediately to the right of those for the 10s number system base 60 *the greatest mathematical achievement was a sexagesima place value number system

Answer 14

*used by babylonians employed two symbols: a thin vertical wedge | for one and a broad horizontal wedge < for ten. Numbers up to 59 were written as the approximate number of 10 symbols, followed closely by that of unit symbols. It's extended to sexagesimal fractions, but lacked a 0 and their sexagesima point. *it was SEXAGESIMAL which means it was to the base 60 and employed POSITIONAL NOTATION

Answer 15

early Babylonians had no 0 symbol, and did not formally separate sets of decimal places or indicate the sexagesimal decimal point. Today we use: 0, zero , decimal place= ; sexigesimal point when transcribing into Modern sexy decimal notation. since the Babylonians had no sexagesimal point, they operated a floating sexagesimal point, which gave them unprecedented flexibility in their calculations, because there was no symbolic distinction between whole numbers and fractions. Thus << could be 20, 1200=(20x60), 72,000=(20x60^2), (1/3)=(20 x (1/60)) or (1/180)=(20x(1/60^2)) etc *from about 300 BC a 0 symbol, two small oblique wedges, came into use for intermediate positions only

Answer 16

This is found on YBC 7289 ( Yale) |<

Answer 17

the Babylonians divided by a number by multiplying by its reciprocal, found from a table text, which usually only included reciprocals of Regular sexagesimals.

Answer 18

a regular sexagesimal is a positive integer whose reciprocal has finite sexagesimal expansion or a positive integer of the form 2^x 3^y 5^z for some non-negative integers x,y,z. 125 is a regular sexagesima l: because 1/125 has finite sexagesimal expansion 0;0,28,48 alternatively 125=5^3 can be expressed in the form 2^x 3^y 5^z 7 is not: 1/7 has infinite recurring sexagesimal expansion 0;(8, 34, 17) overlined() 7 cannot be expressed in the form 2^x 3^y 5^z

Answer 19

* every number, large, small, integer, fraction can be written using the two cuneiform symbols | and < * computations with sexagesimal fractions are performed in an identical way to those with integers * 9 of the first 10 positive integers are regular sexagesimals * allowed any number however large or small to be expressed and computations with fractions could be performed in the same way as those with whole numbers

Answer 20

* ambiguity caused by lack of a 0 symbol and, to a lesser extent, a sexagesimal point * Poor cipherization- 59 needs 14 symbols same number for x60 we denote ; for where sexagesimal point used to be

Answer 21

BM13901 YBC7289 plimpton 322 obverse and reverse 7% egyptian maths 16% babylonian * must be able to write a paragraph about all three tablets * regular sexagesimal and by either definition in plimpton 322 as they only use regular sexagesimals to be exact 16 marks 1800 bc old babylonian period

Answer 22

Babylonian tablet. Columbia University, NEW YORK 12.7 cm by 8.8 cm;right half of larger tablet;4 columns, each of 15 numerals, top left damaged;right column labels Rows 1 to 15. if x,a,c are the first entries on a given row, a is the shortest, c is the longest side of a right-angled triangle ABC with x =sec^2 A= c^2/(c^2-a^2). Side b is not on the tablet but is sqrt(c^2-a^2). The x's (A's) decrease steadily from top 2.0 (45°) to bottom 1.4, (32 degrees) * Boyer corrected 4 errors:difference in writing * shows that the early Babylonians were aware of Pythagorean triples and knew some number theory.

Answer 23

Babylonian tablet British Museum 12 cm by 20 cm problem text, analysed by Neugerbauer 1937. comprises 24 graded quadratic problems with Solutions, three of the 11 on the reverse missing due to damage becoming more complex;involving several squares *one of the three tablets we must write a paragraph about

Answer 24

Babylonian tablet Yale University;outline of square and diagonals with three numbers marked: 30 by NW side; 1;24,51,10 (root 2 to 6 d.p) on diagonal; 42;25,35 (30 root(2) diagonal to 4dp) below diagonal. YALE BABYLONIAN COLLECTION YBC Neugebauer & Sachs *it is a round geometrical diagram text 7.6 cm in diameter, showing a square with its diagonals drawn, and three inscribed cuneiform numbers: 30 (three broad wedges) along NW side ; 1;24,51,10 on the horizontal diagonal; and an upward sloping 42;25,35 below this. despite superficial damage, all numerals are clearly legible. The 30 is the length of a side of the square. The 1;24,51,10 = 1 + (24/60) + (51/60^2) + (10/60^3) = 1.414213 6dp is an approximation to root 2 to 6 decimal places. The Babylonians knew that the diagonal of a square is root 2 times an edge. Here we note that (The (1;24,51,10 )x 30 = 42;25,35 is the upward sloping number below the diagonal. 6.dp equiv 1;24,51,10 ~1.414213 root 2 ~ 1.414214 30root2 ~42.426407 42;25,35 ~ 42.426389

