Irrational Base Counting

Transcription

1 arxiv:5.79v [math.nt] 5 Apr 7 Department of Mathematics Brandeis University, Waltham MA abourla@brandeis.edu June 3, 8 Abstract We will provide algorithmic implementation with proofs of existence and uniqueness for the absolute and alternating irrational base numeration systems. Introduction We can view a positive integer written in our familiar base numeration system as the dot product of a finite sequence of digits (d k ) l {,,..., 9} and the infinite base vector ( k ) truncated to the l position. For instance when l = 3 and (d k := k) 3, we have l d k k = (,, 3) ( k ) = = 3. k= After taking zero as the vacuous expansion obtained when l = and allowing the infinite base vector to alternate in sign as (( ) k ), we can expand all integers base ( ). For instance, 3 = (9, 3, 7, ) (( ) k ) 3, whereas 3 is now given the new digit representation (, 8, 4). We can similarly obtain integer expansions for all fix radix base n systems. In this paper, we how show how to expand integers as a dot product using an irrational base. The idea behind these expansions date back to Ostrowski [3], who used the continued fraction expansion as a tool in inhomogeneous Diophantine Approximation. After fixing the base α (, )\Q, we expand it as an infinite continued fraction α = a + a + a ,

2 obtaining the unique sequence of partial quotients (a k ) (for details refer to any of the standard introductions [, ]). Truncating the iteration after k steps yields the convergent p k q k := a + a a k We will utilize the sequence of denominators (q k ) as the infinite base (α) vector and the alternating sequence (( ) k q k ) as the base ( α) vector, providing rigorous proofs of existence as well as concrete algorithmic realization and some counting examples. We end this section by quoting the well known recursion equation After we define q :=, q :=, q k = a k q k + q k k. () q k := ( ) k q k, k, () we use this relationship to obtain the new recursion equation q :=, q :=, q k = q k a k q k, k. (3) The Base α Expansion. Algorithm and proof The base α expansion is of the dot product of the sequence of digits (c k ) l, where l N and the infinite sequence (q k ) truncated to the l position. We say that the digit sequence (c k ) N is α admissible when it satisfies the following Markov conditions: c a and c k a k for k, not all zeros. If c k = a k then c k =. Theorem.. For every N N there exists l and a unique α admissible sequence of digits (c k ) l such that N = l k= c kq k. Proof. Apply the algorithm:

3 Algorithm : Natural Expansion input : α (, )\Q, N N output: l N, (c k ) l α admissible set N := N, m = n := ; while N m do 3 let n m be such that q nm N m < q nm ; 4 set c nm := N m /q nm ; 5 set N m+ := N m c nm q nm ; 6 set m := m + ; 7 end 8 set M := m, l := n, c k := for all k / {n m } M ; When N =, we have l = n = and the expansion is vacuous. Whenever N m, we see that since q = by definition (), the assignment of step 3 and the step 4 guarantees that n m and that c nm. (4) After we rewrite the assignment of line 4 as the inequality c nm q nm N m < (c nm + )q nm, (5) we observe that, in tandem with the assignment of line-5, we are applying the euclidean algorithm as the repeated integer division of N m by q nm resulting in a quotient c nm and remainder N m+. Thus we must have N m+ < N m N, that is, this iteration scheme must eventually terminate with a finite positive value M, yielding the sequences = N M < N M <... < N = N, n M <... < n < n = l and (c nm ) M m=. For all k l with k / {n m } M we define c k := and then, using the assignment of step 6, we obtain the desired expansion N = N = c n q n + N = c n q n + c n q n + N =... = M m= c nm q nm = l c k q k. k= Furthermore, the uniqueness of the quotient and the remainder terms in the division algorithm guarantees the uniqueness of this expansion. If M is such that n M then c = and if n M =, we use the fact that q = and the inequality (5) to verify that c = c q N < q = a. Conclude that c a as desired. If for some m we have in step that c nm a nm +, then the recursion formula (), the inequality (5) and the fact that the sequence (q k ) is strictly increasing will lead us to the contradiction N m < q nm = a nm q nm + q nm < (a nm + )q nm c nm q nm N m. 3

