APPENDICES APPENDIX A Some Elementary Number Theory We consider the set of all integers, both non-negative and negative. Lemma A.l. Any subset, S, of the integers containing at least one non-zero element and closed under addition and subtraction contains a least positive element and consists of all multiples of this integer. PROOF. Let a E S, a =I= 0. Then S contains the difference a- a= 0, and also 0 - a = -a. Consequently there is at least one positive element, Ia I, in S, and hence there is a smallest positive element, b, inS. Now, S must contain all integral multiples of b, for if it contains nb, n = 1, 2, ... etc., then it must contain (n + 1 )b = nb + b, and we know it contains b. Moreover, (- n )b = 0 - (nb) is the difference of two elements in S, for n = 1, 2, ... and so S contains all negative multiples of b also. We now show that Scan contain nothing but integral multiples of b. For if c is any element of S, there exist integers q and r such that c = bq + r, 0 ~ r < b (qb is the multiple of b closest to c from below; we say c =r(mod b)). Thus r = c- bq must also be in S, since it is a difference of numbers in S. Since r E S, r ::2: 0, r < b, and b is the least positive integer in S, it follows r = 0, so that c = qb. 0 Definition A.l. Every positive integer which divides all the integers a 1, a 2 , ... , ak is said to be a common divisor of them. The largest of these common divisors, is said to be the greatest common divisor (g.c.d.). This number is a well defined positive number if not all a 1, a 2 , ... , ak are zero, which we assume henceforth. Lemma A.2. The greatest common divisor of a 1 , a2 , ••. , ak, say d, can be expressed as a "linear combination", with integral coefficients, of at. a 2 , ••• , 247 248 Appendices ak, i.e. k d = L b;ai> b; integers. i= 1 PROOF. Consider the setS of all numbers of the form L~= 1 b;a;. For any two such k k k I W>ai ± I W>ai = I (W> ± b! 2>)a;, i=1 i=1 i=1 and hence the setS of such numbers is closed under addition and subtrac­ tion, hence by Lemma A.l consists of all multiples of some minimum posi­ tive number k v = Lb;a;. i=l Thus d, the greatest common divisor of a~> ... , at> must divide v, so that 0 <d.::; v. Now each a; is itself a member of S (choose the bs so that b; = 1 and all other bs zero), so that each a; is a multiple of v, by Lemma A.1. Thus a contradiction arises unless d = v, since d is supposed to be the greatest common divisor of the as. D Definition A.2. Let a;, i = 1, 2, ... be an infinite set of positive integers. If dk is the greatest common divisor of a~> •.. , ak, then the greatest common divisor of a;, i = 1, 2, ... , is defined by d =lim dk. k-+oo The limit d ~ 1 clearly exists (since the sequence {dk} is non-increasing). Moreover d is an integer, and must be attained after a finite number of k, since all the dk s are integers. Lemma A.3. An infinite set ofpositive integers, V = {a;, i ~ 1}, which is closed under addition (i.e. if two numbers are in the set, so is their sum), contains all but a finite number of positive multiples of its greatest common divisor. PRooF. We can first divide all elements in the infinite set V by the greatest common divisor d, and thus reduce the problem to the case d = 1, which we consider henceforth. By the fact that d = 1 must be the greatest common divisor of some finite subset, a1, a2 , ••. , ak, of V, it follows from Lemma A.2 that there is a linear combination of these as such that k La;n; = 1, i=1 A. Some Elementary Number Theory 249 where the n;s are integers. Let us rewrite this as m-n=1 (A.l) where m is the sum of the positive terms, and - n the sum of the negative terms. Clearly both n and mare positive integer linear combinations of the a;s and so belong to the set V. Now, let q be any integer satisfying q :::>: n x (n - 1); and write q =an+ b, Osb<n where a is a positive integer, a:::>: (n- 1). Then using (A.1) q =an+ b(m- n) q = (a- b )n + bm so that q is also in the set V. Hence all sufficiently large integers are in the set V, as required. 