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/Vg/J /Vo.a/2./ ALGEBRAIC NUMBERS AND TOPOLOGICALLY EQUIVALENT MEASURES DISSERTATION Presented to the Graduate Council of the North Texas State University in Partial Fulfillment of the Requirements For the Degree of DOCTOR OF PHILOSOPHY By Kuoduo Huang, B.S., M.S, Denton, Texas December, 1983 y Huang, Kuoduo J., Algebraic Numbers and Topologically Equivalent Measures in the Cantor Set. Doctor of Philosophy (Mathematics), December, 1983, 66 pp., bibliography, 23 titles. A set-theoretical point of view to study algebraic numbers has been introduced. We extend a result of Navarro-Bermudez concerning shift invariant measures in the Cantor space which are topologically equivalent to shift invariant measures which correspond to some algebraic integers. It is known that any transcendental numbers and rational numbers in the unit interval are not binomial. We proved that there are algebraic numbers of degree greater than two so that they are binomial numbers. Algebraic inte- gers of degree 2 are proved not to be binomial numbers. A few compositive relations having to do with algebraic numbers on the unit interval have been studied; for instance, rationally related, integrally related, binomially related, -related relations. A formula between binomial numbers and binomial coefficients has been stated. A generalized algebraic equation related to topologically equivalent measures has also been stated. PREFACE This dissertation consists of three chapters. chapter has its own introduction. Each The topologically equi- valent measures in the Cantor set have been studied in Chapter I. The binomial numbers and Pisot numbers have been discussed in Chapter II. In Chapter III, a set-theo- retic approach to algebraic numbers has been developed. I hope that through this point of view, many nice theorems can be discovered in the future. At least the following theorems in this article are original. rems: Some of them are generalizations of known theo- Theorem 1.2, Theorem 1.7, Theorem 1.11, Theorem 2.2, Theorem 2.3, Theorem 2.4, Theorem 3.1 and its corollary, Theorem 3.3, Theorem 3.5, Theorem 3.6, Theorem 3.7, Theorem 3.8, and a relation between binomial numbers and binomial coefficients is stated. I wish to thank Dr. F.J. Navarro-Bermudez and Dr. Helge Tverberg for generously allowing me to make use of some of their unpublished manuscripts. TABLE OF CONTENTS Page i PREFACE Chapter I. TOPOLOGICALLY EQUIVALENT MEASURES IN THE CANTOR SET §1. §2. §3. II. BINOMIAL NUMBERS IN THE UNIT INTERVAL §1. §2. §3. III. Introduction An Improvement of Navarro-Bermudez1s Theorem C-pairs Introduction Existence of Binomial Numbers Existence of Non-binomially Algebraic Numbers of Degree _> 2 ALGEBRAIC NUMBERS § 1. §2. §3. §4. §5. §6. §7. 23 41 Introduction Equivalence Relations on I Homogeneous Functions in Two Variables Rationally Related Relation on I Integrally Related Relation on I Bi-related Relation Binomial Numbers and Binomial Coefficients BIBLIOGRAPHY 65 11 CHAPTER I TOPOLOGICALLY EQUIVALENT MEASURES IN THE CANTOR SET § 1. Introduction Let X be a topological space. Two Borel measures y and v are said to be topologically equivalent whenever y = vh for some homeomorphism h of X onto itself. This notion sets up an equivalence relation which partitions the family of Borel measures into disjoint classes each one consisting of those measures which are mutually equivalent. By restricting attention, if necessary, to a suitably defined subfamily of measures, one can ask for the number of classes there. One can also try to uncover necessary conditions, as well as sufficient conditions, which intrinsically characterize measures which belong to the same class. Topologically equivalent measures in the n-dimensional unit cube, the space of irrational numbers in the unit interval, and the Hilbert cube have been studied, respectively, by Oxtoby and Ulam [1], Oxtoby [2], and by Oxtoby and Prasad [3]. In [4], F.J. Navarro-Bermridez studied the topologically equivalent measures in the Cantor space. his result. I will be able to extend Before getting into details, I prefer to define a few necessary notations and terms. (X,y) is a P-pair provided X is a space of the form oo n xn n=li x = with the product topology. Each factor X^ is finite or countably infinite and carries discrete topology and each X has at least two elements. ]i is a Borel measure in X. n A P-pair is of type C in case each factor X^ is finite (In [ 5 ], F.J. Navarro-Bermudez called aP-pair of type C a "C-pair"). Every P-pair of type C is homeomorphic to the Cantor space, which can be realized as X with X = {0,1} for n n=l,2,..., and its topology is compatible with the metric d such that for any two points x = ( x n ) a n d X' = ( x ^) is given by the formula 00 1 d(X/X ) = ^ -n (%) d (x , x'), n=l each d n is the metric on X n n n n which takes only the values 0 u or 1. A P-pair is of type F-K0 in case there are infinitely many X n having cardinality ^ , and infinitely many of X^ having finite elements. Denote [i1,i z,...,im] = {(x n)€X: x.=i.€X. j j j for i=l,2,...,m} to be a basic open set. The set of all basic open sets forms a base for the topology of X. These sets are both open and closed and will be referred to as the special closed-open sets of X or a basic clopen set. A P-pair is of type ( p 1 , p 2 , — , P n ) provided that the following three conditions hold: (i) p + p 2 + ... + p n = 1 and p l ,p 2 ,...,P n are non-negative real numbers. (ii) For each k, X is a set consisting of n elements, K. for instance, may be taken to be integers 0,1,2,...,n-l. 00 (iii) y (j) = p. for all n and j, and y = n = y n (j) 3 Pj f° r n an II n=l y with n ^ Similarly a P-pair is of type (p^,p2,...) provided the following conditions hold: oo (a) E p = 1, where p ,p ,... are non-negative real 1 2 n=l n numbers and infinitely many of them are positive. (b) For each k, X is a set consisting of K0 elements, iv for instance, may be taken to be integers 1,2,3,... . 00 (c) y n (j) = y n (j) = Pj f° r Pj f° r a-'-^ n anc a H n an< ^ 3r anc ^ U = n y^ with ^ j* Any two P-pairs of type (p^,p2,...) are topologically equivalent which is an immediate result from [2] by John C. Oxtoby. He proved the following theorem in [2]. Theorem. - A topological measure space (X,y) is homeo- morphic to (J,A) if and only if X is homeomorphic to J and y is an everywhere positive, non-atomic, normalized Borel measure in X. In particular, any such measure in J is topo- logically equivalent to X. Recall: (1) A topological measure space is a pair (X,y), where X is a topological space and y is a measure on the class of Borel subsets of X. (2) A measure y is everywhere positive if y(G) > 0 for every non-empty open set G, non-atomic if y({x}) = 0 for each x £ X , and normalized if y(X) = 1. (3) J denotes the set of irrational numbers in I = [0,1] and A denotes the restriction of Lebesgue measure m to the Borel subsets of J. So we are only interested in a P-pair which is not of type (plfp2,...)« Especially we are interested in a P-pair of type (p^/P2/.../Pn)• From now on we let (X,y) be a fixed P-pair of type (p1,P2,...,p ) so that for each positive integer m, X = {0,1,2,...,n-1}. m basic open sets. Every open set U of X is a union of If U is also closed, hence compact, then only finitely many basic open sets enter into this union. It can be easily seen that the closed-open set U can be expressed as the union of finitely many disjoint basic open sets each of which has the form {i }x{i }x...x{i }xx xx x...(i =0,1,2,...,n-l) 1 2 m m+1 m+2 k (1.1) for k=l,2,...,m. The set (1.1) will be denoted as [i,,i„,...,i ]. Two 1 z m of these sets [i,,...,i ] and [j,,j„,...,j ] are equal or 1 m i / m disjoint, depending on whether the n-tuples (i^,i , ...,i ) and (jifj^/---fj ) 1 2 m are identical or not. The sets (1.1) have been named differently by many other authors to be thin cylinder, intervals, special closed-open sets, basic open sets, etc. The truth of the following theorem is obvious. Theorem 1.1. - Let U be a closed-open set of the Cantor space X. There is a positive integer n such that U can be expressed as the disjoint union of finitely many special closed-open sets of length n. The family f of probability measures in X to which attention will be restricted consists of product measures oo y = II ym subject to the condition m=i 1 y (i) = y (i) for all m and i=0,1,2,...,n-l, m m+1 (1.2) the factors y are all normalized; that is, y (0)+...+y (n) = l . m m m For convenience, we denote y = y(p,,p 0 , 1 2 p.^ = V (i) f° r i=0,1, . . .,n-1. ,p ), where m Among the product measures, there are precisely those which are invariant under the Bernoulli shift transformation T of X defined by T(x,x,...,x 1 z ,...) — ( x , x , . . . ) m 1 5 for each point x = (x^,x^,x ,...) in X. X is invariant under T A Borel measure y on (or T is y-measure preserving) if y ( T _ 1 ( U ) ) = y(U) for every Borel set U of X. n For each n triple (p 1 ,...,p11 ) in n [0,1] with i=l p +...