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# Discrete Mathematics - Relations and Functions

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Relations, relation composition, converse relation, reflexivity, symmetry, transitivity. Functions, function composition, one-to-one, onto, bijective functions, inverse function, pigeonhole principle, recursive functions.

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### Discrete Mathematics - Relations and Functions

1. 1. Discrete Mathematics Relations and Functions H. Turgut Uyar Ay¸seg¨ul Gen¸cata Yayımlı Emre Harmancı 2001-2016
3. 3. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
4. 4. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
5. 5. Relation Deﬁnition relation: α ⊆ A × B × C × · · · × N tuple: element of relation binary relation: α ⊆ A × B aαb : (a, b) ∈ α
6. 6. Relation Deﬁnition relation: α ⊆ A × B × C × · · · × N tuple: element of relation binary relation: α ⊆ A × B aαb : (a, b) ∈ α
7. 7. Relation Example A = {a1, a2, a3, a4}, B = {b1, b2, b3} α = {(a1, b1), (a1, b3), (a2, b2), (a2, b3), (a3, b1), (a3, b3), (a4, b1)} b1 b2 b3 a1 1 0 1 a2 0 1 1 a3 1 0 1 a4 1 0 0 Mα = 1 0 1 0 1 1 1 0 1 1 0 0
8. 8. Relation Example A = {a1, a2, a3, a4}, B = {b1, b2, b3} α = {(a1, b1), (a1, b3), (a2, b2), (a2, b3), (a3, b1), (a3, b3), (a4, b1)} b1 b2 b3 a1 1 0 1 a2 0 1 1 a3 1 0 1 a4 1 0 0 Mα = 1 0 1 0 1 1 1 0 1 1 0 0
9. 9. Relation Composition Deﬁnition relation composition: α ⊆ A × B, β ⊆ B × C αβ = {(a, c) | a ∈ A, c ∈ C, ∃b ∈ B [aαb ∧ bβc]} example
10. 10. Relation Composition Mαβ = Mα × Mβ using logical operations: 1 : T 0 : F · : ∧ + : ∨ example Mα = 1 0 0 0 0 1 0 1 1 0 1 0 1 0 1 Mβ = 1 1 0 0 0 0 1 1 0 1 1 0 Mαβ = 1 1 0 0 0 1 1 0 0 1 1 1 0 0 1 1 1 1 1 0
11. 11. Relation Composition Mαβ = Mα × Mβ using logical operations: 1 : T 0 : F · : ∧ + : ∨ example Mα = 1 0 0 0 0 1 0 1 1 0 1 0 1 0 1 Mβ = 1 1 0 0 0 0 1 1 0 1 1 0 Mαβ = 1 1 0 0 0 1 1 0 0 1 1 1 0 0 1 1 1 1 1 0
12. 12. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
13. 13. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
14. 14. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
15. 15. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
16. 16. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
17. 17. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
18. 18. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
19. 19. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
20. 20. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
21. 21. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
22. 22. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
23. 23. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
24. 24. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
25. 25. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
26. 26. Converse Relation Deﬁnition α−1 = {(b, a) | (a, b) ∈ α} Mα−1 = MT α
