SlideShare a Scribd company logo
1 of 46
Relations and their Properties
Lecture 13, CMSC 56
Allyn Joy D. Calcaben
What is Relations?
Relations
Relationships between elements of sets are represented using the
structure called a relation, which is just a subset of the Cartesian
product of the sets.
It can be used to solve problems such as determining which pairs
of cities are linked by airline flights in a network, finding a viable
order for the different phases of a complicated project, or
producing a useful way to store information in computer
databases.
Relations & their Properties
The most direct way to express a relationship between elements
of two sets is to use ordered pairs made up of two related
elements. For this reason, sets of ordered pairs are called binary
relations.
Relations & their Properties
Definition 1
Let A and B be sets.
A binary relation from A to B is a subset of A × B.
Relations & their Properties
We use the notation a R b to denote that (a, b) ∈ R and a R b to
denote that (a, b) / ∈ R. Moreover, when (a, b) belongs to R, a is said to be
related to b by R.
Example
Let A be the set of cities in the U.S.A., and let B be the set of the
50 states in the U.S.A.
Solution
Let A be the set of cities in the U.S.A., and let B be the set of the
50 states in the U.S.A.
Define the relation R by specifying that (a, b) belongs to R if a city with
name a is in the state b.
Solution
Let A be the set of cities in the U.S.A., and let B be the set of the
50 states in the U.S.A.
Define the relation R by specifying that (a, b) belongs to R if a city with
name a is in the state b.
(Boulder, Colorado), (Bangor, Maine), (Ann Arbor, Michigan), (Middletown,
New Jersey), (Middletown, New York), (Cupertino, California), and (Red
Bank, New Jersey) are in R.
Example
Let A = {0, 1, 2} and B = {a, b}. Then {(0, a), (0, b), (1, a), (2, b)} is a
relation from A to B. This means, for instance, that 0 R a, but that
1 R b. Relations can be represented graphically.
Solution
Let A = {0, 1, 2} and B = {a, b}. Then {(0, a), (0, b), (1, a), (2, b)} is a
relation from A to B. This means, for instance, that 0 R a, but that
1 R b. Relations can be represented graphically.
0 •
• a
1 •
• b
2 •
R a b
0 x x
1 x
2 x
Functions as Relations
Functions as Relations
Definition 2
A relation on a set A is a relation from A to A.
(In other words, a relation on a set A is a subset of A × A.)
Example
Which of these relations contain each of the pairs (1, 1), (1, 2),
(2, 1), (1,−1), and (2, 2)?
Example
Which of these relations contain each of the pairs (1, 1), (1, 2),
(2, 1), (1,−1), and (2, 2)?
The pair
(1, 1) is in R1, R3, R4, and R6;
(1, 2) is in R1 and R6;
(2, 1) is in R2, R5, and R6;
(1,−1) is in R2, R3, and R6; and,
(2, 2) is in R1, R3, and R4.
Properties of Relations
Functions as Relations
Definition 3
A relation R on a set A is called reflexive
if (a, a) ∈ R for every element a ∈ A.
Example
Which of these relations are reflexive?
Solution
Which of these relations are reflexive?
The relations R3 and R5 are reflexive because they both contain all
pairs of the form (a, a), namely, (1, 1), (2, 2), (3, 3), and (4, 4). The
other relations are not reflexive because they do not contain all of
these ordered pairs. In particular, R1, R2, R4, and R6 are not
reflexive.
Example
Which of these relations are reflexive?
Solution
Which of these relations are reflexive?
The reflexive relations are R1 (because a ≤ a for every integer a),
R3, and R4. For each of the other relations in this example it is
easy to find a pair of the form (a, a) that is not in the relation.
Functions as Relations
Definition 4
A relation R on a set A is called symmetric
if (b, a) ∈ R whenever (a, b) ∈ R, for all a, b ∈ A.
A relation R on a set A such that for all a, b ∈ A,
if (a, b) ∈ R and (b, a) ∈ R, then a = b is called antisymmetric.
Example
Which of the relations are symmetric and antisymmetric?
Solution
Which of the relations are symmetric and antisymmetric?
The relations R2 and R3 are symmetric, because in each case (b, a) belongs to the
relation whenever (a, b) does.
For R2, the only thing to check is that both (2, 1) and (1, 2) are in the relation.
For R3, it is necessary to check that both (1, 2) and (2, 1) belong to the relation, and
(1, 4) and (4, 1) belong to the relation.
Solution
Which of the relations are symmetric and antisymmetric?
R4, R5, and R6 are all antisymmetric. For each of these relations there is no pair of
elements a and b with a = b such that both (a, b) and (b, a) belong to the relation.
Take Note:
To verify that the relations is symmetric. This is done by finding
a pair (a, b) with a = b such that (a, b) and (b, a) are both in the
relation.
To verify that the relations is antisymmetric. This is done by
