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# Discrete Mathematics - Algebraic Structures

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

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### 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