0 Corollary. If a 1, a2 , •.• , ak are positive integers with g.c.d. unity, then any sufficiently large positive integer q may be expressed as k q = L:a;p; i= I where the P; are non-negative integers. We conclude that this subsection by an application which is of relevance in Chapter 6, and whose proof involves, in a sense, a tightening of the kind of argument given in Lemma A.3. Lemma A.4. 1 Let u; (i = 0, 1, 2, ... ) be non-negative numbers such that ,for all i,j :::>: 0 Suppose the set V of those integers i :::>: 1 for which u; > 0 is non-empty and has g.c.d., say d, which satisfies d = 1. Then u = lim u~fn exists and satisfies 0 < u s oo; further, for all i :::>: 0, u; s ui. PROOF. The set V is closed under addition, in virtue of u; + i :::>: u; ui, and since d = 1, by Lemma A.3, V contains all sufficiently large integers. Hence for any r E V there exists an s E V, s > r, such that the g.c.d. ofr and sis unity. Thus by Lemma A.2 we have 1 Due to Kingman (1963). This is an analogue of the usual theorems on supermultiplicative or subadditive functions (e.g. Hille & Phillips 1957, Theorem 7.6.1; Khintchine, 1969, ~7). 250 Appendices for some integers b1o b2 , Not both of b1 and b2 can be strictly positive, since r, s ~ 1. Assume for the moment b 1 =a>0, -b 2 =b~O. Let n by any integers such that n ~ rs, and let k = k(n) be the smallest positive integer such that bjr::; k/n::; ajs; such an integer certainly exists since n(ajs- bjr) = n(ar- bs)jrs = njrs ~ 1. Then n = Ar + Bs where A= na- k(n)s, B = k(n)r- nb are non-negative integers. Now, from the assumption of the Lemma, Un = UAr+Bs ~ UArUBs ~ .. ' ~ U~U:, (n ~ rs). Thus u~fn ~ u~-(ks/n)u~kr/n)-b. Letting n-+ oo, and noting k(n)/n-+ b/r, we see that n-+oo This holds for any r E V and so for any sufficiently large integer r. Hence lim inf u~ln ~ lim sup u!'' n-+oo r-+oo which shows u = lim u~fn n-+ oo exists. Further, again from the inequality ui+i ~ uiui, uki ~ u~, i, k ~ 0 (ulfki)i U· so that kl >- l and letting k -+ oo, and since ui > 0 for some i ~ 1 the proof is complete, apart from the case i = 0, which asserts 1 ~ u0 • The truth of this follows trivially by putting i = j = 0 in the equality, to obtain u0 ~ u5 . We need now to return to the possibility that b 1 ::; 0, b2 > 0. If we write b2 =a, -b1 = b, so that 1 =as- br, then the roles of sand r need to be changed in the subsequent argument and, apart from this k = k(n) needs to be chosen as the largest integer such that bjs ::; k/n ::; ajr. A. Some Elementary Number Theory 251 This leads eventually once more to and the rest is as before. 0 Corollary. n APPENDIX B Some General Matrix Lemmas Lemma B.l. If A is a finite n x n matrix with real or complex elements such that A k--+ 0 elementwise as k --+ oo, then (I - At 1 exists and 00 (I-At 1 = IAk k=O convergence being elementwise. (A 0 = I by definition.) PROOF. First note that (I- A)(I +A+ ... + Ak- 1 ) =I- Ak. Now, fork sufficiently large, Ak is uniformly close to the zero matrix, and so I - Ak is to I, and is therefore non-singular. (More specifically, by the con­ tinuity of the eigenvalues of a matrix if its elements are perturbed, the eigen­ values of I - Ak must be close to those of I for large k, the latter being alll; hence I- Ak has no zero eigenvalues, and is therefore non-singular.) Taking determinants det (I - A) det (I + A + .. · + A k- 1) = det (I - A k) =f= 0, therefore det (I- A)# 0. Therefore (I- At 1 exists, and I+ A+ ... + Ak-t =(I- At 1(I- Ak). Letting k --+ oo completes the proof of the assertion. 0 252 B. Some General Matrix Lemmas 253 Corollary. Given an n x n matrix A, for all complex z sufficiently close to 0, (I - zA t 1 exists, and 00 (I- zAt1 = L:ZkAk k=O in the sense of elementwise convergence. PROOF. Define (j by (j = max I aii 1- i, j Then putting Ak = {a!J>}, it follows from matrix multiplication that Ia!J> I s n b 2, and, in general Ia!~> I s nk- 1 b\ all i, j = 1, 2, ..