finding a pair (a, b) such that it is in the relation but (b, a) is not.
Example
Which of the relations are symmetric and antisymmetric?
Solution
Which of the relations are symmetric and antisymmetric?
The relations R3, R4, and R6 are symmetric.
R3 is symmetric, for if a = b or a = −b, then b = a or b = −a.
R4 is symmetric because a = b implies that b = a.
R6 is symmetric because a + b ≤ 3 implies that b + a ≤ 3.
Solution
Which of the relations are symmetric and antisymmetric?
The relations R1, R2, R4, and R5 are antisymmetric.
R1 is antisymmetric because the inequalities a ≤ b and b ≤ a imply that a = b.
R2 is antisymmetric because it is impossible that a > b and b > a.
R4 is antisymmetric, because two elements are related with respect to R4 if and only if
they are equal.
R5 is antisymmetric because it is impossible that a = b + 1 and b = a + 1.
Functions as Relations
Definition 5
A relation R on a set A is called transitive
if whenever (a, b) ∈ R and (b, c) ∈ R,
then (a, c) ∈ R, for all a, b, c ∈ A.
Example
Which of the relations are transitive?
Solution
Which of the relations are transitive?
R4, R5, and R6 are transitive. For each of these relations, we can show that it is transitive
by verifying that if (a, b) and (b, c) belong to this relation, then (a, c) also does. For
instance, R4 is transitive, because (3, 2) and (2, 1), (4, 2) and (2, 1), (4, 3) and (3, 1), and
(4, 3) and (3, 2) are the only such sets of pairs, and (3, 1), (4, 1), and (4, 2) belong to R4.
Solution
Which of the relations are transitive?
R1 is not transitive because (3, 4) and (4, 1) belong to R1, but (3, 1) does not.
R2 is not transitive because (2, 1) and (1, 2) belong to R2, but (2, 2) does not.
R3 is not transitive because (4, 1) and (1, 2) belong to R3, but (4, 2) does not.
Example
Which of the relations are transitive?
Solution
Which of the relations are transitive?
The relations R1, R2, R3, and R4 are transitive.
R1 is transitive because a ≤ b and b ≤ c imply that a ≤ c.
R2 is transitive because a > b and b > c imply that a > c.
R3 is transitive because a = ±b and b = ±c imply that a = ±c.
R4 is clearly transitive, because a = b and b = c imply that a = c.
Solution
Which of the relations are transitive?
R5 is not transitive because (2, 1) and (1, 0) belong to R5, but (2, 0) does not.
R6 is not transitive because (2, 1) and (1, 2) belong to R6, but (2, 2) does not.
Combining Relations
Example
Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations:
R1 = {(1, 1), (2, 2), (3, 3)} , and
R2 = {(1, 1), (1, 2), (1, 3), (1, 4)}
can be combined to obtain:
1. R1 ∪ R2
2. R1 ∩ R2
3. R1 − R2
4. R2 − R1
Solution
Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations:
R1 = {(1, 1), (2, 2), (3, 3)} , and
R2 = {(1, 1), (1, 2), (1, 3), (1, 4)}
can be combined to obtain:
1. R1 ∪ R2 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)},
2. R1 ∩ R2
3. R1 − R2
4. R2 − R1
Solution
Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations:
R1 = {(1, 1), (2, 2), (3, 3)} , and
R2 = {(1, 1), (1, 2), (1, 3), (1, 4)}
can be combined to obtain:
1. R1 ∪ R2 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)},
2. R1 ∩ R2 = {(1, 1)},
3. R1 − R2
4. R2 − R1
Solution
Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations:
R1 = {(1, 1), (2, 2), (3, 3)} , and
R2 = {(1, 1), (1, 2), (1, 3), (1, 4)}
can be combined to obtain:
1. R1 ∪ R2 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)},
2. R1 ∩ R2 = {(1, 1)},
3. R1 − R2 = {(2, 2), (3, 3)},
4. R2 − R1
Solution
Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations:
R1 = {(1, 1), (2, 2), (3, 3)} , and
R2 = {(1, 1), (1, 2), (1, 3), (1, 4)}
can be combined to obtain:
1. R1 ∪ R2 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)},
2. R1 ∩ R2 = {(1, 1)},
3. R1 − R2 = {(2, 2), (3, 3)},
4. R2 − R1 = {(1, 2), (1, 3), (1, 4)}.
Combining Relations
Definition 6
Let R be a relation from a set A to a set B and S a relation from B
to a set C. The composite of R and S is the relation consisting of
ordered pairs (a, c), where a ∈ A, c ∈ C, and for which there exists
an element b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. We denote the
composite of R and S by S ◦ R.
Example
What is the composite of the relations R and S, where
R is the relation from {1, 2, 3} to {1, 2, 3, 4} with
R = {(1, 1), (1, 4), (2, 3), (3, 1), (3, 4)} and
S is the relation from {1, 2, 3, 4} to {0, 1, 2} with
S = {(1, 0), (2, 0), (3, 1), (3, 2), (4, 1)}?
Solution
What is the composite of the relations R and S, where
R is the relation from {1, 2, 3} to {1, 2, 3, 4} with
R = {(1, 1), (1, 4), (2, 3), (3, 1), (3, 4)} and
S is the relation from {1, 2, 3, 4} to {0, 1, 2} with
S = {(1, 0), (2, 0), (3, 1), (3, 2), (4, 1)}?
S ◦ R = { (1, 0), (1, 1), (2, 1), (2, 2), (3, 0), (3, 1) }
Any Question?

