Properties of relations Slides to accompany Section 5.4 of Discrete - - PowerPoint PPT Presentation

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Properties of relations Slides to accompany Section 5.4 of Discrete - - PowerPoint PPT Presentation

Sep 13, 2022 233 likes •744 views

Properties of relations Slides to accompany Section 5.4 of Discrete Mathematics and Functional Programming Thomas VanDrunen Reflexivity A relation R on a set X is reflexive if every element is related to itself: x X , ( x , x ) R

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Properties of relations

Slides to accompany Section 5.4 of Discrete Mathematics and Functional Programming Thomas VanDrunen

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Reflexivity

A relation R on a set X is reflexive if every element is related to itself: ∀ x ∈ X, (x, x) ∈ R

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Symmetry

A relation R on a set X is symmet- ric if for every pair in the relation, the inverse of the pair also exists: ∀ x, y ∈ X, if (x, y) ∈ R then (y, x) ∈ R

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Transitivity

A relation R on a set X is tran- sitive if any time one element is related to a second and that sec-

  • nd is related to a third, then the

first is also related to the third: ∀ x, y, z ∈ X, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R

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Summary

∀ x ∈ X, (x, x) ∈ R ∀ x, y ∈ X, if (x, y) ∈ R then (y, x) ∈ R ∀ x, y, z ∈ X, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R

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Proof patterns

∀ x ∈ X, (x, x) ∈ R ∀ x, y ∈ X, if (x, y) ∈ R then (y, x) ∈ R ∀ x, y, z ∈ X, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R

Suppose x ∈ X. . . . Hence (x, x) ∈ R. Therefore R is re- flexive. Suppose x, y ∈ X. Further suppose (x, y) ∈ R. . . . Hence (y, x) ∈ R. Therefore R is sym- metric. Suppose x, y, z ∈ X. Further sup- pose (x, y) ∈ R and (y, z) ∈ R. . . . Hence (x, z) ∈ R. Therefore R is tran- sitive.

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Proof patterns—short versions

∀ x ∈ X, (x, x) ∈ R ∀ x, y ∈ X, if (x, y) ∈ R then (y, x) ∈ R ∀ x, y, z ∈ X, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R

Suppose x ∈ X. . . . Hence (x, x) ∈ R. Therefore R is re- flexive. Suppose (x, y) ∈ R. . . . Hence (y, x) ∈ R. Therefore R is sym- metric. Suppose (x, y) ∈ R and (y, z) ∈ R. . . . Hence (x, z) ∈ R. Therefore R is tran- sitive.

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Reflexivity example

Proposition 1

The relation | on N is reflexive.

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Reflexivity example

Proposition 1

The relation | on N is reflexive.

  • Proof. Suppose a ∈ N.
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Reflexivity example

Proposition 1

The relation | on N is reflexive.

  • Proof. Suppose a ∈ N.

By arithmetic a · 1 = a

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Reflexivity example

Proposition 1

The relation | on N is reflexive.

  • Proof. Suppose a ∈ N.

By arithmetic a · 1 = a, and so by the definition of divides, a|a.

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Reflexivity example

Proposition 1

The relation | on N is reflexive.

  • Proof. Suppose a ∈ N.

By arithmetic a · 1 = a, and so by the definition of divides, a|a. Hence, by the definition of reflexive, | is reflexive.

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Symmetry example

Proposition 2

The relation “is opposite of” on Z is symmetric.

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Symmetry example

Proposition 2

The relation “is opposite of” on Z is symmetric.

  • Proof. Suppose x, y ∈ Z.
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Symmetry example

Proposition 2

The relation “is opposite of” on Z is symmetric.

  • Proof. Suppose x, y ∈ Z. Further suppose x + y = 0.
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Symmetry example

Proposition 2

The relation “is opposite of” on Z is symmetric.

  • Proof. Suppose x, y ∈ Z. Further suppose x + y = 0.

y + x = x + y by commutativity of addition.

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Symmetry example

Proposition 2

The relation “is opposite of” on Z is symmetric.

  • Proof. Suppose x, y ∈ Z. Further suppose x + y = 0.

y + x = x + y by commutativity of addition. y + x = 0 by substitution.

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Symmetry example

Proposition 2

The relation “is opposite of” on Z is symmetric.

  • Proof. Suppose x, y ∈ Z. Further suppose x + y = 0.

y + x = x + y by commutativity of addition. y + x = 0 by substitution. Therefore “is opposite of” is symmetric.

