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    Tuesday 21 April 2015

    Greek Mathematics

    Alexandria remained an important Greek mathematical center, even under the rule of Rome, beginning in 31 BCE. In Chapter 5, we discussed the work of several prominent "applied" mathematicians who flourished under Roman rule in Egypt. But there were other mathemati­cians in the first centuries of the common era whose "pure" mathematical works also had influence stretching into the Renaissance. This chapter deals with four of them.
    We first discuss the works of Nicomachus of Gerasa, a Greek town in Judaea. He wrote in the late first century an Introduction roArith.-rretic, based on his understanding of Pythagorean number philosophy. Besides Books VII-IX of Euclid's Elements, this is the only extant number theory work from Greek antiquity. However, there was another important work entitled Arithmetica, written by Diophantus of Alexandria in the mid-third century, which was destined to be of far more importance than Nicomachus's book. Despite its title, this was a work in algebra, consisting mostly of an organized collection of problems translatable into what are today called indeterminate equations, a!l to be solved i n rational numbers. Like Heron's Metrics, the style of the Arithmetica is that of an Egyptian or Babylonian problem text rather than a classic Greek geometrical work. The third mathematician to be considered is also from Alexandria, the geometer Pappus of the early fourth century. He is best known not for his original work, but for his commentaries on various aspects of Greek mathematics and in particular for his discussion of the Greek method of geometric analysis. The chapter concludes with a brief discussion of the work of Hypatia, the first woman mathematician of whom any details are known. It was her death at the hands of an enraged mob that marked the effective end of the Greek mathematical tradition in Alexandria.
    NICOMACHUS AND ELEMENTARY NUMBER THEORY
    Almost nothing is known about the life of Nicomachus, but since his work is suffused with Pythagorean ideas, it is likely that he studied in Alexandria, the center of mathematical activity and of neo-Pythagorean philosophy. Two of his works survive, the Introduction to Arithmetic and the Introduction to Harmonics. From other sources it appears that he also wrote introductions to geometry and astronomy, thereby completing a series on Plato's basic curriculum, the so-called quadrivium.
    Nicomachus's Introduction to Arithmetic was probably one of several works written over the years to explain Pythagorean number philosophy, but it is the only one still extant. Since no text exists from the time of Pythagoras, it is the source of some of the ideas about Pythagorean number theory already discussed in Chapter 2. Because the work was written some 600 years after Pythagoras, however, it must be considered in the context of its time and compared with the only other treatise on number theory available, Books VII-IX of Euclid's Elements.
    Nicomachus began this brief work, written in two books, with a philosophical introduction. Like Euclid, he followed the Aristotelian separation of the continuous "magnitude" from the discontinuous "multitude" Like Aristotle, he noted that the latter is infinite by increasing indefinitely, while the former is infinite by division. Continuing the distinction in terms of the four elements of the quadrivium, he distinguished arithmetic and music, which deal with the discrete (the former absolutely, the latter relatively), from geometry and astronomy, which deal with the continuous (the former at rest and the latter in motion). Of these four subjects, the one that must be learned first is arithmetic, "not solely because ... it existed before all the others in the mind of the creating God like some universal and exemplary plan, relying upon which as a design and archetypal example the creator of the universe sets in order his material creations and makes them attain to their proper ends, but also because it is naturally
    prior in birth inasmuch as it abolishes other sciences with itself, but is not abolished together with them."3 In other words, arithmetic is necessary for each of the other three subjects.
    Most of Book I of Nicomachus's Arithmetic is devoted to the classification of integers and their relations. For example, the author divided the even integers into three classes, the even times even (those that are powers of two), the even times odd (those that are doubles of odd numbers), and the odd times even (all the others). The odd numbers are divided into the primes and the composites. Nicomachus took what appears to us as an inordinate amount of space discussing these classes and showing how the various members are formed. But it must be remembered that he was writing an introduction for beginners, not a text for mathematicians.
    Nicomachus discussed the Euclidean algorithm of repeated subtraction to find the greatest common measure of two numbers and to determine if two numbers are relatively prime. He also dealt with the perfect numbers, giving the Euclidean construction (Elements IX-36) and, unlike Euclid, actually calculating the first four: 6, 28, 496, and 8128. However, also unlike Euclid, Nicomachus presented no proofs. He just gave examples.
    The final six chapters of the first book are devoted to an elaborate tenfold classification scheme for naming ratios of unequal numbers, a scheme that probably had its origin in early music theory. The scheme was in common use in medieval and Renaissance arithmetics and is sometimes found in early printed editions of Euclid's Elements. Among the classes in this scheme of naming the ratio A : B, which reduces to lowest terms as a : b, are multiple, when a = nb; superparticular, when a = b + 1; and superpartient, when a = b + k(1 < k< b).
    It is Book II of Nicomachus that is, however, of most interest to us, since there he discussed plane and solid numbers, again in great detail but without proofs. This material is not mentioned at all by Euclid. Nicomachus not only dealt with triangular and square numbers (see Chapter 2) but also considered pentagonal, hexagonal, and heptagonal numbers and showed how to extend this series indefinitely. For example, the pentagonal numbers are the numbers 1, 5, 12, 22, 35, 51.... (although Nicomachus noted here that 1 is only the side of a "potential" pentagon). Each of these numbers can be exhibited, using the dot notation of Chapter 2, as a pentagon with equal sides (Fig. 6.1). Beginning with 5, each is formed from the previous one in the sequence by adding the next number in the related sequence 4, 7, 10, .... So 5= 1+ 4, 12 = 5 + 7, 22 = 12 + 10, and so on. This is in perfect analogy to the series of triangular numbers 1, 3, 6, 10, ..., each of which comes from the previous one by adding numbers of the sequence 2, 3, 4, . . . , and the series of squares 1, 4, 9, 16, . . . , each of which results from the previous one by adding numbers of the sequence 3, 5, 7, .... Nicomachus continued this analogy and displayed the first 10 numbers of each of the polygonal classes mentioned.
