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Relations and Their Properties

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Relations and Their Properties

🎯 Learning Objectives

By the end of this topic, you will be able to:
  1. Compute the Cartesian product of two or more sets
  2. Define a binary relation as a subset of a Cartesian product
  3. Classify relations as reflexive, symmetric, transitive, antisymmetric
  4. Identify equivalence relations and partition sets into equivalence classes
  5. Represent relations using ordered pairs, matrices, and graphs

πŸ“‹ Prerequisites

  • Sets and Set Operations β€” sets, subsets, elements, set-builder notation
  • Arithmetic with integers β€” remainders (mod) and factors

πŸ“– Core Content

2.1 Cartesian Product

2.1.1 Intuition: Combining Sets Systematically

Imagine you have two lists: A={Alice,Bob}A = \{\text{Alice}, \text{Bob}\} and B={Pizza,Pasta}B = \{\text{Pizza}, \text{Pasta}\}. How many possible ways can you pair one person with one dish? Every person can be paired with every dish:
  • (Alice, Pizza), (Alice, Pasta)
  • (Bob, Pizza), (Bob, Pasta) This all-possible-pairings is the Cartesian product AΓ—BA \times B. The order matters: (Alice, Pizza) is different from (Pizza, Alice) β€” the first element always comes from AA, the second from BB.
πŸ’‘ Why this matters: In databases, the CROSS JOIN of two tables is a Cartesian product. In probability, the sample space of rolling two dice is {1,2,3,4,5,6}Γ—{1,2,3,4,5,6}\{1,2,3,4,5,6\} \times \{1,2,3,4,5,6\}.

2.1.2 Formal Definition

The Cartesian product of sets AA and BB is:
AΓ—B={(a,b)∣a∈A,Β b∈B}A \times B = \{(a, b) \mid a \in A,\ b \in B\}
Each element (a,b)(a, b) is an ordered pair β€” the position matters. Key properties:
  • ∣AΓ—B∣=∣Aβˆ£Γ—βˆ£B∣|A \times B| = |A| \times |B|
  • AΓ—Bβ‰ BΓ—AA \times B \neq B \times A (unless A=BA = B or one is empty)
  • AΓ—(BΓ—C)β‰ (AΓ—B)Γ—CA \times (B \times C) \neq (A \times B) \times C (triples vs. nested pairs)
  • βˆ…Γ—A=AΓ—βˆ…=βˆ…\emptyset \times A = A \times \emptyset = \emptyset For three sets:
AΓ—BΓ—C={(a,b,c)∣a∈A,Β b∈B,Β c∈C}A \times B \times C = \{(a, b, c) \mid a \in A,\ b \in B,\ c \in C\}
This generalizes to nn sets (ordered nn-tuples).

2.1.3 Worked Examples

Example 1.1 (Easy): Let A={1,2}A = \{1, 2\}, B={3,4}B = \{3, 4\}. Compute AΓ—BA \times B and BΓ—AB \times A. Step 1: AΓ—BA \times B pairs each element of AA with each element of BB:
AΓ—B={(1,3),(1,4),(2,3),(2,4)}A \times B = \{(1,3), (1,4), (2,3), (2,4)\}
Step 2: BΓ—AB \times A pairs each element of BB with each element of AA:
BΓ—A={(3,1),(3,2),(4,1),(4,2)}B \times A = \{(3,1), (3,2), (4,1), (4,2)\} AΓ—Bβ‰ BΓ—A,Β e.g.Β (1,3)∈AΓ—BΒ butΒ (1,3)βˆ‰BΓ—A\boxed{A \times B \neq B \times A,\ \text{e.g.}\ (1,3) \in A \times B \text{ but } (1,3) \notin B \times A}
Example 1.2 (Medium): How many elements are in AΓ—BΓ—CA \times B \times C if ∣A∣=3|A| = 3, ∣B∣=4|B| = 4, ∣C∣=5|C| = 5? Step 1: ∣AΓ—BΓ—C∣=∣Aβˆ£Γ—βˆ£Bβˆ£Γ—βˆ£C∣=3Γ—4Γ—5=60|A \times B \times C| = |A| \times |B| \times |C| = 3 \times 4 \times 5 = 60
60Β orderedΒ triples\boxed{60 \text{ ordered triples}}
Example 1.3 (Exam-style): Let A={1}A = \{1\}, B={2,3}B = \{2, 3\}. List all elements of AΓ—BΓ—AA \times B \times A. Step 1: First coordinates from AA, second from BB, third from AA. AΓ—BΓ—A={(1,2,1),(1,3,1)}A \times B \times A = \{(1,2,1), (1,3,1)\}
{(1,2,1),(1,3,1)}\boxed{\{(1,2,1), (1,3,1)\}}