4 Therefore, for all k we must have c k a k. Next, suppose by contradiction that c k = a k and c k. Since c k = a k, we see from the inequality (4) that there is some m for which n m = k. The the recursion formula (), the inequality (5) and the assignment of line 5 will now leads us to the contradiction N m < q nm = q k < q k = q k N m+ + N m+ q k c nm+q nm + N m+ q k c k q k + N m+ = q k a k q k + N m+ = q k + N m+ c k q k + N m+ = c nm q nm + N m+ = N m.. Examples When α :=.5(5.5 ) = is the golden section, we have {a k } = {}. We then use formula () to verify that the sequence (q k ) is no other than the Fibonacci Sequence (F k ) := (,,, 3, 5, 8, 3,...). The implication of the proposition to this case is the Zeckendorf Theorem, which states that every positive integer can be uniquely written as the sum of nonconsecutive terms in (F k ). When α := = is the sliver section, we have {a k } = {}. By formula (), we verify that (q k ) 3 = (,, 5, ). The following tables display how the digits behave when we count to twenty four using this base: q 3 = q = 5 q = q = N c 4 c 3 c c q 3 = q = 5 q = q = N c 4 c 3 c c

5 3 The Base (-α) Expansion 3. Algorithm and proof The base ( α) expansion is of the dot product of the sequence of digits (b k ) l, where l N and the infinite sequence (qk ) truncated to the l position. We say that the digit sequence (b k ) N is ( α) admissible when: b k a k not all zeros. If b k = a k then b k+ =. Theorem 3.. For every integer Z there is some l and a unique ( α) admissible sequence of digits (b k ) l such that Z = l k= b kq k. Proof. We let I R be the indicator function for the relationship R and apply the algorithm: Algorithm : Integer Expansion input : Z Z, α (, )\Q output: l N, (b k ) l ( α) admissible set Z := Z, m = b = n := ; while Z m do 3 let n m be such that q n m < Z m + I < (Z m ) q n m ; 4 let n m {n m, n m + } be such that I > (( ) nm Z m ) = ; 5 if n m = n m then 6 set b n m := Z m /q nm ; 7 if Z m b n m q n m + I < (Z m b n m q n m ) > q nm then 8 set b nm := b n m + ; 9 else set b nm := b n m ; end else 3 set b nm := ; 4 end 5 set Z m+ := Z m b nm q n m ; 6 set m := m + ; 7 end 8 set M := m, l := n, b := b + Z m, b k := for all k / {n m } M ; The definition () of qk and the assignment of line 4 provides us with the inequality Z m q nm = ( ) nm Z m q nm, (6) whereas the assignment of line 6 provides us with the inequality b n m q nm Z m < (b n m + )q nm. (7) 5

6 When Z =, we have l = and the expansion is vacuous. Assuming Z, we will first show that the sequence of indexes (n m ) M is strictly decreasing. To do so, we will consider the two cases n m {n m, n m } separately: When n m = n m, the inequality of step 3 yields q nm = q n m < Z m Z m + I < (Z m ) q n m = q nm, so when we define Z m+ using b nm = in step 5, we will have by the inequalities (6) and (7) that Z m b n m q nm = Z m b n m q nm (8) and that Z m Z m+, hence Z m+ = q nm Z m q nm q nm. (9) Since n m = n m +, we have q nm, so that Z m+ + q nm and Z m+ + I < (Z m+ ) Z m+ + q nm. Then in step 3 of the next iteration, we will have n m+ n m. If this inequality is strict then we have n m+ n m+ + < n m. If n m+ = n m +, then in step 4 we use the fact that Z m and Z m+ are of opposite sign to obtain that is, I > (( ) nm Z m+ ) = I > ( ( ) n m Z m ) = = I> ( ( ) n m+ Z m+ ), n m+ n m n m+ (mod ). Since n m+ n m+ + n m, we conclude that for this case we have n m+ = n m+ < n m. When n m = n m and Z m >, we have by the inequalities (6), (7), line 5 and the fact that b nm b n m that and Z m+ = Z m b nm q n m = Z m b nm q nm < (b n m + )q nm b n m q nm = q nm q nm = b n m q nm (b n m + )q nm b n m q nm b nm q nm Z m b nm q nm = Z m b nm q n m = Z m+. Similarly, when n m = n m and Z m <, we have by the inequalities (6), (7), line 5 and the fact that b nm b n m that and Z m+ = Z m b nm q n m = Z m + b nm q nm b n m q nm + (b n m + )q nm = q nm q nm = (b n m + )q nm + b n m q nm < Z m + b n m q nm Z m b nm q n m = Z m+ 6