    Nicomachus further explored the solid numbers. A pyramidal number, on a given polyg­onal base of side n, is formed by adding together the first n polygonal numbers of that shape. For example, the pyramidal numbers with triangular base are 1, i+ 3= 4, 1+ 3+ 6= 10, 1+ 3+ 6+ 10 = 20, ..., while those with square base are 1, 1+ 4= 5, 1+ 4+ 9= 14, 1+ 4 + 9 + 16 = 30, .... One can similarly construct pyramidal numbers on any polygonal base.
    Another form of solid number is the cubic number. Nicomachus noted, again without proof, that the cubes are formed from odd numbers, not even. Thus, the first (potential) cube, 1, equals the first odd number, the second cube, 8, equals the sum of the next two odd numbers, the third cube, 27, equals the sum of the next three dd numbers, and so on. Thus, the cubes are closely related to the squares, which are also formed by adding odd numbers. And, Nicomachus concluded, these two facts show that the odd numbers, not the even, are the cause of "sameness."
    The final topic of the treatise is proportion. Nicomachus, referring to pre-Euclidean terminology, used the word "proportion" in a different sense from Euclid's definition 2 of Elements, Book VII. For Euclid, three numbers are in proportion if the first is the same multiple (or part or parts) of the second that the second is of the third. Nicomachus noted that "the ancients" considered not only this type (the type he calls geometric), but also two others, the arithmetic and the harmonic. For Nicomachus, an arithmetic proportion of three terms is a series in which each consecutive pair of terms differs by the same quantity. For example, 3, 7, 11, are in arithmetic proportion. Among the properties of such a proportion are that the product of the extremes is smaller than the square of the mean by the square of the difference. In a geometric proportion, "the only one in the strict sense of the word to be called a proportion;4 the greatest term is to the next greatest as that one is to the next. For example, 3, 9, 27, are in geometric proportion. Among the properties of such a proportion is that the product of the extremes equals the square of the mean. Nicomachus quoted two results of Euclid in this regard, namely, that only one mean term lies between two squares while two lie between two cubes.
    The third type of proportion among three terms, the harmonic, is that in which the greatest term is to the smallest as the difference between the greatest and mean terms is to the difference between the mean and the smallest terms. For example, 3, 4, 6, are in harmonic proportion because 6: 3=(6 - 4) :(4 - 3). Among the properties of this proportion is that when the extremes are added together and multiplied by the mean, the result is twice the product of the extremes. Nicomachus gave as a possible reason for the term "harmonic" that 6, 4, 3, come from the most elementary harmonies. The ratio 6:4 = 3 :2 gives the musical fifth; the ratio 4: 3 gives the fourth, and the ratio 6: 3=(4 : 3)(3: 2) = 2: 1 gives the octave. Today, it is more common to use the names "arithmetic," "geometric;' and "harmonic" for
    means rather than for proportions. Thus, 7 is the arithmetic mean of 3 and 11, 9, is the geometric mean of 3 and 27, and 4 is the harmonic mean of 3 and 6.
    The Introduction to Arithmetic was obviously just that, a basic introduction to elementary ideas about the positive integers. Although it has some points in common with Euclid's Elements, it was written at a much lower level. There are no proofs at all, just a large number of examples. The book was therefore suitable for use by beginners in schools. It was in fact used extensively during ancient times, was translated into Arabic in the ninth century, and was used, in a Latin paraphrase by Boethius (c. 480-524) throughout the early Middle Ages in Europe. For these reasons, copies still exist. That it was so popular and that no more advanced work on the subject, including Euclid's Elements, was studied during much of the period in Europe, shows the level to which mathematical study there fell from its Greek heights. These elementary number properties were for many centuries the summit of the arithmetic i,iiiTiCIiiilrn.
    DIOPHANTUS AND GREEK ALGEBRA
    Little is known about Diophantus's life, other than what is found in the epigram at the beginning of the chapter, except that he lived in Alexandria. It is through his major work, the Arithmetica, that his influence has reached modern times. Diophantus wrote in his introduction that the Arithmetica is divided into thirteen books. Only six have survived in Greek. Four others were recently discovered in an Arabic version. From internal references it appears that these form the fourth through seventh books of the complete work, while the final three Greek books come later. We will refer to the Greek books as I-VI and the Arabic ones as A, B, C, D. The style of the Arabic books is somewhat different from that of the Greek in that each step in the solution of a problem is explained more fully. It is quite possible, therefore, that the Arabic work is a translation not of Diophantus's original, but of a commentary on the Arithmetica, written by Hypatia around 400 CE.
    Before dealing with the problems of the Arithmetica, it is worthwhile to discuss Dio­phantus's major advance in the solution of equations, his introduction of symbolism. The Egyptians and Babylonians wrote out equations and solutions in words. Diophantus, on the other hand, introduced symbolic abbreviations for the various terms involved in equations (Sidebar 6.1). And in a clear break with traditional Greek usage, he dealt with powers higher than the third.
    Note that all of Diophantus's symbols are abbreviations, including the final two: S is a contraction of the first two letters of aptBµos (arithmos, or number), while t1f stands for µovas (monas, or unit). Thus, the manuscripts contain expressions such as O7yStPMB, which stands for 3 squares, 12 numbers, and 9 units, or, as we will write it, 3x2 + 12x + 9. (Recall that the Greeks used an alphabetic cipher for representing numbers in which, for example, y = 3, LP = 12, and B= 9.) Diophantus further used the symbols above with the mark X to designate reciprocals. For example, LTs represented ~. In addition, the symbol
    A, perhaps coming from an abbreviation for.letipts (lepsis, or wanting, or negation), is used for "minus," as in KTasy.AAT ytifn for x3 - 3x2 + 3x - 1. (Negative terms are always collected, so a single A suffices for all terms following it.) In the discussion of Diophantus's problems, however, we use modem notation.