27. 27. Converse Relation Theorems (α−1)−1 = α (α ∪ β)−1 = α−1 ∪ β−1 (α ∩ β)−1 = α−1 ∩ β−1 α−1 = α−1 (α − β)−1 = α−1 − β−1
28. 28. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
29. 29. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
30. 30. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
31. 31. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
32. 32. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
33. 33. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
34. 34. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
35. 35. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
36. 36. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
37. 37. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
38. 38. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
39. 39. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
40. 40. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
41. 41. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
42. 42. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
43. 43. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
44. 44. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
45. 45. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
46. 46. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
47. 47. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
48. 48. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
49. 49. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
50. 50. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
51. 51. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
52. 52. Relation Properties α ⊆ A × A αn: αα · · · α identity relation: E = {(a, a) | a ∈ A} reﬂexivity symmetry transitivity
53. 53. Relation Properties α ⊆ A × A αn: αα · · · α identity relation: E = {(a, a) | a ∈ A} reﬂexivity symmetry transitivity
54. 54. Reﬂexivity reﬂexive α ⊆ A × A ∀a ∈ A [aαa] E ⊆ α nonreﬂexive: ∃a ∈ A [¬(aαa)] irreﬂexive: ∀a ∈ A [¬(aαa)]
55. 55. Reﬂexivity reﬂexive α ⊆ A × A ∀a ∈ A [aαa] E ⊆ α nonreﬂexive: ∃a ∈ A [¬(aαa)] irreﬂexive: ∀a ∈ A [¬(aαa)]
56. 56. Reﬂexivity reﬂexive α ⊆ A × A ∀a ∈ A [aαa] E ⊆ α nonreﬂexive: ∃a ∈ A [¬(aαa)] irreﬂexive: ∀a ∈ A [¬(aαa)]
57. 57. Reﬂexivity reﬂexive α ⊆ A × A ∀a ∈ A [aαa] E ⊆ α nonreﬂexive: ∃a ∈ A [¬(aαa)] irreﬂexive: ∀a ∈ A [¬(aαa)]
58. 58. Reﬂexivity Examples R1 ⊆ {1, 2} × {1, 2} R1 = {(1, 1), (1, 2), (2, 2)} R1 is reﬂexive R2 ⊆ {1, 2, 3} × {1, 2, 3} R2 = {(1, 1), (1, 2), (2, 2)} R2 is nonreﬂexive
59. 59. Reﬂexivity Examples R1 ⊆ {1, 2} × {1, 2} R1 = {(1, 1), (1, 2), (2, 2)} R1 is reﬂexive R2 ⊆ {1, 2, 3} × {1, 2, 3} R2 = {(1, 1), (1, 2), (2, 2)} R2 is nonreﬂexive
60. 60. Reﬂexivity Examples R ⊆ {1, 2, 3} × {1, 2, 3} R = {(1, 2), (2, 1), (2, 3)} R is irreﬂexive
61. 61. Reﬂexivity Examples R ⊆ Z × Z R = {(a, b) | ab ≥ 0} R is reﬂexive
