Discrete Mathematics - Algebraic Structures

16,817 views

Published on

Algebraic structures, groups, rings, Partially ordered sets, lattices, Boolean algebras.

Published in: Education, Technology
0 Comments
12 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
16,817
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
Downloads
755
Comments
0
Likes
12
Embeds 0
No embeds

No notes for slide

Discrete Mathematics - Algebraic Structures

  1. 1. Discrete Mathematics Algebraic StructuresH. Turgut Uyar Ay¸eg¨l Gen¸ata Yayımlı s u c Emre Harmancı 2001-2012
  2. 2. License c 2001-2012 T. Uyar, A. Yayımlı, E. Harmancı You are free: to Share – to copy, distribute and transmit the work to Remix – to adapt the work Under the following conditions: Attribution – You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work). Noncommercial – You may not use this work for commercial purposes. Share Alike – If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one. Legal code (the full license): http://creativecommons.org/licenses/by-nc-sa/3.0/
  3. 3. Topics 1 Algebraic Structures Introduction Groups Rings 2 Lattices Partially Ordered Sets Lattices Boolean Algebra
  4. 4. Topics 1 Algebraic Structures Introduction Groups Rings 2 Lattices Partially Ordered Sets Lattices Boolean Algebra
  5. 5. Algebraic Structure algebraic structure: <set, operations, constants> carrier set operations: binary, unary constants: identity, zero
  6. 6. Operation every operation is a function binary operation: ◦:S ×S →T unary operation: ∆:S →T closed: T ⊆ S
  7. 7. Operation every operation is a function binary operation: ◦:S ×S →T unary operation: ∆:S →T closed: T ⊆ S
  8. 8. Closed Operation Examples Example subtraction is closed on Z subtraction is not closed on Z+
  9. 9. Closed Operation Examples Example subtraction is closed on Z subtraction is not closed on Z+
  10. 10. Binary Operation Properties Definition commutativity: ∀a, b ∈ S a ◦ b = b ◦ a Definition associativity: ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c)
  11. 11. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  12. 12. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  13. 13. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  14. 14. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  15. 15. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  16. 16. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  17. 17. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  18. 18. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  19. 19. Binary Operation Example Example ◦:Z×Z→Z a ◦ b = a + b − 3ab commutative: a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a associative: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)
  20. 20. ConstantsDefinition Definitionidentity: zero:x ◦1=1◦x =x x ◦0=0◦x =0 left identity: 1l ◦ x = x left zero: 0l ◦ x = 0 right identity: x ◦ 1r = x right zero: x ◦ 0r = 0
  21. 21. ConstantsDefinition Definitionidentity: zero:x ◦1=1◦x =x x ◦0=0◦x =0 left identity: 1l ◦ x = x left zero: 0l ◦ x = 0 right identity: x ◦ 1r = x right zero: x ◦ 0r = 0
  22. 22. Examples of Constants Example identity for < N, max > is 0 zero for < N, min > is 0 zero for < Z+ , min > is 1 Example ◦ a b c a a b b b is a left identity b a b c a and b are right zeros c a b a
  23. 23. Examples of Constants Example identity for < N, max > is 0 zero for < N, min > is 0 zero for < Z+ , min > is 1 Example ◦ a b c a a b b b is a left identity b a b c a and b are right zeros c a b a
  24. 24. ConstantsTheorem Theorem∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0rProof. Proof.1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r
  25. 25. ConstantsTheorem Theorem∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0rProof. Proof.1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r
  26. 26. ConstantsTheorem Theorem∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0rProof. Proof.1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r