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Transitivity example

Proposition 3

The relation | on Z is transitive.

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Transitivity example

Proposition 3

The relation | on Z is transitive.

  • Proof. Suppose a, b, c ∈ Z, and suppose a|b and b|c.
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Transitivity example

Proposition 3

The relation | on Z is transitive.

  • Proof. Suppose a|b and b|c.
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Transitivity example

Proposition 3

The relation | on Z is transitive.

  • Proof. Suppose a|b and b|c.

By the definition of divides, there exist d, e ∈ Z such that a · d = b and b · e = c.

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Transitivity example

Proposition 3

The relation | on Z is transitive.

  • Proof. Suppose a|b and b|c.

By the definition of divides, there exist d, e ∈ Z such that a · d = b and b · e = c. By substitution and associativity, a(d · e) = c.

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Transitivity example

Proposition 3

The relation | on Z is transitive.

  • Proof. Suppose a|b and b|c.

By the definition of divides, there exist d, e ∈ Z such that a · d = b and b · e = c. By substitution and associativity, a(d · e) = c. By the definition of divides, a|c.

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Transitivity example

Proposition 3

The relation | on Z is transitive.

  • Proof. Suppose a|b and b|c.

By the definition of divides, there exist d, e ∈ Z such that a · d = b and b · e = c. By substitution and associativity, a(d · e) = c. By the definition of divides, a|c. Hence | is transitive.

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Reflexivity example

Proposition 4

If R is reflexive, then iA ⊆ R.

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Reflexivity example

Proposition 4

If R is reflexive, then iA ⊆ R.

  • Proof. Suppose R is reflexive.
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Reflexivity example

Proposition 4

If R is reflexive, then iA ⊆ R.

  • Proof. Suppose R is reflexive. Further suppose that

(a, b) ∈ iA.

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Reflexivity example

Proposition 4

If R is reflexive, then iA ⊆ R.

  • Proof. Suppose R is reflexive. Further suppose that

(a, b) ∈ iA. By definition of identity relation, a = b.

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Reflexivity example

Proposition 4

If R is reflexive, then iA ⊆ R.

  • Proof. Suppose R is reflexive. Further suppose that

(a, b) ∈ iA. By definition of identity relation, a = b. By definition of reflexivity, since R is reflexive, (a, b) ∈ R.

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Reflexivity example

Proposition 4

If R is reflexive, then iA ⊆ R.

  • Proof. Suppose R is reflexive. Further suppose that

(a, b) ∈ iA. By definition of identity relation, a = b. By definition of reflexivity, since R is reflexive, (a, b) ∈ R. Therefore, by definition of subset, iA ⊆ R.

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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

  • Proof. Suppose R is a relation on a set A.
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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

  • Proof. Suppose R is a relation on a set A. Next,

suppose a, b ∈ A.

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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

  • Proof. Suppose R is a relation on a set A. Next,

suppose a, b ∈ A. Finally, suppose (a, b) ∈ R ∩ R−1.

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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

  • Proof. Suppose R is a relation on a set A. Further

suppose (a, b) ∈ R ∩ R−1.

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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

  • Proof. Suppose R is a relation on a set A. Further

suppose (a, b) ∈ R ∩ R−1. By definition of intersection, (a, b) ∈ R and (a, b) ∈ R−1.

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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

  • Proof. Suppose R is a relation on a set A. Further

suppose (a, b) ∈ R ∩ R−1. By definition of intersection, (a, b) ∈ R and (a, b) ∈ R−1. Since (a, b) ∈ R, the definition of inverse tells us that (b, a) ∈ R−1. Similarly, since (a, b) ∈ R−1, by definition

  • f inverse it is also the case that (b, a) ∈ R.
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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

  • Proof. Suppose R is a relation on a set A. Further

suppose (a, b) ∈ R ∩ R−1. By definition of intersection, (a, b) ∈ R and (a, b) ∈ R−1. Since (a, b) ∈ R, the definition of inverse tells us that (b, a) ∈ R−1. Similarly, since (a, b) ∈ R−1, by definition

  • f inverse it is also the case that (b, a) ∈ R.

By definition of intersection, (b, a) ∈ R ∩ R−1.

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Symmetry example

Proposition 5

If R is a relation on a set A, then R ∩ R−1 is symmetric.