62. 62. Symmetry symmetric α ⊆ A × A ∀a, b ∈ A [(a = b) ∨ (aαb ∧ bαa) ∨ (¬(aαb) ∧ ¬(bαa))] ∀a, b ∈ A [(a = b) ∨ (aαb ↔ bαa)] α−1 = α asymmetric: ∃a, b ∈ A [(a = b) ∧ ((aαb ∧ ¬(bαa)) ∨ (¬(aαb) ∧ bαa))]
63. 63. Symmetry symmetric α ⊆ A × A ∀a, b ∈ A [(a = b) ∨ (aαb ∧ bαa) ∨ (¬(aαb) ∧ ¬(bαa))] ∀a, b ∈ A [(a = b) ∨ (aαb ↔ bαa)] α−1 = α asymmetric: ∃a, b ∈ A [(a = b) ∧ ((aαb ∧ ¬(bαa)) ∨ (¬(aαb) ∧ bαa))]
64. 64. Symmetry symmetric α ⊆ A × A ∀a, b ∈ A [(a = b) ∨ (aαb ∧ bαa) ∨ (¬(aαb) ∧ ¬(bαa))] ∀a, b ∈ A [(a = b) ∨ (aαb ↔ bαa)] α−1 = α asymmetric: ∃a, b ∈ A [(a = b) ∧ ((aαb ∧ ¬(bαa)) ∨ (¬(aαb) ∧ bαa))]
65. 65. Symmetry symmetric α ⊆ A × A ∀a, b ∈ A [(a = b) ∨ (aαb ∧ bαa) ∨ (¬(aαb) ∧ ¬(bαa))] ∀a, b ∈ A [(a = b) ∨ (aαb ↔ bαa)] α−1 = α asymmetric: ∃a, b ∈ A [(a = b) ∧ ((aαb ∧ ¬(bαa)) ∨ (¬(aαb) ∧ bαa))]
66. 66. Antisymmetry antisymmetric: ∀a, b ∈ A [(a = b) ∨ (aαb → ¬(bαa))] ⇔ ∀a, b ∈ A [(a = b) ∨ ¬(aαb) ∨ ¬(bαa)] ⇔ ∀a, b ∈ A [¬(aαb ∧ bαa) ∨ (a = b)] ⇔ ∀a, b ∈ A [(aαb ∧ bαa) → (a = b)]
67. 67. Antisymmetry antisymmetric: ∀a, b ∈ A [(a = b) ∨ (aαb → ¬(bαa))] ⇔ ∀a, b ∈ A [(a = b) ∨ ¬(aαb) ∨ ¬(bαa)] ⇔ ∀a, b ∈ A [¬(aαb ∧ bαa) ∨ (a = b)] ⇔ ∀a, b ∈ A [(aαb ∧ bαa) → (a = b)]
68. 68. Antisymmetry antisymmetric: ∀a, b ∈ A [(a = b) ∨ (aαb → ¬(bαa))] ⇔ ∀a, b ∈ A [(a = b) ∨ ¬(aαb) ∨ ¬(bαa)] ⇔ ∀a, b ∈ A [¬(aαb ∧ bαa) ∨ (a = b)] ⇔ ∀a, b ∈ A [(aαb ∧ bαa) → (a = b)]
69. 69. Antisymmetry antisymmetric: ∀a, b ∈ A [(a = b) ∨ (aαb → ¬(bαa))] ⇔ ∀a, b ∈ A [(a = b) ∨ ¬(aαb) ∨ ¬(bαa)] ⇔ ∀a, b ∈ A [¬(aαb ∧ bαa) ∨ (a = b)] ⇔ ∀a, b ∈ A [(aαb ∧ bαa) → (a = b)]
70. 70. Symmetry Examples R ⊆ {1, 2, 3} × {1, 2, 3} R = {(1, 2), (2, 1), (2, 3)} R is asymmetric
71. 71. Symmetry Examples R ⊆ Z × Z R = {(a, b) | ab ≥ 0} R is symmetric
72. 72. Symmetry Examples R ⊆ {1, 2, 3} × {1, 2, 3} R = {(1, 1), (2, 2)} R is symmetric and antisymmetric
73. 73. Transitivity transitive α ⊆ A × A ∀a, b, c ∈ A [(aαb ∧ bαc) → (aαc)] α2 ⊆ α nontransitive: ∃a, b, c ∈ A [(aαb ∧ bαc) ∧ ¬(aαc)] antitransitive: ∀a, b, c ∈ A [(aαb ∧ bαc) → ¬(aαc)]
74. 74. Transitivity transitive α ⊆ A × A ∀a, b, c ∈ A [(aαb ∧ bαc) → (aαc)] α2 ⊆ α nontransitive: ∃a, b, c ∈ A [(aαb ∧ bαc) ∧ ¬(aαc)] antitransitive: ∀a, b, c ∈ A [(aαb ∧ bαc) → ¬(aαc)]
75. 75. Transitivity transitive α ⊆ A × A ∀a, b, c ∈ A [(aαb ∧ bαc) → (aαc)] α2 ⊆ α nontransitive: ∃a, b, c ∈ A [(aαb ∧ bαc) ∧ ¬(aαc)] antitransitive: ∀a, b, c ∈ A [(aαb ∧ bαc) → ¬(aαc)]
76. 76. Transitivity transitive α ⊆ A × A ∀a, b, c ∈ A [(aαb ∧ bαc) → (aαc)] α2 ⊆ α nontransitive: ∃a, b, c ∈ A [(aαb ∧ bαc) ∧ ¬(aαc)] antitransitive: ∀a, b, c ∈ A [(aαb ∧ bαc) → ¬(aαc)]
77. 77. Transitivity Examples R ⊆ {1, 2, 3} × {1, 2, 3} R = {(1, 2), (2, 1), (2, 3)} R is antitransitive
78. 78. Transitivity Examples R ⊆ Z × Z R = {(a, b) | ab ≥ 0} R is nontransitive
79. 79. Converse Relation Properties Theorem Reﬂexivity, symmetry and transitivity are preserved in the converse relation.