  27. 27. ConstantsTheorem Theorem∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0rProof. Proof.1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r
  28. 28. ConstantsTheorem Theorem∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0rProof. Proof.1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r
  29. 29. ConstantsTheorem Theorem∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0rProof. Proof.1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r
  30. 30. ConstantsTheorem Theorem∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0rProof. Proof.1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r
  31. 31. ConstantsTheorem Theorem∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0rProof. Proof.1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r
  32. 32. Inverse if x ◦ y = 1: x is a left inverse of y y is a right inverse of x if x ◦ y = y ◦ x = 1 x and y are inverse
  33. 33. Inverse Theorem if the operation ◦ is associative: w ◦x =x ◦y =1⇒w =y Proof. w = w ◦1 = w ◦ (x ◦ y ) = (w ◦ x) ◦ y = 1◦y = y
  34. 34. Inverse Theorem if the operation ◦ is associative: w ◦x =x ◦y =1⇒w =y Proof. w = w ◦1 = w ◦ (x ◦ y ) = (w ◦ x) ◦ y = 1◦y = y
  35. 35. Inverse Theorem if the operation ◦ is associative: w ◦x =x ◦y =1⇒w =y Proof. w = w ◦1 = w ◦ (x ◦ y ) = (w ◦ x) ◦ y = 1◦y = y
  36. 36. Inverse Theorem if the operation ◦ is associative: w ◦x =x ◦y =1⇒w =y Proof. w = w ◦1 = w ◦ (x ◦ y ) = (w ◦ x) ◦ y = 1◦y = y
  37. 37. Inverse Theorem if the operation ◦ is associative: w ◦x =x ◦y =1⇒w =y Proof. w = w ◦1 = w ◦ (x ◦ y ) = (w ◦ x) ◦ y = 1◦y = y
  38. 38. Inverse Theorem if the operation ◦ is associative: w ◦x =x ◦y =1⇒w =y Proof. w = w ◦1 = w ◦ (x ◦ y ) = (w ◦ x) ◦ y = 1◦y = y
  39. 39. Algebraic Families algebraic family: algebraic structure, axioms commutativity, associativity inverse elements
  40. 40. Algebraic Family Examples Example axioms: x ◦y =y ◦x (x ◦ y ) ◦ z = x ◦ (y ◦ z) x ◦1=x structures for which these axioms hold: < Z, +, 0 > < Z, ·, 1 > < P(S), ∪, ∅ >
  41. 41. Algebraic Family Examples Example axioms: x ◦y =y ◦x (x ◦ y ) ◦ z = x ◦ (y ◦ z) x ◦1=x structures for which these axioms hold: < Z, +, 0 > < Z, ·, 1 > < P(S), ∪, ∅ >
  42. 42. Subalgebra Definition subalgebra: let A =< S, ◦, ∆, k > ∧ A =< S , ◦ , ∆ , k > A is a subalgebra of A if: S ⊆S ∀a, b ∈ S a ◦ b = a ◦ b ∈ S ∀a ∈ S ∆ a = ∆a ∈ S k =k
  43. 43. Subalgebra Definition subalgebra: let A =< S, ◦, ∆, k > ∧ A =< S , ◦ , ∆ , k > A is a subalgebra of A if: S ⊆S ∀a, b ∈ S a ◦ b = a ◦ b ∈ S ∀a ∈ S ∆ a = ∆a ∈ S k =k
  44. 44. Subalgebra Examples Example < Z+ , +, 0 > is a subalgebra of < Z, +, 0 >. < N, −, 0 > is not a subalgebra of < Z, −, 0 >.
  45. 45. Subalgebra Examples Example < Z+ , +, 0 > is a subalgebra of < Z, +, 0 >. < N, −, 0 > is not a subalgebra of < Z, −, 0 >.
  46. 46. Topics 1 Algebraic Structures Introduction Groups Rings 2 Lattices Partially Ordered Sets Lattices Boolean Algebra
  47. 47. Semigroups Definition semigroup: < S, ◦ > ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c)
  48. 48. Semigroup Examples Example < Σ+ , & > Σ: alphabet, Σ+ : strings of length at least 1 &: string concatenation
  49. 49. Monoids Definition monoid: < S, ◦, 1 > ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c) ∀a ∈ S a ◦ 1 = 1 ◦ a = a
  50. 50. Monoid Examples Example < Σ∗ , &, > Σ: alphabet, Σ∗ : strings of any length &: string concatenation : empty string
  51. 51. Groups Definition group: < S, ◦, 1 > ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c) ∀a ∈ S a ◦ 1 = 1 ◦ a = a ∀a ∈ S ∃a−1 ∈ S a ◦ a−1 = a−1 ◦ a = 1 Abelian group: ∀a, b ∈ S a ◦ b = b ◦ a
  52. 52. Groups Definition group: < S, ◦, 1 > ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c) ∀a ∈ S a ◦ 1 = 1 ◦ a = a ∀a ∈ S ∃a−1 ∈ S a ◦ a−1 = a−1 ◦ a = 1 Abelian group: ∀a, b ∈ S a ◦ b = b ◦ a
  53. 53. Group Examples Example < Z, +, 0 > is a group. < Q, ·, 1 > is not a group. < Q − {0}, ·, 1 > is a group.
