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A Course on Number Theory. Peter J. Cameron
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1 A Course on Number Theory Peter J. Cameron
2 ii
3 Preface These are the notes of the course MTH6128, Number Theory, which I taught at Queen Mary, University of London, in the spring semester of There is nothing original to me in the notes. The course was designed by Susan McKay, and developed by Stephen Donkin, Ian Chiswell, Charles Leedham- Green, and Thomas Müller; I have benefited greatly from Ian Chiswell s notes, which I have followed closely. I am grateful to Mark Walters who stood in for me in the first six lectures of the course, and whose comments have been very helpful; also to the class tutors, markers, and most of all the students who took the course, for their comments and support. The original course was largely based on continued fractions: this technique is very amenable to hand calculation, and can be used to solve Pell s equation, to write an integer as a sum of squares where this is possible, and to classify the indefinite binary quadratic forms. This is still the centrepiece of the course, but I have given alternate treatment of sums of squares. The syllabus for the course reads (a) Continued fractions: finite and infinite continued fractions, approximation by rationals, order of approximation. (b) Continued fractions of quadratic surds: applications to the solution of Pell s equation and the sum of two squares. (c) Binary quadratic forms: equivalence, unimodular transformations, reduced form, class number. Use of continued fractions in the indefinite case. (d) Modular arithmetic: primitive roots, quadratic residues, Legendre symbol, quadratic reciprocity. Applications to quadratic forms. The learning outcomes state Students will be able to use continued fractions to develop arbitrarily accurate rational approximations to rational and irrational numbers. iii
4 iv They will be able to work with Diophantine equations, i.e. polynomial equations with integer solutions. They will know some of the famous classical theorems and conjectures in number theory, such as Fermat s Last Theorem and Goldbach s Conjecture, and be aware of some of the tools used to investigate such problems. The recommended books are [1] H Davenport, The Higher Arithmetic, Cambridge University Press (1999) [2] Allenby & Redfern, Introduction to Number Theory with Computing, Edward Arnold (1989)
5 Contents 1 Overview and revision Overview Euclid s algorithm Primes and factorisation Congruences and modular arithmetic The Chinese Remainder Theorem And finally Algebraic numbers Algebraic numbers and algebraic integers Quadratic irrationals Appendix: Sums, products and quotients Finite continued fractions Introduction The [ ] functions The convergents of a finite continued fraction A party trick Infinite continued fractions An example The definition Approximation by convergents Order of approximation Proof of Lemma Periodic continued fractions Periodic and purely periodic continued fractions Quadratic irrationals The main theorem v
6 vi CONTENTS 6 Lagrange and Pell Introduction The continued fraction for n Sums of two squares The equations x 2 ny 2 = ± Euler s totient function Euler s totient function Evaluation of φ(n) Orders of elements Primitive roots The Möbius function Appendix: An algebraic view Appendix: Cryptography Quadratic residues and non-residues Definition and basic properties The Legendre symbol A Euclid-type theorem Proofs of the first two rules Proofs of Rules 3 and Sums of squares Sums of two squares Sums of four squares Two squares revisited Sums of three squares Where do these identities come from? Pythagoras and Fermat Open problems Appendix: an algebraic proof Quadratic forms Linear forms and degenerate quadratic forms Matrix, discriminant, equivalence Positive definite forms Indefinite quadratic forms Revision problems and solutions Problems Solutions