    Diophantus was also aware of the rules for multiplying with the minus: "A minus mul­tiplied by a minus makes a plus, a minus multiplied by a plus makes a ininus."6 Of course, Diophantus was not here dealing with negative numbers, which did not exist for him. He was simply stating the rules necessary for multiplying algebraic expressions involving subtrac­
    tions. But he did not explicitly state the rules for adding and subtracting with positive and negative terms, simply assuming they were known. Near the conclusion of his introduction, he stated the basic rules for solving equations:
    If a problem leads to an equation in which certain terms are equal to terms of the same species but with different coefficients, it will be necessary to subtract like from like on both sides, until one term is found equal to one term. If by chance there are on either side or on both sides any negative terms, it will be necessary to add the negative terms on both sides, until the terms on both sides are positive, and then again to subtract like from like until one term only is left on each side. This should be the object aimed at in framing the hypotheses of propositions, that is to say, to reduce the equations, if possible, until one term is left equal to one term; but I will show you later how, in the case also where two terms are left equal to one term, such a problem is solved.7
    In other words, Diophantus's general method of solving equations was designed to lead to an equation of the form ax" = bx', where, in the first three books at least, m and n are no greater than 2. On the other hand, he did know how to solve quadratic equations, for example, of the form axZ + c = bx.
    "All numbers are made up of some multitude of units.... Among them are­
    squares, which are formed when any number is multiplied by itself; the number itself is called the side of the square;
    cubes, which are formed when squares are multiplied by their sides;
    square-squares, which are formed when squares are mul­tiplied by themselves;
    square-cubes, which are formed when squares are multi­plied by the cubes formed from the same side;
    cube-cubes, which are formed when cubes are multiplied by themselves;
    and it is from the addition, subtraction, or multiplication of these numbers, or from the ratio which they bear one to an­other or to their own sides, that most arithmetical problems are formed; you will be able to solve them if you follow the method shown below.
    "Now each of these numbers, which have been given abbrevi­ated names, is recognized as an clement in arithmetical science; the square (of the unknown quantity) is called dynamis and its sign is A with the index T, that is, AT; the cube is called ku­bos and has for its sign K with the index T, that is, K7the square multiplied by itself is called dynamo-dynamis and its sign is two deltas with the index T, that is, AT A; the square multiplied by the cube formed from the same root is called dynamo-kubos and its sign is AK with the index T, that is, OK' ; the cube multiplied by itself is called kubo-kubos and its sign is two kappas with the index T, KT K.
    -The number which has none of these characteristics, but merely has in it an undetermined multitude of units, is called arithmos, and its sign is S. There is also another sign denoting the invariable element in determinate numbers, the unit, and its sign is M with the index 0, that is, hf." (From Thomas, Selections, lI, pp. 519-523.)
    6.2.1 Linear and Quadratic Equations
    Most of Diophantus's problems are indeterminate; that is, they can be written as a set of k equations in more than k unknowns. Often there are infinitely many solutions. For these problems, Diophantus generally gave only one solution explicitly, but one can easily extend the method to give other solutions. For determinate problems, once certain quantities are made explicit, there is only one solution. Examples of both of these types are described in what follows .8
    PROBLEM I-1 To divide a given number into two having a given difference.
    Biophantus presented the solution for the case where the given number is 100 and the given difference is 40. If x is the smaller of the two numbers of the solution, then 2x + 40 = 100, so x = 30, and the required numbers are 30 and 70. This problem is determinate, once the
    "given" numbers                                       specified, but :n.,phW t,~s's method works for any ~                            nair T__f _n ig the given
    at-, sp°,^..          , r--.
    number and b< a the given difference, then the equation would be 2x + b= a, and the required numbers would be 2(a - b) and 2(a } b).
    PROBLEM I-5 To divide a given number into two numbers such that given fractions (not the same) of each number when added together produce a given number
    In modern notation, we are given a, b, r, s (r < s) and asked to find u, v, such that u + v= a, ~ u+ s v= b. (Diophantus here, and usually, took his fractions to be unit fractions.)
    Diophantus noted that for this problem to be solvable, it is necessary that Sa < b < !a. He then presented the solution in the case where a = 100, b = 30, r = 3, and s = 5: Let the second part (of 100) be 5x. Therefore, the first part is 3(30 - x). Hence, 90 + 2x = 100 and x = 5. The required parts are then 75 and 25.
    Like Problem I-l, once the "given" numbers are specified, this problem is determinate, and the method works for any choice of the "givens" meeting the required condition. In the present case, Diophantus took 1/5 of the second part for his unknown. This allowed him to avoid fractions in the rest of his calculation because 1/3 of the first part must then equal 30 - x and the first part must be 3(30 - x). The remainder of the solution is clear. To check the generality, let sx represent the second part of a and r(b - x) the first. The equation becomes sx + r(b - x) = a or br +(s - r)x = a. Then x = s b' is a perfectly general solution. Since
    x must be positive, a - br > 0 or b < ~ a, the first half of Diophantus's necessary condition. The second half, that S a< b, or a< sb, comes from the necessity that sx < a or s( S_br )< a In this particular problem, as in most of the problems in Book I, the given values are picked to ensure that the answers are integers. But in the other books, the only general condition on solutions is that they be positive rational numbers. Evidently, Diophantus began with integers merely to make these introductory problems easier. In what follows, then, the word "number" should always be interpreted as "rational number."
    PROBLEM I-28 To find two numbers such that their sum and the sum of their squares are given numbers.
    It is a necessary condition that double the sum of the squares exceeds the square of the sum by a square number. In the problem presented, the given sum is 20 and the sum of the squares is 208.
    This problem is of the general form x+ y= a, s2 + y2 = b, a type solved by the Babyloni­ans. Three other Babylonian types appear in I-27, I-29, and 1-30; namely, x + Y= a, xy = b; x} y= a, X2- yz = 6; and x- y= a, xy = b, respectively. A.e we have seen, results giving methods of solutions of these problem are also found in Euclid, Book II. Diophan­tus's solution to the present problem, although presented strictly algebraically, uses the same basic procedure as the Babylonians. Namely, he took as his "unknown" z half the difference between the two desired numbers. Therefore, since 10 is half the sum of the two numbers, the two numbers themselves are x = 10 } z and y = 10 - z. The Babylonian result tells us that the sum of the squares, here 208, is twice the sum of the squares on half the sum and half the difference. In this case, then, we get 200 + 2z2 = 208. It follows that z = 2 and the required two numbers are 12 and 8. Diophantus's method, applicable to any system of the given form, can be translated into the modem formula

    a                                    26 - aZ    a ~/26 - a2
    x=2+                                2 ,         Y=2-    2 .
