Saturday, August 31, 2019
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This page intentionally left blank Now into its eighth edition and with additional material on primality testing, written by J. H. Davenport, The Higher Arithmetic introduces concepts and theorems in a way that does not require the reader to have an in-depth knowledge of the theory of numbers but also touches upon matters of deep mathematical signi? cance. A companion website (www. cambridge. org/davenport) provides more details of the latest advances and sample code for important algorithms. Reviews of earlier editions: Ã¢â¬Ë. . . the well-known and charming introduction to number theory . . can be recommended both for independent study and as a reference text for a general mathematical audience. Ã¢â¬â¢ European Maths Society Journal Ã¢â¬ËAlthough this book is not written as a textbook but rather as a work for the general reader, it could certainly be used as a textbook for an undergraduate course in number theory and, in the reviewerÃ¢â¬â¢s opinion, is far superior for this purpose to any other book in English. Ã¢â¬â¢ Bulletin of the American Mathematical Society THE HIGHER ARITHMETIC AN INTRODUCTION TO THE THEORY OF NUMBERS Eighth edition H. Davenport M. A. , SC. D. F. R. S. late Rouse Ball Professor of Mathematics in the University of Cambridge and Fellow of Trinity College Editing and additional material by James H. Davenport CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www. cambridge. org Information on this title: www. cambridge. org/9780521722360 Ã © The estate of H. Davenport 2008 This publication is in copyright.Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2008 ISBN-13 ISBN-13 978-0-511-45555-1 978-0-521-72236-0 eBook (EBL) paperback Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. CONTENTS Introduction I Factorization and the Primes 1. 2. 3. 4. . 6. 7. 8. 9. 10. The laws of arithmetic Proof by induction Prime numbers The fundamental theorem of arithmetic Consequences of the fundamental theorem EuclidÃ¢â¬â¢s algorithm Another proof of the fundamental theorem A property of the H. C. F Factorizing a number The series of primes page viii 1 1 6 8 9 12 16 18 19 22 25 31 31 33 35 37 40 41 42 45 46 II Congruences 1. 2. 3. 4. 5. 6. 7. 8. 9. The congruence notation Linear congruences FermatÃ¢â¬â¢s theorem EulerÃ¢â¬â¢s function ? (m) WilsonÃ¢â¬â¢s theorem Algebraic congruences Congruences to a prime modulus Congr uences in several unknowns Congruences covering all numbers v vi III Quadratic Residues 1. 2. 3. 4. . 6. Primitive roots Indices Quadratic residues GaussÃ¢â¬â¢s lemma The law of reciprocity The distribution of the quadratic residues Contents 49 49 53 55 58 59 63 68 68 70 72 74 77 78 82 83 86 92 94 99 103 103 104 108 111 114 116 116 117 120 122 124 126 128 131 133 IV Continued Fractions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Introduction The general continued fraction EulerÃ¢â¬â¢s rule The convergents to a continued fraction The equation ax ? by = 1 In? nite continued fractions Diophantine approximation Quadratic irrationals Purely periodic continued fractions LagrangeÃ¢â¬â¢s theorem PellÃ¢â¬â¢s equation A geometrical interpretation of continued fractionsV Sums of Squares 1. 2. 3. 4. 5. Numbers representable by two squares Primes of the form 4k + 1 Constructions for x and y Representation by four squares Representation by three squares VI Quadratic Forms 1. 2. 3. 4. 5. 6. 7. 8. 9. Introduction Equivalent forms The discriminant The representation of a number by a form Three examples The reduction of positive de? nite forms The reduced forms The number of representations The class-number Contents VII Some Diophantine Equations 1. Introduction 2. The equation x 2 + y 2 = z 2 3. The equation ax 2 + by 2 = z 2 4. Elliptic equations and curves 5.Elliptic equations modulo primes 6. FermatÃ¢â¬â¢s Last Theorem 7. The equation x 3 + y 3 = z 3 + w 3 8. Further developments vii 137 137 138 140 145 151 154 157 159 165 165 168 173 179 185 188 194 199 200 209 222 225 235 237 VIII Computers and Number Theory 1. 2. 3. 4. 5. 6. 7. 8. 9. Introduction Testing for primality Ã¢â¬ËRandomÃ¢â¬â¢ number generators PollardÃ¢â¬â¢s factoring methods Factoring and primality via elliptic curves Factoring large numbers The Dif? eÃ¢â¬âHellman cryptographic method The RSA cryptographic method Primality testing revisited Exercises Hints Answers Bibliography IndexINTRODUCTION T he higher arithmetic, or the theory of numbers, is concerned with the properties of the natural numbers 1, 2, 3, . . . . These numbers must have exercised human curiosity from a very early period; and in all the records of ancient civilizations there is evidence of some preoccupation with arithmetic over and above the needs of everyday life. But as a systematic and independent science, the higher arithmetic is entirely a creation of modern times, and can be said to date from the discoveries of Fermat (1601Ã¢â¬â1665).A peculiarity of the higher arithmetic is the great dif? culty which has often been experienced in proving simple general theorems which had been suggested quite naturally by numerical evidence. Ã¢â¬ËIt is just this,Ã¢â¬â¢ said Gauss, Ã¢â¬Ëwhich gives the higher arithmetic that magical charm which has made it the favourite science of the greatest mathematicians, not to mention its inexhaustible wealth, wherein it so greatly surpasses other parts of mathematics. Ã¢â¬â¢ The theory of numbers is generally considered to be the Ã¢â¬ËpurestÃ¢â¬â¢ branch of pure mathematics.It certainly has very few direct applications to other sciences, but it has one feature in common with them, namely the inspiration which it derives from experiment, which takes the form of testing possible general theorems by numerical examples. Such experiment, though necessary in some form to progress in every part of mathematics, has played a greater part in the development of the theory of numbers than elsewhere; for in other branches of mathematics the evidence found in this way is too often fragmentary and misleading.As regards the present book, the author is well aware that it will not be read without effort by those who are not, in some sense at least, mathematicians. But the dif? culty is partly that of the subject itself. It cannot be evaded by using imperfect analogies, or by presenting the proofs in a way viii Introduction ix which may convey the main idea o f the argument, but is inaccurate in detail. The theory of numbers is by its nature the most exact of all the sciences, and demands exactness of thought and exposition from its devotees. The theorems and their proofs are often illustrated by numerical examples.These are generally of a very simple kind, and may be despised by those who enjoy numerical calculation. But the function of these examples is solely to illustrate the general theory, and the question of how arithmetical calculations can most effectively be carried out is beyond the scope of this book. The author is indebted to many friends, and most of all to Professor o Erd? s, Professor Mordell and Professor Rogers, for suggestions and corrections. He is also indebted to Captain Draim for permission to include an account of his algorithm.The material for the ? fth edition was prepared by Professor D. J. Lewis and Dr J. H. Davenport. The problems and answers are based on the suggestions of Professor R. K. Guy. Chapter VIII a nd the associated exercises were written for the sixth edition by Professor J. H. Davenport. For the seventh edition, he updated Chapter VII to mention WilesÃ¢â¬â¢ proof of FermatÃ¢â¬â¢s Last Theorem, and is grateful to Professor J. H. Silverman for his comments. For the eighth edition, many people contributed suggestions, notably Dr J. F. McKee and Dr G. K. Sankaran.Cambridge University Press kindly re-typeset the book for the eighth edition, which has allowed a few corrections and the preparation of an electronic complement: www. cambridge. org/davenport. References to further material in the electronic complement, when known at the time this book went to print, are marked thus: Ã ¦:0. I FACTORIZATION AND THE PRIMES 1. The laws of arithmetic The object of the higher arithmetic is to discover and to establish general propositions concerning the natural numbers 1, 2, 3, . . . of ordinary arithmetic. Examples of such propositions are the fundamental theorem (I. 4)? hat every nat ural number can be factorized into prime numbers in one and only one way, and LagrangeÃ¢â¬â¢s theorem (V. 4) that every natural number can be expressed as a sum of four or fewer perfect squares. We are not concerned with numerical calculations, except as illustrative examples, nor are we much concerned with numerical curiosities except where they are relevant to general propositions. We learn arithmetic experimentally in early childhood by playing with objects such as beads or marbles. We ? rst learn addition by combining two sets of objects into a single set, and later we learn multiplication, in the form of repeated addition.Gradually we learn how to calculate with numbers, and we become familiar with the laws of arithmetic: laws which probably carry more conviction to our minds than any other propositions in the whole range of human knowledge. The higher arithmetic is a deductive science, based on the laws of arithmetic which we all know, though we may never have seen them form ulated in general terms. They can be expressed as follows. ? References in this form are to chapters and sections of chapters of this book. 1 2 The Higher Arithmetic Addition.Any two natural numbers a and b have a sum, denoted by a + b, which is itself a natural number. The operation of addition satis? es the two laws: a+b =b+a (commutative law of addition), (associative law of addition), a + (b + c) = (a + b) + c the brackets in the last formula serving to indicate the way in which the operations are carried out. Multiplication. Any two natural numbers a and b have a product, denoted by a ? b or ab, which is itself a natural number. The operation of multiplication satis? es the two laws ab = ba a(bc) = (ab)c (commutative law of multiplication), (associative law of multiplication).There is also a law which involves operations both of addition and of multiplication: a(b + c) = ab + ac (the distributive law). Order. If a and b are any two natural numbers, then either a is equal to b o r a is less than b or b is less than a, and of these three possibilities exactly one must occur. The statement that a is less than b is expressed symbolically by a < b, and when this is the case we also say that b is greater than a, expressed by b > a. The fundamental law governing this notion of order is that if a b. We propose to investigate the common divisors of a and b.If a is divisible by b, then the common divisors of a and b consist simply of all divisors of b, and there is no more to be said. If a is not divisible by b, we can express a as a multiple of b together with a remainder less than b, that is a = qb + c, where c < b. (2) This is the process of Ã¢â¬Ëdivision with a remainderÃ¢â¬â¢, and expresses the fact that a, not being a multiple of b, must occur somewhere between two consecutive multiples of b. If a comes between qb and (q + 1)b, then a = qb + c, where 0 < c < b. It follows from the equation (2) that any common divisor of b and c is also a divisor of a.Moreo ver, any common divisor of a and b is also a divisor of c, since c = a ? qb. It follows that the common divisors of a and b, whatever they