7 Chapter 1 Overview and revision In this section we will meet some of the concerns of Number Theory, and have a brief revision of some of the relevant material from Introduction to Algebra. 1.1 Overview Number theory is about properties of the natural numbers, integers, or rational numbers, such as the following: Given a natural number n, is it prime or composite? If it is composite, how can we factorise it? How many solutions do equations like x 2 + y 2 = n or x n + y n = z n have for fixed n, where the variables are required to be natural numbers? How closely can we approximate a given irrational number by rational numbers which are not too complicated? How many primes are there less than (or any other bound we might choose? Are more primes of the form 4k + 1 than 4k 1, or vice versa? Some of these questions are interesting because properties of numbers have fascinated humans for thousands of years. On the other hand, some of them (such as primality testing and factorisation) are of very great practical importance: the secret codes that keep internet commerce secure depend on properties of numbers such as primality, factorisation, and modular arithmetic. Not all these questions will be covered in the course. But here are some problems, which turn out to be closely related to one another, which we will consider. Let p be an odd prime number. 1
8 2 CHAPTER 1. OVERVIEW AND REVISION Can we express p in the form x 2 + y 2 for some natural numbers x and y? (For example, 13 = , but 19 cannot be written in this form, as you can check.) Given a natural number a, is it congruent to the square of a number x modulo p? How do we tell? (For example, mod 13, but there is no solution to 1 x 2 mod 19.) Does the equation x 2 py 2 = 1 have a solution? What about x 2 py 2 = 1? For example, = 1, but there is no solution to x 2 19y 2 = 1. How closely can p be approximated by a rational number? For example, 2 is approximately equal to /100000, but 1393/985 is an even better approximation, and has much smaller numerator and denominator. How does one find such good approximations? 1.2 Euclid s algorithm We will always count 0 as being a natural number. We recall that, if a and b are natural numbers and b > 0, then there exist unique natural numbers q and r such that a = bq + r, with 0 r < b. The numbers q and r are the quotient and remainder when a is divided by b. We sometimes write q = adivb and r = amodb. If amodb = 0, we say that b divides a and write b a. (Note: a/b but b a.) The division algorithm finds q and r from a and b. Euclid s algorithm is a procedure for finding the greatest common divisor of two natural numbers a and b. It can be written as a function gcd(a,b), defined recursively as follows: { a if b = 0, gcd(a,b) = gcd(b,a mod b) if b 0. The greatest common divisor d = gcd(a, b) is characterised by the following properties: d a and d b; if e is a natural number satisfying e a and e b, then e d. Example Find gcd(225,157). Here is the calculation: 225 = =
9 1.3. PRIMES AND FACTORISATION 3 So gcd(225, 157) = = = = The Euclidean algorithm also finds integers u and v such that gcd(a,b) = ua + vb. In the above example, we can work back up the chain: 1 = So we have u = 30, v = 43. = 21 ( ) 4 = = ( ) = = ( ) 30 = Remark Sometimes we will modify the division algorithm as follows. Given integers a and b, with b > 0, there exist unique integers q and r such that a = bq+r and b/2 < r b/2. In other words, we allow the remainder to be negative, but choose it to have modulus at most b/2. For example, the classical division algorithm gives 12 = , but the modified version gives 12 = Primes and factorisation A natural number p is said to be prime if p > 1 and, whenever p = ab holds for some natural numbers a and b, we have either a = p, b = 1, or a = 1, b = p. In other words, p is prime if its only factors in the natural numbers are itself and 1, and these factors are different. The fact that 1 is not counted as being prime is a convention, but is needed in order for unique factorisation to hold. (If we allowed 1 to be prime, then 6 = 2 3 = = = would have infinitely many prime factorisations! Lemma 1.1 Let p be prime. If p ab, then p a or p b. Proof Suppose that p does not divide a. Since the only divisors of p are 1 and p, and p doesn t divide a, we must have gcd(a, p) = 1, so there exist integers u and v with ua + vp = 1. Now b = uab + vpb; and p divides uab (since it divides ab by assumption) and p divides vpb; so p divides their sum, which is b.
