ABSOLUTELY ABNORMAL NUMBERS GREG MARTIN arxiv:math/0006089v2 [math.nt] 7 Nov 2000. Introduction A normal number is one whose decimal exansion or exansion to some base other than 0 contains all ossible finite configurations of digits with roughly their exected frequencies. More formally, when b 2 is an integer, let Nα; b, a, x = #{ n x: the nth digit in the base-b exansion of α is a} denote the counting function of the occurrences of the digit a 0 a < b in the b-ary exansion of the real number α, and define the corresonding limiting frequency δα; b, a = lim x x Nα; b, a, x, 2 if the limit exists. The number α is simly normal to the base b if the limit defining δα; b, a exists and equals /b for each 0 a < b. When α is a b-adic fraction a/b n, which has one b-ary exansion with all but finitely many digits equaling zero and another b-ary exansion with all but finitely many digits equaling b, these limiting frequencies are not uniquely defined; however, such an α will not be simly normal to the base b in either case. A number is normal to the base b if it is simly normal to each of the bases b, b 2, b 3,.... This is equivalent see [6, Chater 8] to demanding that for any finite string b b 2...b k of base-b digits, the limiting frequency of occurrences of this string in the b-ary exansion of α defined analogously to equation 2 above exists and equals /b k. For instance, it was shown by Chamernowne [] that the number 0.23456789023... formed by concatenating all of the ositive integers together into a single decimal is normal to base 0 the analogous construction works for any base b 2, and this sort of examle has been generalized [2, 3]. It is known that almost all real numbers are normal to any given base b see for instance [6, Theorem 8.], and consequently almost all real numbers are absolutely normal, i.e., normal to all bases b 2 simultaneously. On the other hand, we have not roven a single naturally occurring real number to be absolutely normal. Let us call a number abnormal to the base b if it is not normal to the base b, and absolutely abnormal if it is abnormal to all bases b 2 simultaneously. For instance, every rational number r is absolutely abnormal: any b-ary exansion of r will eventually reeat, say with eriod k, in which case r is about as far from being simly normal to the base b k as it can be. Even though the set of absolutely abnormal numbers is the intersection of countably many sets of measure zero, it was ointed out by Maxfield [5] that the set of numbers normal to a given base b is uncountable and dense; later, Schmidt [7] gave a comlicated constructive roof of this fact. In this aer we exhibit a simle construction of a secific irrational in fact, transcendental real number that is absolutely abnormal: Theorem. The number α defined in equation 8 below is irrational and absolutely abnormal. 99 Mathematics Subject Classification. K6 A63.
2 GREG MARTIN In fact, our construction easily generalizes to roduce concretely an uncountable set of absolutely abnormal numbers in any oen interval. It is instructive to consider why constructing an irrational, absolutely abnormal number is even difficult. Since we already know that rational numbers are absolutely abnormal, our first thought might be to choose an irrational number whose b-ary exansions mimic those of rational numbers for long stretches, i.e., an irrational number with very good rational aroximations. Thus a natural class to consider is the Liouville numbers, defined to be those real numbers β such that for every ositive integer m, there exists a rational number q not necessarily in lowest terms satisfying 0 < β q < qm. 3 These Liouville numbers are all transcendcental see Lemma 6 below in fact Liouville introduced these numbers recisely to exhibit secific transcendental numbers, and the offcited examle β = 0 n! = 0.0000000000000000000000... n= is usually the first number that students see roven transcendental. Clearly β is abnormal to the base 0; how would we go about showing, for examle, that β is abnormal to the base 2? We would try to argue that the binary exansion of β agrees with that of each of the rational numbers k β k = 0 n! 4 n= through about the n +! log 2 0-th binary digit. Since each β k is rational and thus abnormal to the base 2, can we conclude that β itself is abnormal to the base 2? Not quite: it seems that we would have to show that there is a fixed ower 2 n such that infinitely many of the β k were not simly normal to the base 2 n. For each β k there is some ower 2 n k such that βk is not simly normal to the base 2 n k, but these exonents nk might very well grow with k. In fact, it is not hard to show using the fact that 2 is a rimitive root modulo every ower of 5 that any 0-adic fraction that is not a 2-adic fraction including each β k is simly normal to the base 2! In general, without actually comuting binary exansions of secific fractions, it seems imossible to rule out the incredible ossibility that the β k are accidentally simly normal to bases that are high owers of 2. In summary, any Liouville number we write down is almost certain morally to be absolutely abnormal, but actually roving its absolute abnormality is another matter. To circumvent this difficulty, we construct a Liouville number whose successive rational aroximations are b-adic fractions with b varying, rather than all being 0-adic fractions as in equation 4. The existence of such Liouville numbers can certainly be roven using just the fact that the b-adic fractions are dense for any integer b 2; however, our construction is comletely exlicit. We first give the comlete construction of our irrational, absolutely abnormal number α and then show afterwards that α has the required roerties.