    His condition is then necessary to ensure that the solution is rational_ Interestingly, the answers to problems 1-27, 1-29, and 1-30 are also- 12 and 8, reminding us of the common Babylonian practice of having the same answers to a series of related problems.
    Did Diophantus have access to the Babylonian material? Or did he learn his methods from a careful study of Euclid's Elements or Data? These questions cannot be answered. It is, however, apparent that there is no geometric methodology in Diophantus's procedures. Perhaps by this time the Babylonian algebraic methods, stripped of their geometric origins, were known in the Greek world.
    PROBLEM 11-8 To divide a given square number into two squares. Here we quote Diophantus exactly:
    Let it be required to divide 16 into two squares. And let the first square = x2; then the other will be 16 - x2; it shall be required therefore to make 16 - sZ = a square- I take a square of the form (ax - 4)Z, a being any integer and 4 the root of 16; for example, let the side be 2x - 4, and the square itself 4x2 + 16 - 16x. Then 4s2 + 16 - l6x = 16 - xZ. Add to both sides the negative terms and take like from like. Then Sx2 = 16x, and x=~. One number will therefore be 256, the other u, and their sum is u° or 216, and each is a square9 [Fig. 6.21.
    This is an example of an indeterminate problem. It translates into one equation in two unknowns, x2 + y2 = 16. This problem also demonstrates one of Diophantus's most common methods. In many problems from Book 11 onward, Diophantus required a solution, expressed in the form of a quadratic polynomial, which must be a square. To ensure a rational solution, he chose his square in the form (ax ± b)2, with a and b selected so that either the quadratic term or the constant term is eliminated from the equation. In this case, where the quadratic polynomial is 16 - x2, he used b = 4 and the negative sign, so the constant term is eliminated and the resulting solution is positive. The rest of the solution is then obvious. The method can be used to generate as many solutions as desired to x2 + y2 = 16, or, in general, to
    z2 + y2 = b2. Take any value for a and set y= ax - b. Then bZ - z2 = a2x2 - 2abx } bz
    or 2abx =(aZ + 1)x2, so x=~.
    As another example where Diophantus needed a square, consider
    PROBLEM II-19 To find three squares such that the difference between the greatest and the middle has a given ratio to the difference between the middle and the least.
    Diophantus assumed that the given ratio is 3 : 1. If the least square is x2, then he took (z + 1)2 = x2 + 2x + 1 as the middle square. Because the difference between these two squares is 2x + 1, the largest square must be z2 + 2x + 1+ 3(2x + 1) = xZ + 8x + 4. To make that quantity a square, Diophantus set it equal to (x + 3)2, in this case choosing the coefficient of x so that the x2 terms cancel. Then 8x + 4 = 6x + 9, so x = 2 2 and the desired
    squares are 6y , 124, 304. One notices, however, that given his initial choice of (x+ 1)2 as the middle square, 3 is the only integer b Diophantus could use in (x + b)2 that would give him a solution. Of course, with other values of the initial ratio, there would be more possibilities as there would with a different choice for the second square. In any case, in this problem as in all of Diophantus's problems, only one solution is required.
    Problem 11-1 1 introduces another general method, that of the double equation.
    PROBLEM II-11 To add the same (required) number to two given numbers so as to make each of them a square.
    Diophantus took the given numbers as 2 and 3. If his required number is x, he needed both x + 2 and x + 3 to be squares. He therefore had to solve z+ 3 = u2, x + 2 = v2, for z, u, v. Again, this is an indeterminate problem. Diophantus described his method as follows: "rake the difference between the two expressions and resolve it into factors. Then take either (a) the
    square of half the difference between these factors and equate it to the lesser expression or (b) the square of half the sum and equate it to the greater." 10
    Since the difference between the expressions is uZ - u2 and this factors as (u + v) (u - v), the difference of the two factors is 2v while the sum is 2u. What Diophantus did not mention explicitly is that the initial factoring must be carefully chosen so that the solution for x is a positive rational number. In the present case, the difference between the two expressions is 1. Diophantus factored that as 4 x 1/4. Thus, u+ v=4 and u - u= 1/4, so 2v = 15/4, z+ 2= v2 = 225/64. and x = 97/64. Note, for example, that the factorization 2 x 1/2 would not give a positive solution, nor would the factorization 3 x 1/3. The factorization 1= at/a
    needs to be chosen so that [Z (a - Q))2 -2.
    6.2.2 Higher-Degree Equations
    Because the problems in Book A involve cubes and even higher powers, Diophantus began with a new introduction in which he described the rules for multiplying such powers. For
    example, since x2, s3, x4, x5, and zb are represented by AT, KT, AT A, OKr, and KTK,
    respectively, Diophantus wrote, forexample, that 4K7 multiplied by g equals K7 multiplied by itself, equals AT multiplied by A70, and all equal K7K. Similarly, if KYK is divided by ATA, the result is AT. Thus, although Diophantus's results are equivalent to our laws of exponents, his notation did not allow him to express it in our familiar way of "add the exponents" when you multiply powers and "subtract the exponents" when you divide.
    Diophantus did, however, explain that, as before, his equations end up with a term in one power equaling a term in another, that is, as" = bs' (n < m), where now m may be any number up to 6. To solve, one must use the rules to divide both sides by the lesser power and end up with one "species" equal to a number, that is, in our notation, a = bz'-". The latter equation is easily solved. Speaking to the reader, he concluded, "when you are acquainted with what I have presented, you will be able to find the answer to many problems which I have not presented, since I shall have shown to you the procedure for solving a great many problems and shall have explained to you an example of each of their types."t t
    As an example of Diophantus's use of higher powers of x, consider
    PROBLEM A-25 To futd two numbers, one a square and the other a cube, such that the sum of their squares is a square.