10 4 CHAPTER 1. OVERVIEW AND REVISION Theorem 1.2 Any natural number greater than 1 can be written as a product of prime numbers, and this expression is unique apart from re-ordering the factors. Proof We show the existence of a factorisation into primes by induction. Given a natural number n, if n is prime, then it is the product of just one prime. (This starts the induction at n = 2, and is also part of the inductive step.) Otherwise, n has a factorisation n = ab with a,b < n. By the induction hypothesis (since both a and b are greater than 1 but smaller than n), they have factorisations into primes; putting them together we have a factorisation of n. For the uniqueness, we use the lemma. Suppose that n = p 1 p 2 p r = q 1 q 2 q s, where p 1, p 2,..., p r,q 1,q 2,...,q s are primes. Clearly p 1 divides q 1 q 2 q s ; by the lemma, either p 1 divides q 1 or p 1 divides q 2 q s. Continuing, we find that p 1 divides one of the primes q 1,...,q s. By re-ordering them if necessary, we can assume that p 1 divides q 1, whence p 1 = q 1 since q 1 is prime. Now we can cancel off the first factor from both sides and continue the process, until we have shown that the two factorisations are the same. 1.4 Congruences and modular arithmetic Let n be a natural number. We say that two integers a and b are congruent modulo n if n divides a b. We write this as a b mod n. Note that this is a slightly different use of the word mod from the one we used earlier to denote the remainder. But it is closely connected; two numbers are congruent modulo n if and only if they leave the same remainder when they are divided by n. Congruence modulo n is an equivalence relation; the equivalence classes are called congruence classes modulo n. There are exactly n congruence classes, corresponding to the n possible remainders (0, 1,..., n 1) we could obtain when we divide a number by n. We denote by [a] n the congruence class modulo n containing a, and by Z n the set of congruence classes modulo n. The set Z n is a ring, in fact a commutative ring with identity; this means that congruence classes can be added or multiplied, by the rules [a] n + [b] n = [a + b] n, [a] n [b] n = [ab] n,
11 1.4. CONGRUENCES AND MODULAR ARITHMETIC 5 and the usual laws (commutative, associative, distributive, identity, and additive inverse laws hold. See the Introduction to Algebra lecture notes if you need a reminder about this. Here are the addition and multiplication tables of Z 4. I have written the entries in the tables as a rather than [a] 4 to save clutter Proposition 1.3 If p is prime, then Z p is a field; that is, all non-zero elements have multiplicative inverses. Proof Suppose that [a] p is a non-zero element of Z p. This means [a] p [0] p, so p does not divide a. Since p is prime, gcd(a, p) = 1. By Euclid s algorithm, there are integers u and v satisfying ua + vp = 1. This means that ua 1 mod p, so that [u] p [a] p = [1] p. So [u] p is the inverse of [a] p. For example, take p = 157. What is the inverse of [225] 157? Our earlier calculation showed that = 1, so that the required inverse is [ 30] 157 = [127] 157. As a consequence we prove Fermat s Little Theorem: Theorem 1.4 Let p be a prime number. Then n p n mod p for any natural number n. Proof If n 0 mod p, then the conclusion is certainly true; so suppose not. Then [n] p is an element of the multiplicative group of non-zero elements of Z p. By Lagrange s Theorem (see the Introduction to Algebra notes), the order of this element divides the order of the group, which is p 1. So ([n] p ) p 1 = [1] p, or in other words, n p 1 1 mod p. Multiplying both sides by n gives the result. Exercise Prove Fermat s Little Theorem by induction on n. (Hint: Use the Binomial ( ) Theorem and the fact (which you should prove) that the binomial coefficients p are divisible by p for 1 k p 1. k Fermat s Little Theorem shows that it is possible to show that a number n is composite without finding any factors of n. If we calculate a n mod n and the answer comes out to be different from a, then we know that n is composite.