  • Proof. Suppose R is a relation on a set A. Further

suppose (a, b) ∈ R ∩ R−1. By definition of intersection, (a, b) ∈ R and (a, b) ∈ R−1. Since (a, b) ∈ R, the definition of inverse tells us that (b, a) ∈ R−1. Similarly, since (a, b) ∈ R−1, by definition

  • f inverse it is also the case that (b, a) ∈ R.

By definition of intersection, (b, a) ∈ R ∩ R−1. Therefore R ∩ R−1 is symmetric by definition.

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Transitivity example

Proposition 6

If R is a relation on A and for all a ∈ A, IR(IR(a)) ⊆ IR(a)), then R is transitive.

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Transitivity example

Proposition 6

If R is a relation on A and for all a ∈ A, IR(IR(a)) ⊆ IR(a)), then R is transitive.

  • Proof. Suppose R is a relation on A and for all a ∈ A,

IR(IR(a)) ⊆ IR(a)).

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Transitivity example

Proposition 6

If R is a relation on A and for all a ∈ A, IR(IR(a)) ⊆ IR(a)), then R is transitive.

  • Proof. Suppose R is a relation on A and for all a ∈ A,

IR(IR(a)) ⊆ IR(a)). Further suppose that (b, c), (c, d) ∈ R.

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Transitivity example

Proposition 6

If R is a relation on A and for all a ∈ A, IR(IR(a)) ⊆ IR(a)), then R is transitive.

  • Proof. Suppose R is a relation on A and for all a ∈ A,

IR(IR(a)) ⊆ IR(a)). Further suppose that (b, c), (c, d) ∈ R. By definition of image, c ∈ IR(b).

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Transitivity example

Proposition 6

If R is a relation on A and for all a ∈ A, IR(IR(a)) ⊆ IR(a)), then R is transitive.

  • Proof. Suppose R is a relation on A and for all a ∈ A,

IR(IR(a)) ⊆ IR(a)). Further suppose that (b, c), (c, d) ∈ R. By definition of image, c ∈ IR(b). By definition of image, d ∈ IR(IR(b))

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Transitivity example

Proposition 6

If R is a relation on A and for all a ∈ A, IR(IR(a)) ⊆ IR(a)), then R is transitive.

  • Proof. Suppose R is a relation on A and for all a ∈ A,

IR(IR(a)) ⊆ IR(a)). Further suppose that (b, c), (c, d) ∈ R. By definition of image, c ∈ IR(b). By definition of image, d ∈ IR(IR(b)) By definition of subset, d ∈ IR(b).

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Transitivity example

Proposition 6

If R is a relation on A and for all a ∈ A, IR(IR(a)) ⊆ IR(a)), then R is transitive.

  • Proof. Suppose R is a relation on A and for all a ∈ A,

IR(IR(a)) ⊆ IR(a)). Further suppose that (b, c), (c, d) ∈ R. By definition of image, c ∈ IR(b). By definition of image, d ∈ IR(IR(b)) By definition of subset, d ∈ IR(b). By definition of image, (b, d) ∈ R.

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Transitivity example

Proposition 6

If R is a relation on A and for all a ∈ A, IR(IR(a)) ⊆ IR(a)), then R is transitive.

  • Proof. Suppose R is a relation on A and for all a ∈ A,

IR(IR(a)) ⊆ IR(a)). Further suppose that (b, c), (c, d) ∈ R. By definition of image, c ∈ IR(b). By definition of image, d ∈ IR(IR(b)) By definition of subset, d ∈ IR(b). By definition of image, (b, d) ∈ R. Therefore R is transitive by definition.

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Proof patterns

∀ x ∈ X, (x, x) ∈ R ∀ x, y ∈ X, if (x, y) ∈ R then (y, x) ∈ R ∀ x, y, z ∈ X, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R

Suppose x ∈ X. . . . Hence (x, x) ∈ R. Therefore R is re- flexive. Suppose x, y ∈ X. Further suppose (x, y) ∈ R. . . . Hence (y, x) ∈ R. Therefore R is sym- metric. Suppose x, y, z ∈ X. Further sup- pose (x, y) ∈ R and (y, z) ∈ R. . . . Hence (x, z) ∈ R. Therefore R is tran- sitive.

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Properties of relations

Slides to accompany Section 5.4 of Discrete Mathematics and Functional Programming Thomas VanDrunen