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25 Figure 5.7 where a, b, and c are positive integers in each case. 5 Arab mathematics The solutions of one equation, 'one square and ten roots of the same equal thirty nine (dirhems)' or x2 + lOx = 39, is interesting historically since it recurs in later Arab and medieval European texts. Al-Khwarizmi's numerical and one of the geometric solutions 'completing the square' are quoted in full. They now constitute a standard repertoire in algebra. Here is al-Khwarizmi's solution for the equation x 2 + lOx = 39. You halve the number (i.e. the coefficient) of the roots, which in the present instance yields five. This you multiply by itself; the product is twenty-five. Add this to thirty-nine; the sum is sixty four. Now take the root of this, which is eight, and subtract from it half the number of the roots, which is five; the remainder is three. This is the root of the square which you sought for; the square itself is nine. Refer to Figure 5.7 for al-Khwarizmi's geometric solution. We proceed from the quadrate A B, which represents the square. It is our business to add to it the ten roots of the same. We halve for this purpose the ten, so that it becomes five, and construct two quadrangles (rectangles) on two sides of the quadrate A B, namely, G and D, the length of each of them being fives,... whilst the breadth of each is equal to a side of the quadrate A B.Then a quadrate remains opposite the corner of the quadrate A B. This is equal to five multiplied by five; this five being half of the number of the roots which we have added to each of the two sides of the first quadrate. Thus we know that the first quadrate, which is the square, and the two quadrangles on its sides, which are the ten roots, make together thirty-nine. In order to complete the great quadrates, there wants only a square of five multiplied by five, or twenty-five. This we add to thirty-nine, in order to complete the great square S H. The sum is sixty- four. We extract its root, eight, which is one of the sides of the great quadrangle. By subtracting from this the same quantity which we have before added, namely five, we obtain three as the remainder. This is the side of the quadrangle A B, which represents the square; it is the root of this square, and the square itself is nine. Activity 5.8 Solutions of quadratic equations 1 Explain in modern notation both solutions for the equation x + \Qx = 39 offered by al-Khwarizmi. Notice that al-Khwarizmi ignores the negative solution. The geometric solution became more abstract and rigorous as Arab mathematicians mastered the works of Euclid and Apollonius. This is evident if you compare al- Khwarizmi's work with that of a later Arab mathematician, Abu Kamil. 2 Study the two mathematicians' geometric solutions, translated into modern notation, to the problem | 'A square and twenty one (dirhems) equal ten roots.' That is, solve x 2 +21 = 10x. 65
D B P N M K Figure 5,9a The Arabs D p B w S S L G K A H Figure 5.9b 66 a Here is a description of al-Khwarizmi's solution (see Figure 5. (5-*) N w M D Figure 5.8 S (5-*) (5-Jt) Beginning with the square ABEF of side x units, extend BE to H and then C and AF to I and then D such that ID = HC = 5 units and FI = EH = 5 - x units. Notice that it follows that H is the mid-point of BC. Therefore, the area of CDFE = *(5 - x) + 5x = \0x - x 2 = 21. Next complete the square HMNC, which has an area of 25 units, by extending HI to M and CD to N such that DN = IM = EH = FI = 5 - x. Now drop a perpendicular from W to S such that the area of square ISWM = (5 — x) Since the area of SDNW = the area of EHIF = x(5 - x), it follows that the area of square HMNC = the area of CDFE + the area of ISWM = 25. Or the area of ISWM = 25 - 21 = 4 so that IS = FI = 2. Subtracting FI from AI will give the root of the square ABEF, that is, x = 3. Adding FI to AI, that is, 5 + (5 - x), will give AS = 7. Hence, x = 3 or 7. L b Here is Abu Kamil's solution (see Figures 5.9a and 5.9b). Figure 5.9a is constructed in the following way. Beginning with the square ABDG of side x units, extend DB to L such that the area of ABLH is 21 units. By construc tion let BL be greater than DB. Also let DL be 10 so that the area of DLHG is 10*. Let P be the mid-point of DL. You need to show that BL x DB + (BP) 2 = (DP) 2 . Figure 5.9b is constructed the following way. Beginning with the square DBAG of side x units, extend DB to W, with BW less than DB. Then let P be the mid-point of DW and draw the square PWLN. Abu Kamil does not show this result but quotes the result from Euclid's Elements, Book II, proposition 5. If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half. The proof given below uses a demonstration of Heron of Alexandria which Abu Kamil must have known.
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Nuffield Advanced Mathematics Histo
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Contents Introduction 7 Unit 1 The
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Introduction This book, about the h
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The Babylonians Chapter 1 Introduct
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Introduction to the Babylonians On
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Figure 1.2 Stele inscribed with Ham
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Token V Figure 1.5 7 Introduction t
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There are no practice exercises for
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A sexagesimal number system is a sy
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Page 24 and 25: Activity 2.6 Babylonian arithmetic
Page 26 and 27: Figure 2.10 Activity 2.9 2 Babyloni
Page 28 and 29: Type Quadratic Table 2.1 x +ax = b
Page 30 and 31: Activity 2.12 Geometry Figure 2.13a
Page 32: Practice exercises for this chapter
Page 35 and 36: 3 An introduction to Euclid fthese
Page 37 and 38: The Greeks Activity 3.3 B Figure 3.
Page 39 and 40: 34 The Greeks Activity 3.4 g Now us
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Page 43 and 44: 38 The Greeks Activity 3.7 therefor
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Page 47 and 48: 42 More Greek mathematics 4.1 Some
Page 49 and 50: The Greeks Q.E.D. stands for 'quod
Page 51 and 52: Figure 4.3 46 The Greeks • treat
Page 53 and 54: Figure 4.6 48 The Greeks together w
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Page 59 and 60: 5 Arab mathematics You should think
Page 61 and 62: The Arabs Western Arabic, or Gobar,
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Page 65 and 66: 7776 Arabs Activity 5.4 This proble
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Calculus including Descartes, under
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118 Calculus Activity 9.7 Cavalieri
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Figure 9. 7 / 120 Calculus dy 1 Wri
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10.1 Visualising negative numbers 1
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126 Searching for the abstract For
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752 12 Summaries and exercises Two
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156 12 Summaries and exercises usin
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158 Hints Activity 2.3, page 15 2 F
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13 Hints Activity 7.3, page 89 1 Wr
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outcome 1 step to the right Figure
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14 Answers 3 0;6,40 and 0;2,13,20 4
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14 Answers 3 If AD intersects the l
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74 Answers 2 Suppose that you can f
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14 Answers 4 If he had a daughter t
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14 Answers Activity 6.5, page 79 1
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14 Answers 6 The equation z 2 + az
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14 Answers b This meets the ellipse
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14 Answers To get an interpretation
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14 Answers of multiplication by neg
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14 Answers b The required number is
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14 Answers operator has the effect
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14 Answers A i- B 5 E %2 5)-25 = 39
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Index Ptolemy 49, 55, 58 Pythagoras
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