12 6 CHAPTER 1. OVERVIEW AND REVISION Example mod 2047, so 2047 is not prime. The computation is not as bad as it might appear. Since 2048 = 2 11, we can work out mod 2047 by successive squaring. All congruences mod 2047; so no number occurring in the calculation is larger than , and the whole thing can easily be done on a calculator. 3 1 = = = 9 2 = = 81 2 = = So mod 2047, on dividing by 3 (equivalently, multiplying by the inverse of 3 in Z 2047, which is 1365). Note that the successive squaring method avoids having to compute very large numbers. We can evaluate by just eleven squaring operations of numbers smaller than 2047 together with taking the remainder mod Unfortunately, it doesn t always work. If we had used 2 rather than 3, we would have found that mod The converse of Fermat s Little Theorem is false! (See the exercise on Carmichael numbers below.) 1.5 The Chinese Remainder Theorem The Chinese Remainder Theorem is about solving simultaneous congruences to different moduli. Theorem 1.5 Let a and b be natural numbers with gcd(a,b) = 1, and let c and d be arbitrary integers. Then there is a solution to the simultaneous congruences x c mod a, x d mod b.
13 1.6. AND FINALLY... 7 Moreover, the solution is unique modulo ab; that is, if x 1 and x 2 are two solutions, then x 1 x 2 mod ab. Proof Since gcd(a,b) = 1, there are integers u and v with ua + bv = 1. Now let x = dau + cbv. We have bv 1 mod a, and au 1 mod b. So x cbv c mod a, and x dau d mod b, as required. If x 1 and x 2 are two solutions, then x 1 c x 2 mod a and x 1 d x 2 mod b. So both a and b divide x 1 x 2. Since a and b are coprime, ab divides x 1 x 2, so that x 1 x 2 mod ab as required. This can be extended to an arbitrary number of congruences to pairwise coprime moduli. Example Find all numbers congruent to 2 mod 3, 1 mod 4 and 3 mod 5. The theorem shows that there is a unique solution mod 60, which can be found by trial and error, or systematically as in the proof, which we do here. Since = 1, the number = 5 satisfies the first two congruences. Now we look for a number congruent to 5 mod 12 and 3 mod 5. We have = 1, so the solution is = 53. So the general solution is the congruence class [53] 60 (all numbers congruent to 53 mod 60). 1.6 And finally... Remember Euclid s famous proof of the existence of infinitely many primes, which you will find in the Introduction to Algebra notes. It is possible to adapt Euclid s method for other purposes. Here is an example. Note that, apart from 2, all primes are odd, and so are of one or other of the forms 4k + 1 and 4k + 3 for some natural number k. Theorem 1.6 There are infinitely many primes of the form 4k+3 for natural numbers k. Proof Suppose that there are only finitely many such primes, say p 1,..., p r. Consider the number n = 4p 1 p r 1. Clearly n is of the form 4k + 3, and so it must be divisible by some prime of this form. (A number with a factor 2 is even, while a product of factors of the form 4k + 1 is itself of this form, since
14 8 CHAPTER 1. OVERVIEW AND REVISION (4k + 1)(4l + 1) = 4(4kl + k + l) + 1.) So one of the primes p 1,..., p r must be a factor of n, since these are all primes congruent to 3 mod 4. But by assumption, n 1 mod p for p = p 1,..., p r, so none of p 1,..., p r can divide n. So we have a contradiction to our assumption, and there must be infinitely many primes of this form. It is also true that there are infinitely many primes of the form 4k + 1 (and indeed, roughly equal numbers of the two forms below any given bound), but these things are more difficult to prove. Exercises 1.1 Prove that a number with a periodic decimal expansion a 1...a k.a k+1...a k+l a k+l+1...a k+l+m is rational. (This notation means that the digits from a k+l+1 to a k+l+m repeat infinitely; for example, = ) 1.2 Find gcd(245,43) and express it in the form 245u + 43v. 1.3 Find all integer solutions of the congruence x 2 2 mod Let Z n denote the ring of integers modulo n. How many solutions does the equation x 2 = 1 have (a) in Z 8, (b) in Z 9, (c) in Z 11? 1.5 List the prime numbers less than 100. Which of them can be written in the form x 2 + y 2 for integers x and y? 1.6 A natural number q is said to be a Carmichael number if q is not prime but satisfies the conclusion of Fermat s Little Theorem, that is, n q n mod q for all integers n. (a) Let p be a prime number, and suppose that p 1 divides q 1. Show that n q n mod p. (b) Hence show that, if q is a product of distinct primes, and every prime p which divides q has the property that also p 1 divides q 1, then q is a Carmichael number.