7 In either case we have Z m+ q nm. () If one of the last inequalities is an equality, then the iteration will terminate at the next step with n m+ = n m, b nm+ = and Z m+ =. Otherwise, we have Z m+ +I < (Z m+ ) q nm so that by line 3 we will have n m+ n m. When n m+ = n m+, we have n m+ < n m and when n m+ = n m+ we use the previous paragraph to conclude that n m+ < n m+. In either case we have n m+ n m+ n m and n m+ < n m. We have just proved that the sequence (n m ) M is non-constant and decreasing and thus conclude that this iteration process will eventually terminate with a finite value M, for which n M and Z M+ =. After we define b k := whenever k / {n m } M, we use the assignment of line 5 to obtain the desired expansion Z = b n q n + Z = b n q n + b n q n + Z =... = l b k qk. To prove uniqueness, we split an expansion of Z into its positive and negative parts and invoke the uniqueness of the absolute irrational expansion. More precisely, if Z = l k= b kqk, then we define Z + := l/ k= b k+ q k = l/ k= l/ b k+ q k, Z := k= k= b k q k = l/ k= b k q k, so that Z = Z + Z. If we also have Z = l b k= k qk then, without changing the representation, we set b k = b k := for all min{l, l} < k max{l, l} and write l/ k= b k q k = Z = Z + Z = = max{l, l}/ k= l/ k= b k+ q k l k= l/ (b k+ b k+ )q k + bk q k. k= b k q k Then theorem. guarantees that l = l and that b k = b k for all k l. To prove that for all k we have b k a k, we will show that for all m M we have b nm a nm. This is clear whenever n m = n m + for by the assignment of line 3, we have b nm =. When n m = n m, we use the inequality of line 3 and the assignments of line 6, line 8 and line, we see that b nm b n m. Furthermore, we cannot have b n m a nm +, for then we would use the recursion relationship () and the inequalities of line 3 and (7) to obtain the contradiction Z m q nm I < (Z m ) q nm = a nm q nm + q nm (b n m )q nm + q nm = b n m q nm (q nm q nm ) < b n m q nm Z m. 7

8 Finally, when b n m = a nm, we will show that we must also have b nm = a nm. If Z m >, then from line 4 and the definition () of q k we have ( )nm = and q n m = q nm so that by the inequality (7) we obtain Z m b n m q n m = Z m b n m q nm. Then the the recursion relationship () and the inequality of line 3 will now yield the inequality Z m b n m q n m + I < (Z m b n m q n m ) = Z m b n m q n m = Z m b nm q nm = Z m a nm q nm q nm a nm q nm = q nm. Similarly, if b n m = a nm and Z m <, then from line 4 we have ( ) nm <, hence q n m = q nm so that, by the inequality (7), we have Z m b n m q n m = Z m + b n m q nm. Then the recursion relationship () and the inequality of line 3 will yield the inequality Z m b n m q n m + I < (Z m b n m q n m ) (Z m b n m q n m ) + = Z m + b n m q n m + q nm I < (Z m ) + b n m q n m + = q nm a nm q nm + = q nm a nm q nm = q nm. In both cases, b n m would not satisfy the condition in line 7, hence we would have b nm = b n m = a nm. Since b k = whenever k / {n m } M, we conclude that for all k we have b k a k. To prove that b k = a k implies that b k+ =, we let k and m are such that n m = k +. If n m+ k then k / {n m } M+, hence b k = a k so that we may assume that n m+ = n m = k. Again we will consider the two cases n m {n m, n m } separately: When n m = n m, we assume that b k+ b k+ and will prove that b k a k. We use the recursion formula (), the fact that the sequence (q k ) is increasing and the inequality (9) to obtain Z m+ < q nm q nm = q k q k = (a k )q k + q k < a k q k. so when we assign b k = b n m = b n m+ using the inequality (7), we will have b k a k. Furthermore, from formula (8), we obtain Z m+ b kq k + I < (Z m+ b kq k ) Z m+ b kq k Z m+ (a k )q k + < q k q k (a k )q k + = q k a k q k + = q k + so that the condition of line 7 is not satisfied and b k := b k a k as desired. When n m = n m, we have n m+ n m = k < n m+ = n m = n m = k < k + = n m. 8