80. 80. Closures reﬂexive closure: rα = α ∪ E symmetric closure: sα = α ∪ α−1 transitive closure: tα = i=1,2,3,... αi = α ∪ α2 ∪ α3 ∪ · · ·
81. 81. Closures reﬂexive closure: rα = α ∪ E symmetric closure: sα = α ∪ α−1 transitive closure: tα = i=1,2,3,... αi = α ∪ α2 ∪ α3 ∪ · · ·
82. 82. Closures reﬂexive closure: rα = α ∪ E symmetric closure: sα = α ∪ α−1 transitive closure: tα = i=1,2,3,... αi = α ∪ α2 ∪ α3 ∪ · · ·
83. 83. Special Relations predecessor - successor R ⊆ Z × Z R = {(a, b) | a − b = 1} irreﬂexive antisymmetric antitransitive
84. 84. Special Relations adjacency R ⊆ Z × Z R = {(a, b) | |a − b| = 1} irreﬂexive symmetric antitransitive
85. 85. Special Relations strict order R ⊆ Z × Z R = {(a, b) | a < b} irreﬂexive antisymmetric transitive
86. 86. Special Relations partial order R ⊆ Z × Z R = {(a, b) | a ≤ b} reﬂexive antisymmetric transitive
87. 87. Special Relations preorder R ⊆ Z × Z R = {(a, b) | |a| ≤ |b|} reﬂexive asymmetric transitive
88. 88. Special Relations limited diﬀerence R ⊆ Z × Z, m ∈ Z+ R = {(a, b) | |a − b| ≤ m} reﬂexive symmetric nontransitive
89. 89. Special Relations comparability R ⊆ U × U R = {(a, b) | (a ⊆ b) ∨ (b ⊆ a)} reﬂexive symmetric nontransitive
90. 90. Special Relations siblings? irreﬂexive symmetric transitive can a relation be symmetric, transitive and irreﬂexive?
91. 91. Special Relations siblings? irreﬂexive symmetric transitive can a relation be symmetric, transitive and irreﬂexive?
92. 92. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
93. 93. Compatibility Relations Deﬁnition compatibility relation: γ reﬂexive symmetric when drawing, lines instead of arrows matrix representation as a triangle matrix αα−1 is a compatibility relation
94. 94. Compatibility Relations Deﬁnition compatibility relation: γ reﬂexive symmetric when drawing, lines instead of arrows matrix representation as a triangle matrix αα−1 is a compatibility relation
95. 95. Compatibility Relations Deﬁnition compatibility relation: γ reﬂexive symmetric when drawing, lines instead of arrows matrix representation as a triangle matrix αα−1 is a compatibility relation
96. 96. Compatibility Relation Example A = {a1, a2, a3, a4} R = {(a1, a1), (a2, a2), (a3, a3), (a4, a4), (a1, a2), (a2, a1), (a2, a4), (a4, a2), (a3, a4), (a4, a3)} 1 1 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 1
97. 97. Compatibility Relation Example A = {a1, a2, a3, a4} R = {(a1, a1), (a2, a2), (a3, a3), (a4, a4), (a1, a2), (a2, a1), (a2, a4), (a4, a2), (a3, a4), (a4, a3)} 1 1 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 1
98. 98. Compatibility Relation Example P: persons, L: languages P = {p1, p2, p3, p4, p5, p6} L = {l1, l2, l3, l4, l5} α ⊆ P × L Mα = 1 1 0 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 0 1 0 0 1 1 0 0 Mα−1 = 1 1 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 0 0
99. 99. Compatibility Relation Example P: persons, L: languages P = {p1, p2, p3, p4, p5, p6} L = {l1, l2, l3, l4, l5} α ⊆ P × L Mα = 1 1 0 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 0 1 0 0 1 1 0 0 Mα−1 = 1 1 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 0 0
100. 100. Compatibility Relation Example αα−1 ⊆ P × P Mαα−1 = 1 1 0 1 0 1 1 1 0 1 0 1 0 0 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 0 1
101. 101. Compatibility Block Deﬁnition compatibility block: C ⊆ A ∀a, b [a ∈ C ∧ b ∈ C → aγb] maximal compatibility block: not a subset of another compatibility block an element can be a member of more than one MCB complete cover: Cγ set of all MCBs