ABSOLUTELY ABNORMAL NUMBERS 3 2. The construction and roof We begin by defining a sequence of integers with the recursive rule d 2 = 2 2, d 3 = 3 2, d 4 = 4 3, d 5 = 5 6, d 6 = 6 30,57,578,25,... This sequence exhibits the attern = j /j j 3. 5 d 4 = 4 32, d 5 = 5 432, d6 = 6 5432,... which in general gives the tyesetting nightmare = j j j 2 j 3... 432... Using these integers, we define the sequence of rational numbers α k = so that α 2 =, α 4 3 = 2, α 3 4 = 2, α 32 5 = 00,35,803,222, and so on. 52,587,890,625 We now nominate α = lim α k = k k j=2. 6, 7 as our candidate for an irrational, absolutely abnormal number. The first few digits in the decimal exansion of α are j=2 α = 0.65624999999569999999... }{{ 99999} 852840420690728..., 9 23,747,29,559 9s from which we can get an inkling of the extreme abnormality of α at least to the base 0. We need to rove three things concerning this number α: first, that the infinite roduct 8 defining α actually converges; second, that α is irrational; and finally, that α is absolutely abnormal. It is aarent from the exressions 5 and 6 that the grow ridiculously raidly and hence that the infinite roduct 8 should indeed converge. The following lemma rovides a crude inequality relating the integers that we can use to rove this assertion rigorously. Lemma. For j 5 we have > 2d 2 j. Proof: We roceed by induction, the case j = 5 being true by insection. For j > 5 we surely have Notice that from the definition 5 of the, = j /j > 5 /j. dj 2 j = j j 2 > j 2 2j 2 =, 8
4 GREG MARTIN and therefore > 5 dj. Now using the fact easily roven by your favorite calculus student that 5 x x 5 for x 5, we conclude that 5 > > 2d 2 j, as desired. Equied with this inequality, we can now show that the infinite roduct 8 defining α converges. Moreover, we can show that the number α is well aroximated by the rational numbers α k. Notice that α 4 is exactly 0.65625 and α 5 is exactly 0.6562499999956992 cf. the decimal exansion 9 of α. Lemma 2. The roduct 8 defining α converges. Moreover, for k 2 we have α k > α > α k 2 d k+. 0 Proof: To show that the roduct 8 converges, we must show that the corresonding sum j=2 / converges. But by Lemma we certainly have > 2 for j 5, and therefore j=2 d 2 + d 3 + j=4 2 j 4 d 4 = d 2 + d 3 + 2 d 4 <. Similarly, using the fact that x j x j for any real numbers 0 x j, we see that for k 3 α k > α = α k αk j=k+ > α k j=k+ j=k+ 2 j k d k+ = α k 2 d k+ > αk 2 d k+. The inequalities 0 for k = 2 follow from those for k = 3, as it is easily verified by hand that α 2 > α 3 > α 3 2 d 4 > α 2 2 d 3. It turns out that both the roof that α is irrational and the roof that α is absolutely abnormal hinge on the fact that each rational aroximation α k is in fact a k-adic fraction that is, when α k is exressed in lowest terms, its denominator divides a ower of k. In other to obtain α k, the numerator of the latter fraction comletely cancels out the denominator of α k, so that all that can remain in the denominator of α k are the owers of k resent in d k. Proving that this always haens is an exercise in elementary number theory, which we resent in the next three lemmas. words, each time we multily α k by d k = d k d k Lemma 3. Let k and r be ositive integers, and let be a rime. If k is divisible by r, then k + is divisible by r+.
