On b-adic Representation of Irrational Numbers

Transcription

1 Welcome! On b-adic Representation of Irrational Numbers Lars Kristiansen Department of Mathematics, University of Oslo Department of Informatics, University of Oslo

2 Introduction

3 Irrational numbers are infinite objects... π = We cannot write them up. When we compute with irrationals, we need to represent them by some kind of program (procedure).

4 Irrational numbers are infinite objects... π = We cannot write them up. When we compute with irrationals, we need to represent them by some kind of program (procedure). Some possible representations: Cauchy sequences base-b expansions (b-adic representations) Dedekind cuts

5 We may represent π by (fast converging) Cauchy sequences 2.9, , , , ,...

6 We may represent π by (fast converging) Cauchy sequences 2.9, , , , ,... A function C : N Q is a Cauchy sequence for the real number α when α C(n) < 1 2 n.

7 We may represent π by a function f : N + {0, 1,... 9} that yields digits of the base-10 expansion of π one by one... f (1) = 1, f (2) = 4, f (3) = 1, f (4) = 5, f (5) = 9, or the digits of the expansion in any other base.

8 We may represent π by a function f : N + {0, 1,... 9} that yields digits of the base-10 expansion of π one by one... f (1) = 1, f (2) = 4, f (3) = 1, f (4) = 5, f (5) = 9, or the digits of the expansion in any other base. A function E : N Z is the base-b expansion of the real number α when E(n) {0, 1,..., b 1} for n 1 E(i) α = E(0) + b i i=1

9 We may represent π by its Dedekind cut D... D(3.14) = below D(3.15) = above... a predicate that tells if a rational numbers is smaller or greater that π.

10 We may represent π by its Dedekind cut D... D(3.14) = below D(3.15) = above... a predicate that tells if a rational numbers is smaller or greater that π. A function D : Q {0, 1} is a Dedekind cut of the real number α when D(q) = 0 iff q < α.

11 Does it matter how we represent the irrational numbers?

12 Does it matter how we represent the irrational numbers? If we have full Turing computability, it does not matter.

13 Does it matter how we represent the irrational numbers? If we have full Turing computability, it does not matter. Let S be a class of computable functions, let S C denote the set of irrationals that can be represented Cauchy sequences in S S be denote the set of irrationals that can be represented base-b expansions in S S D denote the set of irrationals that can be represented Dedekind cuts in S.

14 Does it matter how we represent the irrational numbers? If we have full Turing computability, it does not matter. Let S be a class of computable functions, let S C denote the set of irrationals that can be represented Cauchy sequences in S S be denote the set of irrationals that can be represented base-b expansions in S S D denote the set of irrationals that can be represented Dedekind cuts in S. When S is the class of all Turing computable functions, we have S D = S be = S C.

15 If we assume full Turing computability, we end up with the same class of computable irrational numbers no matter which (reasonable) representation we chose to work with (and there are many other ways to represent irrationals than those I have mentioned so far).

16 If we assume full Turing computability, we end up with the same class of computable irrational numbers no matter which (reasonable) representation we chose to work with (and there are many other ways to represent irrationals than those I have mentioned so far). If we do not assume full Turing computability, we are in a very very different situation.

17 Theorems of Specker and Mostowski.

18 Theorem [Specker 1949]. Let S is the class of primitive recursive functions. Then we have S D S 10E S C.

19 The proof of Specker s theorem.

20 The proof of Specker s theorem. The proof of S D S 10E. Idea:

21 The proof of Specker s theorem. The proof of S D S 10E. Idea: Assume you determined an initial part of the base-10 expansion of π, e.g

22 The proof of Specker s theorem. The proof of S D S 10E. Idea: Assume you determined an initial part of the base-10 expansion of π, e.g Use the Dedekind cut to search for the next digit.

23 The proof of Specker s theorem. The proof of S D S 10E. Idea: Assume you determined an initial part of the base-10 expansion of π, e.g Use the Dedekind cut to search for the next digit. The Dedekind cut will tell you that < π < Thus, the next digit is 5.

24 If we have access to the Dedekind cut of α, then we can primitive recursively compute the base-10 expansion of α. Thus we have S D S 10E.