may be, are the same as the common divisors of b and c. The problem of ? nding the common divisors of a and b is reduced to the same problem for the numbers b and c, which are respectively less than a and b. The essence of the algorithm lies in the repetition of this argument. If b is divisible by c, the common divisors of b and c consist of all divisors of c. If not, we express b as b = r c + d, where d < c. (3)Again, the common divisors of b and c are the same as those of c and d. The process goes on until it terminates, and this can only happen when exact divisibility occurs, that is, when we come to a number in the sequence a, b, c, . . . , which is a divisor of the preceding number. It is plain that the process must terminate, for the decreasing sequence a, b, c, . . . of natural numbers cannot go on for ever. Factorization and the Primes 17 Let us suppose, for the sake of de? niteness, that the process terminates when we reach the number h, which is a divisor of the preceding number g.Then the last two equations of the series (2), (3), . . . are f = vg + h, g = wh. (4) (5) The common divisors of a and b are the same as those of b and c, or of c and d, and so on until we reach g and h. Since h divides g, the common divisors of g and h consist simply of all divisors of h. The number h can be identi? ed as being the last remainder in EuclidÃ¢â¬â¢s algorithm before exact divisibility occurs, i. e. the last non-zero remainder. We have therefore proved that the common divisors of two given natural numbers a and b consist of all divisors of a certain number h (the H. C. F. f a and b), and this number is the last non-zero remainder when EuclidÃ¢â¬â¢s algorithm is applied to a and b. As a numerical illustration, take the numbers 3132 and 7200 which were used in Ã §5. The algorithm runs as follows: 7200 = 2 ? 3132 + 936, 3 132 = 3 ? 936 + 324, 936 = 2 ? 324 + 288, 324 = 1 ? 288 + 36, 288 = 8 ? 36; and the H. C. F. is 36, the last remainder. It is often possible to shorten the working a little by using a negative remainder whenever this is numerically less than the corresponding positive remainder. In the above example, the last three steps could be replaced by 936 = 3 ? 324 ? 6, 324 = 9 ? 36. The reason why it is permissible to use negative remainders is that the argument that was applied to the equation (2) would be equally valid if that equation were a = qb ? c instead of a = qb + c. Two numbers are said to be relatively prime? if they have no common divisor except 1, or in other words if their H. C. F. is 1. This will be the case if and only if the last remainder, when EuclidÃ¢â¬â¢s algorithm is applied to the two numbers, is 1. ? This is, of course, the same de? nition as in Ã §5, but is repeated here because the present treatment is independent of that given previously. 8 7. Another proof of t he fundamental theorem The Higher Arithmetic We shall now use EuclidÃ¢â¬â¢s algorithm to give another proof of the fundamental theorem of arithmetic, independent of that given in Ã §4. We begin with a very simple remark, which may be thought to be too obvious to be worth making. Let a, b, n be any natural numbers. The highest common factor of na and nb is n times the highest common factor of a and b. However obvious this may seem, the reader will ? nd that it is not easy to give a proof of it without using either EuclidÃ¢â¬â¢s algorithm or the fundamental theorem of arithmetic.In fact the result follows at once from EuclidÃ¢â¬â¢s algorithm. We can suppose a > b. If we divide na by nb, the quotient is the same as before (namely q) and the remainder is nc instead of c. The equation (2) is replaced by na = q. nb + nc. The same applies to the later equations; they are all simply multiplied throughout by n. Finally, the last remainder, giving the H. C. F. of na and nb, is nh, wher e h is the H. C. F. of a and b. We apply this simple fact to prove the following theorem, often called EuclidÃ¢â¬â¢s theorem, since it occurs as Prop. 30 of Book VII.If a prime divides the product of two numbers, it must divide one of the numbers (or possibly both of them). Suppose the prime p divides the product na of two numbers, and does not divide a. The only factors of p are 1 and p, and therefore the only common factor of p and a is 1. Hence, by the theorem just proved, the H. C. F. of np and na is n. Now p divides np obviously, and divides na by hypothesis. Hence p is a common factor of np and na, and so is a factor of n, since we know that every common factor of two numbers is necessarily a factor of their H. C. F.We have therefore proved that if p divides na, and does not divide a, it must divide n; and this is EuclidÃ¢â¬â¢s theorem. The uniqueness of factorization into primes now follows. For suppose a number n has two factorizations, say n = pqr . . . = p q r . . . , where all the numbers p, q, r, . . . , p , q , r , . . . are primes. Since p divides the product p (q r . . . ) it must divide either p or q r . . . . If p divides p then p = p since both numbers are primes. If p divides q r . . . we repeat the argument, and ultimately reach the conclusion that p must equal one of the primes p , q , r , . . . We can cancel the common prime p from the two representations, and start again with one of those left, say q. Eventually it follows that all the primes on the left are the same as those on the right, and the two representations are the same. Factorization and the Primes 19 This is the alternative proof of the uniqueness of factorization into primes, which was referred to in Ã §4. It has the merit of resting on a general theory (that of EuclidÃ¢â¬â¢s algorithm) rather than on a special device such as that used in Ã §4. On the other hand, it is longer and less direct. 8. A property of the H. C.F From EuclidÃ¢â¬â¢s algorithm one can deduce a remarkable property of the H. C. F. , which is not at all apparent from the original construction for the H. C. F. by factorization into primes (Ã §5). The property is that the highest common factor h of two natural numbers a and b is representable as the difference between a multiple of a and a multiple of b, that is h = ax ? by where x and y are natural numbers. Since a and b are both multiples of h, any number of the form ax ? by is necessarily a multiple of h; and what the result asserts is that there are some values of x and y for which ax ? y is actually equal to h. Before giving the proof, it is convenient to note some properties of numbers representable as ax ? by. In the ? rst place, a number so representable can also be represented as by ? ax , where x and y are natural numbers. For the two expressions will be equal if a(x + x ) = b(y + y ); and this can be ensured by taking any number m and de? ning x and y by x + x = mb, y + y = ma. These numbers x and y will be natura l numbers provided m is suf? ciently large, so that mb > x and ma > y. If x and y are de? ned in this way, then ax ? by = by ? x . We say that a number is linearly dependent on a and b if it is representable as ax ? by. The result just proved shows that linear dependence on a and b is not affected by interchanging a and b. There are two further simple facts about linear dependence. If a number is linearly dependent on a and b, then so is any multiple of that number, for k(ax ? by) = a. kx ? b. ky. Also the sum of two numbers that are each linearly dependent on a and b is itself linearly dependent on a and b, since (ax1 ? by1 ) + (ax2 ? by2 ) = a(x1 + x2 ) ? b(y1 + y2 ). 20 The Higher ArithmeticThe same applies to the difference of two numbers: to see this, write the second number as by2 ? ax2 , in accordance with the earlier remark, before subtracting it. Then we get (ax1 ? by1 ) ? (by2 ? ax2 ) = a(x1 + x2 ) ? b(y1 + y2 ). So the property of linear dependence on a and b is preserved by addition and subtraction, and by multiplication by any number. We now examine the steps in EuclidÃ¢â¬â¢s algorithm, in the light of this concept. The numbers a and b themselves are certainly linearly dependent on a and b, since a = a(b + 1) ? b(a), b = a(b) ? b(a ? 1). The ? rst equation of the algorithm was a = qb + c.Since b is linearly dependent on a and b, so is qb, and since a is also linearly dependent on a and b, so is a ? qb, that is c. Now the next equation of the algorithm allows us to deduce in the same way that d is linearly dependent on a and b, and so on until we come to the last remainder, which is h. This proves that h is linearly dependent on a and b, as asserted. As an illustration, take the same example as was used in Ã §6, namely a = 7200 and b = 3132. We work through the equations one at a time, using them to express each remainder in terms of a and b. The ? rst equation was 7200 = 2 ? 3132 + 936, which tells s that 936 = a ? 2b. The second equation was 3 132 = 3 ? 936 + 324, which gives 324 = b ? 3(a ? 2b) = 7b ? 3a. The third equation was 936 = 2 ? 324 + 288, which gives 288 = (a ? 2b) ? 2(7b ? 3a) = 7a ? 16b. The fourth equation was 324 = 1 ? 288 + 36, Factorization and the Primes which gives 36 = (7b ? 3a) ? (7a ? 16b) = 23b ? 10a. 21 This expresses the highest common factor, 36, as the difference of two multiples of the numbers a and b. If one prefers an expression in which the multiple of a comes ? rst, this can be obtained by arguing that 23b ? 10a = (M ? 10)a ? (N ? 23)b, provided that Ma = N b.Since a and b have the common factor 36, this factor can be removed from both of them, and the condition on M and N becomes 200M = 87N . The simplest choice for M and N is M = 87, N = 200, which on substitution gives 36 = 77a ? 177b. Returning to the general theory, we can express the result in another form. Suppose a, b, n are given natural numbers, and it is desired to ? nd natural numbers x and y such that ax ? by = n. (6) Such an e quation is called an indeterminate equation since it does not determine x and y completely, or a Diophantine equation after Diophantus of Alexandria (third century A . D . , who wrote a famous treatise on arithmetic. The equation (6) cannot be soluble unless n is a multiple of the highest common factor h of a and b; for this highest common factor divides ax ? by, whatever values x and y may have. Now suppose that n is a multiple of h, say, n = mh. Then we can solve the equation; for all we have to do is ? rst solve the equation ax1 ? by1 = h, as we have seen how to do above, and then multiply throughout by m, getting the solution x = mx1 , y = my1 for the equation (6). Hence the linear indeterminate equation (6) is soluble in natural numbers x, y if and only if n is a multiple of h.In particular, if a and b are relatively prime, so that h = 1, the equation is soluble whatever value n may have. As regards the linear indeterminate equation ax + by = n, we have found the condition for it to be soluble, not in natural numbers, but in integers of opposite signs: one positive and one negative. The question of when this equation is soluble in natural numbers is a more dif? cult one, and one that cannot well be completely answered in any simple way. Certainly 22 The Higher Arithmetic n must be a multiple of h, but also n must not be too small in relation to a and b.It can be proved quite easily that the equation is soluble in natural numbers if n is a multiple of h and n > ab. 9. Factorizing a number The obvious way of factorizing a number is to test whether it is divisible by 2 or by 3 or by 5, and so on, using the series of primes. If a number N v is not divisible by any prime up to N , it must be itself a prime; for any composite number has at least two prime factors, and they cannot both be v greater than N . The process is a very laborious one if the number is at all large, and for this reason factor tables have been computed.The most extensive one which is gener ally accessible is that of D. N. Lehmer (Carnegie Institute, Washington, Pub. No. 105. 