More Related Content

What's hot

Mathematical Induction
Mathematical InductionMathematical Induction
Mathematical Induction
Edelyn Cagas
 
Discrete Mathematics Lecture
Discrete Mathematics LectureDiscrete Mathematics Lecture
Discrete Mathematics Lecture
Genie Rose Santos
 
Mathematical induction
Mathematical inductionMathematical induction
Mathematical induction
Sman Abbasi
 
Discrete Structures. Lecture 1
 Discrete Structures. Lecture 1  Discrete Structures. Lecture 1
Discrete Structures. Lecture 1
Ali Usman
 
Pigeonhole Principle
Pigeonhole PrinciplePigeonhole Principle
Pigeonhole Principle
nielsoli
 

What's hot (20)

Eucledian algorithm for gcd of integers and polynomials
Eucledian algorithm for gcd of integers and polynomialsEucledian algorithm for gcd of integers and polynomials
Eucledian algorithm for gcd of integers and polynomials
 
Discrete Mathematics - Mathematics For Computer Science
Discrete Mathematics -  Mathematics For Computer ScienceDiscrete Mathematics -  Mathematics For Computer Science
Discrete Mathematics - Mathematics For Computer Science
 
Mathematical Induction
Mathematical InductionMathematical Induction
Mathematical Induction
 
Relations
RelationsRelations
Relations
 
Propositional logic
Propositional logicPropositional logic
Propositional logic
 
21 monotone sequences x
21 monotone sequences x21 monotone sequences x
21 monotone sequences x
 
Discrete Math Lecture 01: Propositional Logic
Discrete Math Lecture 01: Propositional LogicDiscrete Math Lecture 01: Propositional Logic
Discrete Math Lecture 01: Propositional Logic
 
Discrete Mathematics Lecture
Discrete Mathematics LectureDiscrete Mathematics Lecture
Discrete Mathematics Lecture
 
Predicates and Quantifiers
Predicates and QuantifiersPredicates and Quantifiers
Predicates and Quantifiers
 
Chapter 2: Relations
Chapter 2: RelationsChapter 2: Relations
Chapter 2: Relations
 
Conditional Probability
Conditional ProbabilityConditional Probability
Conditional Probability
 
number theory.ppt
number theory.pptnumber theory.ppt
number theory.ppt
 
Mathematical induction
Mathematical inductionMathematical induction
Mathematical induction
 
Discrete Structures. Lecture 1
 Discrete Structures. Lecture 1  Discrete Structures. Lecture 1
Discrete Structures. Lecture 1
 
Set Operations
Set OperationsSet Operations
Set Operations
 
Relations
RelationsRelations
Relations
 
Pigeonhole Principle
Pigeonhole PrinciplePigeonhole Principle
Pigeonhole Principle
 
Dm2021 binary operations
Dm2021 binary operationsDm2021 binary operations
Dm2021 binary operations
 
Mathematical induction
Mathematical inductionMathematical induction
Mathematical induction
 
Number theory
Number theoryNumber theory
Number theory
 

Similar to CMSC 56 | Lecture 13: Relations and their Properties

dm_13_RelationsAndTheirProperties (1).pdf
dm_13_RelationsAndTheirProperties (1).pdfdm_13_RelationsAndTheirProperties (1).pdf
dm_13_RelationsAndTheirProperties (1).pdf
SanjanaAdri
 