Answer 25

problem: I add the area and 4/3 of the side of my Square: 0;55. what is the side? solution: 4/3 1;20. Halve 0;40. Square 0;26;40. Add on 0;55:1;21;40. Square root: 1;10. Subtract 0;40 (that was squared), 0;30 is the side. this problem and its solution show the Babylonian sexagesima place value arithmetic coping admirably with fractions, addition, subtraction, multiplication, division and square roots, and exhibits that rhetorical algebra sentences not symbols at work in finding the solution. This is to the quadratic equation x^2+(4/3)x = 11/12. The sol x=0.5 is reached by an accepted, but unexplained verbal algorism, typical of Babylonian recipe methods. The Babylonians knew not of negative numbers, so only the positive square root of 49/36 is given. resulting in just one root of the equation the positive one. The solution of general quadratic equations marks the high point of the Babylonian algebra

Answer 26

Here we note that (The (1;24,51,10 )x 30 = 42;25,35 is the upward sloping number below the diagonal. multiplication by 30 is easily performed in sexagesimal arithmetic, being essentially division by 2. the tablet shows the effectiveness of the Babylonians number system in obtaining this excellent approximation of root 2 perhaps by a square root algorithm it suggests that they knew a special case of Pythagoras's theorem, where are where are irrational numbers and had great interest in geometry per se. *this tablet gives an approximation to route to accurate to almost 6 decimal places

Answer 27

first column entries decrease Steadily. heading of the second contains the word width and third column diagonal. let x a c be first 3 numbers in a row of table text plimpton 322 not 11th. Then x,a,c(b) can be written in terms of Regular sexagesima with p and q, p>q. thus: x=(p^2+q^2)^2/(2pq)^2 a=p^2-q^2 c=p^2+q^2 b=2pq Given a and c, then p and q can be found from eq p= root( 0.5(c+a)) and q= root( 0.5(c-a)), and x calculated; x can also be found directly from a and c using equation x=c^2/(c^2-a^2) ``` On line 1, a=119, c=169 FInd p,q,x: we have c+a= 288=2p^2 and c-a=50=2q^2 , from which we deduce that p = 12 and q = 5, both regular sexagesimals. Thus ``` x=(p^2+q^2)^2/(2pq)^2 = (169^2)/(2(12)(5))^2 = (169/120)^2=28561/14400=1;59,0,15 . In theory, all first entries x on plimpton 322 are exact, since P and Q are regular sexagesimals. In practice, all that can be read of the first row x in the damage top left corner is the final 15.

Answer 28

a positive integer is a regular sexagesima if it's reciprocal has a finite sexagesima expansion or it has the form 2^x3^y5^z, where x,y,z are non-negative integers.

Answer 29

the sexagesimal expansion of 1/7: 1 divided by 7 is 0, remainder 1; 1x60 divided by 7 is 8 remainder 4; 4x60 divided by 7 is 34 remainder 2; 2x60 divided by 7 is 17 remainder 1. From here the process repeats, showing 1/7 has infinite repeating decimal 0;(8,34,17) overlined and by the first criterion it is not a regular sexagesima clearly 7 is not have the required form 2^x3^y5^z for non negative integers x,y,z and so by the second criterion is not a regular sexagesima

Answer 30

The sexagesimal expansion of 1/51840; 1 divided by 51840 is 0, reminder 51840; 1x60 divided by 51840 is 0 remainder 51840; 60x60 divided by 51840 is 0, remainder 51840; 60x60x60 divided by 51840 is 4 remainder 8640; 60 x 8640 divided by 51840 is 10. Thus 1/51840 is 0;0,0,4,10 and is a finite sexagesimal expansion. and by the first criterion is a regular sexagesimal. Also 51840=2^7 3^4 5^1

Answer 31

216=2^3 3^3 by listing in order one primefactor of each after 200 other than 2,3,5,

Answer 32

1/216 regular sexagesimal expansion 1 divided by 216 is 0, remainder 216; 60 divided by 216 is 0 remainder 216; 60x60 divided by 216 is 16 remainder 144; 144x60 divided by 216 is 40. Thus it is0;0,0,16,40.