+P = 1 , there is a measure y(p_,...,p ). Since the i n I n correspondence ( P w . - w P ) -> y(p_,...,p ) is one-to-one, the i n I n family has continuum many measures. Each y(p , . . . , p ) is a 1 n normalized Borel measure which is everywhere positive and n non-atomic for (p ,...,p ) € n (0,1). n i=l If U = [i.,i0,...,i ], then 1 1 m y(px,p2,...,pn)(u) = y 1 (i T )y 2 (i 2 )...y m (i m ). It follows that m. m m y(P1,. . .,Pn) (U) = P x P 2 •••PI1n (1*3) where m. is the number of times of i,=j ( l £ k £ n ) . It is evident that the measure y(px,p2,...,pn) and y(p^(l) f . . . 'Pfi(n)) are always topologically equivalent, where 7T is any permutation of l,2,...,n. Because y(pj,...,p ) = oo y(p p . x)h, where h is the homeomorphism h = n h m=l m with h : X -* X given by h (i) = tt (i) i=l,2,... . m m m m Theorem 1.2. - If the measure y(p^,...,p ) and y(qj/.../q ) are topologically equivalent, then there are positive integers r and s, and integers . . . 3 Ji»'••']o b " . . . i l l , with 0 < a inJ . y • • • 9 in , x i . . 2 Ji >•••>] n and 0 < b < m , 5 • • • jm , i Ql '/ <i 1 Z t 2 3 1 l i < . . . < i £ n, l_<j < j K. x < k m, 1 , m k */ m . 5 x• • • > in # . . m, x x i i where k • • • • m. • • • • •in, • J l£i a J Z £ < . . . < j £ n , m. +...+m. = r 36 1X J. k and nu 1 +...+m_.Z = s, and l £ i , j £ n , such that (1.4) and (1.5) 3 J hold: . . q. = 1 I i1=l a r 1 p. X 1 + . . . Z a 2 lli1<i2<n m m i1' i2 m. p. X 1 m. p. 2 +... (1.4) x 1 2 m. -\ s • s a s s i* / —. m • jfli. * • • • 5 m. 1—<1,<i n<. . .<i- <n 1 2 k— 1x- 2 x0 kxn 1 m. i-i k I-. 1 m1»* p. = J J JlI T b q. Z 3 Jj-1 + 1 £ -J1 Proof: 3 b ^ J 2 V i<^<i2<n . + m. m. 3 Ji * J o ] i Jo b q. q. + . . . y ^ £-n . . (1.5) 32 m. Ji q m. . . .m. 3 J I £ m. - i n Jn J 1 ...q. +...+ b i, l. m . i >11 1, q....q m x n Let h be a homeomorphism of X onto itself such that y(p,, . . . ,p ) = y(q,,...,q 1 ii 1 n set % J l [j] where 1<. j<_n. )h, and let V be the clopen So we have y (P L , . . . ,p n ) (V) = p . On the other hand, h(V), being a closed-open set, is, according to Theorem 1.1, a disjoint union U of finitely many thin cylinders of length s. By (1.3) the y(q,,...,q )-measure of 1 II [ t / . - . / t ] is 1 s m. m. m. 31 J o ^k q. q. ...q. where 1 < i < j 2n < J J 1 2 k "" 1 ... < j, < n, k - {t , . . . ,t } = { j w . - . / j , } / the number of times of m. +...+m. 3 3 1 k If b k is a positive integer and m. 3 i repeating in {tj,...,t }, thus = s. is the number of clopen sets in U in « ) m , $ • • • ) m , 3 3 3 1 2 k m. m. J1 Jv whose y(q,/...,q )-measure is q. ...q (hence 1 ] j i k 3 jljjoj'-'jji cI z k 0< b 1 < •/ , _ ,), then y(q ,...,q )(h(V)) m. • • • m. — m. I . . • m. ! 1 n J 3 3 3 1 k 1 k is equal to m n j j l r E b q + ... + 3 ji =1 1 Z j J,-•-j h 1 U v k b 1 q. . . .q. J <m m j •""j, J 1 k k j 1>2 nm m + ... + b q. ...q n . m, , . . . ,m 1 n 1 n Formula (1.5) follows by substituting into the equation y(P1,...,Pn) = y(qx,...,qn)(h(V)), the values obtained for y(Pj,...,Pn)(V) and y(qx,...,q )(h(V)). Formula (1.4) can be proven similarly by considering 1 h instead of h. Q.E.D. This elementary theorem will be a very useful tool in proving other interesting theorems. The following corollary will be obtained immediately from Theorem 1.2. Corollary 1.3. - If y(pj,...,p ) is topologically equivalent to y(q^,...,q ), then the following statements hold: (i) If p ,...,p i n 9 # *3^ ' ^2 ' * * * (ii) q r are all rationals, then so are If p ^ . . . ^ are all algebraic numbers, so are If p^,...,p are all algebraic integers, so are . ..,qn. (iii) ^3 j f • • • f • The following corollary 1.4 is a special case of Theorem 1.2, but plays an important role in this paper. It was first proved by F.J. Navarro-Bermrtdez in [4]. Corollary 1.4. - Given r,s€ [0,1]. If y(r,l-r), y(s,l-s) are topologically equivalent, then there are positive integers m,n and integers a Q/ a 1 ,... f a n ; b ^ b ^ . . . , ^ with 0 £ a < (J) and 0 < b. < (?) such that ~ J - J s = a r m +a rm_1(l-r)+...+a r(l-r) m_1 +a (l-r)m u m-1 m r = b (1.6) 0sn+bisn_1(1-s)+*••+bn-lS(1~S)n~1+bn(1~S)n * (1 *7) Let C(p1/...,p ) be the class of measures in f which are topologically equivalent to y(p,,...,p ). I We have mentioned n previously that C(pj,...,p ) always contains the measures is any permutation on {l,2,...,n}. Corollary 1.5. - Each class C(p ,...,p ) contains at l n most countably many measures of the family f and the cardinality °f classes C(pj,...,p ) is of power the continuum. Proof: If y(q , — , q ) belongs to the class C(p., ,p ), i II i n then the numbers q ^ . . . ^ must satisfy formula (1.4) for some choice of integers r, • • • i ii > • • • »in a with 1 < i < n m. 5 • • • *ni. — — ^i i ' r! < ^m. ! . . .m. !' 1 - """1 < i 2 < "*"£ — n ' X X x 1 £ 1 £ i i i m. +...+m = r. For each fixed r , i a s O < a ^ ^ m 1 £ ,1 £ 1i i , r!i , th »• •i • j-'-n by var in m I...m !' Y g e integers a * , the n ,•..,n. X 1 * 1 right hand side of (1*4) generates only finitely many differ— ent numbers . It follows that the number of measures - a m. — y (c[•,/•••/q ) is at most countable, j11) in the class C(p i,...,p n 10 Since there are continuum many measures in f and each class C(pj,...,p ) has only countably many measures, the cardinality of classes C(p^,...,p ) is of continuum. Q.E.D. § 2. An Improvement of Navarro-Bermudez 1 s Theorem In [4], F.J. Navarro-Bermudez has proved the following interesting theorem. Theorem 1.6. - If y = y(r,l-r) is topologically equiv- alent to y g = y(s,l-s), and if s is rational or transcendental in the unit interval, then either r = s or r = l - s . In other words, there are only two (except s = h ) elements in the equivalent class C(s,l-s). In this section, we are going to show the same result if s is a real algebraic integer of degree 2 in the unit interval (There are infinitely many algebraic integers of degree 2 in the unit interval). Actually we prove slightly more in Theorem 1.7. Theorem 1.7. - If r is an algebraic integer of degree n ^ 2 , and y is topologically equivalent to y , and s = p+qr S XT for some rational numbers p,q, then s = r or s = 1-r. We denote irr(0) to be the irreducible polynomial of an algebraic number 0 in Z[x] with relatively prime integer coefficients and positive leading coefficient. In this chapter, r and s are always in the unit interval. 11 Theorem 1.8. - If y is topologically equivalent to y TC S and s is a real algebraic number of degree n, then there exists a unique representation of r in terms of s. r = £ n-1 + £ 0 s n - 2 + ...+£, s + n-2 1 0 for some rationals Furthermore, if s is an algebraic integer, then , are integers. U 1 n-1 Let irr(s) = x 4 - a ^ x + . . . + a 1 x + a Q . Upon using n 1 n s = -a n_^ s -...-a^s-a and (1.6) of Corollary 1.4, it follows that there is a representation for (2.1) . Since l,s,...,sn 1 form a basis of Q(s) over a, so the represen- tation of (2.1) is unique. Theorem 1.9. - The sum and product of two algebraic integers are algebraic integers. Also, the set of all algebraic numbers forms a field. Refer to [ 8 ]. Theorem 1.10. - If r is an algebraic number of degree n and y is topologically equivalent to y , then s is an algeS "JC braic number of degree n. Furthermore, in addition, if r is an algebraic integer, then s is an algebraic integer. Proof: degree of r. We are going to prove this by recursion on the If n = l , then this theorem holds according to Theorem 1.6 and Theorem 1.9. Assume this theorem is true for all n_<k and r is an algebraic number of degree k + 1 , and y is topologically equivalent to y . According to s x* (2.1) 12 •n Theorem 1.8, s = ^0''—'^n-1' 1 + ... + £^r + £Q for some rationals and this representation is unique. It follows that s€Q(r), where Q(r) is the minimal field extension over Q so that r€Q(r). It is obvious Q(s) £ Q(r), and it follows that the dimension of Q(s) over Q is less than or equal to the dimension of Q(r) over Q. = k+1. This implies deg(s) £ deg(r) But deg(s) cannot be less than k + 1 ; otherwise, by our inductive hypothesis, the degree of r will be less than k+1. This contradicts deg(r) = k + 1 . the algebraic number s is k + 1 . Hence the degree of In addition, if r is an algebraic integer then, according to Theorem 1.8 and Theorem 1.9, s is an algebraic integer. Recall Theorem 1.7. Q.E.D. If r is an algebraic integer of degree n>_2, and yg is topologically equivalent to y r , and s = p+qr for some rational numbers p,q, then s = r or s = 1-r. Proof: As r is an algebraic integer of degree n> 2 and y s is topologically equivalent to y^, according to Theorem 1.10, it follows that s is an algebraic integer of degree n. Furthermore, owing to Theorem 1.8, we have s = £ r11"1 + ... + £ r + £ n-1 1 0 for some integers unique. ,£^,...,£n_^ and this representation is This forces p and q to be integers and q ^ 0. r = (l/q)s-p/q. So, using the same argument as above, we can have 