ABSOLUTELY ABNORMAL NUMBERS 5 Proof: Writing k = r n, we have from the binomial theorem k + = r n + = { r n + r n + + 2 r n 2 + r n + } = r n + r n + + 2 2r n 2 + r n. Since all of these binomial coefficients k are integers, each term in this last sum is visibly divisible by r+. Lemma 4. For any ositive integers k and m, the integer k + km is divisible by k m+. Proof: If r is any rime ower dividing k, an rm-fold alication of Lemma 3 shows us that k + rm is divisible by r+rm. Then, since k + km = k + rm k + km rm + k + km 2 rm + + k + rm +, we see that k + km is also divisible by rm+. In articular, since r was an arbitrary rime ower dividing k, we see that k+ km is divisible by every rime ower that divides k m+. This is enough to verify that k + km is divisible by k m+ itself. Lemma 5. For each k 2, the roduct d k α k is an integer. In articular, since d k is a ower of k, we see that α k is a k-adic fraction. Proof: We roceed by induction on k, the cases k = 2 and k = 3 being evident by insection. For the inductive ste, suose as our induction hyothesis that d k α k is indeed an integer for a given k 3. We may write d k+ α k+ = d k+ α k = k + d k/k α k = k + d k/k d k d k α k by the definitions 5 and 7 of d k and α k, resectively. The second factor d k α k is an integer by the induction hyothesis. On the other hand, we may rewrite k + d k/k = k + kd k /k. Alying Lemma 4 with m = d k /k, we see that this exression is divisible by k d k /k = d k. Therefore the fraction on the right-hand side of equation is in fact an integer, and so d k+ α k+ is itself an integer, which comletes the roof. As mentioned in the introduction, the key to roving that α is irrational is to show that it is in fact a Liouville number. It is a standard fact that any Liouville number is transcendental see for instance [6, Theorem 7.9]; for the sake of keeing this aer self-contained, we include a roof. Lemma 6. Every Liouville number is transcendental. Proof: We shall rove the contraositive, that no algebraic number can satisfy the Liouville roerty 3 for all ositive m. Suose that β is algebraic. Without loss of generality, we may suose that β by adding an aroriate integer. Let 2 m β x = c d x d + c d x d + + c 2 x 2 + c x + c 0
6 GREG MARTIN be the minimal olynomial for β, where the coefficients c i are integers. Now suose that q is a rational aroximation to β, say β <. Then q 2 m β = mβ mβ β q q d = c d β d 2 + + c 2 β 2 + c q q q β = q β d c d + q q d 2β + + β d + + c 2 q + β + c. Since neither β nor exceeds in absolute value, we see that q m β q q β Cβ, 2 where we have defined the constant Cβ = d c d + d c d + + 2 c 2 + c. On the other hand, m β q is a rational number with denominator at most q d, and it is nonzero since m β is irreducible. Therefore m β q qd. 3 Together, the inequalities 2 and 3 imly that β q Cβ, q d which recludes the inequality 3 from holding when m is large enough. To show that α is indeed a Liouville number, we will need an inequality somewhat stronger than the one given in Lemma. The following lemma furnishes a simle inequality that is strong enough for this urose. Lemma 7. For j 5 we have + > d j. Proof: It is immediate that + = j + dj/j > j dj/j > j 2d2 j /j = j dj /j dj 2j /j = d 2j /j j where we have used Lemma for the second inequality. > d j, Lemma 8. α is a Liouville number; in articular, α is transcendental. Proof: We can easily show show that the α k rovide the very close rational aroximations needed in equation 3 to make α a Liouville number. Indeed, α k can be written as a fraction whose denominator is d k by Lemma 5, while Lemma 2 tells us that for k 5 0 < α α k < 2 < 2, d k+ d d k k where the last inequality is by Lemma 7. Since d k tends to infinity with k, this shows that α is a Liouville number and hence transcendental by Lemma 6.