15 1.6. AND FINALLY... 9 (c) Hence show that 561 is a Carmichael number. 1.7 There is an unknown number of objects. When counted in threes, the remainder is 2; when counted in fives, the remainder is 3; when counted in sevens, the remainder is 2. How many objects are there? This problem is taken from the fourth-century Chinese text Sun Zi suanjing (Master Sun s Arithmetic Manual). He gives the following formula for its solution: Not in every third person is there one aged three score and ten, On five plum trees only twenty-one boughs remain, The seven learned men meet every fifteen days, We get our answer by subtracting one hundred and five over and over again. Can you explain this? Note: Sun Zi (formerly transliterated as Sun Tsu), is not to be confused with the military strategist of the same name. 1.8 (a) Let p be a prime number, not 2 or 5. Show that there exists a positive integer k such that 10 k 1 mod p. (b) Let k p be the smallest positive integer k with this property. Show that k p divides p 1. (c) Show that the digits in the infinite decimal expression for 1/p are periodic with period k p. (d) Find a prime p for which k p = p What goes wrong with the argument in the last section if you try to prove that there are infinitely many primes congruent to 1 mod 4?
17 Chapter 2 Algebraic numbers An algebraic number is one which satisfies a polynomial with integer coefficients. From Pythagoras to the present day, a lot of number theory hae been concerned with these numbers, and in particular in trying to decide whether particular numbers of interest to mathematics are algebraic or not. 2.1 Algebraic numbers and algebraic integers Pythagoras and his school discovered that the square root of 2 is not a rational number. However, it is an easy number to describe geometrically: it is the ratio of the diagonal of a square to its side. The number π has a more complicated geometric description: it is the ratio of the circumference of a circle to its diameter, but there is no simple method to construct a straight line which is equal to the circumference of a given circle. (We know now, for example, that such a line cannot be constructed with the traditional geometric instruments of ruler and compass.) We make a distinction between algebraic numbers (which are roots of polynomials with integer coefficients) and transcendental numbers (which are not): Definition Let u be a complex number. We say that u is an algebraic number if there is a non-zero polynomial f with integer coefficients such that f (u) = 0; and u is a transcendental number otherwise. Moreover, u is an algebraic integer if it is the root of a non-zero monic polynomial (one with leading coefficient 1) over the integers. Note that, if we have any non-zero polynomial over the integers, we can divide by the leading coefficient to get a monic polynomial over the rationals. Conversely, given a monic polynomial over the rationals, we can multiply by the least common multiple of the denominators of the coefficients to obtain a non-zero 11
18 12 CHAPTER 2. ALGEBRAIC NUMBERS polynomial over the integers. So an equivalent definition is: u is an algebraic number if there is a non-zero monic polynomial f with rational coefficients such that f (u) = 0. For example, u = 2 is an algebraic integer since it satisfies the polynomial u 2 2 = 0. The complex number i is an algebraic integer; so is the golden ratio φ = (1+ 5)/2 (it satisfies φ 2 φ 1 = 0). It is known that π and e are transcendental numbers, but we will not give the proof here. (You can find these proofs in Ian Stewart s book Galois Theory. Any integer is an algebraic integer; the integer n satisfies the polynomial x n = 0. Similarly, any rational number is an algebraic number. In the other direction, we have: Proposition 2.1 A rational number is an algebraic integer if and only if it is an integer. For this reason, we sometimes call the ordinary integers rational integers. Proof We have seen that integers are algebraic integers; we have to prove that a rational number which is an algebraic integer is an integer. Let q = a/b be a rational number in its lowest terms (so that gcd(a,b) = 1). Suppose that q