1/q and p/q are integers. or -1. Notice Thus q can be either 1 13 Case 1. q = 1. As 0 < s = p + qr = p + r < 1 and p is an integer and 0 < r <1, it follows that, p = 0; that is, s = r. Case 2. q = -1. Since 0 < s = p + qr = p - r < 1 and p is an integer and 0 < r < 1, so p can only be 1; that is, s = 1 - r. Q.E.D. Theorem 1.11. - If r is an algebraic integer of degree 2 and y is topologically equivalent to y / then s = r or s x* s = 1 - r. Proof: As r is an algebraic integer of degree 2 and y S is topologically equivalent to y , it follows from Theorem 1.10 and Theorem 1.8, that s = p + q r for some rationals p and q. Hence, according to Theorem 1.7, s = r or s = 1 - r . Q.E.D. §3. C-pairs This section consists of some results which have to do with Topologically Equivalent Measures in the Cantor Space. They were done by F.J. Navarro-Bermudez in [5 ] which is an unpublished article. In order to make a complete exposure of the topologically equivalent measures in the Cantor Space, 1 decided to put it here as a section. By a C-pair, we mean a pair (X,y) where X is a space of the form x = n sn n =il 14 with the product topology. Each factor S n carries the discrete topology. is finite and y is a Borel measure in X. It is well known, of course, that whenever (X,y) is a C-pair then X is homeomorphic to the Cantor space of infinite sequences of zeros and ones, and its topology is compatible with the metric d which, for any two points x = (x ) and n x' = (x'n), is given by the formula 00 d(x,x') = r» i (%) d (x ,x1 ) . , n=l By d n n is meant the metric on S n 0 or 1. n n which takes only the values Furthermore, it can readily be seen that a countable basis for the topology of X consists of sets of the form [i ,i ,...,i ] = { (x ) £ X: x.=i. for j=l, 2, . . . ,m} . I z ni n 3 3 These sets are both open and closed and will be referred to as the special closed-open sets of X. They are obtained by fixing the first coordinate, the second coordinate, and so on, up to a finite number of coordinates. Definition. - Let t be an integer, t>2, and let P1,P2/...,Pt be non-negative real numbers such that P 1 + P2 + '-- + P t ~ A C-pair (X,y) is said to be of type (t?P1/P2'•••'Pt) if the following two conditions hold: (i) For each n, is a set consisting of t elements which, for convenience, may be taken to be the integers 1,2,...,t. 15 (ii) p is a shift invariant product measure y = n n u ,pn =l with y (j) = cp. for all n and j. n 2 Let (X,y) be a C-pair of type (t;p x ,p 2 ,...,p ) . x can be expressed as a disjoint union X = U [j] j=l where y([j]) — p_. and diam([j]) = Each one of the special closed-open sets [i ] can in turn be expressed as a disjoint union [i,] = U [i,,j] j=l where y ( [ i p j ] ) = p ± p and diamffi^j]) = %. this process, a sequence of covers ^ By continuing ^ , % ,... of X can be constructed with the following properties: n ^ U iji ...i : l<i.<t, l<j<n^* 12 The members of ^ n — j— (3.1) -J- are mutually disjoint non-empty closed- open sets of diameter less than 1/n. For fixed i , i , — , i , 1 I II • . = U {U. . . .€ # 1 1 2",;Ln j=l 1 1 1 2 " * " 1n:' } » = Pi Pi •••Pi • J- l n 1 2 (3.2) <3-3) n Theorem 1.12. - Let X be a complete metric space and y a Borel measure in X. If there exists an integer t ^ 2 , non- negative real numbers p 1 ,p 2 ,...,p t with p 1 +p 2 +...+p t = 1, and 16 a sequence of covers of X with properties (3.1), (3.2) and (3.3), then (X,y) is isomorphic with a C-pair of type (t; , p^ , . . . , p^ ) . 00 Proof: Put Y = n , where S = {l,2,...,t}, and let n oo n =l v = II v be the product measure in Y given by v (j) = p n n=l n i for all n and j. For y = ( ^ , , i , . . .) in Y, define h(y) = o n o 1 nu n... 1 12 <3.4) 1 2 3 Since X is a complete space and lim (diam U, ) = 1 1 n->oo l 2*'' : L n o, the intersection on the right hand side of (2.4) consists of exactly one point. to X. \ Thus, h is a well defined function from Y It is onto on account of the fact that the families are covers of x ' it: i s one-to-one because each cover %n consists of disjoint sets. and h~ 1 (U =u 12* * * n 12' ' family of all sets U . 12 _ Clearly, h([i ,i ,...,i ]) 1 2 n ) = [i ,i ,...,i ]. As the 1 2 n n is a basis for the topology of n X, and the family of special closed-open sets is a basis for the topology of Y, both functions h and h ^ are continuous. Finally, for U = [i i 1 2 ..,i ], it is the case that v(U) = n . ) = y(h(U)). But every open set P. P* •••?. = y( u . . 1 2 n 1 2"'Xn of Y is a disjoint union of countably many special closedopen sets. Hence, the equation v(U) = y(h(U)) holds for every open set U of Y, and, consequently, for every Borel set as well. 17 The purely topological content of Theorem 1.12, namely, that every compact metric space is the continuous image of the Cantor space, is a well known result. Theorem 1.12 was inspired by the proof that A.H. Schoenfeld [ 9 ] has supplied for that well known result. It is obvious to observe that the converse statement of Theorem 2.1 is nearly true. isomorphic with a C-pair Y to X. (3.2) and Indeed, suppose that (X,y) is (Y,v) through a homeomorphism h from As Y has a sequence of covers with properties (3.1), (3.3), h carries these covers to covers of X with the same properties, except that there is no guarantee that a set of diameter less than 1/n in Y is carried by h to a set of diameter also less than 1/n in X. Some Consequences of Theorem 1.12 A C-pair of a given type may well be isomorphic with a C-pair of a different type. In particular instances, this can be established by appealing to Theorem 1.12. For example, 00 the pair II {1,2} , y ) n n= 1 (X= of type (2;k,h) and the pair 00 (Y = n {1,2,3} ,v) of type n_i (3;h,h,h) are isomorphic. X prove this, a sequence of covers ^ ' ^ 3 ' * * * constructed satisfying properties t=3, P1=h, P = 2 P = 3 ^' L e t the special closed-open sets U3 = [2,2]. t (3.1), (3.2) and ^ le f i r s t cover % = [1], = To will be (3.3) with consist of [2,1] and These sets are mutually disjoint, of diameter less than 1, and u(U^) = h, y ( U 2 ) = y(U^) = In general, 18 if covers \ > " t \ have been constructed satisfying properties (3.1), (3.2) and (3.3), then the cover ^ is n+1 constructed as follows. Let U = H i i 1 ho i i .i LJ 12 n any of the special closed-open sets in % . Put U. n i i . . .i i 12 n J 2 , • • •, js/l] / U ± ± 2 - [ jx , j 2 , . . ., jg ,2,1] and 12* * * iin JU / j»2,2]. i i ...i 3 = j2/« • j is defined s / 2, 2 The cover U 1 2 n n+i to consist of the special closed-open sets U 1 1 1 2 * * * """n+l These sets are clearly disjoint and of diameter less than l/n+l« Property (3.1) is satisfied, and the same is easily seen to be true of properties (3.2) and (3.3). The function h from Y to X defined by (3.4) using the covers establishes an isomorphism between (X,y) and (Y,v). In fact, it is even possible to describe the action of h on the points (i^i^...) Of Y. Put f (1) = 1; f (2) = 2,1; f (3) = 2,2 Then h(i1,i2,...) = ( f d ^ ,f(i2) . Theorem 1.13. - Let (X,y) be a C-pair of type ' • • • '*3^ • -*-n or der for (X,y) to be isomorphic with a C-pair of type (t;Pl,p2,...,pt), it is sufficient that there exist disjoint special closed-open sets U ,U ,...,u J. that x = U U U U... U U such t and y (U.) = p. for all j. t Proof: z. J J Write U = [i ,i . i 1 fnr -i—i o +• j 1 J2 -"nCj) D-J-f • • • ,t. Then p = y(U ) = g q ...q . Let ^ be the cover of X J J 1 . . 1 J2 n (j) consisting of the special closed-open sets U ,U , . . . ,u 1 2 t Suppose that covers , \ of X have been constructed satisfying properties (3.1), (3.2) and (3.3), each cover 19 consisting of special closed-open sets. Let U be l±2 • • • i any member of % say U = [k ,k ,...,k ]. Fo? fixed 11 12*** x k put " i i l 2 . . . v - l V 2 kr,jl>j2 jn(j)1, i and define the cover % ,, to consist of the sets U n+1 i1 i2 i n+l These sets are disjoint special closed—open sets of diameter less than 1/n+l. a y „ d s i „ ~ ^ Properties (3.1) and (3.2) are satisfied, l l V - V " V V " \ V V V « • (Ui i property (3.3) is satisfied i )P 1 = P. P. ...p. p., 1 2* * n 3 2 n3 as well. Thus, it is possible to construct a sequence of covers of X with the requirements of Theorem 2.1. Hence, (X,u) is isomorphic with a C-pair of type (t;p ,p ,...,p ). 