2.2 Binary Relations

2.2.1 Intuition: Connections Between Elements

A relation is simply a rule that connects elements from one set to another. "Alice teaches Mathematics" defines a relation between people and subjects. "5 is less than 10" defines a relation between numbers. A relation doesn't have to connect every element β€” it's a subset of all possible pairings (the Cartesian product).
πŸ’‘ Why this matters: Relations are the mathematical foundation for database tables (each row is a tuple), social networks (edges between users), and dependency graphs (prerequisite relationships).

2.2.2 Formal Definition

A binary relation RR from set AA to set BB is a subset of AΓ—BA \times B:
RβŠ†AΓ—BR \subseteq A \times B
If (a,b)∈R(a, b) \in R, we write a R ba\,R\,b (read: "aa is related to bb").
  • Domain of RR: {a∈Aβˆ£βˆƒb∈B,Β (a,b)∈R}\{a \in A \mid \exists b \in B,\ (a,b) \in R\}
  • Range of RR: {b∈Bβˆ£βˆƒa∈A,Β (a,b)∈R}\{b \in B \mid \exists a \in A,\ (a,b) \in R\}
  • If A=BA = B, we say RR is a relation on AA. (Diagram)

2.2.3 Examples of Relations

Set AASet BBRelation RRRβŠ†AΓ—BR \subseteq A \times B
TeachersCoursestt teaches cc{(t,c)∣t teaches c}\{(t,c) \mid t \text{ teaches } c\}
N\mathbb{N}N\mathbb{N}a∣ba \mid b ( aa divides bb ){(a,b)∣a∣b}\{(a,b) \mid a \mid b\}
R\mathbb{R}R\mathbb{R}a<ba < b{(a,b)∣a<b}\{(a,b) \mid a < b\}
StudentsGPAs≀8.5\leq 8.5 GPA{(s,g)∣GPA(s)≀8.5}\{(s,g) \mid \text{GPA}(s) \leq 8.5\}
Pythagorean triples as a relation:
R={(a,b,c)∣a,b,c∈N, a,b,c>0, a2+b2=c2}R = \{(a,b,c) \mid a,b,c \in \mathbb{N},\ a,b,c > 0,\ a^2 + b^2 = c^2\}
Example triples: (3,4,5)(3,4,5), (5,12,13)(5,12,13), (6,8,10)(6,8,10).

2.3 Properties of Relations (on a Set)

When RR is a relation on a set AA (i.e., RβŠ†AΓ—AR \subseteq A \times A), we classify it by these properties:

2.3.1 Reflexive

Intuition: Every element is related to itself. "Every number is equal to itself." Definition: RR is reflexive if for every a∈Aa \in A, (a,a)∈R(a, a) \in R. Examples:
  • ≀\leq on R\mathbb{R}: a≀aa \leq a for all aa β†’ reflexive βœ“
  • divides on N\mathbb{N}: a∣aa \mid a for all a>0a > 0 β†’ reflexive βœ“
  • << on R\mathbb{R}: a<aa < a is false β†’ NOT reflexive βœ—