102. 102. Compatibility Block Deﬁnition compatibility block: C ⊆ A ∀a, b [a ∈ C ∧ b ∈ C → aγb] maximal compatibility block: not a subset of another compatibility block an element can be a member of more than one MCB complete cover: Cγ set of all MCBs
103. 103. Compatibility Block Deﬁnition compatibility block: C ⊆ A ∀a, b [a ∈ C ∧ b ∈ C → aγb] maximal compatibility block: not a subset of another compatibility block an element can be a member of more than one MCB complete cover: Cγ set of all MCBs
104. 104. Compatibility Block Example C1 = {p4, p6} C2 = {p2, p4, p6} C3 = {p1, p2, p4, p6} (MCB) Cγ = {{p1, p2, p4, p6}, {p3, p4, p6}, {p4, p5}}
105. 105. Compatibility Block Example C1 = {p4, p6} C2 = {p2, p4, p6} C3 = {p1, p2, p4, p6} (MCB) Cγ = {{p1, p2, p4, p6}, {p3, p4, p6}, {p4, p5}}
106. 106. Compatibility Block Example C1 = {p4, p6} C2 = {p2, p4, p6} C3 = {p1, p2, p4, p6} (MCB) Cγ = {{p1, p2, p4, p6}, {p3, p4, p6}, {p4, p5}}
107. 107. Equivalence Relations Deﬁnition equivalence relation: reﬂexive symmetric transitive equivalence classes (partitions) every element is a member of exactly one equivalence class complete cover: C
108. 108. Equivalence Relations Deﬁnition equivalence relation: reﬂexive symmetric transitive equivalence classes (partitions) every element is a member of exactly one equivalence class complete cover: C
109. 109. Equivalence Relation Example R ⊆ Z × Z R = {(a, b) | ∃m ∈ Z [a − b = 5m]} R partitions Z into 5 equivalence classes
110. 110. References Required Reading: Grimaldi Chapter 5: Relations and Functions 5.1. Cartesian Products and Relations Chapter 7: Relations: The Second Time Around 7.1. Relations Revisited: Properties of Relations 7.4. Equivalence Relations and Partitions
111. 111. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
112. 112. Functions Deﬁnition function: f : X → Y ∀x ∈ X ∀y1, y2 ∈ Y [(x, y1), (x, y2) ∈ f → y1 = y2] X: domain, Y : codomain (or range) y = f (x) : (x, y) ∈ f y: image of x under f f : X → Y , X ⊆ X subset image: f (X ) = {f (x) | x ∈ X }
113. 113. Functions Deﬁnition function: f : X → Y ∀x ∈ X ∀y1, y2 ∈ Y [(x, y1), (x, y2) ∈ f → y1 = y2] X: domain, Y : codomain (or range) y = f (x) : (x, y) ∈ f y: image of x under f f : X → Y , X ⊆ X subset image: f (X ) = {f (x) | x ∈ X }
114. 114. Functions Deﬁnition function: f : X → Y ∀x ∈ X ∀y1, y2 ∈ Y [(x, y1), (x, y2) ∈ f → y1 = y2] X: domain, Y : codomain (or range) y = f (x) : (x, y) ∈ f y: image of x under f f : X → Y , X ⊆ X subset image: f (X ) = {f (x) | x ∈ X }
115. 115. Subset Image Examples f : R → R f (x) = x2 f (Z) = {0, 1, 4, 9, 16, . . . } f ({−2, 1}) = {1, 4}
116. 116. Subset Image Examples f : R → R f (x) = x2 f (Z) = {0, 1, 4, 9, 16, . . . } f ({−2, 1}) = {1, 4}
117. 117. One-to-One Functions Deﬁnition f : X → Y is one-to-one (or injective): ∀x1, x2 ∈ X [f (x1) = f (x2) → x1 = x2]
118. 118. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
119. 119. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
120. 120. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
121. 121. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
122. 122. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
123. 123. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
124. 124. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
125. 125. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
126. 126. Onto Functions Deﬁnition f : X → Y is onto (or surjective): ∀y ∈ Y ∃x ∈ X [f (x) = y] f (X) = Y