1 2 t To illustrate Theorem 1.13, consider a C—pair (X,y) of type (2; 1/3,2/3) . Put ^ = [1], U £ = [2,1], U 3 = 3*2,2%. X is the disjoint union of these special sets and since y(U^) = 1/3, y(U2) = 2/9, li(U^) = 4/9, (X,y) is isomorphic with a C-pair of type (3;1/3,2/9,4/9). put Vj = [1,1], V 2 = [1,2], u(^2) = 2/9, y(V^) = 2/3. = [2]. On the other hand, This time yfV^) = 1/9, Thus, (X,y) is also isomorphic with a C-pair of type (3;1/9,2/9,2/3). Therefore, two C-pairs of types (3;1/3,2/9,4/9) and (3;1/9,2/9/2/3), respectively, are always isomorphic. Note that the condition in Theorem 1.13 that the closedopen sets U_. be special cannot be dropped. be a C-pair of type (4jh,h,k,h). u = 2 [2] U [3] U [4]. U 2 , and y (U^) = Put Indeed, let (X,y) = [1] and Then X is the disjoint union of u y(U2) = 3/4. However, (X,y) cannot be and 20 isomorphic with a C-pair of type (2;l/4,3/4), for indeed, by an easy application of Theorem 1.13, (X,y) can be seen to be isomorphic with a C-pair of type (2;h,h). But, by Theorem 3.3 of [4], two C-pairs of types (2;l/4,3/4) and (2;h,h), respectively, cannot be isomorphic, and this in spite of the fact that both measures take values on closed-open sets which are dyadic rationals. Theorem 1.14. - If the requirement that the closed-open sets Uj be special is dropped, then the sufficient condition for a C—pair to be isomorphic with a C—pair of a given type as stated in Theorem 1.13 is also necessary. Proof: Let h: X •+ Y establish an isomorphism between (X,y) and a C-pair (Y,v) of type (t;p 1 ,p 2 ,...,p ). V j = [j3 for j = l,2,...,t. These are mutually disjoint closed-open sets of Y such that Y = V 1 and v (Vj) = p_. . The sets Put U V 2 U U V t = h ^ (V ) are mutually disjoint closed-open sets of X such that X = U y(U ) = v(V.) = p.. J J 3 U U U ... U U and •L Z t Observe that while the sets V are j special, the same is not necessarily true of the sets U . 3 Let (X,y) and (Y,v) be two C-pairs. Denote by C the Cantor space of infinite sequences of zeros and ones, and let f and g be homeomorphisms from C to X and Y, respectively. Put y^ (B) = y(f(B)) and B of C. Clearly, y^ and (B) = v(g(B)) for every Borel set are Borel measures which are topologically equivalent in C if and only if the C-pairs 21 (X,y) and (Y,v) are isomorphic. Thus, the results of Theorem 1.13 and Theorem 1.14 can be carried to measures in C of the form yf where f is a homeomorphism from C to some product space X and y is a shift invariant product measure in X. CHAPTER BIBLIOGRAPHY Oxtoby, J.C. and S.M. Ulaiti, "Measure preserving Homeomorphisms and Metrical Transitivity," Annals of Mathematics (2) 42 (1941), 874-920. Oxtoby, J.C., "Homeomorphic Measures in Metric Space," Proceedings of the American Mathematical Society 24 (1970), 419-423. Oxtoby, J.C. and V. Prasad, "Homeomorphic Measures in the Hilbert cube," Pacific Journal of Mathematics 77 (1978), 483-497. Navarro-Bermudez, F.J., "Topologically equivalent measures in the Cantor space," Proceedings of the American Mathematical Society (2) 77 (1979), 229-236. , "Topologically equivalent measures in the Cantor space II," preprint. Prasad, V.S., "A survey of Homeomorphic Measures," Measure Theory Conference, Oberwolfach, 1981, Springer-Verlag Lecture Notes Series. Von Neumann, J., "Comparison of cells," Collected Works, Vol. II, 558. Hungerford, T.W., Algebra, New York, Holt, Rinehart and Winston, Inc., 1973. Schoenfeld, A.H., "Continuous Surjections from Cantor Sets to Compact Metric Space," Proceedings of the American Mathematical Society 46 (1974), 141-142. 22 CHAPTER II BINOMIAL NUMBERS IN THE UNIT INTERVAL § 1. Introduction In view of Theorem 1.2 and Corollary 1.4 of Chapter I, we will say that two numbers r and s in the unit interval [0,1] are binomially related whenever (1.6) and (1.7) of Corollary 1.4 hold. Furthermore (p1,...,pn) and (q1,...fq ) are said to be multinomially related in case (1.4) and (1.5) of Theorem 1.4 hold. In this chapter, we are only interested in binomial relationship. The following definition plays an important role in this chapter. Definition. — A number r in the unit interval is said to be a binomial number provided that there exists s£ [0,1] such that s is binomially related to r and s ^ r, s#l-r. One can ask whether there are any binomial numbers. this chapter we give the affirmative answer. In According to [1], the rational and transcendental numbers are not binomial numbers. I will give a delicate proof to show that the alge- braic integers of degree 2 are not binomial numbers. It is evident that binomially related relation is an equivalence relation in the unit interval. In other words, a binomial number is a number whose equivalence class has more than two elements. 23 24 Definition. - A Selmer's number r is an algebraic integer of degree n_>3 so that r11 + r-1 = 0 or r11 - r-1 = 0. In this chapter, we are going to show any Selmer's number in the unit interval is a binomial number. Let us have a brief review of Selmer's number. Through a study of generalized continued fractions, Ernst S. Selmer has been led to certain questions concerning the irreducibility of polynomials. Let us first consider the general problem: The ordinary algorithm for a continued fraction may be described as a systematic replacement of two non-negative numbers, a^ ^ and (k+1) _ a (k) a n d ,a (k)= a l ~ 2 2 division a x (k) /a^ k) . by two other numbers, usually the (positive) remainder by the Formulated in this way, a generalization to n dimensions is immediate. Such a division algorithm has been extensively studied by Perron [2]. An alternative pro- cedure, using subtraction instead of division, was introduced by Brun [3]: a At each step, with the n numbers ik) - a2k) - - a n k ) - °' (1-D we replace a^ k ) by the difference a* k ) - a ^ k ) , and rearrange the numbers according to magnitude. Of particular interest are periodic expansions, characterized by a^ k+£ a (k> 1 ) a (k+£) = a(k) 2 a (k + £) = ••• = V ) a „ = A " d-2) 25 In any procedure for generalized continued fractions, the ratio X is determined by an equation of degree n, and this equation is irreducible if and only if the given n numbers are linearly independent. Irreducibility and independence are then, of course, related to the same basic field of rationality. In what follows, we shall always assume this to be the ordinary field of rational numbers. The simplest period consists of one step only; that is, 1 = 1 in (1.2). a ik^ ~ a 2 ^ = a In Brun's algorithm, we must then have nk+1^' since otherwise the smallest number would not be involved at all in the process. The equations (1.2) then take the form £ £ (k) ,(k) a< k) -a 2 (k) n (k) a<k> n = A. n Showing that A is a characteristic root of the n x n matrix 0 1 0 0 0 1 0 0 0 0 1 - 1 0 0 0 0 0 0 0 0 1 0 The corresponding equation for A takes the form An + A-l = 0, (1.3) 26 This naturally led E.S. Selmer to study the irreducibility of the polynomials x n + x ± 1 which have very small discriminants. Note for discriminants. Let K be a field. If f€ K[x] is a monic polynomial of degree n, with roots x, ,...,x we 1 'n define the discriminant of f by: d(f) = 1)/2 n (x -x ) = (-1) 1 i<j J n (x.-x.). . / . 1J J If J Thus D(f) € K and D(f) ^0 if and only if f has distinct roots. It is important to compute the discriminant of an irreducible polynomial f = xn+a1xI1"1 + . . .+an without knowing a priori its roots. = x^+...+xk (where x . , . . . , x If p k. i n i are the n roots of f) for k=0,1,2,..., then pQ = n, pl = -ax and ?2'?3'"* * m a ^ be com P u t e d recursively by the well-known Newton formulas: (a) P If k_<n, then k-Pk-l S l + Pk-2 S 2---- +( - 1)k " 1 Pl s k -l +( - 1)kkS k (b) = °" If k > n, then P k-Pk-lSl+---+<-1)nPk-nSn = °' where Sj,...^^ are the elementary symmetric polynomials on the indeterminants x. ,x I ,. Then n-1 n-1 n D(f) = n-1 n 2n- 2 27 In some cases, the computation of the above determinant is rather awkward. However, it is possible to compute the discriminant of f by the more direct formula: D ,£, = < - D n < n - l ) / 2 N k ( x ) / k ( f ' ( x ) ) where x is any root of f and f 1 is the derivative of the polynomial f. N k(x)/k is the norm of f 1 (relative to k). In [4], E.S. Selmer obtained D (x n -x-l) = ( - l ) ^ ^ " ^ ( n - 2 ) [n n +(-l) n (n-l) n - 1 ] D (x n +x+l) = (_i ) i 2 n ( n- 1 ) [ n ^ - D 1 1 - 1 (n-l) n _ 1 ] . Furthermore, he proved that x n ± x - 1 are irreducible for all n. It happens that there exists a real number y that y ii is a zero of x +x-l. numbers in this chapter. We can see y n e [0,1], such are binomial That is an interesting result. Let us state the following theorem since we will use it in this article. Theorem 2.1. x n - x - 1 are irreducible for all positive integers n>_2. By a trinomial is meant a polynomial of the form x n ± x ± 1. Due to the fast development of computer science, the irreducibility of trinomials over theory. play some role in the coding There are quite a few articles studying irreducibility or irreducible factors of trinomials. would be an interesting one. For instance, [5] 28 We need Gauss' Lemma, in this article. Before stating the lemma, let D be a unique factorization domain and f = . E a.x1 1=0 1 . a non-zero polynomial in D[x]. A greatest common divisor of the coefficients aQ,a^,...,a^ is called a content of f and is denoted by C(f). Strictly speaking, the notation C(f) is ambiguous since greatest common divisors are not unique. an But y two contents of f are necessarily associates and any associate of a content of f is also a content of f. write b ~ c whenever b and c are associates in D. We shall Now ~ is an equivalence relation on D and since D is an integral domain, b ~ c if and only if b = cu for some unit u £ D by Theorem 3.2(vi) of Chapter III in [12]. C(af) ~ aC(f). If f€ D[x] and C(f) is a unit in D, then f is said to be primitive. 