2.3.2 Symmetric

Intuition: If aa relates to bb, then bb relates back to aa. Like a two-way street. Definition: RR is symmetric if whenever (a,b)∈R(a,b) \in R, then (b,a)∈R(b,a) \in R. Examples:
  • = on R\mathbb{R}: a=bβ€…β€ŠβŸΉβ€…β€Šb=aa = b \implies b = a β†’ symmetric βœ“
  • gcd(a,b) = 1 on N\mathbb{N}: gcd⁑(a,b)=1β€…β€ŠβŸΊβ€…β€Šgcd⁑(b,a)=1\gcd(a,b) = 1 \iff \gcd(b,a) = 1 β†’ symmetric βœ“
  • $\leq$ on R\mathbb{R}: 2≀32 \leq 3 does NOT imply 3≀23 \leq 2 β†’ NOT symmetric βœ—

2.3.3 Transitive

Intuition: If aa relates to bb and bb relates to cc, then aa relates to cc. Like a chain. Definition: RR is transitive if whenever (a,b)∈R(a,b) \in R and (b,c)∈R(b,c) \in R, then (a,c)∈R(a,c) \in R. Examples:
  • << on R\mathbb{R}: a<ba < b and b<cb < c implies a<ca < c β†’ transitive βœ“
  • divides on N\mathbb{N}: a∣ba \mid b and b∣cb \mid c implies a∣ca \mid c β†’ transitive βœ“
  • "is a parent of": NOT transitive (my parent's parent is my grandparent, not my parent) βœ—

2.3.4 Antisymmetric

Intuition: If aa relates to bb and bb relates to aa, then aa and bb must be the same. No distinct two-way relationships. Definition: RR is antisymmetric if whenever (a,b)∈R(a,b) \in R and (b,a)∈R(b,a) \in R, then a=ba = b. Examples:
  • ≀\leq on R\mathbb{R}: if a≀ba \leq b and b≀ab \leq a, then a=ba = b β†’ antisymmetric βœ“
  • divides on N\mathbb{N}: if a∣ba \mid b and b∣ab \mid a, then a=ba = b β†’ antisymmetric βœ“
  • a - b = 2 on N\mathbb{N}: neither symmetric nor antisymmetric βœ— (Diagram)

2.3.5 Worked Example: Classifying Relations

Example 2.1: R={(a,b)∣a,b∈N,Β a≀b}R = \{(a,b) \mid a,b \in \mathbb{N},\ a \leq b\}. Classify RR.
PropertyCheckResult
Reflexivea≀aa \leq a for all aa ? βœ“ Yesβœ“
SymmetricIf 2≀32 \leq 3 , does 3≀23 \leq 2 ? Noβœ—
TransitiveIf a≀ba \leq b and b≀cb \leq c , then a≀ca \leq c ? βœ“ Yesβœ“
AntisymmetricIf a≀ba \leq b and b≀ab \leq a , then a=ba = b ? βœ“ Yesβœ“
Reflexive,Β Transitive,Β Antisymmetric\boxed{\text{Reflexive, Transitive, Antisymmetric}}
Example 2.2: R={(a,b)∣a,b∈Z, ∣aβˆ’b∣=2}R = \{(a,b) \mid a,b \in \mathbb{Z},\ |a - b| = 2\}. Classify RR.
PropertyCheckResult
Reflexive$a - a
Symmetric$a - b
Transitive(0,2)∈R(0,2) \in R and (2,4)∈R(2,4) \in R , but (0,4)βˆ‰R(0,4) \notin R since $0-4
OnlyΒ Symmetric\boxed{\text{Only Symmetric}}

2.4 Equivalence Relations

2.4.1 Intuition: Grouping Similar Things

An equivalence relation captures the idea of "sameness" under some criterion. If we group integers by "remainder when divided by 5", all numbers with remainder 2 are equivalent (like 2, 7, 12, 17, ...). They behave the same way modulo 5.
πŸ’‘ Why this matters: Equivalence relations partition data into buckets. In clustering, "same cluster" is an equivalence relation. In cryptography, "same residue class mod n" is fundamental.

