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2 Random Rational Approximations

The one-to-one correspondence between the factional parts of the sequence xn = n log 10 + log π and y = π10n, where n 1, via the map x y = exn clues to n ≥ n −→ n the uniform distribution of both sequences. To explicate the relationship between all the sequences and fractional parts consider the sets X = x = n log 10 + log π : n 1 , (5) { n ≥ } Y = y = π10n : n 1 , { n ≥ } U = x : n 1 , {{ n} ≥ } V = y : n 1 , {{ n} ≥ } Uniform Distribution And Normal Numbers 3 and the correspondence diagram.

exn XY ρ ρ

U V e{xn}

The class function ρ : R [0, 1], refer to the above diagram, assigns the frac- −→ tional parts ρ(x ) = x and ρ(y ) = y of the corresponding sequences x = n { n} n { n} n n log 10 + log π and y = exn = π10n, respectively, where n 1. n ≥

The equidistribution of the sequence xn is established in the Lemma below. Lemma 2.1. The sequence of x = n log 10 + log π, where n 1, is n ≥ uniformly distributed modulo 1.

Proof. A routine calculation based on the Weil criterion, see [12, Theorem 2.1]. 

xn n The conditional proof of the equidistribution of the sequence yn = e = π10 is significantly longer. Some of the required foundational results are established here.

The result below shows that for each n 1, the fractional part π10n is ≥ { } contained in some random subinterval of the form

rn 1 sn cn + µ−1 , + 2 [0, 1], (6) qk qk qk qk  ⊆ where 0 r ,s ,c q . In this application, the lower bound µ(π) 2 of the ≤ n n n ≤ k ≥ irrationality measure of the real number π is sufficient, see [16, p. 556] for the definition of this quantity.

Lemma 2.2. If pk/qk is the kth convergent of the real number π, then, for each n n n qk, the fractional part π10 of the real number π10 satisfies one or both of the≤ following inequalities. { }

n rn 10 n rn +1 n (i) + 2 π10 , if 10 qk, qk 2qk ≤{ } ≤ qk ≤

rn 1 n sn cn n (ii) + µ−1 π10 + 2 , if 10 > qk, qk qk ≤{ } ≤ qk qk where µ = µ(π) 2 is the irrationality measure of π, and 0 c ,r ,s < q . ≥ ≤ n n n k Proof. The verification is split into two cases, depending on the magnitute of the integer 10n. Uniform Distribution And Normal Numbers 4

Case I. Observe that for all sufficiently large q , and any integer 10n q , there is k ≤ k the rational approximation inequality

1 pk 1 1 2 π 2 n , (7) 2qk ≤ − qk ≤ qk ≤ 10 qk for any even index k 1, see [11, Theorem 163]. This leads to the new inequality ≥ n n 10 n 10 pk 1 2 π10 . (8) 2qk ≤ − qk ≤ qk n Now, replace 10 pk = anqk + rn to get the equivalent expression n 10 n rn 1 2 π10 an . (9) 2qk ≤ − − qk ≤ qk Clearly, this implies that the fraction part satisfies n rn 10 n rn +1 + 2 π10 , (10) qk 2qk ≤{ } ≤ qk .

n Case II. For all sufficiently large qk, and any integer 10 > qk, there is the rational approximation inequality 1 pk 1 µ π 2 , (11) qk ≤ − qk ≤ qk for any even index k 1, see [11, Theorem 163]. Here, the quantity µ = µ(π) 2 ≥ ≥ is the irrationality measure of π, see [16, p. 556]. This leads to the new inequality 1 p 1 π k . (12) n µ−1 2 10 qk ≤ − qk ≤ qk Equivalently, this is n n 1 n 10 pk 10 µ−1 π10 2 . (13) qk ≤ − qk ≤ qk Now, rearrange it as

n n pk10 1 n (pkqk + 1)10 + µ−1 π10 2 , (14) qk qk ≤ ≤ qk

n n 2 and replace pk10 = anqk + rn and (pkqk + 1)10 = bnqk + snqk + cn to get the equivalent expression

rn 1 n sn cn an + + µ−1 π10 bn + + 2 , (15) qk qk ≤ ≤ qk qk where 0 a , b ,c ,r ,s < q . This implies that the fraction part satisfies ≤ n n n n n k rn 1 n sn cn + µ−1 π10 + 2 , (16) qk qk ≤{ } ≤ qk qk as claimed.  Uniform Distribution And Normal Numbers 5

Lemma 2.3. Let pk/qk be the kth convergent of the real number π, and let q = n qk + o(qk) be a large prime. Then, for each n q, the fractional part π10 of the real number π10n satisfies one or both of the≤ following inequalities for{ infini}tely many primes q. r 1 r +1 1 (i) n + O π10n n + O , if 10n q , 2q q2  ≤{ } ≤ q q2  ≤ k r 1 1 s 1 (ii) n + + O π10n n + O , if 10n > q , 2q (2q)µ−1 q2  ≤{ } ≤ q q2  k where µ = µ(π) 2 is the irrationality measure of π, and 0 c ,r ,s < q. ≥ ≤ n n n Proof. For a large integer q 2, and a large prime q = q + o(q ), the inequalities k ≥ k k 1 1 1 1 1 + O 2 + O 2 (17) 2q q  ≤ qk ≤ q q  are valid.