  54. 54. Group Examples Example < Z, +, 0 > is a group. < Q, ·, 1 > is not a group. < Q − {0}, ·, 1 > is a group.
  55. 55. Group Example: Permutation Composition permutation: a bijective function on a set representation: a1 a2 ... an p(a1 ) p(a2 ) . . . p(an )
  56. 56. Permutation Examples Example A = {1, 2, 3} 1 2 3 1 2 3 p1 = p2 = 1 2 3 1 3 2 1 2 3 1 2 3 p3 = p4 = 2 1 3 2 3 1 1 2 3 1 2 3 p5 = p6 = 3 1 2 3 2 1
  57. 57. Group Example: Permutation Composition permutation composition is associative identity permutasyon: 1A a1 a2 . . . an a1 a2 . . . an the set of permutations of the elements of a set, the permutation composition operation and the identity permutation constitute a group
  58. 58. Group Example: Permutation Composition permutation composition is associative identity permutasyon: 1A a1 a2 . . . an a1 a2 . . . an the set of permutations of the elements of a set, the permutation composition operation and the identity permutation constitute a group
  59. 59. Group Example: Permutation Composition Example (permutations on {1, 2, 3, 4}) A 1A p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 p11 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 3 3 4 4 1 1 3 3 4 4 3 3 4 2 4 2 3 3 4 1 4 1 3 4 4 3 4 2 3 2 4 3 4 1 3 1 p12 p13 p14 p15 p16 p17 p18 p19 p20 p21 p22 p23 1 3 3 3 3 3 3 4 4 4 4 4 4 2 1 1 2 2 4 4 1 1 2 2 3 3 3 2 4 1 4 1 2 2 3 1 3 1 2 4 4 2 4 1 2 1 3 2 3 1 2 1
  60. 60. Group Example: Permutation Composition Example p8 p12 = p12 p8 = 1A : −1 −1 p12 = p8 , p8 = p12 p14 p14 = 1A : −1 p14 = p14 G1 =< {1A , p1 , . . . , p23 }, , 1A > is a group
  61. 61. Group Example: Permutation Composition Example p8 p12 = p12 p8 = 1A : −1 −1 p12 = p8 , p8 = p12 p14 p14 = 1A : −1 p14 = p14 G1 =< {1A , p1 , . . . , p23 }, , 1A > is a group
  62. 62. Group Example: Permutation Composition Example 1A p2 p6 p8 p12 p14 1A 1A p2 p6 p8 p12 p14 p2 p2 1A p8 p6 p14 p12 p6 p6 p12 1A p14 p2 p8 p8 p8 p14 p2 p12 1A p6 p12 p12 p6 p14 1A p8 p2 p14 p14 p8 p12 p2 p6 1A < {1A , p2 , p6 , p8 , p12 , p14 }, , 1A > is a subgroup of G1
  63. 63. Left and Right Cancellation Theorem a◦c =b◦c ⇒a=b c ◦a=c ◦b ⇒a=b Proof. a◦c = b◦c ⇒ (a ◦ c) ◦ c −1 = (b ◦ c) ◦ c −1 ⇒ a ◦ (c ◦ c −1 ) = b ◦ (c ◦ c −1 ) ⇒ a◦1 = b◦1 ⇒ a = b
  64. 64. Left and Right Cancellation Theorem a◦c =b◦c ⇒a=b c ◦a=c ◦b ⇒a=b Proof. a◦c = b◦c ⇒ (a ◦ c) ◦ c −1 = (b ◦ c) ◦ c −1 ⇒ a ◦ (c ◦ c −1 ) = b ◦ (c ◦ c −1 ) ⇒ a◦1 = b◦1 ⇒ a = b
  65. 65. Left and Right Cancellation Theorem a◦c =b◦c ⇒a=b c ◦a=c ◦b ⇒a=b Proof. a◦c = b◦c ⇒ (a ◦ c) ◦ c −1 = (b ◦ c) ◦ c −1 ⇒ a ◦ (c ◦ c −1 ) = b ◦ (c ◦ c −1 ) ⇒ a◦1 = b◦1 ⇒ a = b
  66. 66. Left and Right Cancellation Theorem a◦c =b◦c ⇒a=b c ◦a=c ◦b ⇒a=b Proof. a◦c = b◦c ⇒ (a ◦ c) ◦ c −1 = (b ◦ c) ◦ c −1 ⇒ a ◦ (c ◦ c −1 ) = b ◦ (c ◦ c −1 ) ⇒ a◦1 = b◦1 ⇒ a = b
  67. 67. Left and Right Cancellation Theorem a◦c =b◦c ⇒a=b c ◦a=c ◦b ⇒a=b Proof. a◦c = b◦c ⇒ (a ◦ c) ◦ c −1 = (b ◦ c) ◦ c −1 ⇒ a ◦ (c ◦ c −1 ) = b ◦ (c ◦ c −1 ) ⇒ a◦1 = b◦1 ⇒ a = b