Discrete-Chapter 08 Relations
Discrete-Chapter 08 RelationsDiscrete-Chapter 08 Relations
Discrete-Chapter 08 Relations
Wongyos Keardsri
 
Relations
RelationsRelations
Relations
Gaditek
 

Similar to CMSC 56 | Lecture 13: Relations and their Properties (20)

dm_13_RelationsAndTheirProperties (1).pdf
dm_13_RelationsAndTheirProperties (1).pdfdm_13_RelationsAndTheirProperties (1).pdf
dm_13_RelationsAndTheirProperties (1).pdf
 
dm_13_RelationsAndTheirProperties (1).pptx
dm_13_RelationsAndTheirProperties (1).pptxdm_13_RelationsAndTheirProperties (1).pptx
dm_13_RelationsAndTheirProperties (1).pptx
 
Per5 relasi
Per5 relasiPer5 relasi
Per5 relasi
 
Relations
RelationsRelations
Relations
 
Properties of relations
Properties of relationsProperties of relations
Properties of relations
 
Discrete-Chapter 08 Relations
Discrete-Chapter 08 RelationsDiscrete-Chapter 08 Relations
Discrete-Chapter 08 Relations
 
Relations
RelationsRelations
Relations
 
Relation in Discrete Mathematics
Relation in Discrete Mathematics Relation in Discrete Mathematics
Relation in Discrete Mathematics
 
Final relation1 m_tech(cse)
Final relation1 m_tech(cse)Final relation1 m_tech(cse)
Final relation1 m_tech(cse)
 
Final relation1 m_tech(cse)
Final relation1 m_tech(cse)Final relation1 m_tech(cse)
Final relation1 m_tech(cse)
 
Final relation1 m_tech(cse)
Final relation1 m_tech(cse)Final relation1 m_tech(cse)
Final relation1 m_tech(cse)
 
Relations
RelationsRelations
Relations
 
Inverse of Relations Presentation.pptx
Inverse of Relations   Presentation.pptxInverse of Relations   Presentation.pptx
Inverse of Relations Presentation.pptx
 
Introductions to Relations
Introductions to RelationsIntroductions to Relations
Introductions to Relations
 
Relations
RelationsRelations
Relations
 
Relations
RelationsRelations
Relations
 
Sadat sumon
Sadat sumonSadat sumon
Sadat sumon
 
Relations between two sets
Relations between two setsRelations between two sets
Relations between two sets
 
Lemh101
Lemh101Lemh101
Lemh101
 
Week 5 ( basic concept of relation )
Week 5 ( basic concept of relation )Week 5 ( basic concept of relation )
Week 5 ( basic concept of relation )
 

More from allyn joy calcaben

More from allyn joy calcaben (16)

CMSC 56 | Lecture 17: Matrices
CMSC 56 | Lecture 17: MatricesCMSC 56 | Lecture 17: Matrices
CMSC 56 | Lecture 17: Matrices
 
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program CorrectnessCMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness
 
CMSC 56 | Lecture 11: Mathematical Induction
CMSC 56 | Lecture 11: Mathematical InductionCMSC 56 | Lecture 11: Mathematical Induction
CMSC 56 | Lecture 11: Mathematical Induction
 
CMSC 56 | Lecture 10: Integer Representations & Algorithms
CMSC 56 | Lecture 10: Integer Representations & AlgorithmsCMSC 56 | Lecture 10: Integer Representations & Algorithms
CMSC 56 | Lecture 10: Integer Representations & Algorithms
 
CMSC 56 | Lecture 9: Functions Representations
CMSC 56 | Lecture 9: Functions RepresentationsCMSC 56 | Lecture 9: Functions Representations
CMSC 56 | Lecture 9: Functions Representations
 
CMSC 56 | Lecture 8: Growth of Functions
CMSC 56 | Lecture 8: Growth of FunctionsCMSC 56 | Lecture 8: Growth of Functions
CMSC 56 | Lecture 8: Growth of Functions
 
CMSC 56 | Lecture 4: Rules of Inference
CMSC 56 | Lecture 4: Rules of InferenceCMSC 56 | Lecture 4: Rules of Inference
CMSC 56 | Lecture 4: Rules of Inference
 
CMSC 56 | Lecture 6: Sets & Set Operations
CMSC 56 | Lecture 6: Sets & Set OperationsCMSC 56 | Lecture 6: Sets & Set Operations
CMSC 56 | Lecture 6: Sets & Set Operations
 