2.4.2 Formal Definition

A relation RR on a set AA is an equivalence relation if it is:
  1. Reflexive: (a,a)∈R(a,a) \in R for all a∈Aa \in A
  2. Symmetric: (a,b)∈Rβ€…β€ŠβŸΊβ€…β€Š(b,a)∈R(a,b) \in R \iff (b,a) \in R
  3. Transitive: (a,b)∈R(a,b) \in R and (b,c)∈Rβ€…β€ŠβŸΉβ€…β€Š(a,c)∈R(b,c) \in R \implies (a,c) \in R

2.4.3 Equivalence Classes

An equivalence relation partitions AA into disjoint subsets called equivalence classes. The equivalence class of a∈Aa \in A is:
[a]={b∈A∣(a,b)∈R}[a] = \{b \in A \mid (a,b) \in R\}
Properties:
  • Every element belongs to exactly one equivalence class
  • [a]=[b][a] = [b] iff (a,b)∈R(a,b) \in R
  • Different equivalence classes are disjoint
  • The union of all equivalence classes equals AA

2.4.4 Worked Examples

Example 3.1 (Easy): Define RR on Z\mathbb{Z} by: a R ba\,R\,b if aβ€Šmodβ€Š5=bβ€Šmodβ€Š5a \bmod 5 = b \bmod 5. Check reflexive: aβ€Šmodβ€Š5=aβ€Šmodβ€Š5a \bmod 5 = a \bmod 5 β†’ βœ“ Check symmetric: If aβ€Šmodβ€Š5=bβ€Šmodβ€Š5a \bmod 5 = b \bmod 5, then bβ€Šmodβ€Š5=aβ€Šmodβ€Š5b \bmod 5 = a \bmod 5 β†’ βœ“ Check transitive: If aβ€Šmodβ€Š5=bβ€Šmodβ€Š5a \bmod 5 = b \bmod 5 and bβ€Šmodβ€Š5=cβ€Šmodβ€Š5b \bmod 5 = c \bmod 5, then aβ€Šmodβ€Š5=cβ€Šmodβ€Š5a \bmod 5 = c \bmod 5 β†’ βœ“ Equivalence classes (remainders 0, 1, 2, 3, 4):
  • [0]={…,βˆ’10,βˆ’5,0,5,10,…}[0] = \{\ldots, -10, -5, 0, 5, 10, \ldots\}
  • [1]={…,βˆ’9,βˆ’4,1,6,11,…}[1] = \{\ldots, -9, -4, 1, 6, 11, \ldots\}
  • [2]={…,βˆ’8,βˆ’3,2,7,12,…}[2] = \{\ldots, -8, -3, 2, 7, 12, \ldots\}
  • [3]={…,βˆ’7,βˆ’2,3,8,13,…}[3] = \{\ldots, -7, -2, 3, 8, 13, \ldots\}
  • [4]={…,βˆ’6,βˆ’1,4,9,14,…}[4] = \{\ldots, -6, -1, 4, 9, 14, \ldots\}
EquivalenceΒ relationΒ withΒ 5Β classes\boxed{\text{Equivalence relation with 5 classes}}
Example 3.2 (Medium): Let A={1,2,3,4,5,6}A = \{1,2,3,4,5,6\}. Define R={(a,b)∣a,b∈A,Β a≑b(mod3)}R = \{(a,b) \mid a,b \in A,\ a \equiv b \pmod{3}\}. Find the equivalence classes. Step 1: Numbers with same remainder modulo 3:
  • Remainder 0: {3,6}\{3,6\}
  • Remainder 1: {1,4}\{1,4\}
  • Remainder 2: {2,5}\{2,5\} Step 2: Verify: every element of AA appears in exactly one class. βœ“
{[3,6],[1,4],[2,5]}\boxed{\{[3,6], [1,4], [2,5]\}}
Example 3.3 (Hard): Is "same parity" (both even or both odd) an equivalence relation on Z\mathbb{Z}? Step 1 β€” Reflexive: Every integer has the same parity as itself (evenΓ—even or oddΓ—odd). βœ“ Step 2 β€” Symmetric: If aa and bb have same parity, then bb and aa do too. βœ“ Step 3 β€” Transitive: If a,ba,b have same parity and b,cb,c have same parity, then a,ca,c have same parity. βœ“ Equivalence classes:
  • Even numbers: {…,βˆ’4,βˆ’2,0,2,4,…}\{\ldots, -4, -2, 0, 2, 4, \ldots\}
  • Odd numbers: {…,βˆ’3,βˆ’1,1,3,5,…}\{\ldots, -3, -1, 1, 3, 5, \ldots\}
Yes,Β itΒ isΒ anΒ equivalenceΒ relationΒ withΒ 2Β classesΒ (evensΒ andΒ odds)\boxed{\text{Yes, it is an equivalence relation with 2 classes (evens and odds)}}