satisfies a monic polynomial with integer coefficients, say f (q) = q n + c n 1 q n c 1 q + c 0 = 0. Putting q = a/b, and multiplying this equation by b n, we obtain a n + c n 1 a n 1 b + + c 1 ab n 1 + c 0 b n = 0. Now every term in this equation except the first is divisible by b, so b divides a n. Applying Lemma 1.1 repeatedly, we find that b a. But also b b, so b gcd(a,b) = 1. This means that b = 1, so that q is an integer. There is a result, which I will not prove, which makes things like this much easier. This is known as Gauss s Lemma. It can be stated in many different ways. (You might like to look at p.258 in my book Introduction to Algebra.) But the following will do for our purposes. We define the minimal polynomial of an algebraic number α to be the monic polynomial with rational coefficients of smallest possible degree satisfied by α. (Any algebraic number satisfies a monic polynomial with rational coefficients, and we can certainly choose one of smallest degree. Why is it unique? Suppose that f 1 (x) and f 2 (x) were two different polynomials of the same (smallest) degree satisfied by α, and let g(x) = f 1 (x) f 2 (x). Then f 1 (α) = f 2 (α) = 0, so g(α) = 0, but g has smaller degree than f 1 and f 2.)
19 2.2. QUADRATIC IRRATIONALS 13 Theorem 2.2 The algebraic number α is an algebraic integer if and only if its minimal polynomial has integer coefficients. Now let q be a rational number. It satisfies the polynomial x q = 0, and clearly this is monic and has smallest possible degree, so it is the minimal polynomial of q. So q is an algebraic integer if and only if the coefficients 1 and q of this polynomial are both integers, i.e. if and only if q is an integer. One of the most important properties of algebraic numbers is the following: Theorem 2.3 (a) Let a and b be algebraic integers. Then a b and ab are algebraic integers. (b) Let a and b be algebraic numbers. Then a b, ab, and (if a 0) 1/a are algebraic numbers. I do not expect you to memorise the proof of this theorem. But it uses ideas from linear algebra, and may be useful revision of linear algebra for you. I have given the proof in an appendix to this section. The theorem can be expressed in the language of algebraic structures as follows: Corollary 2.4 (a) The algebraic numbers form a field. (b) The algebraic integers form a commutative ring with identity. Proof The Theorem above says that they satisfy the conditions of the subfield and subring tests as subsets of the complex numbers. 2.2 Quadratic irrationals In this course we will be particularly interested in numbers of the form a + b d, where a and b are rational numbers and d is a squarefree integer not equal to 1. (An integer d is squarefree if n 2 d implies n = 1 for positive integer n. Clearly, if d were not squarefree, we could write d = cn 2, and then a + b d = a + bn c.) A number of this form is called a quadratic irrational. There will be much more about quadratic irrationals later in the notes! The number u = a+b d is an algebraic number, since it satisfies the quadratic equation u 2 2au+(a 2 db 2 ) = 0. (This is a quadratic with rational coefficients; we obtain one with integer coefficients by multiplying up by the denominators of the coefficients.) In fact, the polynomial f (x) = x 2 2ax + (a 2 db 2 )
20 14 CHAPTER 2. ALGEBRAIC NUMBERS is the minimal polynomial of u. For it is a monic rational polynomial satisfied by u, and has degree 2; a polynomial of smaller degree would have to have degree 1, and have the form x q, but if u satisfied this polynomial, then u = q would be a rational number. Using Gauss s Lemma, we can now decide when a quadratic irrational is an algebraic integer. Proposition 2.5 Let a, b be rational numbers and d a squarefree integer. Then a + b d is an algebraic integer if and only if either (a) a,b are integers; or (b) d 1 mod 4 and a 1 2 and b 1 2 are integers. Example (1 + 5)/2 (the golden ratio) and ( 1 + 3)/2 (a complex cube root of unity) are algebraic integers but (1 + 3)/2 is not. Proof By Gauss s Lemma, we just have to show that the monic quadratic equation satisfied by u = a + b d has integer coefficients precisely in the cases given. The quadratic