Case I. Replacing (17) into (10) yields r 1 r +1 1 n + O π10n n + O (18) 2q q2  ≤{ } ≤ q q2  for any even index k 1, and all sufficiently large q = q + o(q ). ≥ k k Case II. Replacing (17) into (16) yields r 1 1 s 1 n + + O π10n n + O , (19) 2q (2q)µ−1 q2  ≤{ } ≤ q q2  for any even index k 1, see [11, Theorem 163], and the irrationality measure µ = µ(π) 2 of π, see≥ [16, p. 556] for additional information.  ≥ 3 Multiplicative Subgroups And Exponential Sums

The cardinality of a subset of integers H Z is denoted by #H 0. The multiplicative order of an element r = 0 in⊂ a finite ring (Z/mZ)× is≥ defined by ord r = min n 1 : rn 1 mod m ,6 where m 1 is an integer. The multiplica- m { ≥ ≡ } ≥ tive order is a divisor N ϕ(m) of the totient function ϕ(m) = (1 1/p), see | p|m − [1, Theorem 2.4]. Q Lemma 3.1. Let p /q : k 1 be the sequence of convergents of the real number { k k ≥ } π. Assume that gcd(10, qk)=1. Then,

n (i) G = pk10 rn mod qk : n 1 is a large multiplicative subgroup of the { ≡ × ≥ } finite ring (Z/qkZ) .

n (ii) H = (pkqk + 1)10 sn mod qk : n 1 is a large multiplicative subgroup of { ≡× ≥ } the finite ring (Z/qkZ) . Uniform Distribution And Normal Numbers 6

Proof. (ii) The hypothesis gcd(10, qk) = 1 implies that the integer 10 generates a multiplicative subgroup

= 100, 101, 102,..., 10n,... (Z/q Z)× (20) H { } ⊆ k of cardinality N = # . Moreover, since gcd(p q +1, q ) = 1, the map H k k k n (p q +1)10n s mod q (21) −→ k k ≡ n k × is 1 to 1 on the finite ring (Z/qkZ) , it permutes the finite ring Z/qkZ. Therefore, the− two− subsets are equal, that is, H = .  H Trivially, N = # > log qk, but it requires considerable more works to show that ε H N > qk, where ε> 0 is a small number. The ideal case has a large prime q = qk 2, F× ≥ and the integer 10 generates the maximal multiplicative group q of cardinality N = ϕ(q).

Lemma 3.2. Assume the Artin primitive root conjecture is valid. Let p /q : k { k k ≥ 1 be the sequence of convergents of the real number π, and let q qk be a large prime.} Then, ∼

n (i) = pk10 rn mod q : n 0 is the multiplicative group of the finite field GF× { ≡ ≥ } q ,

n (ii) = (pkqk +1)10 sn mod q : n 0 is the multiplicative group of the H { F× ≡ ≥ } finite field q , for infinitely many large primes q = qk + o(qk) such that gcd(pkqk +1, q)=1, as q . k →∞ Proof. (ii) The conditional proof of the Artin primitive root conjecture, states that F× 10 generates the multiplicative group of the finite field q of a subset of primes = prime q 2 : ord 10 = q 1 of density δ( )=0.3739558 ..., see [10, p. Q { ≥ q − } Q 220]. In particular, the interval [qk, qk + O(qk/log qk)] contains approximately

π(q + O(q /log q )) π(q ) q /(log q )2, (22) k k k − k ≫ k k large primes q = q + o(q ) such that < 10 > = F×, as , confer [9, p. 113] or k k q →∞ similar reference. Now, proceed to use the same analysis as in the previous Lemma to complete the proof. 