2.5 Representing Relations

2.5.1 Matrix Representation

For a relation RR on a finite set A={a1,a2,…,an}A = \{a_1, a_2, \ldots, a_n\}, build an nΓ—nn \times n matrix MRM_R where:
MR[i][j]={1if (ai,aj)∈R0otherwiseM_R[i][j] = \begin{cases} 1 & \text{if } (a_i, a_j) \in R \\ 0 & \text{otherwise} \end{cases}
Example: A={1,2,3}A = \{1,2,3\}, R={(1,1),(1,2),(2,3),(3,1)}R = \{(1,1), (1,2), (2,3), (3,1)\}
MR=[110001100]M_R = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}
Properties visible from matrix:
  • Reflexive: All diagonal entries are 1
  • Symmetric: Matrix is symmetric (MR=MRTM_R = M_R^T)
  • Antisymmetric: If MR[i][j]=1M_R[i][j] = 1 and iβ‰ ji \neq j, then MR[j][i]=0M_R[j][i] = 0

2.5.2 Graph Representation (Directed Graphs)

A relation RR on AA can be drawn as a directed graph where:
  • Elements of AA are vertices
  • Each (a,b)∈R(a,b) \in R is a directed edge aβ†’ba \rightarrow b (Diagram) This graph corresponds to the matrix above.

πŸ“ Key Formulas β€” Summary Table

ConceptFormula/DefinitionWhen to Use
Cartesian productAΓ—B={(a,b)∣a∈A,b∈B}A \times B = \{(a,b) \mid a \in A, b \in B\}Generating all ordered pairs
Size of product$\A \times B\
Binary relationRβŠ†AΓ—BR \subseteq A \times BModeling connections
Reflexiveβˆ€a∈A:(a,a)∈R\forall a \in A: (a,a) \in RCheck diagonal in matrix
Symmetric(a,b)∈Rβ€…β€ŠβŸΉβ€…β€Š(b,a)∈R(a,b) \in R \implies (b,a) \in RTwo-way relationships
Transitive(a,b),(b,c)∈Rβ€…β€ŠβŸΉβ€…β€Š(a,c)∈R(a,b),(b,c) \in R \implies (a,c) \in RChain relationships
Antisymmetric(a,b),(b,a)∈Rβ€…β€ŠβŸΉβ€…β€Ša=b(a,b),(b,a) \in R \implies a = bPartial orders
Equivalence relationReflexive + Symmetric + TransitiveGrouping/classification
Equivalence class[a]={b∣(a,b)∈R}[a] = \{b \mid (a,b) \in R\}Partition of a set
Matrix entryMR[i][j]=1M_R[i][j] = 1 if (ai,aj)∈R(a_i,a_j) \in RComputing properties

⚠️ Common Pitfalls

Pitfall 1: Confusing Symmetric with Antisymmetric

Mistake: Thinking a relation cannot be both symmetric and antisymmetric. Why: A relation CAN be both β€” but only if it's a subset of the identity relation. Example: R={(1,1),(2,2)}R = \{(1,1), (2,2)\} on {1,2}\{1,2\} is both symmetric (each pair has its reverse, which is itself) and antisymmetric (if (a,b)(a,b) and (b,a)(b,a) are in RR, then a=ba=b). Most relations are neither symmetric nor antisymmetric. Eg: R={(1,2),(1,3)}R = \{(1,2), (1,3)\} is neither.