Answer 33

*15 rows, 4 rows, broken off corner? the first 3 entries on a general row of plimpton 322, eleventh excepted, are for some regular hexadecimals p an q: (p^2+q^2)^2/(4p^2q^2) p^2-q^2 p^2+q^2 the first of these is c^2/(c^2-a^2), where a and c are respectively the second and third entries on the row. Given p^2-q^2=319 and p^2+q^2=481. adding and subtracting these equations we find that 2p^2=800, 2q^2=162, p=20 and q=9. Both regular sexagesimals.

Answer 34

(481)^2/(2x400x81) (481)^2/(360^2)=1, remainder 101,761; (101,761x60)/(360)^2 =47, remainder 14,460; (14,460)x60/(360)^2 =6, remainder 90,000; 90,000 x60/(360)^2=41, remainder 86,400; (86,400) x 60/(360)^2 =40 the first entry on row 6 of Plimpton 322 is 1;47,6,41,40

Answer 35

Subtract side of square from the area: 14,30 ``` write down 1 Divide by 2 0;30 Multiply 0;30 and 0;30: 0;15 Add 14,30 : 14,30;15 This is 29;30 squared Add 0;30 (which was squared) to 29;30:30, The side of square ``` ``` MODERN x^2-1x=870 Subtract side of square from area: 870 Write down 1 Divide by two ½ Multiply ½ and ½ : ¼ Add 870 : 870¼ This is 29½ squared Add ½ (which was squared) to 29½ : 30, (The) side of square. ```

Answer 36

The key to reading Babylonic cuneiform was found in 1846 by Henry Creswicke Rawlinson, British Consul in Baghdad, who deciphered the Old Persian part of a trilingual cuneiform text on the Behistun Cliff. Later he deciphered the Elamite and Babylonian parts of the text, and noticed the sexagesimal place-value number system. THE BEHISTUN CLIFF This monumental, thirteen panel inscription of 516 BC, cut on a sheer rock face, high above an ancient caravan route in west Iran, records the victories of Darius the Great and depicts him addressing a group of prisoners.