ABSOLUTELY ABNORMAL NUMBERS 7 At last we have all the tools we need to establish the theorem stated in the introduction: Theorem. The number α defined in equation 8 is irrational and absolutely abnormal. Proof: We have just shown in Lemma 8 that α is irrational. As for roving that α is absolutely abnormal, the idea is that for every integer base b 2, the number α is just a tiny bit less than the b-adic fraction α b. Since the b-ary exansion of α b terminates in an infinite string of zeros, the slightly smaller number α will have a long string of digits equal to b before resuming a more random behavior. This is evident in the decimal exansion 9 of α, as α 5 is a 0-adic fraction as well as a 5-adic fraction. This haens more than once, as each of α b, α b 2, α b 3, and so on is a b-adic fraction. Consequently, the b-ary exansion of α will have increasingly long strings consisting solely of the digit b, which will revent it from being even simly normal to the base b. More quantitatively, let b 2 and r be ositive integers. Since d b rα b r is an integer by Lemma 5, and since d b r = b r d b r /b r by definition, the b-ary exansion of α b r terminates after at most rd b r /b r nonzero digits. On the other hand, by Lemma 2 we know that α is less than α b r but by no more than 2/d b r + = 2/b r + d b r/br < 2/b rd b r /br. Therefore, when we subtract this small difference from α b r, the resulting b-ary exansion will have occurrences of the digit b beginning at the rd b r /b r + -th digit at the latest, and continuing through at least the rd b r/b r -th digit since the difference will start to show only in the rd b r/b r -th digit at the soonest. Using the notation defined in equation, this imlies that N α; b, b, rd b r rd br b r b r At this oint we can calculate that lim su x Using Lemma, we see that lim su x rdbr rdbr > b r b r x Nα; b, b, x lim su b r r rd b r x Nα; b, b, x lim su r lim su r 2d b r 2rd br b r. N α; b, b, rd b r b r 2d b r. 2d 2 b r d b r = lim su =. r d b r In articular, the frequency δα; b, b defined in equation 2 either does not exist or else equals, either of which recludes α from being simly normal to the base b. Since b 2 was arbitrary, this shows that α is absolutely abnormal. 3. Generalizations and further questions We mentioned in the introduction that our construction of an irrational, absolutely abnormal number can be generalized to exhibit an uncountable set of absolutely abnormal numbers in any oen interval, and we now describe that extension. Clearly we may limit our attention to subintervals of [0, ], since the set of normal numbers to any base is invariant under translation by an integer. In the original construction at the beginning of Section 2, we began with d 2 = 2 2 and α 2 = 3 4 ; to be more general, let α 2 be any 2-adic fraction a d 2, where d 2 = 2 n 2 for some ositive integer n 2. Next fix any sequence n 3, n 4,... of ositive
8 GREG MARTIN integers and modify the recursive definition 5 of the to If we now set = j n j /j α k = α 2 k j=3 j 3. 4 where of course the numbers α k now deend on α 2 and the n j, then the resence of the integers n j will not hinder the roof of Lemma 5 that d k α k is always a k-adic fraction this is a consequence of the fact that x always divides x n. Therefore the new limit α = lim k α k can be shown to be a transcendental, absolutely abnormal number in exactly the same way, the modifications only accelerating the convergence of the infinite roduct and enhancing the ease with which the various inequalities in the Lemmas are satisfied. When we make this modification, then the one case we must avoid is α 2 = and n 2 2 = n 3 = =, for in this case it will haen that = j for every j 2. Then each roduct α k is a telescoing roduct with value, and their limit α = 0, while certainly absolutely abnormal, will be k uninterestingly so. In articular, Lemma 2 alied with k = 2 would show in this context that a > α > a 2; 2 n 2 2 n 2 thus by choosing α 2 = a aroriately, we can ensure that the resulting number α lies in 2 n 2 any rescribed oen subinterval of [0, ]. Moreover, the various choices of the integers n 3, n 4,... will give rise to distinct limits α; one can show this by considering the first index j 3 at which the choices of n j differ, say, and then alying Lemma 2 with k = j to each resulting α. This generalization thus ermits us to construct uncountably many transcendental, absolutely abnormal numbers in any rescribed oen interval. One interesting secial case of this generalized construction arises from the choices α 2 = 2 and n j = φj for all j 3, where φ is the Euler totient function; these choices give