127. 127. Onto Function Examples onto f : R → R f (x) = x3 not onto f : Z → Z f (x) = 3x + 1
128. 128. Onto Function Examples onto f : R → R f (x) = x3 not onto f : Z → Z f (x) = 3x + 1
129. 129. Bijective Functions Deﬁnition f : X → Y is bijective: f is one-to-one and onto
130. 130. Function Composition Deﬁnition f : X → Y , g : Y → Z g ◦ f : X → Z (g ◦ f )(x) = g(f (x)) not commutative associative: f ◦ (g ◦ h) = (f ◦ g) ◦ h
131. 131. Function Composition Deﬁnition f : X → Y , g : Y → Z g ◦ f : X → Z (g ◦ f )(x) = g(f (x)) not commutative associative: f ◦ (g ◦ h) = (f ◦ g) ◦ h
132. 132. Composition Commutativity Example f : R → R f (x) = x2 g : R → R g(x) = x + 5 g ◦ f : R → R (g ◦ f )(x) = x2 + 5 f ◦ g : R → R (f ◦ g)(x) = (x + 5)2
133. 133. Composition Commutativity Example f : R → R f (x) = x2 g : R → R g(x) = x + 5 g ◦ f : R → R (g ◦ f )(x) = x2 + 5 f ◦ g : R → R (f ◦ g)(x) = (x + 5)2
134. 134. Composition Commutativity Example f : R → R f (x) = x2 g : R → R g(x) = x + 5 g ◦ f : R → R (g ◦ f )(x) = x2 + 5 f ◦ g : R → R (f ◦ g)(x) = (x + 5)2
135. 135. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
136. 136. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
137. 137. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
138. 138. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
139. 139. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
140. 140. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is onto ∧ g is onto ⇒ g ◦ f is onto Proof. ∀z ∈ Z ∃y ∈ Y g(y) = z ∀y ∈ Y ∃x ∈ X f (x) = y ⇒ ∀z ∈ Z ∃x ∈ X g(f (x)) = z
141. 141. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is onto ∧ g is onto ⇒ g ◦ f is onto Proof. ∀z ∈ Z ∃y ∈ Y g(y) = z ∀y ∈ Y ∃x ∈ X f (x) = y ⇒ ∀z ∈ Z ∃x ∈ X g(f (x)) = z
142. 142. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is onto ∧ g is onto ⇒ g ◦ f is onto Proof. ∀z ∈ Z ∃y ∈ Y g(y) = z ∀y ∈ Y ∃x ∈ X f (x) = y ⇒ ∀z ∈ Z ∃x ∈ X g(f (x)) = z
143. 143. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is onto ∧ g is onto ⇒ g ◦ f is onto Proof. ∀z ∈ Z ∃y ∈ Y g(y) = z ∀y ∈ Y ∃x ∈ X f (x) = y ⇒ ∀z ∈ Z ∃x ∈ X g(f (x)) = z
144. 144. Identity Function Deﬁnition identity function: 1X 1X : X → X 1X (x) = x
145. 145. Inverse Function Deﬁnition f : X → Y is invertible: ∃f −1 : Y → X [f −1 ◦ f = 1X ∧ f ◦ f −1 = 1Y ] f −1: inverse of function f
146. 146. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
147. 147. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
148. 148. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
149. 149. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
150. 150. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
151. 151. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
152. 152. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
153. 153. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
154. 154. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
155. 155. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
156. 156. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
157. 157. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
158. 158. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
159. 159. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
160. 160. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
161. 161. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
162. 162. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
163. 163. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
164. 164. Invertible Function Theorem A function is invertible if and only if it is one-to-one and onto.