9 = If a€ D and f€ D[x], then Clearly for any polynomial g6 D[x], C(g)g^ with g^ primitive. (Gauss' Lemma). If D is a unique factorization domain and g,f€D[x], then C(fg) s C(f)C(g). In particular, the product of primitive polynomials is primitive. Eisenstein s Criterion. — Let D be a unique factorization domain with quotient field F. If f = E a. x* € D[x], deg f> 1 i = 0 1 and p is an irreducible element of D, such that PVV P|a. for i=0,1, . . .,n-1; p then f is irreducible in F[x], irreducible in D[x], For a proof, refer to [5]. 2 ^, If f is primitive, then f is 29 §2. Existence of Binomial Numbers In view of [1,6], one may ask the existence of binomial numbers. tively. The following theorem answers the question affirmaMoreover, the theorem gives binomial numbers for each degree n > 3. Theorem 2.2. - For each positive integer n>^3, there are algebraic integer x and algebraic fractional y of degree n such that x,y are binomial numbers. Proof: (I) For any positive integers n^3, First show that there are algebraic integers of degree n which are binomial numbers. Case 1. n is even. Let f(x) = x n + x - l . As f(X) is irreducible (according to Theorem 2.1) and f(0) = -1 < 0, f(i) = 1 > 0, it follows that there exists an algebraic integer r of degree n such that r11 + r - 1 = 0 that is, r = l-r 11 = 1- (r2 ) n ^2 which is equivalent to r m = 1 - sm _ = T K ,c -,m m-k.,1 V( l - s* ) Kk k k=l vz (2.1) where m = n/2 and s = r2 (2.2) 30 As s = r or s = 1-r will imply that r is an algebraic number of degree <2 which contradicts r is an algebraic integer of degree nj>3, it follows that s ^ r and s^l-r. Furthermore, (2.1) and (2.2) imply that r is binomially related to s. Case 2. Hence r is a binomial number of degree n. n is odd, say n=2m+l, m>_l. Considering f(x) = x2m+1 +x2m - 1, which is irreducible (according to Theorem 2.1, Gauss' Lemma, and substituting — for x). Moreover f(0) = —1 < 0 and f(1) = 1 > o. Therefore, there is an algebraic integer r of degree n in the unit interval so that 2m+l 2m r + r - 1 = n0. (2.3) Multiplying both sides of equation (2.3) by (r-1), we obtain r = l - r2m + r 2 m + 2 (2.4) s = r2. (2.5) Set Then we have y~ — i I m m+1 m n N s +s = 1 - s (1-s); . (2.6) that is, m+1 _ / y /-.m+1 m-k+1 .1 ,k, . m,, . r - ( h C s (1-s) ) +ms (1-s). k=0 k#l r (2.7) 31 Since s = r or s = 1-r implies the degree of r is less than or equal to 2, we have a contradiction. s^l-r. Therefore s # r and Equations (2.5) and (2.7) imply that r is a binomial number. (II) Next we must prove that there are algebraic frac- tional of degree n ^ 3 , which are binomial numbers. The following lemma will simplify the original proof of (II), which is also done by a Norwegian mathematician, Helge Tverberg, at the University of Bergen in Norway. Lemma. 2x 2n + 2x 2 n _ 1 - x 2 - x - 1 is irreducible over Z[x] for all positive integers n>_2. Proof: According to Gauss' Lemma and substituting ^ for x, it suffices to prove f(x) = x 2 n + x 2 n _ 1 + x 2 n ~ 2 - 2x - 2 to be irreducible. Assume the contrary that f(x) = g(x)h(x) where g,h€ Z [x] and deg g >_ 1, deg h 1. As f(x) 5 x 2 n + x 2 n _ 1 + x 2 n " 2 E x 2n " 2 (x 2 +x+l) (mod 2), and 2 x +x+l is irreducible in 7>1 [x] , it follows that g(x) = (x2+x+l)xr+ 2gx(x) h(x) = x s + 2h (x) where r + 2 = deg(g) > degfg^, s = deg(h) > degd^) and gj,h e z[x]. Thus 32 „2n x 2n-l +x 2n-2 +x ~ „ - 2x - 2 (2.8) = x 2 n + x 2 n _ 1 + x 2 n " 2 + 2[ (x2+x+l)xrh1+xsg1] +4g 1 h 1> Now s> 0. If also r> 0, then the right hand side of (2.8) would have a constant term divisible by r. Thus r=0, and so g(x) = (x2+x+l) +2(ax+b), where a,b€ Z. As g(1) divides f(l), g(-1) divides f(-l), g(0) divides f(0), it follows that either g(x) = x2+x-l or g(x) =x2-x-l. (2.9) If g(x) = x 2 + x - l , then f(x) = ( x 2 n + x 2 n _ 1 - x 2 n - 2 ) + 2(x2n_2-x-l) which is equivalent to (x2+x-l)h(x) = (x2+x-l)x2n_1 + 2(x2n-2-x-l). This implies that x 2 n " 2 - x - 1 has factor x 2 + x - 1 which 2 n~ 2 contradicts x — x — 1 is irreducible for n > 2 (according to Theorem 2.1). If g(x) = x 2 + x - l , then f(x) = (-x 2 n +x 2 n _ 1 +x 2 n ~ 1 ) +2(x2n-x-l). A similar argument shows that x 2 n - x - 1 has a factor x 2 - x - 1, which contradicts x 2 n - x -1 is irreducible. Q.E.D. 33 Continuing to prove Theorem 2.2: (II) To show that there are algebraic fractionals of degree n>_3 (for each n) , which are binomial numbers. Case 3. n is odd, say n = 2m+1. f(x) = 2x 2 m + 1 - 2x 2m + 2xm - 1 is irreducible over ]R , according to Eisenstein's Criterion. Using the fact that f(0) = -1 < 0, f(l) = 1 > 0, we find that there exists an algebraic fractional r of degree 2m + 1 in the unit interval so that 2r 2 m + 1 - 2r 2m + 2rm - 1 = 0 which is equivalent to (2.10), if we multiply (1-r) to both sides of the previous equation. r = 1 - 2rm(1-r) + 2[rm(1-r)]2 (2.10) s = rm(1-r). (2.11) Set Then (2.10) can be transformed into r = 1 - 2s + 2s2 which is equivalent to r = s 2 + (1-s)1. Using the fact that r ^ s , r^l-s, (2.11) and (2.12), we conclude that r is a binomial number of degree 2m +1. (2.12) 34 Case 4. n is even, say n = 2m, m> 1. Considering f(x) = 2x 2m + 2x 2 m _ 1 - x 2 - x - 1, according to the previous lemma, f(x) is irreducible over Z[x]. Upon using the fact that f(0) < 0, f(1) > 0, we find an algebraic fractional r of degree n in the unit interval such that (2.13) holds. o„2m 0 2m-l 2 . 2r + 2r - r - r - 1 = 0; (2.13) that is, 1+ r + r 2 = 2r 2m_1 (1+r). Multiplying both sides of the above equation by (1-r), we obtain 1 - r 3 = 2r 2m ~ 1 (1-r 2 ) which is equivalent to 1 = r 3 + 2r 2m - 1 (l-r 2 ). MultiplyinQ" both sid.Gs of th.B cibovG scjiicition by ITF WG obtain r = r 4 + 2r2nl(l-r2). (2 . 1 4) Set s = rl - (2.15) Then (2.15) becomes r = s 2 + 2sm(1-s) (2.16) 35 which is equivalent to r = (m+1)sm(1-s) + m-1 z ( m ~ )s k=0 k k^l (l-s)k. Similarly reasoning, we have s / r , s ^ 1 - r . (2 17) Therefore (2.17), (2.15), s ^ r, and s i1 1-r imply that r is a binomial number. Q.E.D. §3. Existence of Non-binomially Algebraic Numbers of Degree > 2 Iri this section, we are going to prove that any algebraic integer of degree 2 is not a binomial number. Let [x] be the largest the fractional part of x. integer £ x and <x) = x - [x] be Then (3.1), (3.2), (3.3), (3.4) are evidently true. If 0 € (0,1) and a,b€ B , then either or (3.1) <(a+b)0) = <a0} + <b0), ((a+b)0) = <a0) + <b0) - 1. If s > r, s,r are positive integers, O < 0 < 1 , then <(s-r)6) = f<se)-<r0) < |^<s0) - (l-<r0)) ((q(r0)) = l - < q e ) if <S0> > <r0) if <r0) > <s0) if q € JSF , O < 0 < 1 (-n0) = 1 - <n0) (3.2) if n€ u . The following theorem is very useful. (3.3) (3.4) 36 Theorem 2.3. - Let s be an irrational number and m an integer with |m| > 1 . If r = <ms>, then s^<kr> for any integer k. Proofs It is obvious that k 5^ 0. Assume the contrary that s = <kr) and r = <ms) with |m| > 1, k # 0. Notice that kms = k ([ms] + <ms >) = (k[ms] + [k<ms>]) + <k<ms>) = (k[ms] + [k<ms)]) + s. Therefore, we have (km-l)s as an integer. By using the fact that s is irrational and (km - l)s is an integer. have km - 1 = 0. This contradicts |km| > 1. So we must Hence s#<kr) for any non-zero integer k. Q.E.D. Theorem 2.4. — If r is an algebraic integer of degree 2 and s is binomially related to r, then either s= r or s=1-r. Proof: Since s is binomially related to r so that there are non-negative integers m,n; a , a a ; bn,b1f...,b so u i n u i m that 0 < a. < (^), 0 < b < ( m ) for i=0,1,2,...,n; •J J = j 0/1/2,...,m, and (3.5) holds. s = a rI1 0 + a 1 rI1 ~ 1 d-r) + ... + a^ (1-r) n r = b s m + b sm_1(1—s) + ... + b (l-s)m 1 Jm Using the fact that r is an algebraic integer of degree 2, there exist integers a,b so that 2 r + ar + b = 0; (3 ' 5) 37 that is, (3.6) holds. r2 = - a r - b . (3.6) Repeatedly applying (3.5) and (3.6), we obtain s = <pr) and r = (qs) for some integers p,q. According to Theorem 2.3, we must have p = 0,1,-1 and q = 0,1,-1. These possible combi- nations of values of p and q force either s = <r) or s = <-r). In other words, according to (3.4), s = r or s = 1-r. Q.E.D. It follows from Theorem 2.4 that any algebraic integer of degree 2 is not a binomial number. §4. Binomial Numbers and Pisot Numbers We shall denote by ||a|| the absolute value of the difference between a and the nearest integer. Thus, I I a| | = min |a-n|, n=0,±l,±2,... . If m is the integer nearest to a, we shall also write a = m ±e where 0 < e £ so that ||a|| = e. If n is an integer, then ||a|| = ||a+n||. Definition. — Let 0 > 0 be a real algebraic integer such that all its conjugates (not 0 itself) have moduli strictly less than 1. Then we shall say that 0 is a Pisot number. Refer to [7, 8, 9, 10]. 