the simle recursive rule = j φdj for j 3. In this secial case, the crucial roerty that d k α k is always an integer is in fact a direct consequence of Euler s theorem that a φn is always congruent to modulo n as long as a and n have no common factors. In general, the smallest exonent e k we can take in the recursive rule d k = k e k so that this crucial roerty is satisfied is the multilicative order of k modulo n, which might be smaller than d k /k ; however, our construction given in Section 2 has the advantage of being more exlicit, as it is not necessary to wait and see the exact value of d k before knowing how we will construct d k. We remark that Schmidt s construction [7] mentioned in the introduction actually gives the following more owerful result: given any set S of integers exceeding with the roerty that an integer b is in S if and only if every erfect ower of b is in S, Schmidt constructed real numbers that are normal to every base b S and abnormal to every base b / S. The roblem considered herein is the secial case where S =. It would be interesting to see if the construction in this aer could be modified to roduce these selectively normal numbers as well. We conclude with a few remarks about absolutely simly abnormal numbers, numbers that are simly normal to no base whatsoever. As we saw in the roof of our Theorem, the number α does in fact meet this stronger criterion of abnormality. On the other hand, while all rational numbers are absolutely abnormal, many of them are in fact simly normal to various bases. For examle, /3 is simly normal to the base 2, as its binary reresentation is
ABSOLUTELY ABNORMAL NUMBERS 9 0.000.... In fact, one can check that every fraction in reduced form whose denominator is 3, 5, 6, 9, 0,, 2, 3, 7, 8, 9, 20,... is simly normal to the base 2 resumably there are infinitely many such odd denominators can it be roven?; every fraction whose denominator is 7, 4, 9, 2, 3,... is simly normal to the base 3; and so on. Somewhat generally, if is a rime such that one of the divisors b of is a rimitive root for, then every fraction whose denominator is is simly normal to the base b although this is not a necessary condition, as the normality of fractions with denominator 7 to base 2 shows. For a fraction with denominator q to be simly normal to the base b we can assume, by multilying by b a few times if necessary, that b and q are relatively rime, it is necessary for b to divide the multilicative order of b modulo q, and hence b must certainly divide φq by Euler s theorem. Therefore, we immediately see that every fraction whose denominator is a ower of 2 is absolutely simly abnormal. One can also verify by this criterion that every fraction whose denominator in reduced form is 5 or 28, for examle, is absolutely simly abnormal. It seems to be a nontrivial roblem to classify, in general, which rational numbers are absolutely simly abnormal. Since some fractions with denominator 63 are simly normal to the base 2 while others are not, as one can check, absolute simle abnormality robably deends in general on the numerator as well as the denominator of the fraction. Acknowledgements: Glyn Harman gave a survey talk on normal numbers at the Millennial Conference on Number Theory at the University of Illinois in May 2000 see [4], at the end of which Andrew Granville asked the question about a secific absolutely abnormal number that surred this aer. Carl Pomerance suggested the number n= n! n!, a Liouville number which again is almost surely absolutely abnormal, but a roof of this seems hoeless. The author would like to thank all three for their interest in this construction, and also an anonymous referee for several helful comments on a reliminary version of this aer. References [] D.G. Chamernowne, The construction of decimals normal in the scale of ten, J. London Math. Soc. 8 933, 254 260. [2] A.H. Coeland and P. Erdős, Note on normal numbers, Bull. Amer. Math. Soc. 52 946, 857 860. [3] H. Davenort and P. Erdős, Note on normal decimals, Canadian J. Math. 4 952, 58 63. [4] G. Harman, 00 years of normal numbers, Proceedings of the Millennial Conference on Number Theory Urbana, IL, A K Peters, to aear. [5] J.E. Maxfield, Normal k-tules, Pacific J. Math. 3 953, 89 96. [6] I. Niven, Irrational numbers, The Mathematical Association of America. Distributed by John Wiley and Sons, Inc., New York, N.Y., 956, The Carus Mathematical Monograhs, No.. [7] W.M. Schmidt, Über die Normalität von Zahlen zu verschiedenen Basen, Acta Arith. 7 96/962, 299 309. Deartment of Mathematics, University of Toronto, Canada M5S 3G3 E-mail address: gerg@math.toronto.edu