  68. 68. Left and Right Cancellation Theorem a◦c =b◦c ⇒a=b c ◦a=c ◦b ⇒a=b Proof. a◦c = b◦c ⇒ (a ◦ c) ◦ c −1 = (b ◦ c) ◦ c −1 ⇒ a ◦ (c ◦ c −1 ) = b ◦ (c ◦ c −1 ) ⇒ a◦1 = b◦1 ⇒ a = b
  69. 69. Basic Theorem of Groups Theorem The unique solution of the equation a ◦ x = b is: x = a−1 ◦ b. Proof. a◦c = b ⇒ a−1 ◦ (a ◦ c) = a−1 ◦ b ⇒ 1◦c = a−1 ◦ b ⇒ c = a−1 ◦ b
  70. 70. Basic Theorem of Groups Theorem The unique solution of the equation a ◦ x = b is: x = a−1 ◦ b. Proof. a◦c = b ⇒ a−1 ◦ (a ◦ c) = a−1 ◦ b ⇒ 1◦c = a−1 ◦ b ⇒ c = a−1 ◦ b
  71. 71. Basic Theorem of Groups Theorem The unique solution of the equation a ◦ x = b is: x = a−1 ◦ b. Proof. a◦c = b ⇒ a−1 ◦ (a ◦ c) = a−1 ◦ b ⇒ 1◦c = a−1 ◦ b ⇒ c = a−1 ◦ b
  72. 72. Basic Theorem of Groups Theorem The unique solution of the equation a ◦ x = b is: x = a−1 ◦ b. Proof. a◦c = b ⇒ a−1 ◦ (a ◦ c) = a−1 ◦ b ⇒ 1◦c = a−1 ◦ b ⇒ c = a−1 ◦ b
  73. 73. Basic Theorem of Groups Theorem The unique solution of the equation a ◦ x = b is: x = a−1 ◦ b. Proof. a◦c = b ⇒ a−1 ◦ (a ◦ c) = a−1 ◦ b ⇒ 1◦c = a−1 ◦ b ⇒ c = a−1 ◦ b
  74. 74. Topics 1 Algebraic Structures Introduction Groups Rings 2 Lattices Partially Ordered Sets Lattices Boolean Algebra
  75. 75. Ring Definition ring: < S, +, ·, 0 > ∀a, b, c ∈ S (a + b) + c = a + (b + c) ∀a ∈ S a + 0 = 0 + a = a ∀a ∈ S ∃(−a) ∈ S a + (−a) = (−a) + a = 0 ∀a, b ∈ S a + b = b + a ∀a, b, c ∈ S (a · b) · c = a · (b · c) ∀a, b, c ∈ S a · (b + c) = a · b + a · c (b + c) · a = b · a + c · a
  76. 76. Ring Definition ring: < S, +, ·, 0 > ∀a, b, c ∈ S (a + b) + c = a + (b + c) ∀a ∈ S a + 0 = 0 + a = a ∀a ∈ S ∃(−a) ∈ S a + (−a) = (−a) + a = 0 ∀a, b ∈ S a + b = b + a ∀a, b, c ∈ S (a · b) · c = a · (b · c) ∀a, b, c ∈ S a · (b + c) = a · b + a · c (b + c) · a = b · a + c · a
  77. 77. Ring Definition ring: < S, +, ·, 0 > ∀a, b, c ∈ S (a + b) + c = a + (b + c) ∀a ∈ S a + 0 = 0 + a = a ∀a ∈ S ∃(−a) ∈ S a + (−a) = (−a) + a = 0 ∀a, b ∈ S a + b = b + a ∀a, b, c ∈ S (a · b) · c = a · (b · c) ∀a, b, c ∈ S a · (b + c) = a · b + a · c (b + c) · a = b · a + c · a
  78. 78. Field Definition field: < S, +, ·, 0, 1 > all properties of a ring ∀a, b ∈ S a · b = b · a ∀a ∈ S a · 1 = 1 · a = a ∀a ∈ S ∃a−1 ∈ S a · a−1 = a−1 · a = 1
  79. 79. Field Definition field: < S, +, ·, 0, 1 > all properties of a ring ∀a, b ∈ S a · b = b · a ∀a ∈ S a · 1 = 1 · a = a ∀a ∈ S ∃a−1 ∈ S a · a−1 = a−1 · a = 1
  80. 80. References Grimaldi Chapter 5: Relations and Functions 5.4. Special Functions Chapter 16: Groups, Coding Theory, and Polya’s Method of Enumeration 16.1. Definitions, Examples, and Elementary Properties Chapter 14: Rings and Modular Arithmetic 14.1. The Ring Structure: Definition and Examples
  81. 81. Topics 1 Algebraic Structures Introduction Groups Rings 2 Lattices Partially Ordered Sets Lattices Boolean Algebra
  82. 82. Partially Ordered Set Definition partial order relation: reflexive anti-symmetric transitive partially ordered set (poset): a set with a partial order relation defined on its elements