38 Theorem 2.5. - If 0 is a Pisot number, then | | 0 n | | -> o as n oo. For a proof, refer to [7, 8]. It is worthy to mention an important unsolved problem here. Suppose that 0 > 1 is a number such that | |0n| | o as n + oo (or, more generally, that 0 is such that there exists a real number A such that | |X0n| | + 0 as n -* oo) . C an we assert that 0 is a Pisot number? The following theorem is a well-known good theorem. Theorem 2.6. - The set of all Pisot numbers is a closed set. For a proof, refer to [8]. In view of Theorem 2.6, we have the following definition. Definition. - A real number 0 > 1 is a binomial number if 1/0 is a binomial number in [0,1]. It is very interesting to determine the relationship between Pisot numbers and binomial numbers or other similar questions. According to the previous definition and [1], we know that the positive integers 2,3,4,— are trivially non-binomial Pisot numbers. Using Theorem 2.4, we can determine some of the Pisot numbers of degree 2 that are non-binomial. instance, if a is an integer and a <-2, then (-1 + /aJ^4) /2 is a non-binomial Pisot number. For 39 According to Theorem 2.2 and Case 3, the binomially algebraic integer 0 of degree 3 such that I 3 - 2 0 2 + 20 - 2 = and it is obvious 1 < 0 < 1.5. 0ab = 0 Let a,b be its conjugates. 2 and 1 < 0 < 1.5, it follows that |ab| > 1. words, at least one of a,b has modulus > 1 . and non-Pisot number. m As other So 0 is a binomial For more details of binomial numbers and Pisot numbers, refer to [11]. CHAPTER BIBLIOGRAPHY 1. Navarro-Bermudez, F.J., "Topologically equivalent measures in the Cantor space," Proceedings of the American Mathematical Society (2) 77 (1979), 229-236^ 2. Perron, 0., "Grundlagen fur eine Theorie des Jacubischen Kettenbruchalgorithms," Mathematische Annalen 64 (1907), 1-76. 3. Brun, V. "Engeneralisation av kjedebroken I, II" (avec des resumes en fransais), Vid, Selsk, Skrifter, Mat.-Nat. Kl., Kristiania 1919, Nr. 6 1-29 1920, 1-24. 4. Selmer, Ernst S., "On the irreducibility of certain trinomials," Mathematica Scandinavica 4 M9R61. 287-302. 5. Marsh, Richard W., W.H. Mills, Robert L. Ward, Howard Ramsey, Jr., and Lloyd R. Welch, "Round Trinomials," Pacific Journal of Mathematics, Vol. 96. 1 ' 1981, 175-192. 6. Navarro-Bermudez, F.J., "A Number Theoretic Result with Applications to Measure Theory," preprint. 7. Meyer, Yves, Algebraic Numbers and Harmonic Analysis, Amsterdam, North-Holland Publishing Co., 1972. 8. Salem^ Raphael, Algebraic Numbers and Fourier Analysis, Belmont, California, Wadsworth Mathematics Series, 1963. ' Pisot, C. and J. Dufresnoy, Annales Scientifigues de l'Ecole Normale Superieure, Vol. 70 (1953), 105-133. 10. Pisot, C., Annali di Pisa, Vol. 7 (1938), 205-248. 11. Huang, K., "Binomial Numbers and Pisot Numbers," to appear. 12. Hungerford, T.W., Algebra, New York, Holt, Rinehart and Winston, Inc., 1973. 40 CHAPTER III ALGEBRAIC NUMBERS §1. Introduction There are many points of view from which one may approach algebraic number theory. The most favorite point of view, possibly, is the algebraic point of view. Study- ing algebraic numbers through algebraic point of view has to do with groups, rings, fields, ideals, quotient fields, homeomorphisms and isomorphisms, modules and vector spaces, field extensions, Galois theory, Dedkind rings, localization, the ideal function, ideles and adeles, class field theory, etc. There is another favorite point of view; namely, the analytic point of view. For instance, there are Liouville's Theorem, Functional Equation ([1]), Diophantine Approximation ([2]), Fourier Analysis ([3]), Harmonic Analysis ([4]), etc. There are some interrelations between these two viewpoints. There are numerously unsolved problems related to algebraic numbers. Neither algebraic point of view nor analytic point of view can solve all the problems. There are many popular theorems having to do with algebraic numbers. Both points of view have developed a lot of concepts and techniques to solve problems. They also have developed their own theories and have proved many nice theorems, respectively. 41 42 The study of topologically equivalent measures has not only led us to study the binomial numbers but also intrinsically led us to have a different point of view for looking at algebraic numbers. §2. Equivalence Relations on I Let I be the unit interval on the real line, and S be the set of all the functions from I into I. If T is a subset of S, we give the following definition. Definition. - Let r and s be two elements of I. We say that r is T-related to s provided that there exist f,g in T so that r = f(s) , s = g(r). If T is closed under the operation of composition of two functions, then T-related relation is an equivalence relation on I. From now on, we are only interested in this type of relation. We shall call it compositive relation. In other words, if T-related relation is a compositive relation, then it is an equivalence relation on I. For instance, if T = {id}, where id is the identity function on I, then each equivalence class has only one element. itself, If T = S, then there is only one equivalent class - I if T = {f: f(x) = 1-x or f = {id}}, then each equi- valent class is of the form {r,l-r} for r£ I. It is obvious that the three relations appearing above are compositive relations on I. 43 Let Z[x] - the set of all polynomials with integer coefficients, Q[x] = the set of all polynomials with rational coefficients, £ ak x k (1-x) n - k k=0 a ° ± k l (£) for k-0,1,... ,n}. B [x] = {f 6 Z[x]: f(x) = Bq[x]-related equivalence relation has been called binomially related in Chapter II. Let us call Z[x]-related relation the integrally related relation and Q[x]-related relation the rationally related relation. It is obvious that integrally related and rationally related relations are compositive relations. It is not triv- ial that Bq[X]-related relation is a compositive relation. The next section will prove that Bq[x]-related relation is a compositive relation. §3. Homogeneous Functions in Two Variables Let H[x,y] be the set of all homogeneous functions in two variables x,y with non-negative coefficients. that n f(x / Y ) = Z a x iy n-i, i=0 g(x,y) = are two arbitrary elements of H[x,y]. f <g 11 n Z b.x j y n ~ j j=0 J We define if a. <b. for all i=0,l,2, .. . ,n. Assume 44 One can easily prove the following two statements (3.1), (3.2): If f,g,h are in H[x,y] and if f < g, then f«h < g-h. (3.1) If f < g, then f n •< g11 for n=l,2, (3.2) Let B [x,y] = {f 6 H[x,y] : f(x,y) = n O i a . l t . ) for i=0,1,2,..., n} . . E a.x1y11 i= 0 1 1 and Lemma. - Let A — a x m + a x m ^y+ ... + a y m and u i m B = b x m + b 1 x m _ 1 y + ... + b y m U I ill where °-ak^(k}' ° < b k ^O' and °^ak+bk^(k)' If f(A,B) = C q A £ + C 1 A A _ 1 B + ... + C £ _ 1 A B £ _ 1 + C £ B £ = e xp + e x p _ 1 y + • • • + e u I p-1 1 u x y p - 1 + e2 y p p- and 0 £ C i £ ( £ ) , then p = m£ and 0 < e± <_ ( p ) for i=0,l, . P. .., Proof: First, it is easy to see p = m£, and e x p + e x p _ 1 y + . . . + e .1x y p _ 1 + e yp u i p-1 pJ = f(A,B) = C Q A £ + C 1 A £ _ 1 B + ... + C £ _ 1 AB £ ~ 1 + C^B £ < (£)A^ + ( £ ) A £ _ 1 B + . . . + ( * )AB £ _ 1 + ( £ )B £ = (A + B) £ = [ (a n + b n ) x m + ( a i + b i )xin 1y + • • • + (a +b )y m ] Z u u 1 1 m m "< [(™)x m + ("1)xm~1y+ ... + ( m )y m ] £ u 1 m = (x + y) = ( um n £ )x m £ + (n,8-)xm<!-1y+ ... + ( m J)y m £ ; 1 m£ 45 that is, e 0 X ^ + e ^x p y + ... + e _. xy p ^ + e y p P p •< ( m 0 £ ) x m £ + ( m 1 A )x m A - 1 y+ ... + ( ^ ) y m £ . Uisng the fact that p = m£ / we find ° - 8 i - * Q.E.D. If f(x,y) 6 B[x,y], and f(x,y) = a 0 x n + a i x n _ 1 y + ... + a n y n where 0 < a. < (J) for i=0,1,2,...,n. Then we define the conjugate of f, denoted by f, by f(x,y) = (l-a 0 )x n + ((J)-a )xn"1y+ ... + ( ( n ) -a )y n . n n From the previous lemma, we obtain the following theorem. Theorem 3.1. - if f(x,y) 6 B[x,y], then f(g,g) 6 B[x,y] for any g 6 B[x,y]. Corollary. - B q [x]-related relation is a compositive relation on I. Proof: and g(x) = Suppose g , f 6 B [x] and f(x) = m z bk x k (l-x) m ~ k . k=0 E a x k (l-x) n ~ k , f -K.= rj k Define F(x,y) = E akxkyn~k, k=0 G (x,y) = ? b xkym"k. k=0 k According to Theorem 3.1, F(G,G) €B[x,y] and 46 F(G(x,1-x),G (x,1-x)) n = k=0 = aK lG(x,l-x))k(G(x,l-x))n-k , l a k ( j „ V £ ( 1 - x ) m " £ ) k ( ? ((®)-b£)xil(l-x)in-£)n-k. k=0 £=0 £=o Using the fact that 1 = (x+(l-x))m = in /Hmix)x __kk(l-x) , ^ vm m-k £ ( ~k, we k= o k find that h(x) = F(G(x,l-x),G(x,l-x)) n = Z a (g(x)) (l-g(x)) n_k = f(g(x)). k=0 Since F(G,G) € B[x,y], so h(x) 6 B q [x]. Hence f(g(x)) is an element in B Q [X], that is, Bq[x]-related relation is a compositive relation on I. §4. Rationally Related Relation on I In this section, we are going to investigate the rationally related relation on I. Although it does little classi- fication of algebraic numbers, it certainly gives a very clear-cut equivalence classes for the unit interval. Let Q(r) be the field obtained by adjoining r to Q. Q(r) a = deg r. vector space over Q and the dimension [Q(r): r] The following theorem is needed in this section. Theorem 3.2. - Let W be a subspace of a vector space V over a division ring D. (i) (ii) Then the following statements hold. dim W £ dim V; if dim W = dim V and dim V is finite, then W = V; 47 (iii) dim V = dim W + dim (V/W). The following theorem is the main theorem of this section. Theorem 3.3. - The equivalence classes for rationally related relation on I are either of the form (4.1) or of the form (4.2). n-1 A(r) = {s: s 6 i , s = a„+a 1r+...