    The goal is to find x, y, and z such that (xZ)2 + (y3)2 = zz Thus this is an indeterminate problem with one equation in three unknowns. Diophantus set z equal to 2y (the 2 is arbitrary) and performed the exponentiation to conclude that 16y4 + y6 must be a square, which he took to be the square of ky2. So l6y° + y6 = k2 y4, y6 =(kz - 16)y4, and yz = k2 - 16. It follows that k2 - 16 must be a square. Diophantus chose the easiest value, namely, k2 = 25, so y = 3. Therefore, the desired numbers are y3 = 27 and (2y)2 = 36. This solution is easily generalized. Take x = ay for any positive a. Then k and y must be found so that kz - a4 = yz or so that k2 - yz = a4. Diophantus had, however, already demonstrated in Problem 11-10 that one can always find two squares whose difference is given.
    Problem B-7 shows that Diophantus knew the expansion of (x + y)3. As he put it, "whenever we wish to form a cube from some side made up of the sum of, say, two different terms-so that a multitude of terms does not make us commit a mistake-we have to take the
    cubes of the two different terms, and add to them three times the results of the multiplication of the square of each term by the other." 12
    PROBLEM B-7 To find two numbers such that their sum and the sum of their cubes are equal to two given numbers.
    The problem asks to solve x+ y = a, x3 + y3 = b. This system of two equations in two unknowns is determinate. It is a generalization of the "Babylonian" problem 1-28, z+ y= a, x2 + y2 = b, and Diophantus's method of solution generalized his method there. Letting a = 20 and b = 2240, he began as before by letting the two numbers be 10 + z and 10 - z. The second equation then becomes (10 + z)3 + (10 - z)3 = 2240 or, using the expansion already discussed, 2000 + 60z2 = 2240, or 60z2 = 240, z2 = 4, and z = 2. Diophantus gave, of course, a condition for a rational solution, namely, that (4b - a3)/3a i_s a_ square (equivalent to the more natural condition that f_b - 2(° )3]/3a is a square). It is interesting that the answers here are the same as in 1-28, namely, 12 and 8.
    When reading through the Arithmetica, one never quite knows what to expect next. There are a great variety of problems. Often there are several similar problems grouped together, one involving a subtraction where the previous one involved an addition, for example. But then one wonders why other similar ones were not included. For example, the first four problems of Book A ask for (1) two cubes whose sum is a square, (2) two cubes whose difference is a square, (3) two squares whose sum is a cube, and (4) two squares whose difference is a cube. What is missing from this list is, first, to find two squares whose sum is a square-but that had been solved in 11-8-and second, to find two cubes whose sum is a cube. This latter problem is impossible to solve, and there are records stating this impossibility dating back to the tenth century. Probably Diophantus was also aware of the impossibility. At the very least, he must have tried the problem and failed to solve it. But he did not mention anything about it in his work. A similar problem with fourth powers occurs as V-29: to find three fourth powers whose sum is a square. Although Diophantus solved that problem, he did not mention the impossibility of finding two fourth powers whose sum is a square. Again, one assumes that he tried the latter problem and failed to solve it.
    In his discussion of Problem D-I1, he finally addressed an impossibility. After solving that problem, to divide a given square into two parts such that the addition of one part to the square gives a square and the subtraction of the other part from the square also gives a square, he continued, "since it is not possible to find a square number such that, dividing it into two parts and increasing it by each of the parts, we obtain in both cases a square, we shall now present something which is possible." 13
    PROBLEM D-12 To divide a given square into two parts such that when we subtract each from the given square, the remainder is (in both cases) a square.
    Why is the quoted case impossible? To solve x2 = a + b, x2 +a = c2, z2 + b= d2 would imply that
    3x2=c2+d2 or 3=' X~2+/ d )2.
    It is, in fact, impossible to decompose 3 into two rational squares. One can show this easily by congruence arguments modulo 4. Diophantus himself did not give a proof, nor later, when
    he stated in VI-14 that 15 is not the sum of two squares, did he tell why. The solution of D-12, however, is very easy.
    6.2_3 The Method of False Position
    In Book IV, Diophantus began use of anew technique, a technique reminiscent of the Egyptian "false position." Among many problems he solved using this technique, the following one will be important in our later discussion of elliptic curves.
    PROBLEM IV-24 To divide a given number into two Paris such that their product is a cube minus its side.
    If a is the given number, the problem is to find x and y such that y(a - y) = x3 - x. This is an indeterminate problem. As usual, Diophantus began by choosing a particular value for a, here a= 6. So 6y - yZ must equal a cube minus its side. He chose the side s to be of the form x = my - L The question is, What value should he choose for m? Diophantus picked m= 2 and calculated: 6y - y2 = (2y - 1)3 - (2y - 1), or 6y - y2 = 8y3 - 12y2 + 4y. We note immediately that the "1° in z= my - I was chosen so that there would be no constant term in this equation. Nevertheless, this is still an equation with three separate species, not the type Diophantus could solve most easily. So he noted that if the coefficients of y on each side were the same, then the solution would be simple. Now the "6" on the left is the "given number," so that cannot be changed. But the "4" on the right comes from the calculation 32 - 2, which in turn depends on the choice m= 2 in x= my - 1. Therefore, Diophantus needed to find m so that 3• m- m = 6. Therefore, m= 3. We can then begin again: z= 3y - I and 6y - y2 = (3y - 1)3 - (3y - 1), or 6y - y2 = 27y3 - 27y2 + 6y. Therefore, 27y3 = 26y2 and y = 26. The two parts of 6, therefore, are Z~ and t~, while the product of those two numbers is (L7)3 - 9. The general solution to this problem, for arbitrary a, is then given by
    6°Z - 8 _ 3a2-a
    Y= a3 ,x---~-1_
    In Problem IV-31, Diophantus found again that his original assumption did not work. But here the problem is that a mixed quadratic equation, the first one to appear in the Arithmetica, fails to have a rational solution.
    PROBLEM IV-31 To divide unity into two parts so that, if given numbers are added to them respectively, the product of the two sums is a square.