Answer 37

``` x^2 - bx = c Write down b Divide by two ½b Multiply ½b and ½b : ¼b^2 Add c : ¼b^2 + c This is root(¼b^2 + c) squared Add ½b (which was squared) to root(¼b^2 + c) ``` x = ½ b + root(¼b^2 + c)

Answer 38

first of all explain it: halving squaring it etc what method is being done then comments and explanations: eg problem 2 * Quadratics not uncommon in Babylonian mathematics (unheard in Egyptian) * Specific numerical problem- never general * Sexagesimal place-value number system effective- no trouble performing the operations roots, adding etc * Rhetorical algebra- didnt have a formula but wrote out in words, long-hand * Algorithmic/recipe solution, no explanation * Only positive square root recognized-one solution * Areas and sides subtracted, not practical- wouldn't normally do this, not a practical solution * Exact square root suggests use in scribal training- teaching contexts and exercises from the nice square roots * Verbal commentary possibly given to trainee scribes

Answer 39

BM13901 rectangular tablet with many problems obverse? YBC7289 YALE circlular tablet with diamond shape in middle 1,24,51,10 42,2,5,35 side 30 written plimpton 322 landscape tablet with rows(4? labelled)

Answer 40

even the oldest tablets attest the high level of computational skill arithmetic process as being carried out with the aid of table texts showing multiplication tables, tables of reciprocals, squares, cubes, square roots and exponentials.. exponentials were used together with linear interpolation to solve problems on compound interest while reciprocal tables were used to reduce division to multiplication.

Answer 41

algebra did the most impressive mathematics by 2000 BC they had developed an extensive rhetorical algebra, solve quadratic equations by a method equivalent to substitution in a general formula, solved simple cubic and quartic equations and mustard simultaneous equations with several unknowns. although general algebraic methods were understood they were only illustrated by particular cases. Babylonian geometry was not formal but closely related to Practical menstruation. They found areas and volumes of basic geometric figures, but not all of their rules were correct. They knew Pythagoras's theorem, although most likely not approve. Calculations of circumferences and areas of circles usually implied a value of Pi of 3, but in one example a better approximation was implied.

Answer 42

the overwhelming impression is the study of numbers, and techniques for solving problems involving numbers. Where the numbers arise from where the land measurement, economic questions, geometrical objects, or just abstractly seems a relatively secondary matter.

Answer 43

16 x 60 + 40x1 | .|

Answer 44

keeps deleting but 10*60 + 10

Answer 45

some mistakes made by the scribe titles at the top: ___ width diagonal 15 rows EXAM: we should be able to work out the number which is missing given a row :third side of triangle isn't on the tablet, by pythagoras and then work out the angle. The middle side (adjacent) is missing from the tablet for the right angles triangles. The number that isn't on the tablet is a regular sexagesimal. because reciprocal has finite expansion.

Answer 46

width a = 1 , 59 = 119 diagonal c = 2 , 49 = 169 b^2 = c^2 - a^2 = 169^2 - 119^2 = 120^2 b = 120 = 2^3 . 3^1 . 5^1 A roughly equal to 45 degrees (nearest)

Answer 47

column 1 2 3 4 c^2/(c^2- a^2) a c row c^2/(c^2- a^2) , diagonal c , width, (row number on right) ordered in decreasing angles b^2= c^2-a^2 : b isnt on the tablet, has form 2^x3^y5^z

Answer 48

972 / 722 = 1 ; 48 , 54 , 1 , 40 65 97 5 a~ 42 degrees b=72 = 2^33^25^0 97^2/72^2=1;48,54,1,40 ~ 1.813007716...

Answer 49

Pythagorean triple: a = p^2 - q^2 ; b = 2pq ; c = p^2 + q^2 p,q are regular sexagesimals qith p>q side not on tablet: b=2pq you'll be given the two numbers on the middle of the row and you calculate p and q given a and c. first entry on the row is (p^2+q^2)^2/4p^2q^2 = (c^2/b^2) which is always finite sexagesimal expansion

Answer 50

except eleventh (p^2+q^2)^2/4p^2q^2 (c^2/b^2) p^2- q^2 = a p^2 + q^2 = c row fifth row p=9 q=4 (9^2+4^2)^2/4 .9^2.4^2 9^2-4^2 9^2 + 4^2 5

Answer 51

see booklet | see past exam papers

Answer 52

in MATHEMATICAL CUNEIFORM TEXTS by Neugebauer & Sachs published these tablets

Answer 53

0;1,1,1,1,1,... infinite so irregular sexagesimal

Answer 54

this is solving quadratic equation x^2 +px=q where p = 0;40, q= 0 ;35. Recipe Leads first to (0.5p)^2 thensqrt((0.5p)^2+q), the side.

Answer 55

solution: let z^2=(p/2)^2-q. Then x=(p/2)/+z, y=(p/2)-z. clearly x and y satisfy the equations, which are equivalent to the quadratic equation x^2-(p+q)x+q=0. With p=6;30 and q=7;30, z=7/4 and x=5, y = 1.5 or vice versa

Answer 56

x^2+px+q=0 | is missing as it has no positive roots

Answer 57

0 symbol the Seleucid period 312-63 bc Seleucus, one of Alexander’s leading generals, became satrap (governor) of Babylonia in 321, two years after the death of Alexander.

Answer 58

using Pythagoras's theorem. *shows that the early Babylonians were aware of Pythagorean triples and knew some number theory.

Answer 59

0;57,36 3(1/8) Area=(8/9 diameter)^2 `~256/81 * r^2 6% increase by squaring and doubling

Answer 60

| || Big space in between 1*60 + 2

Answer 61

1,21,11 it is | "less than""less than"| "less than"| 1*60^2 +(21*60) +11

Answer 62

BM 13901: Problem 2 Commentary Quadratics not uncommon in Babylonian mathematics Specific numerical problem Sexagesimal place-value number system effective Rhetorical algebra Algorithmic/recipe solution, no explanation Only positive square root recognized Areas and sides subtracted, not practical Exact square root suggests use in scribal training Verbal commentary possibly given to trainee scribes

Answer 63

One of the most remarkable documents of Old Babylonian mathematics O. Neugebauer on PLIMPTON 322

Egypt and Mesopotamia Flashcards

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