- + a l 0* ret l" * * n-1 r ' where deg s= deg r, a ,a , ...,a are U 1 n-1 rational numbers} if r (4.1) i-s a n algebraic number of degree n and r is in the unit interval. T(r) = {a+br€l: a,b are rationals, b ^ O } (4.2) if r is a transcendental number and r€ I. In order to prove Theorem 3.2, we need the following famous classical theorem in algebra. Theorem 3.4. - Let k be a field, u an element of a larger field, and suppose that u is algebraic over k. Let f be a monic polynomial with coefficients in k of least degree such that f(u) = 0, and let this minimal degree be n. Then (a) f is unique; (b) f is irreducible over k; 2 n__ ^ l,u,u ,...,u form a vector space basis of K(u) (c) over k; 48 (d) [k(u) : k] = n; (e) A polynomial g with coefficients in k satisfies g(u) = 0 if and only if g is a multiple of f. For a proof, refer to [6]. Proof of Theorem 3.3: Case 1. Suppose r€ [0,1]. Suppose r is transcendental. Let s be rationally related to r. f(x) = aQ + a ] [ x+ ... + a x n Then there are f,g, say (a ^0), g(x) = b + b. x + . .. + b x m 11 u i m (bm^0) being two elements in Q[x], such that r = f(s) and s = g(r). It is easy to see that r = f(g(r)), which is equi- valent to r = a + a 0 i ( b 0 + b i r + * • • + b m r m ) + ••• + a n ( b o+...+ b m r m ) n Claim. (4.3) m = l and n = 1. Suppose the contrary - that m ^ l or n # l . Since m = 0 , n - 0 would imply that r is rational, so we can assume m> 1 or n> 1. Then the degree of f(g(x)) is nm and a b n is its n m leading coefficient. As the left hand side and right hand side of (4.3) are equal, it follows that a b n = 0. That n m contradicts S = b 0+bir* a n ^ 0 and I n ot Hence m = l , n = l ; that is, her words, if s is rationally related to r, then s = b + b r. Conversely, if s = a + b r € I and a,b are rational numbers with b ^ O , then we take f(x) = a + bx and g(x) = ^ + I x b Then b s = f(r) and r = g(s); that is, r and s are rationally related. Hence, the ecjuiv&lence eXciss of T is of the form (4.2) 49 Case 2. r is algebraic over Q. Suppose s is rationally related to r. Then there exist f,g£Q[x] so that r = f(s), s = g(r) . Let irr(r) = x n + a ^ x 1 1 - 1 + ...+ a Q , where a. 6 Q, for i=0,1,...,n-l. So we have r n + a ^ r 1 1 " 1 + ... + a Q = 0; that is, n r = _a rn-l n-1 " *'* ~ V (4.4) Applying (4.4) t o s = g ( r ) finitely many times, we obtain s6A(r). Furthermore deg s < deg r. r = f(s) which would imply deg r < n. Conversely, suppose s€A(r). If deg s < deg r, then Therefore, deg s = deg r. We want to show s is rationally related to r. s€A(r) implies there are rational numbers a ,a ,...,a 0' 1' ' n-1 so that (4.5) holds. S = a0 + a l r + • • • + V l 1 " " 1 Let f (x) a Q + ajX + . . . + ^ ^ x 1 1 1 . s€A(r) also implies deg s = deg r. A H.s) So we have s = f (r) . Using the fact that (r) £ Q(r), s^A(r), we find that Q(s) £Q(r). Moreover, Q(r) is a finite dimensional vector space, Q(s) is a subspace of Q(r), and dim Q(s) = dim Q(r). Therefore Q(r) = Q(s). That is, r = g(s) for some g(x) € Q [ X ] . Q.E.D. §5. Integrally Related Relation on I After investigating the rationally related relation on I, the next composition relation on I to be studied is integrally related relation. We have the following theorem for the inte- grally related relation on I. 50 Theorem 3.5. - Let r be any number in the unit interval. Then the following statements are true. (a) If r is transcendental and s is integrally related to r, then either s = r o r s = l - r. (b) If r is an algebraic integer of degree 2 and s is integrally related to r, then either s = r or s = 1 — r. (c) If r is an algebraic integer, then s is integrally related to r if and only if z [ s ] = Z[r]. Proof: (a) Suppose r is transcendental and s is integrally related to r. Then there exist f,g€ Z [x] , say f (x) = a + a.x+ ... + a x n ; u i n g(x) = b + b x + ... + b x n (b #0) , u i n n so that r = f(s) and s = g(r). r = f(g(r)). r = a 0 + a Claim. it is easy to see that r = f(g(r)) is equivalent to l < b + b 0 1 r+. • •+bmrTn) + ... + a n (b Q +. . .+b m r m ) n . (5.1) m = l and n = l . Suppose the contrary that m # l or n # l . Since m = 0 , n — 0 would imply that r is rational, so we can assume m > 1 or n > 1. and a nbm Since a n # 0 , k>m^0, the degree of f(g(x)) is nm is i t s lead i n g coefficient. As the left hand side and the right hand side of (5.1) are equal and l,r,r^,... form a basis of Q[r] over Q, it follows that a b n = 0. This n m a contradicts n ^ 0 and fc>m^0. Hence m = l , n = l; that is, 51 s r = b = a 0 + b^r (5.2) 0 + alS (5.3) From (5.2) and (5.3), we obtain r = a So we have 0 + alb0 + alblr- = 1. Then either = b^ = 1 or and (5.2) becomes either s = b Q + r or s = b Q - r. = b =-1 Using the fact that r € [0,1] and b^ is an integer, we conclude that either s = r o r s = l - r. (b) If r is an algebraic integer of degree 2 and s is integrally related to r, applying Theorem 1.7, Theorem 1.8, we will have r,s satisfy (5.2) and (5.3) for some integers b 0' b l ,a 0' a l* With a similar argument, we reach s = r or s = 1 - r. (c) is obviously true. Z[r] is a larger set than the equivalent class of r under integrally related relation. For instance, s = nr is not integrally related to r for any integer |n| > 1. §6. B^-related Relation Let Bj[x] be the set of all polynomials f(x) of Z[x] so that there are non-negative integers a.,a.,...,a with U 1 n (aQ = 0 or 1) and (an = 0 or 1) , and fix) = a 0 x n + a 1 x n - 1 (l-x) + ... +a n _ 1 x(l-x) n - 1 +a n (l-x) n . It is obvious that B1[x]-related relation is a compositive relation on I. From now on, we will call Bx[x]-related (6.1) 52 relation the B^related relation. Since (s is B -related to r) implies (s is integrally related to r), the following theorem is an immediate result of Theorem 3.5. Theorem 3.6. - Let r be a number in the unit interval and s Bj-related to r. And if r is an algebraic integer of degree 2 or if r is transcendental, then either s = r or s = 1-r. The following Theorem 3.7 was inspired by the proof that F.J. Navarro-Bermudez has supplied for his main theorem in [7]. Lemmas (6.1), (6.2), (6.3), and (6.4) are based on his proof in [7]. Theorem 3.7. - Let r and s be two rational numbers in the unit interval. The numbers r and s are B^-related if and only if s = r or s = 1-r. Proof: The sufficient condition is clear. the necessary condition proceeds as follows. The proof for Let r = a/b and s - c/d, where b and d are positive integers and a and c are non-negative integers. in reduced form. Assume that the fractions are written Using the fact that r and s are B^-related, we find (6.2), (6.3) are true for some non-negative integers a Q ,a 1 ,...,a n ; b Q ,^,...,b m ; a 0 /b 0 ,a n ,b m are either 0 or 1, and n,m are positive integers. s = V 1 r m = V "!''" + 1 ! 1 " ! + ... + a n _ 1 r ( l - r ) n - 1 + a n ( l - r ) n V m " 1 < l - = > + - . . + b B _ 1 s < l - s ) m - 1 + b (i-sl"1. (6.2) (6-3) 53 Since Bj related relation is an equivalence relation, we can interchange the roles of r and s, if necessary, and assume, without loss of generality, that b < d. If r = 0 or 1 and applying (6.2), then s = 0 or 1. now on, we can suppose that neither r nor s is 0 or 1. From in other words b > 1, 0 < a < b and 0 < c < d . Suppose the contrary that s ^ r and s # 1-r. We are going to show that (6.3) cannot be true. Note that if c = d-c, then s = c/d = 1/2. and since l<b<_d, then b = 2 also. Hence d = 2, But then a = l and r = %. Thus c £ d-c and, by interchanging the roles of s and 1-s, if necessary, it may be assumed, without loss of generality, that c < d-c. It is being assumed throughout that b £ d . From now on, the proof splits into two cases; either b = d or b < d. Case 1. b = d. If it were true that a = b-a, then r = a/b = 1/2. Hence, d = b = 2, and since 0 < c < d, it would follow that c = 1 and therefore s=h. Thus a ^ b - a and, by interchanging the roles of r and 1-r, if necessary, it may be assumed that a < b-a. Also, if it were true that d - c < a , the d-c < b-a, and as c< d-c <a, it would follow that d = c + (d-c) < a + (b-a) = b. But b = d. Consequently, a<d-c. Finally, observe that under the assumption b = d, it is not possible to have a = c, as this would force s = r. 54 Lemma 6.1. - I f b - d , c < d - c and a < d - c, there cannot exist non-negative integers b ,b , — , b (m>l) with u i m — b Q = 0 or 1 such that ^ = b (—)m + b f b Vd' + Vd' Proof: (6.4) holds. ad — i + ,i /d-c»m • * • + bm < — ) • (6.4) Assume that, for some choice of integers, line Then, upon clearing denominators, b Q c +b 1 c m 1 (d-c) + .. . + b^ (d-c)m. (6.5) As d e c (mod d-c), it follows from (6.5) that a c m - 1 = b0cm (mod d-c). (6>6) Since c/d is in reduced form, the integers c and d-c are Line (6.6) can be divided by cm-''' to relatively prime. obtain a e bQc (mod d-c). (6m7) However, 0 < a < d-c, so that it is not possible for b Q to be 0. Thus, b Q = 1, but then a E c (mod d-c) is not possible either, as both a and c are less than d-c and a ^ c . This contradiction proves the lemma. Case 2. b < d. Then the inequalities a > c and b-a > d-c cannot be both true. Thus, a < c or b-a < d-c, and there is no loss of gener- ality to assume that a < c, provided that it is not assumed also that c < d-c as was done earlier. 