  83. 83. Partially Ordered Set Definition partial order relation: reflexive anti-symmetric transitive partially ordered set (poset): a set with a partial order relation defined on its elements
  84. 84. Partial Order Examples Example (set of sets, ⊆) A⊆A A⊆B ∧B ⊆A⇒A=B A⊆B ∧B ⊆C ⇒A⊆C
  85. 85. Partial Order Examples Example (Z, ≤) x ≤x x ≤y ∧y ≤x ⇒x =y x ≤y ∧y ≤z ⇒x ≤z
  86. 86. Partial Order Examples Example (Z+ , |) x|x x|y ∧ y |x ⇒ x = y x|y ∧ y |z ⇒ x|z
  87. 87. Comparability a b: a precedes b a b∨b a: a and b are comparable total order (linear order): all elements are comparable with each other
  88. 88. Comparability a b: a precedes b a b∨b a: a and b are comparable total order (linear order): all elements are comparable with each other
  89. 89. Comparability Examples Example Z+ , |: 3 and 5 are not comparable Z, ≤: total order
  90. 90. Comparability Examples Example Z+ , |: 3 and 5 are not comparable Z, ≤: total order
  91. 91. Hasse Diagrams a b: a immediately precedes b ¬∃x a x b Hasse diagram: draw a line between a and b if a b preceding element is below
  92. 92. Hasse Diagrams a b: a immediately precedes b ¬∃x a x b Hasse diagram: draw a line between a and b if a b preceding element is below
  93. 93. Hasse Diagram Examples Example {1, 2, 3, 4, 6, 8, 9, 12, 18, 24} the relation |
  94. 94. Consistent Enumeration consistent enumeration: f :S →N a b ⇒ f (a) ≤ f (b) there can be more than one consistent enumeration
  95. 95. Consistent Enumeration Examples Example {a −→ 5, b −→ 3, c −→ 4, d −→ 1, e −→ 2} {a −→ 5, b −→ 4, c −→ 3, d −→ 2, e −→ 1}
  96. 96. Consistent Enumeration Examples Example {a −→ 5, b −→ 3, c −→ 4, d −→ 1, e −→ 2} {a −→ 5, b −→ 4, c −→ 3, d −→ 2, e −→ 1}
  97. 97. Maximal - Minimal Elements Definition maximal element: max ∀x ∈ S max x ⇒ x = max Definition minimal element: min ∀x ∈ S x min ⇒ x = min
  98. 98. Maximal - Minimal Elements Definition maximal element: max ∀x ∈ S max x ⇒ x = max Definition minimal element: min ∀x ∈ S x min ⇒ x = min
  99. 99. Maximal - Minimal Element Examples Example max : 18, 24 min : 1
  100. 100. BoundsDefinition DefinitionA⊆S A⊆SM is an upper bound of A: m is a lower bound of A:∀x ∈ A x M ∀x ∈ A m xM(A): set of upper bounds of A m(A): set of lower bound of Asup(A) is the supremum of A: inf (A) is the infimum of A:∀M ∈ M(A) sup(A) M ∀m ∈ m(A) m inf (A)
  101. 101. BoundsDefinition DefinitionA⊆S A⊆SM is an upper bound of A: m is a lower bound of A:∀x ∈ A x M ∀x ∈ A m xM(A): set of upper bounds of A m(A): set of lower bound of Asup(A) is the supremum of A: inf (A) is the infimum of A:∀M ∈ M(A) sup(A) M ∀m ∈ m(A) m inf (A)
  102. 102. Bound Example Example (factors of 36) inf = greatest common divisor sup = least common multiple
  103. 103. Topics 1 Algebraic Structures Introduction Groups Rings 2 Lattices Partially Ordered Sets Lattices Boolean Algebra
  104. 104. Lattice Definition lattice: < L, ∧, ∨ > ∧: meet, ∨: join a∧b =b∧a a∨b =b∨a (a ∧ b) ∧ c = a ∧ (b ∧ c) (a ∨ b) ∨ c = a ∨ (b ∨ c) a ∧ (a ∨ b) = a a ∨ (a ∧ b) = a