25 The proof of S 10E S C.

26 The proof of S 10E S C. If we have access to the base-10 of α, then we can primitive recursively compute a Cauchy sequence for α.

27 The proof of S 10E S C. If we have access to the base-10 of α, then we can primitive recursively compute a Cauchy sequence for α. This is very easy to see. Just note that and that π = , 3.1, 3.14, 3.141, , , ,... is a Cauchy sequence for π.

28 The proof of S 10E S C. If we have access to the base-10 of α, then we can primitive recursively compute a Cauchy sequence for α. This is very easy to see. Just note that and that π = , 3.1, 3.14, 3.141, , , ,... is a Cauchy sequence for π. Thus we have S 10E S C.

29 Now we have established that S D S 10E S C. (We want to prove S D S 10E S C. )

30 Now we have established that S D S 10E S C. (We want to prove S D S 10E S C. ) It is obvious that S D is closed under multiplication with natural numbers.

31 Now we have established that S D S 10E S C. (We want to prove S D S 10E S C. ) It is obvious that S D is closed under multiplication with natural numbers. It is obvious that S C is closed under multiplication with natural numbers.

32 Now we have established that S D S 10E S C. (We want to prove S D S 10E S C. ) It is obvious that S D is closed under multiplication with natural numbers. It is obvious that S C is closed under multiplication with natural numbers. Specker proves that S 10E is not closed under multiplication with natural numbers.

33 Now we have established that S D S 10E S C. (We want to prove S D S 10E S C. ) It is obvious that S D is closed under multiplication with natural numbers. It is obvious that S C is closed under multiplication with natural numbers. Specker proves that S 10E is not closed under multiplication with natural numbers. Hence we have S D S 10E S C.

34 Specker constructs α such that α S 10E and 3 α S 10E.

35 Specker constructs α such that α S 10E and 3 α S 10E. α is of the form 0. 3} {{ }? 0 3} {{ }? 0 3} {{ }? 0... a lot of 3 s a lot of 3 s a lot of 3 s where? is 2 or 4.

36 Theorem [Mostowski 1957]. Let S be the class of primitive recursive functions. We have prim(a) prim(b) S be S ae. Notation: We use prim(x) denote the set of prime factors of the base x, that is, prim(x) = {p p is a prime and p x}.

37 Theorem [Mostowski 1957]. Let S be the class of primitive recursive functions. We have prim(a) prim(b) S be S ae. Notation: We use prim(x) denote the set of prime factors of the base x, that is, prim(x) = {p p is a prime and p x}. Open Problem [Mostowski 1957]. prim(a) prim(b) S be S ae.

38 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae.

39 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae. I will discuss the proof of this theorem...

40 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae. I will discuss the proof of this theorem... but first note that we have prim(2) prim(10) and prim(10) prim(2). Thus S 10E S 2E. We have more base-2 reals than base-10 reals!

41 The theorem also yields a nice proof of a slightly stronger version of Specker s Theorem.

42 The theorem also yields a nice proof of a slightly stronger version of Specker s Theorem. Corollary. For any subrecursive class S closed under Kalmar elementary operations and any base b, we have S D S be S C.

43 The proof of the corollary. Let b be any base.

44 The proof of the corollary. Let b be any base. Pick a such that prim(a) prim(b) and prim(b) prim(a).

45 The proof of the corollary. Let b be any base. Pick a such that prim(a) prim(b) and prim(b) prim(a). By the theorem we have S ae S be and S be S ae.

46 The proof of the corollary. Let b be any base. Pick a such that prim(a) prim(b) and prim(b) prim(a). By the theorem we have S ae S be and S be S ae. As explained above, it is straightforward to prove S D S ae S C and S D S be S C.

47 The proof of the corollary. Let b be any base. Pick a such that prim(a) prim(b) and prim(b) prim(a). By the theorem we have S ae S be and S be S ae. As explained above, it is straightforward to prove S D S ae S C and S D S be S C. Hence S D S be S C.

48 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae. Let us discuss the proof.