55 Lemma 6.2. - Let b < d and a < c. If a E b m b (mod c) and a = b Q b (mod d-c) with b m = 0 or 1, bQ = 0 or 1, then (i) b m = 1 ' (ii) a < d - c Proof: k. ' and (iii) b Q = 1. If b^ = 0, then a = kc for some positive integer Hence a >_ c, which contradicts the hypothesis. Therefore b m = 1; thus proving (i). It follows from (i) that b-a = jc for some non-negative integer j. As b ^ a , j must be positive. Then b-a > c. If it were true that a _> d-c, it would follow that b = a + (b-a) >_ (d-c) + c = d, or b > d, contrary to the hypothesis. Therefore, a < d-c, thus proving (ii). then a = p(d-c) for some positive integer p. but this contradicts (ii). If b^ = 0, Hence a > d-c, So, b Q = 1, and the lemma is proved. Lemma 6.3. - Let b < d and a < c. Then it is not possible to have a = b (mod c) and a = b (mod d-c) simultaneously. Proof: Suppose on the contrary that b = a + k c and that b = a + j(d-c) for some non-negative integers k and j. As b # a , both k and j must be larger than or equal to 1. Both inequalities c < d/2 and d-c < d/2 cannot be true simultaneously. Assume that c d/2. If k > 2, then b = a + k c > a + d or b > d, which contradicts the hypothesis. Hence, k = 1. But then, c = kc = b-a = j(d-c), so that d-c divides c. c and d-c are relatively prime and distinct. But Hence, d-c = 1, which is impossible on account of Lemma 6.2(ii) and the fact that a is positive. d, 56 It must then be true that d-c > d/2. b = a + j (d-c) thesis. a + d > d. Thus, j = l. c divides d-c. distinct. If j > 2, then But this contradicts the hypo- Then d-c = j(d-c) = b-a = kc, and Again c and d-c are relatively prime and So, c = 1, which is impossible as a < c by the hypothesis and a is positive. This contradiction proves the lemma. Lemma 6.4. - Let b < d and a < c. Then there cannot exist non-negative integers hQ/bj r • •. fbm (m> 1) with b Q = 0 or 1/ b m = 0 or 1, such that f - V§>" Proof: (6.8) holds. + Vi>'"1<TS> + -"+VT £ >". Assume that, for some choice of integers, line Then, upon clearing denominators, d m a = b 0 bc m + b 1 bc m - 1 (d-c) + ... + bb(d-c) m . As d E c (6-8) (6 .9) (mod d-c) and d E d-c (mod c) , it follows from (6.9) that c " 3. = b be (mod d-c) (6.10) and (d-c) a = b^bfd-c)"1 (mod c) . Since c and d-c are relatively prime, line (6.10) can be divided by c m and line (6.11) by (d-c)m to obtain (6.11) 57 a = b b (mod d-c) a = b b ni (mod c) . (6.12) Then, by Lemma 6.2, b = b = 1 and thus a E b (mod c) and u m a E b (mod d-c). However, by Lemma 6.3, these two congruences cannot hold simultaneously. This contradiction proves the lemma. Lemma 6.1 and Lemma 6.2 show that if s ^ r and s^ 1-r, then Equation (6.3) does not hold. Therefore, Theorem 3.7 has been proved. Q.E.D. As BQ [X] ^ B1[x], it follows that "s and r are binomially related" implies "s and r are Bj[x]-related." We will have the following theorem. Theorem 3.8. - Let n be a positive integer. The follow- ing two statements hold. (a) For n>_2, there are algebraic fractional numbers in the unit interval so that each of their equivalence classes under B -related has more than two elements. (b) For n>_3, there are algebraic fractional numbers in the unit interval such that each of their equivalence classes under B^-related has more than two elements. Proof: Using the fact that "s,r are binomially related" implies "s,r are B -related," and in Theorem 2.2 we find that the only thing we need to prove here is to find the algebraic fractional 0 of degree 2 so that the equivalence class of 0 under B^-related has at least three elements. 58 Look at f(x) = 4x(l-x) = 4x2(1-x) +4x(l-x)2. r = (5 + /5)/8 and s = (5-/5)/8. and f(x) 6 Let Since s = f(r) and r = f(s) [x], so the equivalence class of r has at least four elements, namely r, s, 1-r, 1-s. And it is obvious that the degree of r is 2. Note. Q.E.D. Let A be the set of all polynomials f(x) of Z[x] so that there are integers a^ra^,...,a^ with (aQ = 0 or 1) and (an = 0 or 1) , and 0 _< a^ £ (?)+l for i=2,3,...,n-l. f (x) = a.x11 + a. x n ~ 1 (1-x) + . . . + a U I And . x (1-x) n ~ 1 + a (1-x) n . n-1 n Let B be one smallest subset containing A so that the compositions of two functions B are closed in B. All the theorems true for B^-related relation are true for B-related relation in this section. §7. Binomial Numbers and Binomial Coefficients Binomial coefficients are the combinations of n things, k at a time. The Pascal triangle identity (7.1), the binomial theorem (7.2), and many other combinatorial identities have to do with the magic binomial coefficients. (j) - I";1)+ <;:}> (a + b)n = ? (")a k b n_k . n.u (7.2) k=0 Borel measures on the Cantor set are naturally connected with binomial coefficients through taking the measures on the basic open sets. 59 In this section, we are going to expose an interesting formula related to binomial numbers and binomial coefficients. Let r be a binomial number in the unit interval and s is binomially related to r. So there exists f,gf BQ[x] such that s = f(r) and r = g(s). Hence r = g(f(r)). According to Theorem 3.1, g,f€B^[x]; that is, there exist integers n,aQ,a1,...,an such that n > 1, 0 < a £ (") for i=0,l,...,n; and r = aQrn + (1-r) + . . . + a R (l-r)n. (7.3) Expanding the right hand side by using the binomial theorem, we have r = (ajj-ajt.. , + (-l) n a n )r n + (a1-(jj) a 2 + (J) a3-... + (-1) n+1 (") a^ + ••• + + --- (a + k" ( k k 1 ) a k+l + --- + ( - 1 ) n + k < k ) a „ ) r k ( a n - l " C l ) a n ) r + <7"4» O V Define h(x) as follows: h(x) = ( Z (-l) k a 1 )x I1 +( E (-l) k+1 (J) a. )x n_1 + . . . k k=0 + 1 k=l k (ky-l)k+£(k)ak)x^... + (( Z k=n-1 {?>5) (-l) k + n _ 1 ( k )a )-l)x + ( n )a . n-1 k n n Now let irr(r) = b^ + b, x + . . . + b x m . 0 1 m As h(r) = 0 and b_+b. r + . . . + b r m = 0, it follows that 0 1 irr(r) is a factor of h(x). m Therefore there exist integers d ,d1,...,dn (dn^0) and t(x)€z[x], so that 60 h(x) = t(x) (irr (r)) = d Q + c^x + ... + d n x n , (7.6) Comparing (7.5) with (7.6), we have (7.7). d n = a d n-l 0-al+a2+---+(-1)nan = U 1 " (l)a2+ (l)a3" + <-l>n+1<?>an> (7.7) j t d = a k" 1 ,k+l. , k )ak+l + ••• + , , -i v n+k ,ni k n (_1) n d.1 + 1 = a n-1 - (n-1 )a n d. = a . 0 n In other words, n n-1 (7.7) = A d +1 n where (-1)n 1 - 1 1 0 1 (-l) n+1 (?) (\) A = 0 0 1 0 -Ci' 61 The determinant of A is 1 and the inverse matrix of A exists, say B. Using the orthogonal relation (7.8) (refer to [8]), we obtain (7.9). 6 = (7.8) E ( - l ) ^ + k (?) (J) 3 k j =k where 6 is the kronecker delta. mk 1 1 1 1 0 0 1 0 2 1 3 3 0 0 0 1 (7.9) B = ( nJ n-1 1 (*) . 1 1 1 Hence • • • © n (7.10) = B dl + 1 a n — According to3 (7.7) and i (7.10), we have the following: m ( k ) d , dm = m k k=m k+m(k)a m k where d, = d. for k # l , d = d +1. k k 1 i By using (7.10), we will be able to use the computer to help us to study a specific binomial number with its given irreducible polynomial. 62 In the end of Chapter III, I would like to mention two unsolved problems which were thought to be important by many mathematicians; for instance, Problem 1 by Stan Ulam and Dan Mauldin, Problem 2 by R.D. Mauldin. Problem 1. - Let r,s be any two elements in the unit interval. If r is binomially related to s, is it true that u is topologically equivalent to us ? r Problem 2. - Are there any r£ [0,1] so that the equivalent class of r under binomially related relation (or y r under topologically equivalent relation) has infinitely many elements? Of course, these are not the only two unsolved problems left in this article. If you read this article carefully, you will find many other interesting unsolved problems. CHAPTER BIBLIOGRAPHY 1. Tate, J., Fourier Analysis in Number Fields and Hecke's zeta Function, Thesis, Princeton, 1950. 2. Stolarsky, K.B., Algebraic Numbers and Diophantine Approximation. New York, Marcel Dekker, Inc., 1974. 3. Salem, Raphael, Algebraic Numbers and Fourier Analysis, Wadsworth Mathematics Series, 1963. 4. Meyer, Yves, Algebraic Numbers and Harmonic Analysis, Amsterdam, North-Holland Publishing Co., 1972. 5. Hungerford, T.W., Algebra, New York, Holt, Rinehart and Winston, Inc., 1973. 6. 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