9 Suppose by contradiction that b k = a k and b k+ b k+. Then by the recursion relationship (), the inequalities (7), () and the assignment of line 5, we obtain the contradiction q k b k+q k = b n m q nm Z m = b nm q n m + Z m+ = b nm q nm Z m+ = b nm q nm b nm q nm + Z m+ b nm q nm b nm q nm + Z m+ < b nm q nm b nm q nm + (b nm+ + )q nm+ = b k q k b k q k + (b k + )q k = a k q k b k q k + (b k + )q k = q k (b k + )q k + (b k + )q k = q k. 3. Examples When α is the golden section, we have (qk ) := (,,, 3, 5,...) and are able to extend Zeckendorf s Theorem to the integers. When α is the silver section, we have (qk ) = (,, 5,, 9,...). The following tables displays how the digits behave when counting from -4 to 4 using this base: q =5 q = q = Z b 3 b b q4 =9 q 3 = q =5 q = q = Z b 5 b 4 b 3 b b

10 Irrational Base Counting Z q3 = b4 q =5 b3 q = b q = b Z q3 = b4 q =5 b3 q = b q = b Appendix Mathematica Implementation We use MathematicaTM to implement the algorithm and with the base whose first continued fraction partial quotients are (ak := k)9. The vectors b and c start at position and the vectors q and q start in positions so that we obtain the dot product representation N = c q = Ost(N ) q and Z = b q = AltOst(Z) q. Α = FromContinuedFraction@Prepend@Table@k, 8k, 9<D, DD; q = Denominator@Convergents@Α, DD; Ost@N_D := Module@8n = N, c = Table@, 8i, <D<, While@n >, j := First@Flatten@Position@q, First@Select@q, ð > n &, DDDDD - ; c = ReplacePart@c, j Quotient@n, qpjtdd; n = Mod@n, qpjtdd; cd Α = FromContinuedFraction@Prepend@Table@k, 8k, <D, DD; q = Prepend@Denominator@Convergents@Α, DD, D; q * = q * Table@H-L^n, 8n, <D; AltOst@Z_D := ModuleA8b = Table@, 8i, <D, nm, z = Z<, WhileA z ¹, nm = First@Flatten@Position@q, First@Select@q, ð ³ HAbs@zD + Boole@z < DL &, DDDDD; IfAH-L^nm * z >, bpnmt =, nm -= ; bpnmt = Floor@Abs@zD qpnmtd ; IfAAbsAz - bpnmt q * PnmTE + BooleAz - bpnmt q * PnmT < E > qpnm - T, bpnmt += EE; z -= bpnmt q * PnmTE; be

11 5 Acknowledgments This work could have not been completed without the guidance, encouragement and good company of Robbie Robinson from George Washington University. References [] G.H. Hardy and E.M. Wright, An introduction to the theory of numbers, 5ed, Clarendon Press, New York, 979 [] A. Ya. Khintchine, Continued Fractions, P. Noordhoff Ltd. Groningen, 963. [3] A. Ostrowski, Bemerkungen zur theorie der diophantischen approximationen, Abh. Math. Semin. Hamburg Univ, :7798, 9.