1909; reprinted by Hafner Press, New York, 1956), which gives the least prime factor of each number up to 10,000,000. When the least prime factor of a particular number is known, this can be divided out, and repetition of the process gives eventually the complete factorization of the number into primes. Several mathematicians, among them Fermat and Gauss, have invented methods for reducing the amount of trial that is necessary to factorize a large number.Most of these involve more knowledge of number-theory than we can postulate at this stage; but there is one method of Fermat which is in principle extremely simple and can be explained in a few words. Let N be the given number, and let m be the least number for which m 2 > N . Form the numbers m 2 ? N , (m + 1)2 ? N , (m + 2)2 ? N , . . . . (7) When one of these is reached which is a perfect square, we get x 2 ? N = y 2 , and consequently N = x 2 ? y 2 = (x ? y)(x + y). The calculation of the numbers (7) is facilitated by noting that their successive differences increase at a constant rate. The identi? ation of one of them as a perfect square is most easily made by using BarlowÃ¢â¬â¢s Table of Squares. The method is particularly successful if the number N has a factorization in which the two factors are of about the same magnitude, since then y is small. If N is itself a prime, the process goes on until we reach the solution provided by x + y = N , x ? y = 1. As an illustration, take N = 9271. This comes between 962 and 972 , so that m = 97. The ? rst number in the series (7) is 972 ? 9271 = 138. The Factorization and the Primes 23 subsequent ones are obtained by adding successively 2m + 1, then 2m + 3, and so on, that is, 195, 197, and so on.This gives the series 138, 333, 530, 729, 930, . . . . The fourth of these is a perfect square, namely 272 , and we get 9271 = 1002 ? 272 = 127 ? 73. An interesting algorithm for fact orization has been discovered recently by Captain N. A. Draim, U . S . N. In this, the result of each trial division is used to modify the number in preparation for the next division. There are several forms of the algorithm, but perhaps the simplest is that in which the successive divisors are the odd numbers 3, 5, 7, 9, . . . , whether prime or not. To explain the rules, we work a numerical example, say N = 4511. The ? st step is to divide by 3, the quotient being 1503 and the remainder 2: 4511 = 3 ? 1503 + 2. The next step is to subtract twice the quotient from the given number, and then add the remainder: 4511 ? 2 ? 1503 = 1505, 1505 + 2 = 1507. The last number is the one which is to be divided by the next odd number, 5: 1507 = 5 ? 301 + 2. The next step is to subtract twice the quotient from the ? rst derived number on the previous line (1505 in this case), and then add the remainder from the last line: 1505 ? 2 ? 301 = 903, 903 + 2 = 905. This is the number which is to be divi ded by the next odd number, 7. Now we an continue in exactly the same way, and no further explanation will be needed: 905 = 7 ? 129 + 2, 903 ? 2 ? 129 = 645, 645 ? 2 ? 71 = 503, 503 ? 2 ? 46 = 411, 645 + 2 = 647, 503 + 8 = 511, 411 + 5 = 416, 647 = 9 ? 71 + 8, 511 = 11 ? 46 + 5, 416 = 13 ? 32 + 0. 24 The Higher Arithmetic We have reached a zero remainder, and the algorithm tells us that 13 is a factor of the given number 4511. The complementary factor is found by carrying out the ? rst half of the next step: 411 ? 2 ? 32 = 347. In fact 4511 = 13? 347, and as 347 is a prime the factorization is complete. To justify the algorithm generally is a matter of elementary algebra.Let N1 be the given number; the ? rst step was to express N1 as N1 = 3q1 + r1 . The next step was to form the numbers M2 = N1 ? 2q1 , The number N2 was divided by 5: N2 = 5q2 + r2 , and the next step was to form the numbers M3 = M2 ? 2q2 , N 3 = M3 + r 2 , N 2 = M2 + r 1 . and so the process was continued. It can be deduced from these equations that N2 = 2N1 ? 5q1 , N3 = 3N1 ? 7q1 ? 7q2 , N4 = 4N1 ? 9q1 ? 9q2 ? 9q3 , and so on. Hence N2 is divisible by 5 if and only if 2N1 is divisible by 5, or N1 divisible by 5. Again, N3 is divisible by 7 if and only if 3N1 is divisible by 7, or N1 divisible by 7, and so on.When we reach as divisor the least prime factor of N1 , exact divisibility occurs and there is a zero remainder. The general equation analogous to those given above is Nn = n N1 ? (2n + 1)(q1 + q2 + Ã · Ã · Ã · + qn? 1 ). The general equation for Mn is found to be Mn = N1 ? 2(q1 + q2 + Ã · Ã · Ã · + qn? 1 ). (9) If 2n + 1 is a factor of the given number N1 , then Nn is exactly divisible by 2n + 1, and Nn = (2n + 1)qn , whence n N1 = (2n + 1)(q1 + q2 + Ã · Ã · Ã · + qn ), (8) Factorization and the Primes by (8). Under these circumstances, we have, by (9), Mn+1 = N1 ? 2(q1 + q2 + Ã · Ã · Ã · + qn ) = N1 ? 2 n 2n + 1 N1 = N1 . n + 1 25 Thus the complementary factor to the factor 2n + 1 is Mn+1 , as stated in the example. In the numerical example worked out above, the numbers N1 , N2 , . . . decrease steadily. This is always the case at the beginning of the algorithm, but may not be so later. However, it appears that the later numbers are always considerably less than the original number. 10. The series of primes Although the notion of a prime is a very natural and obvious one, questions concerning the primes are often very dif? cult, and many such questions are quite unanswerable in the present state of mathematical knowledge.We conclude this chapter by mentioning brie? y some results and conjectures about the primes. In Ã §3 we gave EuclidÃ¢â¬â¢s proof that there are in? nitely many primes. The same argument will also serve to prove that there are in? nitely many primes of certain speci? ed forms. Since every prime after 2 is odd, each of them falls into one of the two progressions (a) 1, 5, 9, 13, 17, 21, 25, . . . , (b) 3, 7, 11, 15, 19, 23, 27, . . . ; the progression (a) consisting of all numbers of the form 4x + 1, and the progression (b) of all numbers of the form 4x ? 1 (or 4x + 3, which comes to the same thing).We ? rst prove that there are in? nitely many primes in the progression (b). Let the primes in (b) be enumerated as q1 , q2 , . . . , beginning with q1 = 3. Consider the number N de? ned by N = 4(q1 q2 . . . qn ) ? 1. This is itself a number of the form 4x ? 1. Not every prime factor of N can be of the form 4x + 1, because any product of numbers which are all of the form 4x + 1 is itself of that form, e. g. (4x + 1)(4y + 1) = 4(4x y + x + y) + 1. Hence the number N has some prime factor of the form 4x ? 1. This cannot be any of the primes q1 , q2 , . . . , qn , since N leaves the remainder ? when 26 The Higher Arithmetic divided by any of them. Thus there exists a prime in the series (b) which is different from any of q1 , q2 , . . . , qn ; and this proves the proposition. The same argument cannot be used to prove t hat there are in? nitely many primes in the series (a), because if we construct a number of the form 4x +1 it does not follow that this number will necessarily have a prime factor of that form. However, another argument can be used. Let the primes in the series (a) be enumerated as r1 , r2 , . . . , and consider the number M de? ned by M = (r1 r2 . . rn )2 + 1. We shall see later (III. 3) that any number of the form a 2 + 1 has a prime factor of the form 4x + 1, and is indeed entirely composed of such primes, together possibly with the prime 2. Since M is obviously not divisible by any of the primes r1 , r2 , . . . , rn , it follows as before that there are in? nitely many primes in the progression (a). A similar situation arises with the two progressions 6x + 1 and 6x ? 1. These progressions exhaust all numbers that are not divisible by 2 or 3, and therefore every prime after 3 falls in one of these two progressions.One can prove by methods similar to those used above that there ar e in? nitely many primes in each of them. But such methods cannot cope with the general arithmetical progression. Such a progression consists of all numbers ax +b, where a and b are ? xed and x = 0, 1, 2, . . . , that is, the numbers b, b + a, b + 2a, . . . . If a and b have a common factor, every number of the progression has this factor, and so is not a prime (apart from possibly the ? rst number b). We must therefore suppose that a and b are relatively prime. It then seems plausible that the progression will contain in? itely many primes, i. e. that if a and b are relatively prime, there are in? nitely many primes of the form ax + b. Legendre seems to have been the ? rst to realize the importance of this proposition. At one time he thought he had a proof, but this turned out to be fallacious. The ? rst proof was given by Dirichlet in an important memoir which appeared in 1837. This proof used analytical methods (functions of a continuous variable, limits, and in? nite series), an d was the ? rst really important application of such methods to the theory of numbers.It opened up completely new lines of development; the ideas underlying DirichletÃ¢â¬â¢s argument are of a very general character and have been fundamental for much subsequent work applying analytical methods to the theory of numbers. Factorization and the Primes 27 Not much is known about other forms which represent in? nitely many primes. It is conjectured, for instance, that there are in? nitely many primes of the form x 2 + 1, the ? rst few being 2, 5, 17, 37, 101, 197, 257, . . . . But not the slightest progress has been made towards proving this, and the question seems hopelessly dif? cult.Dirichlet did succeed, however, in proving that any quadratic form in two variables, that is, any form ax 2 + bx y + cy 2 , in which a, b, c are relatively prime, represents in? nitely many primes. A question which has been deeply investigated in modern times is that of the frequency of occurrence of the p rimes, in other words the question of how many primes there are among the numbers 1, 2, . . . , X when X is large. This number, which depends of course on X , is usually denoted by ? (X ). The ? rst conjecture about the magnitude of ? (X ) as a function of X seems to have been made independently by Legendre and Gauss about X 1800.It was that ? (X ) is approximately log X . Here log X denotes the natural (so-called Napierian) logarithm of X , that is, the logarithm of X to the base e. The conjecture seems to have been based on numerical evidence. For example, when X is 1,000,000 it is found that ? (1,000,000) = 78,498, whereas the value of X/ log X (to the nearest integer) is 72,382, the ratio being 1. 