165. 165. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
166. 166. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
167. 167. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
168. 168. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
169. 169. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
170. 170. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
171. 171. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
172. 172. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
173. 173. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
174. 174. Invertible Function If bijective then invertible. f : X → Y f is onto ⇒ ∀y ∈ Y ∃x ∈ X f (x) = y let g : Y → X be deﬁned by x = g(y) is it possible that g(y) = x1 = x2 = g(y) ? this would mean: f (x1) = y = f (x2) but f is one-to-one
175. 175. Invertible Function If bijective then invertible. f : X → Y f is onto ⇒ ∀y ∈ Y ∃x ∈ X f (x) = y let g : Y → X be deﬁned by x = g(y) is it possible that g(y) = x1 = x2 = g(y) ? this would mean: f (x1) = y = f (x2) but f is one-to-one
176. 176. Invertible Function If bijective then invertible. f : X → Y f is onto ⇒ ∀y ∈ Y ∃x ∈ X f (x) = y let g : Y → X be deﬁned by x = g(y) is it possible that g(y) = x1 = x2 = g(y) ? this would mean: f (x1) = y = f (x2) but f is one-to-one
177. 177. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
178. 178. Pigeonhole Principle Deﬁnition pigeonhole principle (Dirichlet drawers): If m pigeons go into n holes and m > n, then at least one hole contains more than one pigeon. f : X → Y |X| > |Y | ⇒ f is not one-to-one ∃x1, x2 ∈ X [x1 = x2 ∧ f (x1) = f (x2)]
179. 179. Pigeonhole Principle Deﬁnition pigeonhole principle (Dirichlet drawers): If m pigeons go into n holes and m > n, then at least one hole contains more than one pigeon. f : X → Y |X| > |Y | ⇒ f is not one-to-one ∃x1, x2 ∈ X [x1 = x2 ∧ f (x1) = f (x2)]
180. 180. Pigeonhole Principle Examples Among 367 people, at least two have the same birthday. In an exam where the grades are integers between 0 and 100, how many students have to take the exam to make sure that at least two students will have the same grade?
181. 181. Pigeonhole Principle Examples Among 367 people, at least two have the same birthday. In an exam where the grades are integers between 0 and 100, how many students have to take the exam to make sure that at least two students will have the same grade?
182. 182. Generalized Pigeonhole Principle Deﬁnition generalized pigeonhole principle: If m objects are distributed to n drawers, then at least one of the drawers contains m/n objects. example Among 100 people, at least 100/12 = 9 were born in the same month.
183. 183. Generalized Pigeonhole Principle Deﬁnition generalized pigeonhole principle: If m objects are distributed to n drawers, then at least one of the drawers contains m/n objects. example Among 100 people, at least 100/12 = 9 were born in the same month.
184. 184. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 9}, T ⊂ S, |T| = 6 ∃s1, s2 ∈ T [s1 + s2 = 10]
185. 185. Pigeonhole Principle Example Theorem S ⊆ Z+, ∀a ∈ S [a ≤ 14], |S| = 6 T = P(S) − ∅ X = {ΣA | A ∈ T}, ΣA : sum of the elements in A |X| < |T| Proof Attempt holes: 1 ≤ ΣA ≤ 9 + · · · + 14 = 69 pigeons: 26 − 1 = 63 Proof. consider T − S holes: 1 ≤ sA ≤ 10 + · · · + 14 = 60 pigeons: 26 − 2 = 62
186. 186. Pigeonhole Principle Example Theorem S ⊆ Z+, ∀a ∈ S [a ≤ 14], |S| = 6 T = P(S) − ∅ X = {ΣA | A ∈ T}, ΣA : sum of the elements in A |X| < |T| Proof Attempt holes: 1 ≤ ΣA ≤ 9 + · · · + 14 = 69 pigeons: 26 − 1 = 63 Proof. consider T − S holes: 1 ≤ sA ≤ 10 + · · · + 14 = 60 pigeons: 26 − 2 = 62
187. 187. Pigeonhole Principle Example Theorem S ⊆ Z+, ∀a ∈ S [a ≤ 14], |S| = 6 T = P(S) − ∅ X = {ΣA | A ∈ T}, ΣA : sum of the elements in A |X| < |T| Proof Attempt holes: 1 ≤ ΣA ≤ 9 + · · · + 14 = 69 pigeons: 26 − 1 = 63 Proof. consider T − S holes: 1 ≤ sA ≤ 10 + · · · + 14 = 60 pigeons: 26 − 2 = 62
188. 188. Pigeonhole Principle Example Theorem S ⊆ Z+, ∀a ∈ S [a ≤ 14], |S| = 6 T = P(S) − ∅ X = {ΣA | A ∈ T}, ΣA : sum of the elements in A |X| < |T| Proof Attempt holes: 1 ≤ ΣA ≤ 9 + · · · + 14 = 69 pigeons: 26 − 1 = 63 Proof. consider T − S holes: 1 ≤ sA ≤ 10 + · · · + 14 = 60 pigeons: 26 − 2 = 62