CMSC 56 | Lecture 5: Proofs Methods and Strategy
CMSC 56 | Lecture 5: Proofs Methods and StrategyCMSC 56 | Lecture 5: Proofs Methods and Strategy
CMSC 56 | Lecture 5: Proofs Methods and Strategy
 
CMSC 56 | Lecture 3: Predicates & Quantifiers
CMSC 56 | Lecture 3: Predicates & QuantifiersCMSC 56 | Lecture 3: Predicates & Quantifiers
CMSC 56 | Lecture 3: Predicates & Quantifiers
 
CMSC 56 | Lecture 2: Propositional Equivalences
CMSC 56 | Lecture 2: Propositional EquivalencesCMSC 56 | Lecture 2: Propositional Equivalences
CMSC 56 | Lecture 2: Propositional Equivalences
 
CMSC 56 | Lecture 1: Propositional Logic
CMSC 56 | Lecture 1: Propositional LogicCMSC 56 | Lecture 1: Propositional Logic
CMSC 56 | Lecture 1: Propositional Logic
 
La Solidaridad and the Propaganda Movement
La Solidaridad and the Propaganda MovementLa Solidaridad and the Propaganda Movement
La Solidaridad and the Propaganda Movement
 
Computer Vision: Feature matching with RANSAC Algorithm
Computer Vision: Feature matching with RANSAC AlgorithmComputer Vision: Feature matching with RANSAC Algorithm
Computer Vision: Feature matching with RANSAC Algorithm
 
Chapter 7 expressions and assignment statements ii
Chapter 7 expressions and assignment statements iiChapter 7 expressions and assignment statements ii
Chapter 7 expressions and assignment statements ii
 
#11 osteoporosis
#11 osteoporosis#11 osteoporosis
#11 osteoporosis
 

Recently uploaded

Spellings Wk 4 and Wk 5 for Grade 4 at CAPS
Spellings Wk 4 and Wk 5 for Grade 4 at CAPSSpellings Wk 4 and Wk 5 for Grade 4 at CAPS
Spellings Wk 4 and Wk 5 for Grade 4 at CAPS
AnaAcapella
 
SURVEY I created for uni project research
SURVEY I created for uni project researchSURVEY I created for uni project research
SURVEY I created for uni project research
CaitlinCummins3
 
SPLICE Working Group: Reusable Code Examples
SPLICE Working Group:Reusable Code ExamplesSPLICE Working Group:Reusable Code Examples
SPLICE Working Group: Reusable Code Examples
Peter Brusilovsky
 

Recently uploaded (20)

OS-operating systems- ch05 (CPU Scheduling) ...
OS-operating systems- ch05 (CPU Scheduling) ...OS-operating systems- ch05 (CPU Scheduling) ...
OS-operating systems- ch05 (CPU Scheduling) ...
 
Observing-Correct-Grammar-in-Making-Definitions.pptx
Observing-Correct-Grammar-in-Making-Definitions.pptxObserving-Correct-Grammar-in-Making-Definitions.pptx
Observing-Correct-Grammar-in-Making-Definitions.pptx
 
OSCM Unit 2_Operations Processes & Systems
OSCM Unit 2_Operations Processes & SystemsOSCM Unit 2_Operations Processes & Systems
OSCM Unit 2_Operations Processes & Systems
 
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
Spring gala 2024 photo slideshow - Celebrating School-Community PartnershipsSpring gala 2024 photo slideshow - Celebrating School-Community Partnerships
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
 
Spellings Wk 4 and Wk 5 for Grade 4 at CAPS
Spellings Wk 4 and Wk 5 for Grade 4 at CAPSSpellings Wk 4 and Wk 5 for Grade 4 at CAPS
Spellings Wk 4 and Wk 5 for Grade 4 at CAPS
 
How to Manage Website in Odoo 17 Studio App.pptx
How to Manage Website in Odoo 17 Studio App.pptxHow to Manage Website in Odoo 17 Studio App.pptx
How to Manage Website in Odoo 17 Studio App.pptx
 
The Liver & Gallbladder (Anatomy & Physiology).pptx
The Liver &  Gallbladder (Anatomy & Physiology).pptxThe Liver &  Gallbladder (Anatomy & Physiology).pptx
The Liver & Gallbladder (Anatomy & Physiology).pptx
 
SURVEY I created for uni project research
SURVEY I created for uni project researchSURVEY I created for uni project research
SURVEY I created for uni project research
 
Analyzing and resolving a communication crisis in Dhaka textiles LTD.pptx
Analyzing and resolving a communication crisis in Dhaka textiles LTD.pptxAnalyzing and resolving a communication crisis in Dhaka textiles LTD.pptx
Analyzing and resolving a communication crisis in Dhaka textiles LTD.pptx
 
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...
 