084 . . . . Numerical evidence of this kind may, of course, be quite misleading. But here the result suggested is true, in the sense that the ratio of ? (X ) to X/ log X tends to the limit 1 as X tends to in? ity. This is the famous Prime Number Theorem, ? rst proved by Hadamard and de la Vall? e e Poussin independently in 1896, by the use of new and powerful analytical methods. It is impossible to give an account here of the many other results which have been proved concerning the distribution of the primes. Those proved in the nineteenth century were mostly in the nature of imperfect approaches towards the Prime Number Theorem; those of the twentieth century included various re? nements of that theorem. There is one recent event to which, however, reference should be made.We have already said that the proof of DirichletÃ¢â¬â¢s Theorem on primes in arithmetical progressions and the proof of the Prime Number Theorem were analytical, and made use of methods which cannot be said to belong properly to the theory of numbers. The propositions themselves relate entirely to the natural numbers, and it seems reasonable that they should be provable without the intervention of such foreign ideas. The search for Ã¢â¬ËelementaryÃ¢â¬â¢ proofs of these two theorems was u nsuccessful until fairly recently. In 1948 A. Selberg found the ? rst elementary proof of DirichletÃ¢â¬â¢s Theorem, and with 28 The Higher Arithmetic he help of P. Erd? s he found the ? rst elementary proof of the Prime Numo ber Theorem. An Ã¢â¬ËelementaryÃ¢â¬â¢ proof, in this connection, means a proof which operates only with natural numbers. Such a proof is not necessarily simple, and indeed both the proofs in question are distinctly dif? cult. Finally, we may mention the famous problem concerning primes which was propounded by Goldbach in a letter to Euler in 1742. Goldbach suggested (in a slightly different wording) that every even number from 6 onwards is representable as the sum of two primes other than 2, e. g. 6 = 3 + 3, 8 = 3 + 5, 10 = 3 + 7 = 5 + 5, 12 = 5 + 7, . . . Any problem like this which relates to additive properties of primes is necessarily dif? cult, since the de? nition of a prime and the natural properties of primes are all expressed in terms of multiplic ation. An important contribution to the subject was made by Hardy and Littlewood in 1923, but it was not until 1930 that anything was rigorously proved that could be considered as even a remote approach towards a solution of GoldbachÃ¢â¬â¢s problem. In that year the Russian mathematician Schnirelmann proved that there is some number N such that every number from some point onwards is representable as the sum of at most N primes.A much nearer approach was made by Vinogradov in 1937. He proved, by analytical methods of extreme subtlety, that every odd number from some point onwards is representable as the sum of three primes. This was the starting point of much new work on the additive theory of primes, in the course of which many problems have been solved which would have been quite beyond the scope of any pre-Vinogradov methods. A recent result in connection with GoldbachÃ¢â¬â¢s problem is that every suf? ciently large even number is representable as the sum of two numbers, one of which is a prime and the other of which has at most two prime factors.Notes Where material is changing more rapidly than print cycles permit, we have chosen to place some of the material on the bookÃ¢â¬â¢s website: www. cambridge. org/davenport. Symbols such as Ã ¦I:0 are used to indicate where there is such additional material. Ã §1. The main dif? culty in giving any account of the laws of arithmetic, such as that given here, lies in deciding which of the various concepts should come ? rst. There are several possible arrangements, and it seems to be a matter of taste which one prefers. It is no part of our purpose to analyse further the concepts and laws of ? rithmetic. We take the commonsense (or na? ve) view that we all Ã¢â¬ËknowÃ¢â¬â¢ Factorization and the Primes 29 the natural numbers, and are satis? ed of the validity of the laws of arithmetic and of the principle of induction. The reader who is interested in the foundations of mathematics may consult Bertrand Russe ll, Introduction to Mathematical Philosophy (Allen and Unwin, London), or M. Black, The Nature of Mathematics (Harcourt, Brace, New York). Russell de? nes the natural numbers by selecting them from numbers of a more general kind. These more general numbers are the (? ite or in? nite) cardinal numbers, which are de? ned by means of the more general notions of Ã¢â¬ËclassÃ¢â¬â¢ and Ã¢â¬Ëone-to-one correspondenceÃ¢â¬â¢. The selection is made by de? ning the natural numbers as those which possess all the inductive properties. (Russell, loc. cit. , p. 27). But whether it is reasonable to base the theory of the natural numbers on such a vague and unsatisfactory concept as that of a class is a matter of opinion. Ã¢â¬ËDolus latet in universalibusÃ¢â¬â¢ as Dr Johnson remarked. Ã §2. The objection to using the principle of induction as a de? ition of the natural numbers is that it involves references to Ã¢â¬Ëany proposition about a natural number nÃ¢â¬â¢. It seems plain the th at Ã¢â¬ËpropositionsÃ¢â¬â¢ envisaged here must be statements which are signi? cant when made about natural numbers. It is not clear how this signi? cance can be tested or appreciated except by one who already knows the natural numbers. Ã §4. I am not aware of having seen this proof of the uniqueness of prime factorization elsewhere, but it is unlikely that it is new. For other direct proofs, see Mathews, p. 2, or Hardy and Wright, p. 21.? Ã §5. It has been shown by (intelligent! computer searches that there is no odd perfect number less than 10300 . If an odd perfect number exists, it has at least eight distinct prime factors, of which the largest exceeds 108 . For references and other information on perfect or Ã¢â¬Ënearly perfectÃ¢â¬â¢ numbers, see Guy, sections A. 3, B. 1 and B. 2. Ã ¦I:1 Ã §6. A critical reader may notice that in two places in this section I have used principles that were not explicitly stated in Ã §Ã §1 and 2. In each place, a proof by induction co uld have been given, but to have done so would have distracted the readerÃ¢â¬â¢s attention from the main issues.The question of the length of EuclidÃ¢â¬â¢s algorithm is discussed in Uspensky and Heaslet, ch. 3, and D. E. KnuthÃ¢â¬â¢s The Art of Computer Programming vol. II: Seminumerical Algorithms (Addison Wesley, Reading, Mass. , 3rd. ed. , 1998) section 4. 5. 3. Ã §9. For an account of early methods of factoring, see DicksonÃ¢â¬â¢s History Vol. I, ch. 14. For a discussion of the subject as it appeared in ? Particulars of books referred to by their authorsÃ¢â¬â¢ names will be found in the Bibliography. 30 The Higher Arithmetic the 1970s see the article by Richard K. Guy, Ã¢â¬ËHow to factor a numberÃ¢â¬â¢, Congressus Numerantium XVI Proc. th Manitoba Conf. Numer. Math. , Winnipeg, 1975, 49Ã¢â¬â89, and at the turn of the millennium see Richard P. Brent, Ã¢â¬ËRecent progress and prospects for integer factorisation algorithmsÃ¢â¬â¢, Springer Lecture Notes in Comp uter Science 1858 Proc. Computing and Combinatorics, 2000, 3Ã¢â¬â22. The subject is discussed further in Chapter VIII. It is doubtful whether D. N. LehmerÃ¢â¬â¢s tables will ever be extended, since with them and a pocket calculator one can easily check whether a 12-digit number is a prime. Primality testing is discussed in VIII. 2 and VIII. 9. For DraimÃ¢â¬â¢s algorithm, see Mathematics Magazine, 25 (1952) 191Ã¢â¬â4. 10. An excellent account of the distribution of primes is given by A. E. Ingham, The Distribution of Prime Numbers (Cambridge Tracts, no. 30, 1932; reprinted by Hafner Press, New York, 1971). For a more recent and extensive account see H. Davenport, Multiplicative Number Theory, 3rd. ed. (Springer, 2000). H. Iwaniec (Inventiones Math. 47 (1978) 171Ã¢â¬â88) has shown that for in? nitely many n the number n 2 + 1 is either prime or the product of at most two primes, and indeed the same is true for any irreducible an 2 + bn + c with c odd. DirichletÃ¢â¬â¢s p roof of his theorem (with a modi? ation due to Mertens) is given as an appendix to DicksonÃ¢â¬â¢s Modern Elementary Theory of Numbers. An elementary proof of the Prime Number Theorem is given in ch. 22 of Hardy and Wright. An elementary proof of the asymptotic formula for the number of primes in an arithmetic progression is given in Gelfond and Linnik, ch. 3. For a survey of early work on GoldbachÃ¢â¬â¢s problem, see James, Bull. American Math. Soc. , 55 (1949) 246Ã¢â¬â60. It has been veri? ed that every even number from 6 to 4 ? 1014 is the sum of two primes, see Richstein, Math. Comp. , 70 (2001) 1745Ã¢â¬â9. For a proof of ChenÃ¢â¬â¢s theorem that every suf? iently large even integer can be represented as p + P2 , where p is a prime, and P2 is either a prime or the product of two primes, see ch. 11 of Sieve Methods by H. Halberstam and H. E. Richert (Academic Press, London, 1974). For a proof of VinogradovÃ¢â¬â¢s result, see T. Estermann, Introduction to Modern Prime Number Theory (Cambridge Tracts, no. 41, 1952) or H. Davenport, Multiplicative Number Theory, 3rd. ed. (Springer, 2000). Ã¢â¬ËSuf? ciently largeÃ¢â¬â¢ in VinogradovÃ¢â¬â¢s result has now been quanti? ed as Ã¢â¬Ëgreater than 2 ? 101346 Ã¢â¬â¢, see M. -C. Liu and T. Wang, Acta Arith. , 105 (2002) 133Ã¢â¬â175.Conversely, we know that it is true up to 1. 13256 ? 1022 (Ramar? and Saouter in J. Number Theory 98 (2003) 10Ã¢â¬â33). e II CONGRUENCES 1. The congruence notation It often happens that for the purposes of a particular calculation, two numbers which differ by a multiple of some ? xed number are equivalent, in the sense that they produce the same result. For example, the value of (? 1)n depends only on whether n is odd or even, so that two values of n which differ by a multiple of 2 give the same result. Or again, if we are concerned only with the last digit of a number, then for that purpose two umbers which differ by a multiple of 10 are effectively the same. The congruence notation, introduced by Gauss, serves to express in a convenient form the fact that two integers a and b differ by a multiple of a ? xed natural number m. We say that a is congruent to b with respect to the modulus m, or, in symbols, a ? b (mod m). The meaning of this, then, is simply that a ? b is divisible by m. The notation facilitates calculations in which numbers differing by a multiple of m are effectively the same, by stressing the analogy between congruence and equality.Congruence, in fact, means Ã¢â¬Ëequality except for the addition of some multiple of mÃ¢â¬â¢. A few examples of valid congruences are: 63 ? 0 (mod 3), 7 ? ?1 (mod 8), 52 ? ?1 (mod 13). A congruence to the modulus 1 is always valid, whatever the two numbers may be, since every number is a multiple of 1. Two numbers are congruent with respect to the modulus 2 if they are of the same parity, that is, both even or both odd. 31 32 The Higher Arithmetic Two congruences can be added, subtracted, or m ultiplied together, in just the same way as two equations, provided all the congruences have the same modulus.If a ? ? (mod m) and b ? ? (mod m) then a + b ? ? + ? (mod m), a ? b ? ? ? ? (mod m), ab ? (mod m). The ? rst two of these statements are immediate; for example (a + b) ? (? + ? ) is a multiple of m because a ? ? and b ? ? are both multiples of m. The third is not quite so immediate and is best proved in two steps. First ab ? ?b because ab ? ?b = (a ? ?)b, and a ? ? is a multiple of m. Next, ? b ? , for a similar reason. Hence ab ? (mod m). A congruence can always be multiplied throughout by any integer: if a ? ? (mod m) then ka ? k? (mod m).Indeed this is a special case of the third result above, where b and ? are both k. But it is not always legitimate to cancel a factor from a congruence. For example 42 ? 12 (mod 10), but it is not permissible to cancel the factor 6 from the numbers 42 and 12, since this would give the false result 7 ? 2 (mod 10). The reason is obvious : the ? rst congruence states that 42 ? 