I P B´ E P C E A Proportionally modular Diophantine inequalities J. C. Rosales Porto 2008
I P B´ E P C E A • J. C. Rosales, P . A. Garc´ ıa-S´ anchez and J. M. Urbano-Blanco, The set of solutions of a proportionally modular diophantine inequality, J. Number Theory, 128 (2008), 453-467 • M. Bullejos, J. C. Rosales, Proportionally modular diophantine inequalities and Stern-Brocot tree, preprint • J. C. Rosales, P . A. Garc´ ıa-S´ anchez and J. M. Urbano-Blanco, Modular diophantine inequalities and numerical semigroups, Pacific J. Math. 218 (2005), 379-398 • J. C. Rosales, P . A. Garc´ ıa-S´ anchez, J. I. Garc´ ıa-Garc´ ıa and J. M. Urbano-Blanco, Proportionally modular diophantine inequalities, J. Number Theory 103(2003), 281-294 • J. M. Urbano-Blanco and P . Vasco PhD theses
I P B´ E P C E A What we intend to study • A proportionally modular Diophantine inequality is an expression of the form ax mod b ≤ cx , with a , b , and c positive integers • The set S = S( a , b , c ) of solutions to such inequality is a numerical semigroup • A numerical semigroup is proportionally modular if it is of this form
I P B´ E P C E A Some basic definitions • If A ⊂ N , we denote by � A � the submonoid of ( N , + ) generated by A , � A � = { λ 1 a 1 + ··· + λ n a n | n ∈ N ,λ 1 ,...,λ n ∈ N , a 1 ,..., a n ∈ A } • � A � is a numerical semigroup if and if gcd ( A ) = 1 • Every numerical semigroup S is finitely generated, and thus there exist positive integers n 1 ,..., n p ∈ S such that S = � n 1 ,..., n p � . If no proper subset of { n 1 ,..., n p } generates S , we say that this set is a minimal generating set of S • Minimal generating sets are unique and its elements are minimal generator of the semigroup
I P B´ E P C E A Example S( 12 , 32 , 3 ) = { x ∈ N | 12 x mod 32 ≤ 3 x } = { 0 , 3 , 6 , 7 , 8 , 9 , 10 , →} = � 3 , 7 , 8 � 12 × 1 mod 32 = 12 > 3 × 1 12 × 7 mod 32 = 20 ≤ 3 × 7 12 × 2 mod 32 = 24 > 3 × 2 12 × 8 mod 32 = 0 ≤ 3 × 8 12 × 3 mod 32 = 4 ≤ 3 × 3 12 × 9 mod 32 = 12 ≤ 3 × 9 12 × 4 mod 32 = 16 > 3 × 4 12 × 10 mod 32 = 24 ≤ 3 × 10 . . 12 × 5 mod 32 = 28 > 3 × 5 . 12 × 6 mod 32 = 8 ≤ 3 × 6
I P B´ E P C E A Some simplifications • The inequality ax mod b ≤ cx has the same solutions as ( a mod b ) x mod b ≤ cx • If c ≥ a , then S( a , b , c ) = N • Thus we can assume that c < a < b
I P B´ E P C E A Lemma If c < a < b are positive integers, S( a , b , c ) = T ∩ N , where T is the submonoid of ( Q , + ) generated by [ b b a , a − c ] . Conversely, if a 1 , a 2 , b 1 , b 2 are positive integers with a 1 b 1 < a 2 b 2 , and T is the submonoid of ( Q , + ) generated by [ a 1 b 1 , a 2 b 2 ] , then T ∩ N = S( a 2 b 1 , a 1 a 2 , a 2 b 1 − a 1 b 2 ) • T ∩ N is the proportionally modular numerical semigroup associated to the interval [ a 1 b 1 , a 2 b 2 ] , and we denote it by S([ a 1 b 1 , a 2 b 2 ])
I P B´ E P C E A Lemma If c < a < b are positive integers, S( a , b , c ) = T ∩ N , where T is the submonoid of ( Q , + ) generated by [ b b a , a − c ] . Conversely, if a 1 , a 2 , b 1 , b 2 are positive integers with a 1 b 1 < a 2 b 2 , and T is the submonoid of ( Q , + ) generated by [ a 1 b 1 , a 2 b 2 ] , then T ∩ N = S( a 2 b 1 , a 1 a 2 , a 2 b 1 − a 1 b 2 ) Example • S( 12 , 32 , 3 ) = S([ 32 12 , 32 9 ]) = S([ 8 3 , 32 9 ]) k ∈ N [ k 8 3 , k 32 9 ] = { 0 }∪ [ 8 3 , 32 9 ] ∪ [ 16 3 , 64 9 ] ∪ [ 8 , 32 • T = � 3 ] ∪··· • S( 12 , 32 , 3 ) = T ∩ N = { 0 , 3 , 6 , 7 , 8 , 9 , 10 ,... }
I P B´ E P C E A First results Lemma A positive integer x belongs to S([ a 1 b 1 , a 2 b 2 ]) if and only if there exists a positive integer y such that a 1 y ≤ a 2 b 1 ≤ x b 2 Lemma If a 2 b 1 − a 1 b 2 = 1, then S([ a 1 b 1 , a 2 b 2 ]) = � a 1 , a 2 �
I P B´ E P C E A B´ ezout sequence a p A sequence a 1 b 1 < a 2 b 2 < ··· < b p of rational numbers is a B´ ezout sequence if a 1 ,..., a p , b 1 ,..., b p are positive integers and a i + 1 b i − a i b i + 1 = 1 for all i ∈ { 1 ,..., p − 1 } a p Its length is p , and a 1 b 1 and b p are its ends Proposition a p a p If a 1 b 1 < a 2 ezout sequence, S([ a 1 b 2 < ··· < b p is a B´ b p ]) = � a 1 ,..., a p � b 1 ,
I P B´ E P C E A Example Let us solve the Diophantine inequality 50 x mod 131 ≤ 3 x We know that the set of solutions is S([ 131 50 , 131 47 ]) As 131 50 < 76 29 < 21 8 < 8 3 < 11 4 < 25 9 < 39 14 < 131 47 is a B´ ezout sequence, � � [ 131 50 , 131 S 47 ] = � 131 , 76 , 21 , 8 , 11 , 25 , 39 � = � 8 , 11 , 21 , 25 , 39 �
I P B´ E P C E A Problems: 1) If a b < c ezout sequence with ends a b and c d , is there a B´ d ? 2) If so, find a procedure to compute such a sequence 3) Characterize proportionally modular numerical semigroups in terms of their minimal generators
I P B´ E P C E A Proposition Let a 1 , a 2 , b 1 and b 2 be positive integers with a 1 b 1 < a 2 b 2 and mcd { a 1 , b 1 } = 1 = mcd { a 2 , b 2 } . Then there exists a B´ ezout sequence of length less than or equal to a 2 b 1 − a 1 b 2 + 1 and with ends a 1 b 1 and a 2 b 2 Corollary S([ a 1 b 1 , a 2 b 2 ]) has embedding dimension less than or equal to a 2 b 1 − a 1 b 2 + 1
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