ESSENTIAL of (CS/IT/IS) class 07 (Networks)
ESSENTIAL of (CS/IT/IS) class 07 (Networks)ESSENTIAL of (CS/IT/IS) class 07 (Networks)
ESSENTIAL of (CS/IT/IS) class 07 (Networks)
 
Improved Approval Flow in Odoo 17 Studio App
Improved Approval Flow in Odoo 17 Studio AppImproved Approval Flow in Odoo 17 Studio App
Improved Approval Flow in Odoo 17 Studio App
 
PSYPACT- Practicing Over State Lines May 2024.pptx
PSYPACT- Practicing Over State Lines May 2024.pptxPSYPACT- Practicing Over State Lines May 2024.pptx
PSYPACT- Practicing Over State Lines May 2024.pptx
 
male presentation...pdf.................
male presentation...pdf.................male presentation...pdf.................
male presentation...pdf.................
 
SPLICE Working Group: Reusable Code Examples
SPLICE Working Group:Reusable Code ExamplesSPLICE Working Group:Reusable Code Examples
SPLICE Working Group: Reusable Code Examples
 
How to Send Pro Forma Invoice to Your Customers in Odoo 17
How to Send Pro Forma Invoice to Your Customers in Odoo 17How to Send Pro Forma Invoice to Your Customers in Odoo 17
How to Send Pro Forma Invoice to Your Customers in Odoo 17
 
Graduate Outcomes Presentation Slides - English (v3).pptx
Graduate Outcomes Presentation Slides - English (v3).pptxGraduate Outcomes Presentation Slides - English (v3).pptx
Graduate Outcomes Presentation Slides - English (v3).pptx
 
When Quality Assurance Meets Innovation in Higher Education - Report launch w...
When Quality Assurance Meets Innovation in Higher Education - Report launch w...When Quality Assurance Meets Innovation in Higher Education - Report launch w...
When Quality Assurance Meets Innovation in Higher Education - Report launch w...
 
Trauma-Informed Leadership - Five Practical Principles
Trauma-Informed Leadership - Five Practical PrinciplesTrauma-Informed Leadership - Five Practical Principles
Trauma-Informed Leadership - Five Practical Principles
 
VAMOS CUIDAR DO NOSSO PLANETA! .
VAMOS CUIDAR DO NOSSO PLANETA!                    .VAMOS CUIDAR DO NOSSO PLANETA!                    .
VAMOS CUIDAR DO NOSSO PLANETA! .
 