    Diophantus set the given numbers at 3, 5, and the parts of unity as x, 1- x. Therefore, (x + 3) (6 - x) = 18 + 3x - x2 must be a square. Since neither of his usual techniques for determining a square will work here (neither 18 nor -1 are squares), he tried (2x)2 = 4x2 as the desired square. But the resulting quadratic equation, 18 + 3x = SxZ "does not give a rational result." He needed to replace 4x2 by a square of the form (MX)2, which does give a rational solution. Thus, since 5 = 2 2 + 1, he noted that the quadratic equation will be solvable if (M2 + 1) 18 +(3/2)Z is a square. This implies that 72m2 -1- 81 is a square, say, (8m + 9)Z. (Here, his usual technique succeeds.) Then m = 18 and, returning to the beginning, he set 18 + 3x - z2 = 324z2. He then simply presented the solution: x = 78/325 = 6/25, and the desired numbers are 6(?.5, 19125.

    Although Diophantus did not give details in IV-31 on the solution of the quadratic, he did give them in Problem IV-39. His words in that problem are easily translated into the formula
    z + Jac+ (z)2
    X =_____________________
    a
    for solving the equation c + bx = ax2. This formula translates correctly the Babylonian procedure, which began by multiplying the equation through by a and solving for ax. Diophantus was sufficiently familiar with this formula and its variants that he used it in various later problems not only to solve quadratic equations but also to solve quadratic inequalities.
    PROBLEM V-10 To divide unity into two parts such that, if we add different given numbers to each, the results will be squares.
    In this problem the manuscripts have, for one of only two times in the entire work, a diagram (Fig. 6.3). Diophantus assumed that the two given numbers are 2 and 6. He represented them, as well as 1, by setting DA = 2, AB = 1, and BE = 6. The point G is chosen so that DG (= AG + DA) and GE (= BG + BE) are both squares. Since DE = 9, the problem is reduced to dividing 9 into two squares such that one of them lies between 2 and 3. If that square is xZ, the other is 9- xZ. Unlike the situation in previous problems, Diophantus could not simply put 9 - x2 equal to (3 - mx)2 with an arbitrary m, for he needed x2 to satisfy the inequality condition. So he set it equal to (3 - mx)2 without specifying m. Then
    6m
    x _
    m2 -} 1.
    Rather than substitute the expression for x into 2< xZ < 3 and attempt to solve a fourth-degree inequality, he picked two squares close to 2 and 3, respectively, namely, 289/144 = (17/[2}Z and 361/144 = (19/12)2, and substituted the expression into the inequality 17/12 < z< 19/12. Therefore,
    17                                      6m   19
    [2<mz+l<12'
    The left inequality becomes 72m >[7m2 + 17. Although the corresponding quadratic equa­tion has no rational solution, Diophantus nevertheless used the quadratic formula and showed that since J(72/2)Z - 172 = 1007 is between 31 and 32, the number m must be chosen so that m< 67/17. The right inequality similarly shows that m>_ 66/19. Diophantus therefore picked the simplest m between these two limits, namely, 3 1/2. So
    /                                             2
    9-x2=/3-3z~ and x=
    2                                                         584
    3 Then x2 = 7056/2809 and the desired segments of i are 1438/2809 and 1371/2809.
    Diophantus's work, the only example of a genuinely algebraic work surviving from ancient Greece, was highly influential. Not only was it commented on in late antiquity, but it was also studied by Islamic authors. Many of its problems were taken over by Rafael Bombelli and published in his Algebra of 1572, while the initial printed Greek edition of Bachet, published in 1621, was carefully studied by Pierre Fermat and led him to numerous general results in number theory, about which Diophantus himself only hinted. Perhaps more important, however, is the fact that this work, as a work of algebra, was in effect a treatise on the analysis of problems. Namely, the solution of each problem began with the assumption that the answer x, for example, had been found. The consequences of this fact were then followed to the point where a numerical value of x could be determined by solving a simple equation. The synthesis, which in this case is the proof that the answer satisfies the desired conditions, was never given by Diophantus because it only amounted to an arithmetic computation. Thus, Diophantus's work is at the opposite end of the spectrum from the purely synthetic work of Euclid.
    PAPPUS AND ANALYSIS
    Although analysis and synthesis had been used by all of the major Greek mathematicians, there was no systematic study of the methodology published, as far as is known, until the work of Pappus, who lived in Alexandria early in the fourth century (Sidebar 6.2). Pappus was one of the last mathematicians in the Greek tradition_ He was familiar with the major and minor works of the men already discussed, and even extended some of their work in certain ways. He is best known for his Collection, a group of eight separate works on various topics in mathematics, probably put together shortly after his death by an editor attempting to preserve Pappus's papers. The books of the collection vary greatly in quality, but most of the material consists of surveys of certain mathematical topics collected from the works of his predecessors.
    The preface to Book 3 provides an interesting sidelight to the work. Pappus addressed the preface to Pandrosian, a woman teacher of geometry. He complained that "some persons professing to have learned mathematics from you lately gave me a wrong enunciation of problems ." td By that Pappus meant that these people attempted to solve problems by methods that could not work, for example, to solve the problem of the two mean proportionals using only circles and straight lines. 'Mere is no indication of how Pappus knew that such a construction was impossible_ From his remark, however, we learn that women were involved in mathematics in Alexandria. 15
    Book 5, the most polished book of the Collection, deals with isoperimetric figures, figures of different shape but with the same perimeter. Pappus's introduction provided a counterpoint to the pure mathematics of the text as he wrote of the intelligence of bees:

    [The bees], believing themselves, no doubt, to be entrusted with the task of bringing from the gods to the more cultured part of mankind a share of ambrosia in this form, .., do not think it proper to pour it carelessly into earth or wood or any other unseemly and irregular material, but, collecting the fairest parts of the sweetest flowers growing on the earth, from them they prepare for the reception of the honey the vessels called honeycombs, [with cells] all equal, similar and adjacent, and hexagonal in form.