Pitfall 2: Forgetting That Transitive Requires ALL Chains

Mistake: Checking only one chain and concluding transitivity. Correct approach: You must check every pair (a,b),(b,c)(a,b),(b,c). If ANY chain exists without its corresponding (a,c)(a,c), the relation is NOT transitive. Example: R={(1,2),(2,3),(1,3),(3,4)}R = \{(1,2), (2,3), (1,3), (3,4)\}. Is (1,2),(2,3)β†’(1,3)(1,2),(2,3) \to (1,3) βœ“. But (1,2),(2,4)(1,2),(2,4)? (2,4)βˆ‰R(2,4) \notin R, so no check needed. However (1,3),(3,4)β†’(1,4)(1,3),(3,4) \to (1,4)? (1,4)βˆ‰R(1,4) \notin R β†’ NOT transitive!

Pitfall 3: Thinking {(1,1),(2,2)}\{(1,1), (2,2)\} is Reflexive on {1,2,3}\{1,2,3\}

Mistake: Reflexive requires (a,a)(a,a) for every a∈Aa \in A. If A={1,2,3}A = \{1,2,3\}, then (3,3)(3,3) must be present. Without it, the relation is not reflexive. Correct: Check each element in the set AA, not just those that appear in RR.

Pitfall 4: Confusing "divides" Relation Domain

Mistake: Thinking 0∣50 \mid 5 (0 divides 5) is valid. Correct: Division by zero is undefined. So the divides relation is defined only on positive integers or non-zero integers. a∣ba \mid b requires aβ‰ 0a \neq 0.