12 is a multiple of 10, but this does not imply that 1 (42 ? 12) is a multiple of 10. The cancellation of 6 a factor from a congruence is legitimate if the factor is relatively prime to the modulus.For let the given congruence be ax ? ay (mod m), where a is the factor to be cancelled, and we suppose that a is relatively prime to m. The congruence states that a(x ? y) is divisible by m, and it follows from the last proposition in I. 5 that x ? y is divisible by m. An illustration of the use of congruences is provided by the well-known rules for the divisibility of a number by 3 or 9 or 11. The usual representation of a number n by digits in the scale of 10 is really a representation of n in the form n = a + 10b + 100c + Ã · Ã · Ã · , where a, b, c, . . . re the digits of the number, read from right to left, so that a is the number of units, b the number of tens, and so on. Since 10 ? 1 (mod 9), we have also 102 ? 1 (mod 9), 103 ? 1 (mod 9), and so on. Hence it follows from the above representation of n that n ? a + b + c + Ã · Ã · Ã · (mod 9). Congruences 33 In other words, any number n differs from the sum of its digits by a multiple of 9, and in particular n is divisible by 9 if and only if the sum of its digits is divisible by 9. The same applies with 3 in place of 9 throughout. The rule for 11 is based on the fact that 10 ? ?1 (mod 11), so that 102 ? +1 (mod 11), 103 ? 1 (mod 11), and so on. Hence n ? a ? b + c ? Ã · Ã · Ã · (mod 11). It follows that n is divisible by 11 if and only if a ? b+c? Ã · Ã · Ã · is divisible by 11. For example, to test the divisibility of 9581 by 11 we form 1? 8+5? 9, or ? 11. Since this is divisible by 11, so is 9581. 2. Linear congruences It is obvious that every integer is congruent (mod m) to exactly one of the numbers 0, 1, 2, . . . , m ? 1. (1) r < m, For we can express the integer in the form qm + r , where 0 and then it is congruent to r (mod m). Obviously there are othe r sets of numbers, besides the set (1), which have the same property, e. . any integer is congruent (mod 5) to exactly one of the numbers 0, 1, ? 1, 2, ? 2. Any such set of numbers is said to constitute a complete set of residues to the modulus m. Another way of expressing the de? nition is to say that a complete set of residues (mod m) is any set of m numbers, no two of which are congruent to one another. A linear congruence, by analogy with a linear equation in elementary algebra, means a congruence of the form ax ? b (mod m). (2) It is an important fact that any such congruence is soluble for x, provided that a is relatively prime to m.The simplest way of proving this is to observe that if x runs through the numbers of a complete set of residues, then the corresponding values of ax also constitute a complete set of residues. For there are m of these numbers, and no two of them are congruent, since ax 1 ? ax2 (mod m) would involve x1 ? x2 (mod m), by the cancellation of the factor a (permissible since a is relatively prime to m). Since the numbers ax form a complete set of residues, there will be exactly one of them congruent to the given number b. As an example, consider the congruence 3x ? 5 (mod 11). 34 The Higher ArithmeticIf we give x the values 0, 1, 2, . . . , 10 (a complete set of residues to the modulus 11), 3x takes the values 0, 3, 6, . . . , 30. These form another complete set of residues (mod 11), and in fact they are congruent respectively to 0, 3, 6, 9, 1, 4, 7, 10, 2, 5, 8. The value 5 occurs when x = 9, and so x = 9 is a solution of the congruence. Naturally any number congruent to 9 (mod 11) will also satisfy the congruence; but nevertheless we say that the congruence has one solution, meaning that there is one solution in any complete set of residues. In other words, all solutions are mutually congruent.The same applies to the general congruence (2); such a congruence (provided a is relatively prime to m) is precisely equivalent to the con gruence x ? x0 (mod m), where x0 is one particular solution. There is another way of looking at the linear congruence (2). It is equivalent to the equation ax = b + my, or ax ? my = b. We proved in I. 8 that such a linear Diophantine equation is soluble for x and y if a and m are relatively prime, and that fact provides another proof of the solubility of the linear congruence. But the proof given above is simpler, and illustrates the advantages gained by using the congruence notation.The fact that the congruence (2) has a unique solution, in the sense explained above, suggests that one may use this solution as an interpretation b for the fraction a to the modulus m. When we do this, we obtain an arithmetic (mod m) in which addition, subtraction and multiplication are always possible, and division is also possible provided that the divisor is relatively prime to m. In this arithmetic there are only a ? nite number of essentially distinct numbers, namely m of them, since two numbers w hich are mutually congruent (mod m) are treated as the same.If we take the modulus m to be 11, as an illustration, a few examples of Ã¢â¬Ëarithmetic mod 11Ã¢â¬â¢ are: 5 ? 9 ? ?2. 3 Any relation connecting integers or fractions in the ordinary sense remains true when interpreted in this arithmetic. For example, the relation 5 + 7 ? 1, 5 ? 6 ? 8, 1 2 7 + = 2 3 6 becomes (mod 11) 6 + 8 ? 3, because the solution of 2x ? 1 is x ? 6, that of 3x ? 2 is x ? 8, and that of 6x ? 7 is x ? 3. Naturally the interpretation given to a fraction depends on the modulus, for instance 2 ? 8 (mod 11), but 2 ? 3 (mod 7). The 3 3 Congruences 35 nly limitation on such calculations is that just mentioned, namely that the denominator of any fraction must be relatively prime to the modulus. If the modulus is a prime (as in the above examples with 11), the limitation takes the very simple form that the denominator must not be congruent to 0 (mod m), and this is exactly analogous to the limitation in ordina ry arithmetic that the denominator must not be equal to 0. We shall return to this point later (Ã §7). 3. FermatÃ¢â¬â¢s theorem The fact that there are only a ? nite number of essentially different numbers in arithmetic to a modulus m means that there are algebraic relations which are satis? d by every number in that arithmetic. There is nothing analogous to these relations in ordinary arithmetic. Suppose we take any number x and consider its powers x, x 2 , x 3 , . . . . Since there are only a ? nite number of possibilities for these to the modulus m, we must eventually come to one which we have met before, say x h ? x k (mod m), where k < h. If x is relatively prime to m, the factor x k can be cancelled, and it follows that x l ? 1 (mod m), where l ? h ? k. Hence every number x which is relatively prime to m satis? es some congruence of this form. The least exponent l for which x l ? (mod m) will be called the order of x to the modulus m. If x is 1, its order is obviously 1. To illustrate the de? nition, let us calculate the orders of a few numbers to the modulus 11. The powers of 2, taken to the modulus 11, are 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, . . . . Each one is twice the preceding one, with 11 or a multiple of 11 subtracted where necessary to make the result less than 11. The ? rst power of 2 which is ? 1 is 210 , and so the order of 2 (mod 11) is 10. As another example, take the powers of 3: 3, 9, 5, 4, 1, 3, 9, . . . . The ? rst power of 3 which is ? 1 is 35 , so the order of 3 (mod 11) is 5.It will be found that the order of 4 is again 5, and so also is that of 5. It will be seen that the successive powers of x are periodic; when we have reached the ? rst number l for which x l ? 1, then x l+1 ? x and the previous cycle is repeated. It is plain that x n ? 1 (mod m) if and only if n is a multiple of the order of x. In the last example, 3n ? 1 (mod 11) if and only if n is a multiple of 5. This remains valid if n is 0 (since 30 = 1), and it remains valid also for negative exponents, provided 3? n , or 1/3n , is interpreted as a fraction (mod 11) in the way explained in Ã §2. 36 The Higher ArithmeticIn fact, the negative powers of 3 (mod 11) are obtained by prolonging the series backwards, and the table of powers of 3 to the modulus 11 is n =Ã¢â¬ ¦ ?3 ? 2 ? 1 0 1 2 3 4 5 6 . . . 9 5 4 1 3 9 5 4 1 3 Ã¢â¬ ¦ . 3n ? . . . Fermat discovered that if the modulus is a prime, say p, then every integer x not congruent to 0 satis? es x p? 1 ? 1 (mod p). (3) In view of what we have seen above, this is equivalent to saying that the order of any number is a divisor of p ? 1. The result (3) was mentioned by Fermat in a letter to Fr? nicle de Bessy of 18 October 1640, in which he e also stated that he had a proof.But as with most of FermatÃ¢â¬â¢s discoveries, the proof was not published or preserved. The ? rst known proof seems to have been given by Leibniz (1646Ã¢â¬â1716). He proved that x p ? x (mod p), which is equivalent to (3), b y writing x as a sum 1 + 1 + Ã · Ã · Ã · + 1 of x units (assuming x positive), and then expanding (1 + 1 + Ã · Ã · Ã · + 1) p by the multinomial theorem. The terms 1 p + 1 p + Ã · Ã · Ã · + 1 p give x, and the coef? cients of all the other terms are easily proved to be divisible by p. Quite a different proof was given by Ivory in 1806. If x ? 0 (mod p), the integers x, 2x, 3x, . . . , ( p ? )x are congruent (in some order) to the numbers 1, 2, 3, . . . , p ? 1. In fact, each of these sets constitutes a complete set of residues except that 0 has been omitted from each. Since the two sets are congruent, their products are congruent, and so (x)(2x)(3x) . . . (( p ? 1)x) ? (1)(2)(3) . . . ( p ? 1)(mod p). Cancelling the factors 2, 3, . . . , p ? 1, as is permissible, we obtain (3). One merit of this proof is that it can be extended so as to apply to the more general case when the modulus is no longer a prime. The generalization of the result (3) to any modulus was ? rst given b y Euler in 1760.To formulate it, we must begin by considering how many numbers in the set 0, 1, 2, . . . , m ? 1 are relatively prime to m. Denote this number by ? (m). When m is a prime, all the numbers in the set except 0 are relatively prime to m, so that ? ( p) = p ? 1 for any prime p. EulerÃ¢â¬â¢s generalization of FermatÃ¢â¬â¢s theorem is that for any modulus m, x ? (m) ? 1 (mod m), provided only that x is relatively prime to m. (4) Congruences 37 To prove this, it is only necessary to modify IvoryÃ¢â¬â¢s method by omitting from the numbers 0, 1, . . . , m ? 1 not only the number 0, but all numbers which are not relatively prime to m.There remain ? (m) numbers, say a 1 , a2 , . . . , a? , Then the numbers a1 x, a2 x, . . . , a? x are congruent, in some order, to the previous numbers, and on multiplying and cancelling a1 , a2 , . . . , a? (as is permissible) we obtain x ? ? 1 (mod m), which is (4). To illustrate this proof, take m = 20. The numbers less than 20 and relati vely prime to 20 are 1, 3, 7, 9, 11, 13, 17, 19, so that ? (20) = 8. If we multiply these by any number x which is relatively prime to 20, the new numbers are congruent to the original numbers in some other order.For example, if x is 3, the new numbers are congruent respectively to 3, 9, 1, 7, 13, 19, 11, 17 (mod 20); and the argument proves that 38 ? 1 (mod 20). In fact, 38 = 6561. where ? = ? (m). 4. EulerÃ¢â¬â¢s function ? (m) As we have just seen, this is the number of numbers up to m that are relatively prime to m. It is natural to ask what relation ? (m) bears to m. We saw that ? ( p) = p ? 