    Ztaphael's painting The School ofAthens depicts Ptolemy as a prince with Italian features, while the most common "portrait" of Hypatia, attributed to an artist named Gasparo, shows her as Italian as well. There is nothing surprising in this; artists usu­ally use their contemporaries as models for figures from long ago. But the more serious question is to what extent the Alexan­drian mathematicians of the period from the first to the fifth centuries CE were Greek. Certainly, all of them wrote in Greek and were part of the Greek intellectual community of Alexan­dria. And most modern studies of Hellenistic Egypt conclude that the Greek community and the native Egyptian community coexisted, with little mutual influence. So do we then conclude that Ptolemy and Diophantus, Pappus and Hypatia were ethni­cally Greek, that their ancestors had come from Greece at some point in the past and had remained effectively isolated from the Egyptians for many centuries?
    The question is, of course, not possible to answer definitively. But the research in papyri dating from the early centuries of
    .{L/LGI14llLLGNdiW; ....... :..

    the common era also demonstrates that there was significant intermarriage between the Greek and Egyptian communities, chiefly by Greek men taking Egyptian wives. And it is known, for example, that Greek marriage contracts increasingly re­sembled Egyptian ones. In addition, even from the founding of Alexandria, small numbers of Egyptians were admitted to the privileged classes in the city to fulfill numerous civic roles. Of course, it was essential in this case for the Egyptians to become "Hellenized," to adopt Greek habits and the Greek lan­guage. Given that the Alexandrian mathematicians mentioned above were active several hundred years after the founding of the city, however, it would seem at least equally possible that they were ethnically Egyptian as that they remained ethnically Greek. In any case, it is unreasonable for us today to portray these mathematicians with pure European features when we have no physical descriptions of them whatsoever.
    That they have contrived this in accordance with a certain geometrical forethought we may thus infer. They would necessarily think that the figures must all be adjacent one to another and have their sides common, in order that nothing else might fall into the interstices and so defile their work- Now there are only three rectilineal figures which would satisfy the condition, I mean regular figures which are equilateral and equiangular, inasmuch as irregular figures would be displeasing to the bees.... [Thcse being] the triangle, the square and the hexagon, the bees in their wisdom chose for their work that which has the most angles, perceiving that it would hold more honey than either of the two others [Fig. 6.4].
    Bees, then, know just this fact which is useful to them, that the hexagon is greater than the square and the triangle and will hold more honey for the same expenditure of material in constructing each. But we, claiming a greater share in wisdom than the bees, will investigate a somewhat wider problem, namely that, of all equilateral and equiangular plane figures having an equal perimeter, that which has the greater number of angles is always greater, and the greatest of them all is the circle having its perimeter equal to them. 16
    The most influential book of Pappus's Collection, however, is Book 7, On the Domain of Analysis, which contains the most explicit discussion from Greek times of the method of analysis, the methodology Greek mathematicians used to solve problems. The central ideas
    are spelled out in the introduction to this book:
    That which is called the Domain of Analysis ... is, taken as a whole, a special resource ... for those who want to acquire a power in geometry that is capable of solving problems set to them; and it is useful for this alone. It was written by three men, Euclid the writer of the Elements, Apollonius of Perga, and Aristaeus the elder, and proceeds by analysis and synthesis.
    More important for Greek mathematics than theorematic analysis is the problematic analysis. We have already discussed several examples of this type of analysis, including the problems of angle trisection and cube duplication and Archimedes' problems on the division of a sphere by a plane. And although Euclid did not present the analysis as such, one can carry out the procedure in solving Elements VI-28, the geometric algebra problem leading to the solution of the quadratic equation x2 + c = bx. The analysis there shows that an additional condition is required for the solution, namely, that c<_ ( Z)2.
    Pappus's Book 7, then, is a companion to the Domain of Analysis, which itself consists of several geometric treatises, all written many centuries before Pappus. These works, Apollo­nius's Conics and six other books (all but one lost), Euclid's Data and two other lost works, and single works (both lost) by Aristaeus and Eratosthenes, even though the last-named au­thor is not mentioned in Pappus's introduction, provided the Greek mathematician with the
    -wit." ~ problems that .b~.,it
    tools necessary to solve problems by analysis. For example, to deal w. pro,.~ ~ u•". •..~~••
    in conic sections, one needs to be familiar with Apollonius's work To deal with problems solvable by "Euclidean" methods, the material in the Data is essential.
    Pappus's work does not include the Domain of Analysis itself. It is designed only to be read along with these treatises. Therefore, it includes a general introduction to most of the individual books along with a large collection of lemmas that are intended to help the reader work through the actual texts. Pappus evidently decided that the texts themselves were too difficult for most readers of his day to understand as they stood. The teaching tradition had been weakened through the centuries, and there were few, like Pappus, who could appreciate these several-hundred-year-old works. Pappus's goal was to increase the numbers who could understand the mathematics in these classical works by helping his readers through the steps where the authors wrote "clearly ...!" He also included various supplementary results as well as additional cases and alternative proofs.
    Among these additional remarks is the generalization of the three- and four-line locus problems discussed by Apollonius. Pappus noted that in that problem itself the locus is a conic section. But, he says, if there are more than four lines, the loci are as yet unknown; that is, "their origins and properties are not yet known." He was disappointed that no one had given the construction of these curves that satisfy the five- and six-line locus. The problem in these cases is, given five (six) straight lines, to find the locus of a point such that the rectangular parallelepiped contained by the lines drawn at given angles to three of these lines has a given ratio to the rectangular parallelepiped contained by the remaining two lines and some given line (remaining three lines). Pappus noted that one can even generalize the problem further to more than six lines, but in that case, "one can no longer say `the ratio is given between some figure contained by four of them to some figure contained by the remainder' since no figure can be contained in more than three dimensions." Nevertheless, according to Pappus, one can express this ratio of products by compounding the ratios that individual lines have to one another, so that one can in fact consider the problem for any number of lines. But, Pappus complained, "[geometers] have by no means solved [the multi-line locus problem] to the extent that the curve can be recognized.... The men who study these matters are not of the same quality as the ancients and the best writers. Seeing that all geometers are occupied with the first principles of mathematics . . . and being ashamed to pursue such topics myself, I have proved propositions of much greater importance and utility." 18
    Pappus concluded Book 7 by stating one of the "important" results he had proved, that "the ratio of solids of complete revolution is compounded of that of the revolved figures and that of the straight lines similarly drawn to the axes from the center of gravity in them." 19 The modem version of this theorem is that the volume of a solid formed by revolving a region 0 around an axis not intersecting SZ is the product of the area of S2 and the circumference of the circle traversed by the center of gravity of SZ. Unfortunately, there is no record of Pappus's proof. There is some indication that it is in one of the books of the Collection now lost.