is x 2 2ax + (a 2 db 2 ). So the question is, when is it true that 2a and a 2 db 2 are integers? If 2a is even then a is an integer; if 2a is odd then a 2 1 is an integer. Suppose that a is an integer. Then db 2 is an integer; since d is squarefree, this implies that b is also an integer, since if b = m/n with gcd(m,n) = 1 then necessarily n 2 d. Suppose that a = k + 1 2, with k Z. Then a2 = k 2 + k + 1 4, so db2 1 4 is an integer. This means that b = l for l Z (so that db2 has denominator 4). Then db = (l2 + l)d + (d 1)/4, so we must have d 1 mod Appendix: Sums, products and quotients In this section we prove Theorem 2.3. If a and b satisfy monic polynomials over the integers or rationals, we have to show that their difference and product do also. The direct approach is quite difficult; to convince yourself of this, try writing down a monic polynomial over the integers which has as a root. So we need a different strategy. First, we give an equivalent characterisation of algebraic numbers and algebraic integers, using the concept of eigenvalues from linear algebra. Proposition 2.6 Let u be a complex number. Then
21 2.3. APPENDIX: SUMS, PRODUCTS AND QUOTIENTS 15 (a) u is an algebraic number if and only if it is an eigenvalue of a matrix over Q; (b) u is an algebraic integer if and only if it is an eigenvalue of a matrix over Z. Proof The eigenvalues of a matrix A are the roots of the characteristic polynomial of A, the monic polynomial det(xi A). If A is a rational (resp. integer) matrix, this polynomial has rational (resp. integer) coefficients. Conversely, given any monic polynomial f (x) of degree n, there is a matrix C( f ), called the companion matrix of f, whose characteristic polynomial is f (and hence whose eigenvalues are the roots of f ). It is the n n matrix which has entries 1 immediately above the diagonal in the first n 1 rows, 0 in all other positions in these rowss, and the coefficients of f (other than the coefficient of x n ) in reverse order with the signs changed in the nth row. So C( f ) is a matrix of integers (resp. rational numbers) if and only if the coefficients of f are integers (resp. rational numbers). Now we define the Kronecker product of matrices. Let A and B be two matrices, possibly of different sizes. Suppose that A is n m, with (i, j) entry a i j. Then the matrix A B is defined to be the matrix in block form with m n blocks, the (i, j) block being a i j B. For example, if then A = ( ) 1 2, B = 3 4 ( ) A B = The following result is what is needed. The proof is just a rather boring linear algebra argument involving playing with subscripts. Proposition 2.7 Let A, B,C, D be matrices of the appropriate sizes so that AC and BD are defined. Then (A B)(C D) is defined, and (A B)(C D) = AC BD. In particular, if A and B are matrices and v,w are column vectors such that Av and Bw are defined, then (A B)(v w) = Av Bw.
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text:
As a result of customer acquisition since its launch in October 2007, Lebara Mobile has virtually doubled its revenue during the last year.The UK business specialises in providing cheap, overseas calls for more than one million customers and the company overall has a total customer base of three million customers across the Nordic countries, Spain and the UK. It also recently ventured into Australia.Yoganathan Ratheesan, the Chief Executive Officer and a founder said, "the UK business benefited from continued interest in Lebara's proposition from the millions of UK residents with family in a vast range of different countries".The company is looking to target areas with a high ethnic population in order to increase growth even further. Cities such as Birmingham, Manchester, Newcastle, Leeds and Glasgow will be targeted first.Offering free Lebara-to-Lebara calls and texts is another part of the strategy to grow its business although more approaches such as postpay or subsidised handset bundles have been decided against at this point in time.Ratheesan said: 'Growth is pretty strong in the current format and we want to support that growth for the coming year.'
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