1 for any prime p. It is also easy to evaluate ? ( p a ) for any prime power pa . The only numbers in the set 0, 1, 2, . . . , pa ? 1 which are not relatively prime to p are those that are divisible by p. These are the numbers pt, where t = 0, 1, . . , pa? 1 ? 1. The number of them is pa? 1 , and when we subtract this from the total number pa , we obtain ? ( pa ) = pa ? pa? 1 = pa? 1 ( p ? 1). (5) The determination of ? (m) for general values of m is effected by proving that this function is multiplicative. By this is meant that if a and b are any two relatively prime numbers, then ? (ab) = ? (a)? (b). (6) 38 The Higher Arithmetic To prove this, we begin by observing a general principle: if a and b are relatively prime, then two simultaneous congruences of the form x ? ? (mod a), x ? ? (mod b) (7) are precisely equivalent to one congruence to the modulus ab.For the ? rst congruence means that x = ? + at where t is an integer. This satis? es the second congruence if and only if ? + at ? ? (mod b), or at ? ? ? ? (mod b). This, being a linear congruence for t, is soluble. Hence the two congruences (7) are simultaneously soluble. If x and x are two solutions, we have x ? x (mod a) and x ? x (mod b), and therefore x ? x (mod ab). Thus there is exactly one solution to the modulus ab. This principle, which extends at once to several congruences, provided that the moduli ar e relatively prime in pairs, is sometimes called Ã¢â¬Ëthe Chinese remainder theoremÃ¢â¬â¢.It assures us of the existence of numbers which leave prescribed remainders on division by the moduli in question. Let us represent the solution of the two congruences (7) by x ? [? , ? ] (mod ab), so that [? , ? ] is a certain number depending on ? and ? (and also on a and b of course) which is uniquely determined to the modulus ab. Different pairs of values of ? and ? give rise to different values for [? , ? ]. If we give ? the values 0, 1, . . . , a ? 1 (forming a complete set of residues to the modulus a) and similarly give ? the values 0, 1, . . . , b ? 1, the resulting values of [? , ? constitute a complete set of residues to the modulus ab. It is obvious that if ? has a factor in common with a, then x in (7) will also have that factor in common with a, in other words, [? , ? ] will have that factor in common with a. Thus [? , ? ] will only be relatively prime to ab if ? is relatively prime to a and ? is relatively prime to b, and conversely these conditions will ensure that [? , ? ] is relatively prime to ab. It follows that if we give ? the ? (a) possible values that are less than a and prime to a, and give ? the ? (b) values that are less than b and prime to b, there result ? (a)? (b) values of [? ? ], and these comprise all the numbers that are less than ab and relatively prime to ab. Hence ? (ab) = ? (a)? (b), as asserted in (6). To illustrate the situation arising in the above proof, we tabulate below the values of [? , ? ] when a = 5 and b = 8. The possible values for ? are 0, 1, 2, 3, 4, and the possible values for ? are 0, 1, 2, 3, 4, 5, 6, 7. Of these there are four values of ? which are relatively prime to a, corresponding to the fact that ? (5) = 4, and four values of ? that are relatively prime to b, Congruences 39 corresponding to the fact that ? (8) = 4, in accordance with the formula (5).These values are italicized, as also are the corresponding values of [? , ? ]. The latter constitute the sixteen numbers that are relatively prime to 40 and less than 40, thus verifying that ? (40) = ? (5)? (8) = 4 ? 4 = 16. ? ? 0 1 2 3 4 0 0 16 32 8 24 1 25 1 17 33 9 2 10 26 2 18 34 3 35 11 27 3 19 4 20 36 12 28 4 5 5 21 37 13 29 6 30 6 22 38 14 7 15 31 7 23 39 We now return to the original question, that of evaluating ? (m) for any number m. Suppose the factorization of m into prime powers is m = pa q b . . . . Then it follows from (5) and (6) that ? (m) = ( pa ? pa? 1 )(q b ? q b? 1 ) . . . or, more elegantly, ? (m) = m 1 ? For example, ? (40) = 40 1 ? and ? (60) = 60 1 ? 1 2 1 2 1 p 1? 1 q Ã¢â¬ ¦. (8) 1? 1 3 1 5 = 16, 1 5 1? 1? = 16. The function ? (m) has a remarkable property, ? rst given by Gauss in his Disquisitiones. It is that the sum of the numbers ? (d), extended over all the divisors d of a number m, is equal to m itself. For example, if m = 12, the divisors are 1, 2, 3, 4, 6, 12, and we have ? (1) + ? (2) + ? (3) + ? (4) + ? (6) + ? (12) = 1 + 1 + 2 + 2 + 2 + 4 = 12. A general proof can be based either on (8), or directly on the de? nition of the function. 40 The Higher ArithmeticWe have already referred (I. 5) to a table of the values of ? (m) for m 10, 000. The same volume contains a table giving those numbers m for which ? (m) assumes a given value up to 2,500. This table shows that, up to that point at least, every value assumed by ? (m) is assumed at least twice. It seems reasonable to conjecture that this is true generally, in other words that for any number m there is another number m such that ? (m ) = ? (m). This has never been proved, and any attempt at a general proof seems to meet with formidable dif? culties. For some special types of numbers the result is easy, e. g. f m is odd, then ? (m) = ? (2m); or again if m is not divisible by 2 or 3 we have ? (3m) = ? (4m) = ? (6m). 5. WilsonÃ¢â¬â¢s theorem This theorem was ? rst publis
Thursday, August 29, 2019
The executive, Anucha Browne Sanders, has asked for an additional $9. 6 million in compensatory damages, which the judge will decide on in the coming weeks. The Garden and Mr. Thomas said they would appeal. Ms. Browne Sanders accused Mr. Thomas of verbally abusing and sexually harassing her over a two-year period. Less than a month after she formally complained to the Garden, the company chairman, James L. Dolan, fired her. In court, the Garden cited numerous explanations for the dismissal, including poor job performance and the claim that she had interfered with the GardenÃ¢â¬â¢s internal investigation of her accusations.Ms. Browne Sanders, who wept when the decision was read, described her victory as important for Ã¢â¬Å"the women who donÃ¢â¬â¢t have the means and couldnÃ¢â¬â¢t possibly have done what I was able to do,Ã¢â¬ and for Ã¢â¬Å"everybody that cares about working in a civil work environment. Ã¢â¬ Mr. Thomas emerged from the courthouse and said, Ã¢â¬Å"I want to say it as loud as I possibly can: I am innocent; I am very innocent. I did not do the things that she accused me in the courtroom of doing. Ã¢â¬ Patting his chest for emphasis, he added, Ã¢â¬Å"I am extremely disappointed that the jury did not see the facts in this case.I will appeal this. Ã¢â¬ The sordid four-week trial was the latest chapter in the story of a once-respected N. B. A. franchise. During Mr. ThomasÃ¢â¬â¢s nearly four-year tenure as president and now coach, the team has spent millions on free agents without any progress toward a championship. And still to come, the Garden faces a second sexual harassment trial brought by a former Rangers cheerleader against team officials. The Garden was ordered to pay $6 million for subjecting Ms. Browne Sanders to a hostile work environment and another $2. million for firing her in retaliation. The jury ordered that Mr. Dolan pay $3 million for the retaliation. In his testimony, Mr. Dolan said that he alone made the decision to fire her. Mr. Dolan had no comment on the verdict or the award. Although found liable, Mr. Thomas will not have to pay any of the punitive damages for sexually harassing Ms. Browne Sanders with unwanted sexual advances. One holdout on the seven-member jury kept the panel from holding him financially responsible for the harassment. The juror Sally Anne Foster, 49, of Cortlandt Manor, N. Y. said it was just Ã¢â¬Å"different personality traitsÃ¢â¬ among the jurors that led them to hold Mr. Dolan, not Mr. Thomas, financially liable for Ms. Browne SandersÃ¢â¬â¢s claims. Asked if she believed Mr. DolanÃ¢â¬â¢s testimony, Ms. Foster said: Ã¢â¬Å"I canÃ¢â¬â¢t say. IÃ¢â¬â¢m not a psychiatrist. Ã¢â¬ United States District Court Judge Gerard E. Lynch is expected to make a decision as early as next month on compensatory damages for Ms. Browne Sanders, a formerNorthwestern University basketball star and mother of three, who was fired in January 2006 from her $260,000-a-year job as the KnicksÃ¢â¬â¢ senior vice president of marketing.She said that the firing by Mr. Dolan led her to search for more than year to find her current job as the associate athletic director at the State University of New York at Buffalo, at about half her salary with the Knicks. The trial painted Mr. Thomas as the foul-mouthed president of basketball operations who clashed with Ms. Browne Sanders about their executive responsibilities. It featured testimony about sex between the teamÃ¢â¬â¢s star,Stephon Marbury, and a Knicks intern in his truck; a slipshod internal Garden investigation of Ms. Browne SandersÃ¢â¬â¢s claims; and the hiring of Mr.MarburyÃ¢â¬â¢s cousin and a boyfriend of Mr. DolanÃ¢â¬â¢s stepdaughter as Knicks employees. The trial and the verdicts exposed more concern about the state of the Knicks, and the Garden, under Mr. DolanÃ¢â¬â¢s leadership. Mr. Thomas stepped in to coach after Mr. Dolan fired Larry Brown, who received an $18. 5 million contract settlem ent during an arbitration by N. B. A. Commissioner David Stern. The Garden refused to settle the case against Ms. Browne Sanders, exposing the organization to ridicule in newspapers and in television reports. But the verdicts will not lead to sanctions by the league.Although the league penalizes players, coaches and team owners for criminal infractions, said Tim Frank, the leagueÃ¢â¬â¢s vice president for basketball communications, Ã¢â¬Å"Our policies do not encompass civil litigation. Ã¢â¬ The claims that Ms. Browne Sanders made against Mr. Thomas appeared to come down to a test of her credibility as a little-known executive against his celebrity standing and Hall of Fame status. Her claims that he repeatedly referred to her as a Ã¢â¬Å"bitchÃ¢â¬ and made unwanted advances had little corroboration from witnesses. Still, they stood up to a barrage of testimony by witnesses for Mr.Thomas and the Garden that her job performance had dramatically floundered in 2005. Ã¢â¬Å"If th is was something I made up, it would have been a lot juicier,Ã¢â¬ Ms. Browne Sanders said in an interview. After the verdict, Mr. Thomas left for the KnicksÃ¢â¬â¢ training camp in Charleston, S. C. , where he arrived later in the afternoon. Ms. Browne Sanders gathered with her legal team and relatives for a prayer circle outside the courthouse. In its statement, the Garden said, Ã¢â¬Å"We look forward to presenting our arguments to an appeals court and believe they will agree that no sexual harassment took place. Carl Tobias, a professor at the University of Richmond Law School and a former civil litigator, doubted the appeal would succeed. Ã¢â¬Å"The jury heard the facts and it is unusual that an appeals court would overturn a jury finding which was so fact-intensive,Ã¢â¬ he said in a telephone interview. Ã¢â¬Å"There isnÃ¢â¬â¢t much basis unless there was a legal error, and Judge Lynch has a very good reputation for making sure his cases are fairly and properly tried. Ã¢ â¬ Tobias said that the jury not holding Mr. Thomas financially liable Ã¢â¬Å"is not a large enough inconsistency that could have this overturned. Ã¢â¬
Wednesday, August 28, 2019