    Much of the explicit analysis in Greek mathematics has to do with material we generally think of as algebraic. The examples from Elements JffIt-1 and VI-28 are clearly such. The examples using the conic sections are ones that today would be solved using analytic geometry, a familiar application of algebra. It is somewhat surprising, then, that Pappus does not mention the strictly algebraic Arithmetica of Diophantus as a prime example of analysis, because, in effect, every problem in Diophantus's work is solved according to Pappus's model. Perhaps Pappus did not inciude this work because it was not on the level of the classic geometric works. In any case, it was the algebraic analysis of Diophantus and the "quasi­algebraic" analysis of many of the other mentioned works, rather than the pure geometric analysis, that provided the major impetus for sixteenth- and seventeenth-century European mathematicians to expand on the notion of algebra and develop it into a major tool to solve even purely geometric problems.20
    HYPATIA AND THE END OF GREEK MATHEMATICS
    Pappus's aim of reviving Greek mathematics was unsuccessful, probably in part because the increasingly confused political and religious situation affected the stability of the Alexandrian Museum and Library. In his time, Christianity was changing from a persecuted sect into the official religion of the Roman Empire. In 313 the emperor Galerius issued an edict of toleration in the Eastern Empire, and two years later the same was done in the West by Constantine. The latter in fact converted to Christianity before his death in 337. Within 60 years, Christianity became the state religion of the empire and the ancient worship of the Roman gods was banned. Of course, the banning of paganism did not cause everyone to adopt Christianity_ In fact, in the late fourth and early fifth centuries, Hypatia (c. 355-415), the daughter of "Ibeon of Alexandria, was a respected and eminent teacher in that city, not only of mathematics but also of some of the philosophic doctrines dating back to Plato's Academy. And although she maintained her non-Christian religious beliefs, she enjoyed intellectual independence and even had eminent Christians among her students, including Synesius of Cyrene (in present-day Libya), who later became a bishop.
    Although there is some evidence of earlier women being involved in Greek mathematics, it is only about Hypatia that the evidence is substantial enough to give some indication of her mathematical accomplishments. Hypatia was given a very thorough education in mathematics and philosophy by her father. Although the only surviving documents with a clear reference to Hypatia are Synesius's letters to her requesting scientific advice, recent detailed textual studies of Greek, Arabic, and medieval Latin manuscripts lead to the conclusion that she was responsible for many mathematical works. These include several parts of her father's commentary on Ptolemy's Almagest, the edition of Archimedes' Measurement of the Circle
    Why did Greek mathematics decline so dramatically from its height in the fourth and third centuries tsce? Among the several answers to this question, the most important is the change in the sociopolitical scene in the region surrounding the eastern
    Mediterranean.
    A consideration of mathematical development ir, the various ancient societies already studied shows that mathematical cre­ativity requires some sparks of intellectual curiosity, whether or not these are stimulated by practical concerns. But this spark nf rnr;ncity nds a Climate of gwernmenf rm.oura_gement for its flames to spread. The Babylonians used their most advanced techniques, not for everyday purposes, but for solving intellec­tually challenging problems. The government encouraged the use of these mathematical problems to help train the minds of its future leaders. In Greek civilization, the intellectual curios­ity ran even deeper. In the Greek homeland, the sociopolitical system provided philosophy and mathematics with encourage­ment. The Ptolemies continued this encouragement in Egypt after 300 ace.
    Buteven in Greek society, the actual numberofthose who un­derstood theoretical mathematics was small. There were never many who could afford to spend their lives as mathematicians or astronomers and persuade the rulers to provide them with stipends. The best of the mathematicians wrote works that were discussed and commented on in the various mathemati­cal schools, but not everything could be learned from the texts. An oral teaching tradition was necessary to keep mathematics
    progressing because, in general, one could not master Euclid's Elements or Apollonius's Conics on one's own. A break of a generation in this tradition thus meant that the entire process of mathematical research would be severely damaged.
    One factor certainly weakening the teaching tradition, if not breaking it entirely, was the political strife around the east­em Mediterranean in the years surrounding the beginning of the common era. More important, because the Roman imperial government evidently decided that mathematical research was not an important national interest, it did not support it. There was little encouragement of mathematical studies in Rome. Few Greek scholars were imported to teach mathematics to the children of the elite. Soon, no one in Rome could even un­derstand, let alone extend, the works of Euclid or Apollonius.
    The Greek tradition did continue for several centuries, how­ever, under the Roman governors of Egypt, particularly be­cause the Alexandrian Museum and Library remained in exis­tence. Anyone interested could continue to study and interpret the ancient texts. With fewer and fewer teachers, however, less and less new work was accomplished. The virtual destruction of the great library by the late fourth century finally severed the tenuous links with the past. Although there continued to be some limited mathematical activity for a while in Athens and elsewhere-wherever copies of the classic works could be found-by the end of the fifth century, there were too few people devoting their energies to mathematics to continue the tradition, and Greek mathematics ceased to be.
    from which most later Arabic and Latin translations stem, a work on areas and volumes reworking Archimedean material, and a text on isoperimetric figures related to Pappus's Book 5.21 She was also responsible for commentaries on Apollonius's Conics and, as noted earlier, on Diophantus's Arithmetica.
    Unfortunately, although Hypatia had many influential friends in Alexandria, including the Roman prefect Orestes, they were primarily from the upper classes. The populace at large in general supported the patriarch Cyril in his struggle with Orestes for control of the city. So when Cyril spread rumors that the famous woman philosopher in reality practiced sorcery as part of her philosophical, mathematical, and astronomical work, a group emerged that was willing to eliminate this "satanic" figure. Hypatia's life was thus cut short as already described. Her death effectively ended the Greek mathematical tradition of Alexandria (Sidebar 6.3). 
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