  105. 105. Lattice Definition lattice: < L, ∧, ∨ > ∧: meet, ∨: join a∧b =b∧a a∨b =b∨a (a ∧ b) ∧ c = a ∧ (b ∧ c) (a ∨ b) ∨ c = a ∨ (b ∨ c) a ∧ (a ∨ b) = a a ∨ (a ∧ b) = a
  106. 106. Poset - Lattice Relationship If P is a poset, then < P, inf , sup > is a lattice. a ∧ b = inf (a, b) a ∨ b = sup(a, b) Every lattice is a poset where these definitions hold.
  107. 107. Poset - Lattice Relationship If P is a poset, then < P, inf , sup > is a lattice. a ∧ b = inf (a, b) a ∨ b = sup(a, b) Every lattice is a poset where these definitions hold.
  108. 108. Duality Definition dual: ∧ instead of ∨, ∨ instead of ∧ Theorem (Duality Theorem) Every theorem has a dual theorem in lattices.
  109. 109. Duality Definition dual: ∧ instead of ∨, ∨ instead of ∧ Theorem (Duality Theorem) Every theorem has a dual theorem in lattices.
  110. 110. Lattice Theorems Theorem a∧a=a Proof. a ∧ a = a ∧ (a ∨ (a ∧ b))
  111. 111. Lattice Theorems Theorem a∧a=a Proof. a ∧ a = a ∧ (a ∨ (a ∧ b))
  112. 112. Lattice Theorems Theorem a b ⇔a∧b =a⇔a∨b =b
  113. 113. Lattice Examples Example < P{a, b, c}, ∩, ∪ > ⊆ relation
  114. 114. Bounded LatticeDefinition Definitionlower bound of lattice L: 0 upper bound of lattice L: I∀x ∈ L 0 x ∀x ∈ L x I Theorem Every finite lattice is bounded.
  115. 115. Bounded LatticeDefinition Definitionlower bound of lattice L: 0 upper bound of lattice L: I∀x ∈ L 0 x ∀x ∈ L x I Theorem Every finite lattice is bounded.
  116. 116. Bounded LatticeDefinition Definitionlower bound of lattice L: 0 upper bound of lattice L: I∀x ∈ L 0 x ∀x ∈ L x I Theorem Every finite lattice is bounded.
  117. 117. Distributive Lattice distributive lattice: ∀a, b, c ∈ L a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) ∀a, b, c ∈ L a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c)
  118. 118. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ c = c
  119. 119. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ c = c
  120. 120. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ c = c
  121. 121. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ c = c
  122. 122. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ c = c
  123. 123. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ c = c
  124. 124. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ c = c
  125. 125. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ I = I
  126. 126. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ I = I
  127. 127. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ I = I
  128. 128. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ I = I
  129. 129. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ I = I
  130. 130. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ I = I
  131. 131. Counterexamples Example a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ I = I
  132. 132. Distributive Lattice Theorem A lattice is nondistributive if and only if it has a sublattice isomorphic to any of these two structures.