189. 189. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] ﬁrst, show that: ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] then, use this to prove the main theorem
190. 190. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] ﬁrst, show that: ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] then, use this to prove the main theorem
191. 191. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
192. 192. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
193. 193. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
194. 194. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
195. 195. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
196. 196. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
197. 197. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
198. 198. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
199. 199. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
200. 200. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
201. 201. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
202. 202. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
203. 203. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] Proof. P = {p | p ∈ S, ∃i ∈ Z [p = 2i + 1]}, |P| = 100 f : S → P, r ∈ N, s = 2r p → f (s) = p |T| = 101 ⇒ at least two elements have the same image in P: f (s1) = f (s2) ⇒ 2r1 p = 2r2 p s1 s2 = 2r1 p 2r2 p = 2r1−r2
204. 204. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] Proof. P = {p | p ∈ S, ∃i ∈ Z [p = 2i + 1]}, |P| = 100 f : S → P, r ∈ N, s = 2r p → f (s) = p |T| = 101 ⇒ at least two elements have the same image in P: f (s1) = f (s2) ⇒ 2r1 p = 2r2 p s1 s2 = 2r1 p 2r2 p = 2r1−r2
205. 205. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] Proof. P = {p | p ∈ S, ∃i ∈ Z [p = 2i + 1]}, |P| = 100 f : S → P, r ∈ N, s = 2r p → f (s) = p |T| = 101 ⇒ at least two elements have the same image in P: f (s1) = f (s2) ⇒ 2r1 p = 2r2 p s1 s2 = 2r1 p 2r2 p = 2r1−r2
206. 206. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] Proof. P = {p | p ∈ S, ∃i ∈ Z [p = 2i + 1]}, |P| = 100 f : S → P, r ∈ N, s = 2r p → f (s) = p |T| = 101 ⇒ at least two elements have the same image in P: f (s1) = f (s2) ⇒ 2r1 p = 2r2 p s1 s2 = 2r1 p 2r2 p = 2r1−r2
207. 207. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
208. 208. Recursive Functions Deﬁnition recursive function: a function deﬁned in terms of itself f (n) = h(f (m)) inductively deﬁned function: a recursive function where the size is reduced at every step f (n) = k if n = 0 h(f (n − 1)) if n > 0
209. 209. Recursion Examples f 91(n) = n − 10 if n > 100 f 91(f 91(n + 11)) if n ≤ 100 n! = 1 if n = 0 n · (n − 1)! if n > 0
210. 210. Recursion Examples f 91(n) = n − 10 if n > 100 f 91(f 91(n + 11)) if n ≤ 100 n! = 1 if n = 0 n · (n − 1)! if n > 0
211. 211. Greatest Common Divisor gcd(a, b) = b if b | a gcd(b, a mod b) if b a gcd(333, 84) = gcd(84, 333 mod 84) = gcd(84, 81) = gcd(81, 84 mod 81) = gcd(81, 3) = 3
212. 212. Greatest Common Divisor gcd(a, b) = b if b | a gcd(b, a mod b) if b a gcd(333, 84) = gcd(84, 333 mod 84) = gcd(84, 81) = gcd(81, 84 mod 81) = gcd(81, 3) = 3
213. 213. Fibonacci Sequence Fn = ﬁb(n) =    1 if n = 1 1 if n = 2 ﬁb(n − 2) + ﬁb(n − 1) if n > 2 F1 F2 F3 F4 F5 F6 F7 F8 . . . 1 1 2 3 5 8 13 21 . . .
214. 214. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
215. 215. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
216. 216. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
217. 217. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
218. 218. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
219. 219. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
220. 220. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
221. 221. Ackermann Function ack(x, y) =    y + 1 if x = 0 ack(x − 1, 1) if y = 0 ack(x − 1, ack(x, y − 1)) if x > 0 ∧ y > 0
222. 222. References Required Reading: Grimaldi Chapter 5: Relations and Functions 5.2. Functions: Plain and One-to-One 5.3. Onto Functions: Stirling Numbers of the Second Kind 5.5. The Pigeonhole Principle 5.6. Function Composition and Inverse Functions