49 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae. Let us discuss the proof. First I will explain the proof of prim(a) prim(b) S be S ae. This is also proved by Mostowski, but I prove some intermediate results that some might find interesting. Then I will explain the proof of prim(a) prim(b) S be S ae.

50 The Base Transition Factor and the Base Transition Theorem

51 a base-b digit is a natural number in the set {0, 1,..., b 1}

52 a base-b digit is a natural number in the set {0, 1,..., b 1} let M be an integer and let D 1,..., D n be base-b digits, then (M.D 1 D 2... D n ) b = M + n i=1 D i b i

53 a base-b digit is a natural number in the set {0, 1,..., b 1} let M be an integer and let D 1,..., D n be base-b digits, then (M.D 1 D 2... D n ) b = M + n i=1 D i b i let M be an integer, and let D 1, D 2, D 3, D 4... be an infinite sequence of base-b digits, then (M.D 1 D 2...) b is the base-b expansion of the real number α if and only if (by definition)

54 a base-b digit is a natural number in the set {0, 1,..., b 1} let M be an integer and let D 1,..., D n be base-b digits, then (M.D 1 D 2... D n ) b = M + n i=1 D i b i let M be an integer, and let D 1, D 2, D 3, D 4... be an infinite sequence of base-b digits, then (M.D 1 D 2...) b is the base-b expansion of the real number α if and only if (by definition) (M.D 1 D 2... D n ) b α < (M.D 1 D 2... D n ) b + b n for all n 1.

55 Example π = ( ) 10

56 Example π = ( ) 10 We have 3.1 π < = 3.2

57 Example π = ( ) 10 We have 3.1 π < = π < = 3.15

58 Example π = ( ) 10 We have 3.1 π < = π < = π < = 3.142

59 Example π = ( ) 10 We have 3.1 π < = π < = π < = and so on..

60 The base transition factor from base a to base b is

61 The base transition factor from base a to base b is the least natural number k such that

62 The base transition factor from base a to base b is the least natural number k such that for any n 1 and any base-a digits D 1, D 2,..., D n

63 The base transition factor from base a to base b is the least natural number k such that for any n 1 and any base-a digits D 1, D 2,..., D n there exist base-b digits Ḋ 1, Ḋ 2,..., Ḋ kn

64 The base transition factor from base a to base b is the least natural number k such that for any n 1 and any base-a digits D 1, D 2,..., D n there exist base-b digits Ḋ 1, Ḋ 2,..., Ḋ kn such that (0.D 1... D n ) a = (0.Ḋ 1... Ḋ kn ) b.

65 Example The base transition factor from base 16 to base 10 is 4.

66 Example The base transition factor from base 16 to base 10 is 4. We have (0.23) 16 = ( ) 10

67 Example The base transition factor from base 16 to base 10 is 4. We have (0.23) 16 = ( ) 10 (0.42) 16 = ( ) 10

68 Example The base transition factor from base 16 to base 10 is 4. We have (0.23) 16 = ( ) 10 (0.42) 16 = ( ) 10 (0.1) 16 = (0.0625) 10

69 Example The base transition factor from base 16 to base 10 is 4. We have (0.23) 16 = ( ) 10 (0.42) 16 = ( ) 10 (0.1) 16 = (0.0625) 10 Every base-16 representation of the form 0.D 1... D n can converted to a base-10 representation of the form 0.Ḋ 1... Ḋ 4n.

70 The base transition factor from base a to base b exists if and only if prim(a) prim(b).

71 The formal definition of the base transition factor: Let prim(a) prim(b). Let b = p k 1 1 pk pkn n, where p i is a prime and k i N \ {0} (for i = 1,..., n), be the prime factorization of b. Then, a can be written of the form a = p j 1 1 p j pn jn where j i N (for i = 1,..., n). The base transition factor from a to b is the natural number k such that ji k = max{ 1 i n }. k i This definition is hard to read and does not give much intuition. But it shows how to compute k when a and b are given.