  133. 133. Join Irreducible Definition join irreducible element: a=x ∨y ⇒a=x ∨a=y atom: a join irreducible element which immediately succeeds the minimum
  134. 134. Join Irreducible Definition join irreducible element: a=x ∨y ⇒a=x ∨a=y atom: a join irreducible element which immediately succeeds the minimum
  135. 135. Join Irreducible Example Example (divisibility relation) prime numbers and 1 are join irreducible 1 is the minimum, the prime numbers are the atoms
  136. 136. Join Irreducible Example Example (divisibility relation) prime numbers and 1 are join irreducible 1 is the minimum, the prime numbers are the atoms
  137. 137. Join Irreducible Theorem Every element in a lattice can be written as the join of join irreducible elements.
  138. 138. Complement Definition a and x are complements: a ∧ x = 0 and a ∨ x = I
  139. 139. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  140. 140. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  141. 141. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  142. 142. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  143. 143. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  144. 144. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  145. 145. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  146. 146. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  147. 147. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  148. 148. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  149. 149. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  150. 150. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  151. 151. Complemented Lattice Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I x = x ∨ 0 = x ∨ (a ∧ y ) = (x ∨ a) ∧ (x ∨ y ) = I ∧ (x ∨ y ) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y
  152. 152. Topics 1 Algebraic Structures Introduction Groups Rings 2 Lattices Partially Ordered Sets Lattices Boolean Algebra
  153. 153. Boolean Algebra Definition Boolean algebra: < B, +, ·, x, 1, 0 > a+b =b+a a·b =b·a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a+0=a a·1=a a+a=1 a·a=0
  154. 154. Boolean Algebra Definition Boolean algebra: < B, +, ·, x, 1, 0 > a+b =b+a a·b =b·a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a+0=a a·1=a a+a=1 a·a=0
  155. 155. Boolean Algebra Definition Boolean algebra: < B, +, ·, x, 1, 0 > a+b =b+a a·b =b·a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a+0=a a·1=a a+a=1 a·a=0
  156. 156. Boolean Algebra Definition Boolean algebra: < B, +, ·, x, 1, 0 > a+b =b+a a·b =b·a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a+0=a a·1=a a+a=1 a·a=0
  157. 157. Boolean Algebra Definition Boolean algebra: < B, +, ·, x, 1, 0 > a+b =b+a a·b =b·a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a+0=a a·1=a a+a=1 a·a=0
  158. 158. Boolean Algebra - Lattice Relationship Definition A Boolean algebra is a finite, distributive, complemented lattice.
  159. 159. Duality Definition dual: + instead of ·, · instead of + 0 instead of 1, 1 instead of 0 Example (1 + a) · (b + 0) = b dual of the theorem: (0 · a) + (b · 1) = b
  160. 160. Duality Definition dual: + instead of ·, · instead of + 0 instead of 1, 1 instead of 0 Example (1 + a) · (b + 0) = b dual of the theorem: (0 · a) + (b · 1) = b
  161. 161. Boolean Algebra Examples Example B = {0, 1}, + = ∨, · = ∧ Example B = { factors of 70 }, + = lcm, · = gcd
  162. 162. Boolean Algebra Examples Example B = {0, 1}, + = ∨, · = ∧ Example B = { factors of 70 }, + = lcm, · = gcd
  163. 163. Boolean Algebra Theorems a+a=a a·a=a a+1=1 a·0=0 a + (a · b) = a a · (a + b) = a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a=a a+b =a·b a·b =a+b
  164. 164. Boolean Algebra Theorems a+a=a a·a=a a+1=1 a·0=0 a + (a · b) = a a · (a + b) = a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a=a a+b =a·b a·b =a+b
  165. 165. Boolean Algebra Theorems a+a=a a·a=a a+1=1 a·0=0 a + (a · b) = a a · (a + b) = a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a=a a+b =a·b a·b =a+b
  166. 166. Boolean Algebra Theorems a+a=a a·a=a a+1=1 a·0=0 a + (a · b) = a a · (a + b) = a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a=a a+b =a·b a·b =a+b
  167. 167. Boolean Algebra Theorems a+a=a a·a=a a+1=1 a·0=0 a + (a · b) = a a · (a + b) = a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a=a a+b =a·b a·b =a+b
  168. 168. Boolean Algebra Theorems a+a=a a·a=a a+1=1 a·0=0 a + (a · b) = a a · (a + b) = a (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a=a a+b =a·b a·b =a+b
  169. 169. References Required Reading: Grimaldi Chapter 7: Relations: The Second Time Around 7.3. Partial Orders: Hasse Diagrams Chapter 15: Boolean Algebra and Switching Functions 15.4. The Structure of a Boolean Algebra

×