Impact and association of vision impairment from diabetic retinopathy in diabetic patients - Thesis Example Another cause of concern is the increase of the disease among younger age groups, ranging from ages 18 to 69 years, making it a possible leading cause of eye and vision loss at early ages (Teck, 2006). Diabetic retinopathy is one of the reliable tools to assess the status, rate of progress and complications developing in the patient with the disease. Therefore, it can be used for systematic evaluation and checkups to ascertain the efficacy of a certain treatment regarding management of diabetes (Chun et al, 2007). American researches claim diabetic retinopathy to be the leading cause of blindness among people younger than 65 years of age (Knobbe and Hadrill, 2011). Alongside, researches on Australian and American populations also suggest that diabetic retinopathy is associated with higher risks of mortality due to heart complications and stroke respectively (Wu, 2010, pp 263). Diabetes can lead to many types of eye complications, including glaucoma, or the increase in the intraocular pressure, which in turn may increase chances of developing diabetic retinopathy (Sugiura et al, 2011, pp 624 and 625). Again, the type of retinopathy affects the intraocular pressure.
Statistics in Business Decision Making - Term Paper Example As discussed in the introduction that this paper will attempt to describe statistical tools such as mean, standard deviation, variance etc. These statistical tools often come under the heading of descriptive statistics as they are the main tools used to collect data quantitatively and present in the more meaningful manner to draw some logical conclusions from the data collected. Once data is collected, it is nothing more than a raw set of data which may provide no clue about the potential information that they may provide. Thus one meaningful way of manipulating the data will calculate the mean or average of the data. It is also important to note that mean values may provide distorted information because of the outliers effect. One large observation value can distort the results and mean values may become more inflated due to the impact of outliers or larger values in the population.Mean value is considered as one of the most significant and important measures especially in finance. There are various uses of this measure in finance i.e. from measuring the average rate of return on an investment to calculating the weighted average rate of return of a portfolio.Similarly, average values are also calculated for studying the costs also as concepts such as average cost, average variable costs, average fixed costs are important concepts to understand in order to make important business decisions because controlling costs is one of the fundamental responsibilities of the managers.
Tuesday, August 27, 2019
Development Needs analysis (explanation and justification)(personal develop) - Essay Example It is a self-reflective process of metacognition. The evaluation of oneÃ¢â¬â¢s skills in strategic learning is a critical part of the school curricular program. It entails the student reflecting on the abilities, strengths and weakness in the tackling of the everyday work at school. Since historical times at the invention of the formal education system, teachers used academic performance as the driving tool to determine the excellence for the students. Conveniently, this provides a framework and a record of results that reflect on the student is thinking capabilities. However, this makes the students less equipped in handling day-to-day life situations. Time to time, oral presentation in class gives a reflection of the oratory skills of the students and can be harnessed and perfected for good public speaking skills. In addition, make-up, and personal tutorials help to build the culture of critical thinking as the student are engaged in the explanation of the answers. It offers a better platform for the students to assess their personal capabilities in their academics and general mindset. The teacher can capitalize on this by asking the student a one on one question to evaluate how well they answer the questions. My learning culture is very progressive putting into consideration the previous class evaluations. I am not blowing anything out of proportion and being a victim of the Lake Wobegon effect of thinking beyond my capabilitie s. Despite not putting the best of effort due to the co-curricular activities, I still manage to register excellent scores. It has culminated from the fact that I represent the school in much of the outdoor curricular activities. I have attended student-led conferences in which I presented an article before the teachers and the parents. It gave me opportunities to reflect finally on my oratory skills that I have been practicing over a period. I had a well-prepared portfolio prior to the conference day that assisted and guided me
Monday, August 26, 2019
Argument of evaluation - Assignment Example when he should have enjoyed his retirement) and the circumstances under which Reagan conducted his presidency, it is clear how great his influence was, which is still felt today, which most people do not know but which everybody somehow benefited. All previous presidents will be evaluated on how good they were on both domestic and international affairs. On the domestic side, issues like the economy, employment, taxes, and governance will be used as criteria to measure their achievements. On the international front, the main criteria to be used will be conduct of foreign policy and international agenda like globalization, trade issues, and foreign wars (whether limited wars or proxy wars). President Ronald Reagan will be compared to the other great presidents in terms of his accomplishments, how he measures up against public opinion polls, how all other foreign governments view his administration, the long-term effects of his actions while in office, and how history will probably judge him, even in the years to come. An example of his enduring legacy is the adoption of the word Ã¢â¬Å"ReaganomicsÃ¢â¬ which means lowered taxes and less of the government interference in business (Dunn 51) and in peoples lives, and a strong
Sunday, August 25, 2019
Further Exploration assignment - Essay Example The Janice and Melwyn, looking back said that they felt each new baby created an increased bond of closeness between them, but then as the kids became older, Melwyn reports feeling that they were often, Ã¢â¬Å"in the wayÃ¢â¬ . As a result, the couple reports that for nearly a decade, they couldnÃ¢â¬â¢t remember going on a date, seeing a movie together or taking a trip without the children. This eroded the quality of their marriage and relationship. Stevie and Lucas say that they have felt their love grow for one another as they support each other in caring for a toddler and preparing for an infant. They feel that they have had to change very little so far in their lives because of kids. They never were into partying or going out much before kids and that hasnÃ¢â¬â¢t changed. They do miss the quiet that has been lost to the two year-oldÃ¢â¬â¢s toys and constant talking. They know that crying is soon ahead, as are late night feedings and diapers, but all in all they feel as clo se as they ever have. Janice and Melwyn report that when they had their first child, Janice quit her job as a bank teller to stay home with the children. She had a high school diploma and Melwyn was trained as a machinist. The economy was good, so Melwyn took overtime in order to make-up for the loss of JaniceÃ¢â¬â¢s income. Janice did not return to the workforce until her youngest child was in 7th grade. She says she didnÃ¢â¬â¢t want to work once the kids came along, but admitted that being a housewife was boring and isolating at times. Stevie is going to continue her studies at university through her pregnancy and after the birth of the next child. She is studying elementary education. She feels this would provide good additional income at some point and allow her to have time with her kids. Lucas is an engineer. He is just starting his career and wants to be the sole provider for the family for as long as
Saturday, August 24, 2019
Camp Nou Stadium - Research Paper Example Camp Nou Stadium Camp nou is the largest football stadium in Spain and the second largest football stadium in Europe. It is located in Barcelona and seats 99354 people in a football match but in any match organized by the UEFA the stadium is reduced to 96336 people per stadium. The stadium has hosted many matches and was one of the stadiums that hosted the 1992 summer Olympics that was held in Spain. Construction of this amazing football stadium started on the 28th match of 1954. This was because the previous Barcelona stadium camp de les corts could not be expanded anymore and so the then government decided to build another bigger stadium that could accommodate all the crazy soccer fans who were increasing in number daily and increased revenue to the government. This construction began in front of more than 60000 fans of the Barcelona football club team watching the construction. Before the construction began, Archbishop of Barcelona laid the ground with prayers before the civil gov ernor of Barcelona preceded by laying down the first stone to mark the beginning of the construction. The architectures in the team were Francesca Titans, Lorenzo Garcia and Josep soteras. The construction ended on 24 September 1957 and since then it has been the official Barcelona football club stadium. This means that the construction took more than three years. After its construction, the constructor later had many tenders for more construction and the architects behind the design of the stadium were given several awards for being the best designers. Culture and society The stadium was built in a culture that was characterized by mad football support and competition with opponent football club Real Madrid that at the time had the largest soccer stadium. This led to the Barcelona fans to want a stadium that could host more people and help them be proud of it, as it was larger than the opponentÃ¢â¬â¢s stadium. The construction of the stadium was seen as an opportunity for the rep utable Spanish architectures to show off their architectural skills to the population by designing the best stadium ever. Therefore, the panel picked the best stadium that they felt was affordable and could last for a longer period. Foster and collaborates were the contractors they finally won the tender to build the stadium. At the time of its building, many people were fascinated by the way, that football stadiums were being built in Europe starting with the construction of Old Trafford stadium which was one of the state of the art stadiums at the time. After this, many club supporters across Europe wanted to help in the building of similar stadiums with the same capacity or even more. At this time, there were several universities that offered different courses in architecture and there were different competition on who could design and build the most efficient and best clubs. There were also upcoming technologies that were used in the building process that was of importance. Cons truction of the stadium Constructed in 1957, the tools and the methods of construction are not so different from those that are currently used. However, there is a slight difference in the tools, the materials, and the labour that was used. Trucks did transportation and cranes did loading. Since the Portland cement used in the making of concrete was not far from the site off construction, the transportation did not take a lot of time. Cranes were then used to uplift the building materials to greater heights and the major