72 The Base Transition Theorem. Let k be the base transition factor from a to b, and let (M.D 1 D 2...) a and (M.Ḋ 1 Ḋ 2...) b be, respectively, the base-a and base-b expansion of an arbitrary real number α. Then, for all n N, we have (M.D 1... D n ) a (M.Ḋ 1... Ḋ kn ) b α < (M.Ḋ 1... Ḋ kn ) b + b kn (M.D 1... D n ) a + a n.

73 I will not prove the Base Transition Theorem, but what I told you so far gives some intuition why it holds.

74 Let (M.Ḋ 1 Ḋ 2...) b be the base-b expansion of α k be the base transition factor from a to b.

75 Let (M.Ḋ 1 Ḋ 2...) b be the base-b expansion of α k be the base transition factor from a to b. The Base Transition Theorem tells us that the n th digit of the base-a expansion of α is the same as the n th digit of the base-a expansion of (M.Ḋ 1... Ḋ kn ) b.

76 Thus, if the base transition factor from base a to base b exits, then we can easily convert irrational numbers from base b to base a.

77 Thus, if the base transition factor from base a to base b exits, then we can easily convert irrational numbers from base b to base a. We can find the n th digit of the base-a expansion of α by the following procedure: compute the rational number (M.Ḋ 1... Ḋ kn ) b compute the n th digit of the base-a expansion of this number. This procedure is (Kalmar) elementary.

78 If the base transition factor from a to b exists, the base-a expansion of α can be computed elementary in the base-b expansion of α.

79 If the base transition factor from a to b exists, the base-a expansion of α can be computed elementary in the base-b expansion of α. The base transition factor from a to b exists if and only if prim(a) prim(b).

80 If the base transition factor from a to b exists, the base-a expansion of α can be computed elementary in the base-b expansion of α. The base transition factor from a to b exists if and only if prim(a) prim(b). The left-right implication of our theorem follows: Theorem. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae.

81 The Proof of the Converse Implication

82 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae.

83 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae. Assume prim(a) prim(b).

84 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae. Assume prim(a) prim(b). We need to prove S be S ae.

85 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae. Assume prim(a) prim(b). We need to prove S be S ae. I will explain how to construct α such that α S be and α S ae. I will use a = 10 and b = 2.

86 Theorem [Kristiansen]. For any subrecursive class S closed under Kalmar elementary operations, we have prim(a) prim(b) S be S ae. Assume prim(a) prim(b). We need to prove S be S ae. I will explain how to construct α such that α S be and α S ae. I will use a = 10 and b = 2. I need a blackboard.

87 Some relevant papers: L. Kristiansen: On Subrecursive Representability of Irrational Numbers, Computability (the journal of CiE, preprint available from my homepage). L. Kristiansen: On Subrecursive Representability of Irrational Numbers, Part II, Computability (preprint available from my homepage). E. Specker: Nicht Konstruktiv Beweisbare Satze Der Analysis, The Journal of Symbolic Logic 14(3) (1949), A. Mostowski: On Computable Sequences, Fundamenta Mathematica 44 (1957),

88 Base Pairs and Base-(b, l) Expansions

89 A base pair is a pair of natural numbers (b, l) such that b > 1 and 0 l < b.

90 A base pair is a pair of natural numbers (b, l) such that b > 1 and 0 l < b. A base-(b, l) digit is a natural number in the set {l b + 1, l b + 2,..., l 1, l}.

91 Example The base-(10,9) digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The base-(10,0) digits are 9, 8, 7, 6, 5, 4, 3, 2, 1, 0. The base-(10,6) digits are 3, 2, 1, 0, 1, 2, 3, 4, 5, 6. The base-(3,1) digits are 1, 0, 1.

92 Theorem. Let (b, l) be an arbitrary base pair. Any irrational number has a unique base-(b, l) expansion.

93 Theorem. Let (b, l) be an arbitrary base pair. Any irrational number has a unique base-(b, l) expansion. Open Problem. Let S be a subrecursive class closed under elementary operations. Let S (b,l) denote the set of irrational numbers with a base-(b, l) expansion in S. For which base pairs (b, l) and (a, k) do we, or do we not, have S (b,l) S (a,k).

94 Goodbye! Thanks for your attention!

95 ???????????????... well, maybe I have time for a few more...