CMSC 56 | Lecture 13: Relations and their Properties

  • 1. Relations and their Properties Lecture 13, CMSC 56 Allyn Joy D. Calcaben
  • 3. Relations Relationships between elements of sets are represented using the structure called a relation, which is just a subset of the Cartesian product of the sets. It can be used to solve problems such as determining which pairs of cities are linked by airline flights in a network, finding a viable order for the different phases of a complicated project, or producing a useful way to store information in computer databases.
  • 4. Relations & their Properties The most direct way to express a relationship between elements of two sets is to use ordered pairs made up of two related elements. For this reason, sets of ordered pairs are called binary relations.
  • 5. Relations & their Properties Definition 1 Let A and B be sets. A binary relation from A to B is a subset of A × B.
  • 6. Relations & their Properties We use the notation a R b to denote that (a, b) ∈ R and a R b to denote that (a, b) / ∈ R. Moreover, when (a, b) belongs to R, a is said to be related to b by R.
  • 7. Example Let A be the set of cities in the U.S.A., and let B be the set of the 50 states in the U.S.A.
  • 8. Solution Let A be the set of cities in the U.S.A., and let B be the set of the 50 states in the U.S.A. Define the relation R by specifying that (a, b) belongs to R if a city with name a is in the state b.
  • 9. Solution Let A be the set of cities in the U.S.A., and let B be the set of the 50 states in the U.S.A. Define the relation R by specifying that (a, b) belongs to R if a city with name a is in the state b. (Boulder, Colorado), (Bangor, Maine), (Ann Arbor, Michigan), (Middletown, New Jersey), (Middletown, New York), (Cupertino, California), and (Red Bank, New Jersey) are in R.
  • 10. Example Let A = {0, 1, 2} and B = {a, b}. Then {(0, a), (0, b), (1, a), (2, b)} is a relation from A to B. This means, for instance, that 0 R a, but that 1 R b. Relations can be represented graphically.
  • 11. Solution Let A = {0, 1, 2} and B = {a, b}. Then {(0, a), (0, b), (1, a), (2, b)} is a relation from A to B. This means, for instance, that 0 R a, but that 1 R b. Relations can be represented graphically. 0 • • a 1 • • b 2 • R a b 0 x x 1 x 2 x
  • 13. Functions as Relations Definition 2 A relation on a set A is a relation from A to A. (In other words, a relation on a set A is a subset of A × A.)
  • 14. Example Which of these relations contain each of the pairs (1, 1), (1, 2), (2, 1), (1,−1), and (2, 2)?
  • 15. Example Which of these relations contain each of the pairs (1, 1), (1, 2), (2, 1), (1,−1), and (2, 2)? The pair (1, 1) is in R1, R3, R4, and R6; (1, 2) is in R1 and R6; (2, 1) is in R2, R5, and R6; (1,−1) is in R2, R3, and R6; and, (2, 2) is in R1, R3, and R4.
  • 17. Functions as Relations Definition 3 A relation R on a set A is called reflexive if (a, a) ∈ R for every element a ∈ A.
  • 18. Example Which of these relations are reflexive?
  • 19. Solution Which of these relations are reflexive? The relations R3 and R5 are reflexive because they both contain all pairs of the form (a, a), namely, (1, 1), (2, 2), (3, 3), and (4, 4). The other relations are not reflexive because they do not contain all of these ordered pairs. In particular, R1, R2, R4, and R6 are not reflexive.
  • 20. Example Which of these relations are reflexive?
  • 21. Solution Which of these relations are reflexive? The reflexive relations are R1 (because a ≤ a for every integer a), R3, and R4. For each of the other relations in this example it is easy to find a pair of the form (a, a) that is not in the relation.
  • 22. Functions as Relations Definition 4 A relation R on a set A is called symmetric if (b, a) ∈ R whenever (a, b) ∈ R, for all a, b ∈ A. A relation R on a set A such that for all a, b ∈ A, if (a, b) ∈ R and (b, a) ∈ R, then a = b is called antisymmetric.
  • 23. Example Which of the relations are symmetric and antisymmetric?
  • 24. Solution Which of the relations are symmetric and antisymmetric? The relations R2 and R3 are symmetric, because in each case (b, a) belongs to the relation whenever (a, b) does. For R2, the only thing to check is that both (2, 1) and (1, 2) are in the relation. For R3, it is necessary to check that both (1, 2) and (2, 1) belong to the relation, and (1, 4) and (4, 1) belong to the relation.
  • 25. Solution Which of the relations are symmetric and antisymmetric? R4, R5, and R6 are all antisymmetric. For each of these relations there is no pair of elements a and b with a = b such that both (a, b) and (b, a) belong to the relation.
  • 26. Take Note: To verify that the relations is symmetric. This is done by finding a pair (a, b) with a = b such that (a, b) and (b, a) are both in the relation. To verify that the relations is antisymmetric. This is done by finding a pair (a, b) such that it is in the relation but (b, a) is not.
  • 27. Example Which of the relations are symmetric and antisymmetric?
  • 28. Solution Which of the relations are symmetric and antisymmetric? The relations R3, R4, and R6 are symmetric. R3 is symmetric, for if a = b or a = −b, then b = a or b = −a. R4 is symmetric because a = b implies that b = a. R6 is symmetric because a + b ≤ 3 implies that b + a ≤ 3.
