Discrete Mathematics: Relations, Lecture notes of Discrete Mathematics

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Discrete Mathematics

(Relations)

Pramod Ganapathi

Department of Computer Science State University of New York at Stony Brook

January 24, 2021

Are these functions?

Problem Are these functions? − rational p = rational q − m < n − d does not divide n − n leaves a remainder of 5 when divided by d − line 1 is parallel to line 2 − person a is a parent of person b − triangle t 1 is congruent to triangle t 2 − edge e 1 is adjacent to edge e 2 − matrix A is orthogonal to matrix B No! (Because an input is mapped to more than one output.) What are these mappings called? Relations!

Functions vs. relations

-10 -5 5 10

x

20

40

60

80

100

y

y = x^220 40 60 80 100 x

5

10

y x = y^2

y = x^2 y = ± √ x Function? 3 7 Relation? 3 3

Functions vs. relations

-10 -5 5 10

x

5

10

y

-10 -5 5 10

x

5

10

y

y = x y ≥ x Function? 3 7 Relation? 3 3

Example: Marriage relation

M 1

M 2

M 3

M 4

F 1

F 2

F 3

F 4

F 5

Male

Female

Example: Less than

Problem A relation L : R → R as follows. For all real numbers x and y, (x, y) ∈ L ⇔ x L y ⇔ x < y. Draw the graph of L as a subset of the Cartesian plane R × R. Solution L = {(− 10. 678 , 30 .23), (17. 13 , 45 .98), (100/ 9 , 200),.. .} Graph:

Inverse of a relation

Male Male

• Example: Congruence modulo
• M
• M
• M
• M
• F
• F
• F
• F
• F
• M Female
• M
• M
• M
• F
• F
• F
• F
• F

Example: Inverse of a finite relation

Problem Let A = { 2 , 3 , 4 } and B = { 2 , 6 , 8 }. Let R : A to B. For all (a, b) ∈ A × B, a R b ⇔ a | b Determine R and R−^1. Draw arrow diagrams for both. Describe R−^1 in words. Solution R = {(2, 2), (2, 6), (2, 8), (3, 6), (4, 8)} R−^1 = {(2, 2), (6, 2), (8, 2), (6, 3), (8, 4)} For all (b, a) ∈ B × A, (b, a) ∈ R−^1 ⇔ b is a multiple of a

Example: Inverse of an infinite relation

Problem Define a relation R from R to R as follows: For all (u, v) ∈ R × R, u R v ⇔ v = 2|u|. Draw the graphs of R and R−^1 in the Cartesian plane. Is R−^1 a function? Solution R−^1 is not a function. Why?

Example: Relation on a set

Problem Let A = { 3 , 4 , 5 , 6 , 7 , 8 }. Define relation R on A as follows. For all x, y ∈ A, x R y ⇔ 2 |(x − y). Draw the graph of R. Solution

Reflexivity, symmetry, and transitivity

Properties Set A = { 2 , 3 , 4 , 6 , 7 , 9 } Relation R on set A is: ∀x, y ∈ A, x R y ⇔ 3 | (x − y)

Reflexivity. ∀x ∈ A, (x, x) ∈ R. Symmetry. ∀x, y ∈ A, if (x, y) ∈ R, then (y, x) ∈ R. Transitivity. ∀x, y, z ∈ A, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.

Example

Problem A = { 0 , 1 , 2 , 3 }. R = {(0, 0), (0, 2), (0, 3), (2, 3)}. Is R reflexive, symmetric, and transitive? Solution

Not reflexive. e.g.: (1, 1) 6 ∈ R. ∃x ∈ A, (x, x) 6 ∈ R. Not symmetric. e.g.: (0, 3) ∈ R but (3, 0) 6 ∈ R. ∃x, y ∈ A, if (x, y) ∈ R, then (y, x) 6 ∈ R. Transitive. ∀x, y, z ∈ A, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.

Example

Problem A = { 0 , 1 , 2 , 3 }. R = {(0, 1), (2, 3)}. Is R reflexive, symmetric, and transitive? Solution

Not reflexive. e.g.: (0, 0) 6 ∈ R. ∃x ∈ A, (x, x) 6 ∈ R. Not symmetric. e.g.: (0, 1) ∈ R but (1, 0) 6 ∈ R. ∃x, y ∈ A, if (x, y) ∈ R, then (y, x) 6 ∈ R. Transitive. Why? ∀x, y, z ∈ A, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.