Friday, August 23, 2019
What exactly are the consequences of performance appraisal in organisations and how does it contribute to organisational or individual performance - Essay Example Additionally, the process provides the management with information that helps them to identify employee potentials. The organization can desire to affect the employee attitude by the results obtained by the employee through the evaluation cycle. The evaluation cycle provide the employee with clear performance SMART goals. The performance SMART goals are Specific, Measurable, Actionable, Realistic and Time-bound that the employee should work towards achieving. If the employee is performing well, the organization can reward such an employee by job promotion or increase in salary pay. If an employee performance is poor, the organization can choose to train the employee so that the employee can improve in performance (Kerwick 2013). After goal setting in performance appraisal, the supervisor and employee hold a self-appraisal meeting to discuss the employeeÃ¢â¬â¢s performance so that they can identify potential goals for the upcoming appraisal period (Michelle, Douglas & John 2010). Based on the management, the employees perform their job and the supervisor should a keep note of the employeeÃ¢â¬â¢s accomplishments and challenges. When a performance appraisal is carried out, a supervisor needs to show the employee how their performance affects the productivity of the organization. The employees should also understand how their performance affects the ability of others employees to do their jobs, and it helps put his job duties into an overall company context. This helps improve the notion of teamwork among the staff and can also encourage the cooperation to achieve corporate goals. After employee evaluation, employees with high scores become motivated and are likely to perform well and increase organizational productivity (Michelle et al 2010).An employee with a positive feedback makes himself feel like a worthy contributor in the organization hence encouraging them to perform better in the future. This
Thursday, August 22, 2019
Victorian England Essay When Oliver reaches London he falls into the hands of Fagin who is likened to the devil by Dickens who upon Olivers first meeting is described to be wearing a red cloak and holding a fork in front of a fire, he then continues with a more frightening description, whose villainous-looking and repulsive face was obscured by a quantity of matted red hair. Dickens cleverly does this as Fagin like the devil tempts people into a life of sin and crime. Even the colours black and red suggest that it was not a nice place, that it was evil and dangerous. Something that Nancy, in the book struggles to break free from like many of the paupers in those situations. Like Oliver, Nancy has a sense of right and wrong, after playing a part of the re-capturing of Oliver she realises what she has done is wrong but it is her loyalties to Sikes and his gang that keep her from taking action sooner. Although she does save Oliver, Dickens keeps the story realistic, when she is discovered to have alerted Mr Brownlow and Rose, Sikes clubs her to death. The one person that ever truly cared for him, who he constantly abused he eventually killed. Dickens based him on typical hardened criminals in Victorian England. Children in Victorian England were intensely used by adult gang members, as pick pockets, prostitutes and even in organised robberies. Once involved in crime there was no escape unless you were very lucky as Oliver was. Oliver is taught to pickpocket and winds up in trouble when he is thought to have stolen from a wealthy man when in fact it wasnt him. Mr Brownlow shows compassion and takes him into his home even though he has been wronged, this is an ideal that Dickens holds dear and tried to vent it through his writing even if he does make it comical when he ridicules the rich by comparing Olivers terrible illness and suffering to Mr Brownlows worrying for his health when he uses a slightly damp cloth. As the future begins to look bright for Oliver you discover that Fagin is not far and he wants Oliver back as he could be a danger to them, this is an example of the never-ending crime cycle in Victorian England. Oliver is captured and forced to help Sikes in a robbery where he is shot and then left presumed dead. Dickens uses this as an example of how harsh Victorian England could be Sikes and his gang had fled when it had gone wrong leaving anything that would slow them down.
Wednesday, August 21, 2019
Improving Health and Wellness in Students Essay Rising consumerism is a problem that has a profound effect on children today.Ã Children and adults watch television and are inundated with commercials that urge viewers to buy the latest technological gadgets that replace outdoor activities and exercise.Ã As well, the latest candy, ice cream, and other unhealthy products are cast in between cartoons that capture childrenÃ¢â¬â¢s attentions and their parents are pressured to please their children and buy them material objects to satisfy them rather than engaging them in healthy activities and studying much of the time. Many parents are working and have little time to spend with their kids, sitting in front of the television together or encouraging kids to quietly play with their gadgets (play stations, computers, and others), so parents can relax may become the norm.Ã To make up for this lack of involvement many parents guiltily give in to childrenÃ¢â¬â¢s whims and buy them candy and unhealthy food, to save time Ã¢â¬Å"fast-foodÃ¢â¬ replaces a healthy dinner and that much needed time at the dinner table to interact and be involved in their childrens lives. An efficient school program would not only target the youth, but their parents, as well then.Ã Ã¢â¬Å"Operation Pause the PlayStationÃ¢â¬ will be aimed at educating parents and children separately on issues involving obesity and other unhealthy behaviors.Ã It is probable that parents of children, who are not obese, will be less receptive and unlikely to come to these after-school classes. But, it is postulated this involvement program will be less receptive if it were labeled as a program for troubled kids.Ã Therefore this program will have the contingency that children will not be able to pass to the next grade level unless parents attend (this is fitting as the program should be implemented at the end of the school year before children have summer break and may be more sedentary and involved in unhealthy activities without the benefit of healthy school lunches and physical education). Therefore, parents and children should attend at least three classes that tackle these problems.Ã Class one should tackle the Ã¢â¬Å"McDonaldization of SocietyÃ¢â¬ and demonstrate that busy working families do not mean to do their children harm when settling for fast food, but that fast food is unhealthy and can lead to obesity and/or unhealthy learned behaviors that will follow children into adulthood. Ã¢â¬Å"Giving inÃ¢â¬ to children and buying them unhealthy food to fill the void that is left from lack of quality time, should also be addressed.Ã Most importantly, the lack of parental involvement in school activities due to rushed lifestyles should be addressed.Ã This class will be a sort of forum, as well, not meant to single out any parent, but an opportunity for parents facing the same kinds of strain to network with one another and see that they can be part of a healthy solution. Class two should encompass the overuse of technical gadgets (including television) that interfere with healthy activity.Ã Teachers of these classes that can be taken from high school level health classes and may choose films or other forms of media that deal with these issues to show that, in a sense, these parents and their families are Ã¢â¬Å"victimsÃ¢â¬ of consumerism. Outside of class, more and more children are watching more and more TV, to the point that they are watching approximately 40,000 TV commercials annually. (The CEO of Prism Communication notes, Ã¢â¬Å"They arenÃ¢â¬â¢t children as much as what I like to call Ã¢â¬Ëevolving consumersÃ¢â¬â¢Ã¢â¬ (Heiner, 2006). Deconstructing these facts that lead to sedentary and possibly unhealthy behaviors in children from a larger, systemic base will, also, help parents to realize that they are not being singled out, but instead part of a consumer culture that demands this type of behavior. While the parents are involved in the first two classes, high school level physical education and health teachers should teach the children about food pyramid and what different foods do to help the body grow and be strong.Ã They should, also, focus on different exercise techniques that are fun and help to keep children in shape.Ã The two courses should help children to begin to think beyond McDonaldÃ¢â¬â¢s and PlayStation and the children will take what they learn and teach their parents. This is what class three should be revolving around, a fun and light-hearted end to the requirement.Ã Here students will tech their parents what they have learned as far as healthy eating and a better overall lifestyle.Ã The parents will, most likely, appreciate that their children are making an effort to improve their lifestyles and will continue where the classes left off.Ã Additionally, there should be representatives from various summer camps and programs that are inexpensive , so that even children in poverty could attend.Ã The YMCA, Boys and Girls Club, and other organizations should promote what they have to offer at this time and, hopefully the summer will serve as a break from studies, but a beginning to more healthy behavior. In conclusion, problems with obesity and unhealthy behavior are systemic.Ã We live in a consumer culture that causes both the old and young to sometimes believe that having Ã¢â¬Å"thingsÃ¢â¬ is essential.Ã Hurried lifestyles, as well, from long work hours, and especially in single-parent households may lead to turning to Ã¢â¬Å"fast foodÃ¢â¬ and a lesser interest in school activities.Ã Competing with friends to have the most up-to-date technology may lead to parents having pressure put on them to provide these unnecessary gadgets and relaxing may start to take the form of television watching or other technological time.Ã These problems are not unique to any one group, but all parents and all children are at risk.Ã Ã¢â¬Å"Operation Pause the PlaystationÃ¢â¬ , should help change attitudes on this. References Heiner, R. (2006).Ã Social Problems: An Introduction to Critical Constructionism.Ã New York: Oxford University Press. Insidehighered.com.Ã Advanced Placement Still Ascending.Ã (2007). Retrieved February 18, 2007 from Ã Ã http://www.insidehighered.com/news/2007/02/07/ap. Ã Ã Ã Ã Ã Ã Ã Ã Ã Ã Ã Leone, Peter Drakeford, William.Ã Alternative Education: From a Last Chance to a Proactive Mode. (1999).Ã Reprinted with permission of The Clearing House: Volume 3, Number 2, November/December 1999: The Helen Dwight Reid Educational Foundation. Published by Heldref Publications, 1319 18th St. N.W. Washington, D.C. 20036-1802. Copyright 19.Ã Retrieved February 18, 2007 from http://www.edjj.org/Publications/pub_06_13_00_1.html. Payne, R.Ã (1996).Ã A Framework for Understanding Poverty.Ã p. 59.Ã Highlands: aha! Process, Inc.