  • 29. Solution Which of the relations are symmetric and antisymmetric? The relations R1, R2, R4, and R5 are antisymmetric. R1 is antisymmetric because the inequalities a ≤ b and b ≤ a imply that a = b. R2 is antisymmetric because it is impossible that a > b and b > a. R4 is antisymmetric, because two elements are related with respect to R4 if and only if they are equal. R5 is antisymmetric because it is impossible that a = b + 1 and b = a + 1.
  • 30. Functions as Relations Definition 5 A relation R on a set A is called transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R, for all a, b, c ∈ A.
  • 31. Example Which of the relations are transitive?
  • 32. Solution Which of the relations are transitive? R4, R5, and R6 are transitive. For each of these relations, we can show that it is transitive by verifying that if (a, b) and (b, c) belong to this relation, then (a, c) also does. For instance, R4 is transitive, because (3, 2) and (2, 1), (4, 2) and (2, 1), (4, 3) and (3, 1), and (4, 3) and (3, 2) are the only such sets of pairs, and (3, 1), (4, 1), and (4, 2) belong to R4.
  • 33. Solution Which of the relations are transitive? R1 is not transitive because (3, 4) and (4, 1) belong to R1, but (3, 1) does not. R2 is not transitive because (2, 1) and (1, 2) belong to R2, but (2, 2) does not. R3 is not transitive because (4, 1) and (1, 2) belong to R3, but (4, 2) does not.
  • 34. Example Which of the relations are transitive?
  • 35. Solution Which of the relations are transitive? The relations R1, R2, R3, and R4 are transitive. R1 is transitive because a ≤ b and b ≤ c imply that a ≤ c. R2 is transitive because a > b and b > c imply that a > c. R3 is transitive because a = ±b and b = ±c imply that a = ±c. R4 is clearly transitive, because a = b and b = c imply that a = c.
  • 36. Solution Which of the relations are transitive? R5 is not transitive because (2, 1) and (1, 0) belong to R5, but (2, 0) does not. R6 is not transitive because (2, 1) and (1, 2) belong to R6, but (2, 2) does not.
  • 38. Example Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations: R1 = {(1, 1), (2, 2), (3, 3)} , and R2 = {(1, 1), (1, 2), (1, 3), (1, 4)} can be combined to obtain: 1. R1 ∪ R2 2. R1 ∩ R2 3. R1 − R2 4. R2 − R1
  • 39. Solution Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations: R1 = {(1, 1), (2, 2), (3, 3)} , and R2 = {(1, 1), (1, 2), (1, 3), (1, 4)} can be combined to obtain: 1. R1 ∪ R2 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)}, 2. R1 ∩ R2 3. R1 − R2 4. R2 − R1
  • 40. Solution Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations: R1 = {(1, 1), (2, 2), (3, 3)} , and R2 = {(1, 1), (1, 2), (1, 3), (1, 4)} can be combined to obtain: 1. R1 ∪ R2 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)}, 2. R1 ∩ R2 = {(1, 1)}, 3. R1 − R2 4. R2 − R1
  • 41. Solution Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations: R1 = {(1, 1), (2, 2), (3, 3)} , and R2 = {(1, 1), (1, 2), (1, 3), (1, 4)} can be combined to obtain: 1. R1 ∪ R2 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)}, 2. R1 ∩ R2 = {(1, 1)}, 3. R1 − R2 = {(2, 2), (3, 3)}, 4. R2 − R1
  • 42. Solution Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations: R1 = {(1, 1), (2, 2), (3, 3)} , and R2 = {(1, 1), (1, 2), (1, 3), (1, 4)} can be combined to obtain: 1. R1 ∪ R2 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)}, 2. R1 ∩ R2 = {(1, 1)}, 3. R1 − R2 = {(2, 2), (3, 3)}, 4. R2 − R1 = {(1, 2), (1, 3), (1, 4)}.
  • 43. Combining Relations Definition 6 Let R be a relation from a set A to a set B and S a relation from B to a set C. The composite of R and S is the relation consisting of ordered pairs (a, c), where a ∈ A, c ∈ C, and for which there exists an element b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. We denote the composite of R and S by S ◦ R.
  • 44. Example What is the composite of the relations R and S, where R is the relation from {1, 2, 3} to {1, 2, 3, 4} with R = {(1, 1), (1, 4), (2, 3), (3, 1), (3, 4)} and S is the relation from {1, 2, 3, 4} to {0, 1, 2} with S = {(1, 0), (2, 0), (3, 1), (3, 2), (4, 1)}?
  • 45. Solution What is the composite of the relations R and S, where R is the relation from {1, 2, 3} to {1, 2, 3, 4} with R = {(1, 1), (1, 4), (2, 3), (3, 1), (3, 4)} and S is the relation from {1, 2, 3, 4} to {0, 1, 2} with S = {(1, 0), (2, 0), (3, 1), (3, 2), (4, 1)}? S ◦ R = { (1, 0), (1, 1), (2, 1), (2, 2), (3, 0), (3, 1) }

Editor's Notes

  1. S ◦R is constructed using all ordered pairs in R and ordered pairs in S, where the second element of the ordered pair in R agrees with the first element of the ordered pair in S. For example, the ordered pairs (2, 3) in R and (3, 1) in S produce the ordered pair (2, 1) in S ◦R. Computing all the ordered pairs in the composite, we find
  2. S ◦R is constructed using all ordered pairs in R and ordered pairs in S, where the second element of the ordered pair in R agrees with the first element of the ordered pair in S. For example, the ordered pairs (2, 3) in R and (3, 1) in S produce the ordered pair (2, 1) in S ◦R. Computing all the ordered pairs in the composite, we find