πŸ“ Practice Questions

Q1: Let A={x,y}A = \{x, y\}, B={1,2,3}B = \{1, 2, 3\}. Compute AΓ—BA \times B.
Strategy Hint: List all ordered pairs where first coordinate is from A, second from B.
{(x,1),(x,2),(x,3),(y,1),(y,2),(y,3)}\boxed{\{(x,1), (x,2), (x,3), (y,1), (y,2), (y,3)\}} Q2: Does R={(a,b)∣a,b∈N, a<b}R = \{(a,b) \mid a,b \in \mathbb{N},\ a < b\} satisfy reflexivity?
Strategy Hint: Check if any natural number is less than itself.
Solution: a<aa < a is false for all aa. So RR is NOT reflexive.
No\boxed{\text{No}} Q3: Is "is a sibling of" an equivalence relation?
Strategy Hint: Check reflexivity, symmetry, transitivity.
Reflexive: A person is not their own sibling β†’ Not reflexive. Symmetric: If X is sibling of Y, then Y is sibling of X β†’ Yes. Transitive: If X is sibling of Y and Y is sibling of Z, then X is sibling of Z β†’ Yes (assuming full siblings).
But since it's not reflexive, it's NOT an equivalence relation.
NoΒ β€”Β notΒ reflexive\boxed{\text{No β€” not reflexive}} Q4: Find the equivalence classes of R={(a,b)∣a,b∈Z,Β a≑b(mod4)}R = \{(a,b) \mid a,b \in \mathbb{Z},\ a \equiv b \pmod{4}\}.
Strategy Hint: Group by remainder when dividing by 4.
[0]={…,βˆ’8,βˆ’4,0,4,8,…},Β [1]={…,βˆ’7,βˆ’3,1,5,9,…},Β [2]={…,βˆ’6,βˆ’2,2,6,10,…},Β [3]={…,βˆ’5,βˆ’1,3,7,11,…}\boxed{[0] = \{\ldots,-8,-4,0,4,8,\ldots\},\ [1] = \{\ldots,-7,-3,1,5,9,\ldots\},\ [2] = \{\ldots,-6,-2,2,6,10,\ldots\},\ [3] = \{\ldots,-5,-1,3,7,11,\ldots\}} Q5: Is R={(a,b)∣a,b∈R,Β a2=b2}R = \{(a,b) \mid a,b \in \mathbb{R},\ a^2 = b^2\} an equivalence relation?
Strategy Hint: Check the three properties.
Reflexive: a2=a2a^2 = a^2 β†’ βœ“ Symmetric: a2=b2β€…β€ŠβŸΉβ€…β€Šb2=a2a^2 = b^2 \implies b^2 = a^2 β†’ βœ“ Transitive: a2=b2a^2 = b^2 and b2=c2β€…β€ŠβŸΉβ€…β€Ša2=c2b^2 = c^2 \implies a^2 = c^2 β†’ βœ“
Yes! Equivalence classes are {x,βˆ’x}\{x, -x\} for each x∈Rx \in \mathbb{R}.
YesΒ β€”Β equivalenceΒ classesΒ areΒ pairsΒ {x,βˆ’x}\boxed{\text{Yes β€” equivalence classes are pairs } \{x, -x\}} Q6: Let A={1,2,3}A = \{1,2,3\}. How many relations exist on AA?
Strategy Hint: Any relation is a subset of AΓ—AA \times A.
Step 1: ∣AΓ—A∣=3Γ—3=9|A \times A| = 3 \times 3 = 9 Step 2: Number of subsets = 29=5122^9 = 512
512Β relations\boxed{512 \text{ relations}} Q7: Draw the directed graph for R={(1,2),(2,3),(3,1),(1,1)}R = \{(1,2), (2,3), (3,1), (1,1)\} on {1,2,3}\{1,2,3\}.
Strategy Hint: Vertices = 1,2,3. Edges = the pairs.
pseudo
1 β†’ 2 β†’ 3
↑    ↓
β””β”€β”€β”€β”€β”˜
With a self-loop at 1.
GraphΒ representation:Β 1β†’2β†’3β†’1,Β plusΒ 1β†’1\boxed{\text{Graph representation: } 1 \to 2 \to 3 \to 1 \text{, plus } 1 \to 1} Q8: If ∣A∣=3|A| = 3 and ∣B∣=4|B| = 4, what is ∣AΓ—BΓ—A∣|A \times B \times A|?
Strategy Hint: Multiply cardinalities.
∣AΓ—BΓ—A∣=3Γ—4Γ—3=36|A \times B \times A| = 3 \times 4 \times 3 = 36
36\boxed{36} Q9: Let R={(1,1),(1,2),(2,1),(2,2),(3,3)}R = \{(1,1), (1,2), (2,1), (2,2), (3,3)\} on {1,2,3,4}\{1,2,3,4\}. Is RR reflexive?
Strategy Hint: Reflexive requires (4,4)(4,4).
(4,4)βˆ‰R(4,4) \notin R, so NOT reflexive.
No\boxed{\text{No}} Q10: Find the matrix representation of R={(a,b)∣a,b∈{1,2,3}, a∣b}R = \{(a,b) \mid a,b \in \{1,2,3\},\ a \mid b\}.
Strategy Hint: aa divides bb means bb is a multiple of aa.
(a,b)123
11 (1|1)1 (1|2)1 (1|3)
201 (2|2)0
3001 (3|3)
>MR=[111010001]>> M_R = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} >
>MR=[111010001]>> \boxed{M_R = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}} >
Q11: If RR is symmetric and antisymmetric, prove RβŠ†{(a,a)∣a∈A}R \subseteq \{(a,a) \mid a \in A\}.
Strategy Hint: Take any (a,b)∈R(a,b) \in R and deduce a=ba = b.
Proof: Let (a,b)∈R(a,b) \in R. By symmetry, (b,a)∈R(b,a) \in R. Since RR is antisymmetric, (a,b),(b,a)∈Rβ€…β€ŠβŸΉβ€…β€Ša=b(a,b),(b,a) \in R \implies a = b. Thus every pair in RR has the form (a,a)(a,a).
Proved\boxed{\text{Proved}} Q12: Show that "same birthday month" is an equivalence relation on the set of all people. How many equivalence classes are there?
Strategy Hint: The three properties are straightforward. Classes = 12 months.
Reflexive: A person has the same birthday month as themselves. βœ“ Symmetric: If A and B share a month, B and A do too. βœ“ Transitive: If A∼B (same month) and B∼C (same month), then A∼C. βœ“
12 equivalence classes (January through December), possibly with empty classes if no one in the set has that month.
Yes,Β 12Β equivalenceΒ classes\boxed{\text{Yes, 12 equivalence classes}}

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