Theorem 3.1. ([4, Theorem 1]) Let F be a multiplicative subgroup of order H ⊆ p # >pc0/log log p for some sufficiently large constant c > 1. Then H 0 c max ei2πax/p < e−(log p) # , (23) gcd(a,p)=1 H xX∈H where c> 0 is an absolute constant. Uniform Distribution And Normal Numbers 7

Lemma 3.3. Let p /q : k 1 be the sequence of convergents of the real number { k k ≥ } π, and let q = qk + o(qk) be a large integer. If the fractional part has an effective rational approximation

s 1 π10n n , (24) { } − q ≪ q2

where 0 sn < qk, then ≤ s n i2π n 1 ei2π{π10 } e q . (25) − ≪ q2

Proof. Basically, this follows from the Lipschitz property

f(x) f(y) x y (26) | − |≪| − | of the continuous function f(x)= eix of the real variable 0 x < 1. Specifically, ≤| |

n i2π sn D = ei2π{π10 } e q (27) − i2π sn i2π {π10n}− sn = e q e ( q ) 1  −  s s = cos2π π10n n 1+ i sin 2π π10n n { } − q  − { } − q 

s s cos2π π10n n 1 + sin 2π π10n n ≤ { } − q  − { } − q 

s sin 2π π10n n ≤ { } − q 

s π10n n ≤ { } − q

1 , ≪ q2 since cos z =1+ O(z2) and sin z = z + O(z3) for 0 z < 1.  ≤| | 4 Conditional Proof For The Normality Of Pi

The proof is based on the foundational results in Section 2, Section 3, and the Wall criterion stated below. Theorem 4.1. (Wall) An irrational number α R is a normal number in base 10 ∈ if and only if the sequence α10n : n 1 is uniformly distributed modulo 1. { ≥ } The proof of this criterion appears in [13, Theorem 8.15]. Proof. (Theorem 1.1) By Lemma 2.3, almost every fractional part has the random rational approximation r 1 1 s 1 n + + O π10n n + O , (28) 2q (2q)µ−1 q2  ≤{ } ≤ q q2  Uniform Distribution And Normal Numbers 8 where 0 c ,r ,s q , and µ 2. There are at most O(log q) exceptions, see ≤ n n n ≤ k ≥ Lemma 2.2. The nonsymmetric inequalities (28) are rewritten as

r 1 1 π10n n , (29) { } − 2q − (2q)µ−1 ≪ q2

and s 1 π10n n , (30) { } − q ≪ q2

Therefore, by Lemma 3.3, the corresponding exponentials pairing

rn 1 i2πm + − n e  2q (2q)µ 1  ei2πm{π10 }, (31) ≍ and n i2πm sn ei2πm{π10 } e q , (32) ≍ where m = 0, are proportionals. Similarly, the corresponding exponential sums 6 rn 1 i2πm + − n e  2q (2q)µ 1  ei2πm{π10 }, (33) ≍ Xn≤q Xn≤q and n i2πm sn ei2πm{π10 } e q , (34) ≍ Xn≤q Xn≤q where q = qk +o(qk) is a large prime such that gcd(pkqk +1, q) = 1, are proportionals.

By Lemma 3.2 the subsets

G = p 10n r mod q and H = (p q + 1)10n s mod q (35) { k ≡ n } { k k ≡ n } F× are sufficiently large multiplicative subgroups of the finite field q . In particu- lar, conditional on the Artin primitive root conjecture, the cardinalities are #G = q 1 qε, and #H = q 1 qε, where ε> 0. − ≫ − ≫ By Theorem 3.1, the exponential sums have nontrivial upper bounds

rn 1 i2πm + − n e  2q (2q)µ 1  ei2πm{π10 } q1−ε, (36) ≍ ≪ Xn≤q Xn≤q and n i2πm sn ei2πm{π10 } e q q1−ε. (37) ≍ ≪ Xn≤q Xn≤q By the Weil criterion, see [12, Theorem 2.1], any of the expressions (36) or (37) is sufficient to prove the uniform distribution of the sequence π10n : n 1 .  { ≥ } Uniform Distribution And Normal Numbers 9

Establishing the main result as an unconditional does not seem to be difficult, be- cause proving the existence of infinitely many large multiplicative subgroups

= 100, 101, 102,..., 10n,... (Z/q Z)× (38) H { } ⊆ k ε of cardinalities # q as qk , is not a difficult task. The techniques employed appear to be extendableH≫ to other→∞ irrational numbers α of finite irrationality measure µ(α) 2 and base b = 10. But, the generalization to other bases b = 10 seems to ≥ 6 require significant additional works.

5 References References

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[8] Copeland, A. H.; Erdos, P. Note on Normal Numbers. Bulletin of the American Mathematical Society, 52 (10): 857–860, 1946. doi:10.1090/S0002-9904-1946- 08657-7.

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[11] G. H. Hardy and E. M. Wright. An Introduction to the Theory of Numbers. 5th ed., Oxford University Press, Oxford, 1975. Uniform Distribution And Normal Numbers 10

[12] Kuipers, L.; Niederreiter, H. Uniform distribution of sequences. Pure and Applied Mathematics. Wiley-Interscience, New York-London-Sydney, 1974. MR419394.

[13] Niven, I. Irrational Numbers. Am. Math. Assotiation, Weley and Sons Inc., 1956.

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