2.2 Relational Properties: Reflexivity, Symmetry, Transitivity & Antisymmetry

Back to Discrete Mathematics Notes

📚Module 02: Binary Relations & ClosuresTopic 2.2: Relational Properties: Reflexivity, Symmetry & Transitivity⏱️5 min read

🎯High-Yield For:Semester Exams • GATE CSE • Technical Interviews

💡 Core Intuition: The 3-Second Mental Model

🍳 The Everyday Analogy: Social Connections & Family Bloodlines

Imagine analyzing relationships among people in a social community:

1. Reflexive ("Has the same blood type as"): Every single person has the exact same blood type as themselves ($x \, R \, x$). In a network graph, every vertex has a self-loop.
2. Irreflexive ("Is strictly taller than"): No person can be strictly taller than themselves. Self-loops are strictly forbidden!
3. Symmetric ("Is a sibling of"): If Alice is a sibling of Bob, Bob is guaranteed to be a sibling of Alice. Every forward arrow implies a reverse arrow ($a \leftrightarrows b$).
4. Anti-Symmetric ("Is a biological ancestor of"): If Alice is an ancestor of Bob, Bob cannot possibly be an ancestor of Alice unless they are the exact same individual! Bidirectional cycles between distinct people are strictly prohibited.
5. Asymmetric ("Is strictly older than"): If Alice is older than Bob, Bob cannot be older than Alice, and no person is older than themselves (zero self-loops and zero bidirectional arrows).
6. Transitive ("Is an ancestor of"): If Alice is an ancestor of Bob, and Bob is an ancestor of Charlie, Alice is guaranteed to be an ancestor of Charlie ($a \to b \to c \implies a \to c$).

💻 Bridging to Computer Science: Equivalence Partitioning & Partial Orders

In computer science, relations are categorized by these foundational properties to form algebraic structures:

• Equivalence Relations (Reflexive + Symmetric + Transitive): Partition datasets into disjoint equivalence classes, used in SQL GROUP BY, compiler register allocation, and DFA state minimization.
• Partial Orders / Posets (Reflexive + Anti-Symmetric + Transitive): Model dependency graphs, build pipelines (Makefiles, Docker layers), and task scheduling.

📑Table of Contents

• 🔄 Reflexive & Irreflexive Relations
  • Foundational Invariants of Reflexivity
  • Diagnostic Evaluation: Reflexive vs Irreflexive
• ⚖️ Symmetric, Antisymmetric & Asymmetric Relations
  • Diagnostic Decision Matrix on a Three-Element Set
• ⛓️ Transitive Relations: Multi-Hop Reachability Invariants
  • The Vacuous Truth Trap in Transitivity
  • Number of Transitive Relations
• 🏛️ Master Taxonomy & Counting Reference Table
• 🌍 Real-World Engineering Context
• 📝 Step-by-Step Worked Examples
  • Example 1: Property Evaluation of Concrete Relations
  • Example 2: Combinatorial Counting of Relations with Mixed Properties
  • Example 3: Verifying Matrix and Algebraic Invariants
• 🎯 Exam & Interview Pitfall Check

---

🔄 Reflexive & Irreflexive Relations

A binary relation on a set $A$ is evaluated primarily on how it handles diagonal identity pairs $(x, x)$ where both coordinates match.

📖Definition 2.6: Reflexive & Irreflexive Relationsdefinition

Let $R$ be a binary relation on set $A$:

1. Reflexive Relation: $$\forall x \in A, \quad (x, x) \in R \iff \Delta_A \subseteq R$$ Requirement: Every single diagonal pair $(x, x)$ must be present in $R$. Additional non-diagonal pairs are allowed.
2. Irreflexive Relation: $$\forall x \in A, \quad (x, x) \notin R \iff R \cap \Delta_A = \emptyset$$ Requirement: Zero diagonal pairs are permitted. Not even a single self-loop may exist.

📐Formula 2.3: Combinatorial Counting for Reflexivityformula

For a set $A$ containing $|A| = n$ elements, the Cartesian product $A \times A$ contains $n^2$ total pairs:

• Diagonal Pairs: Exactly $n$ pairs of the form $(x, x)$.
• Non-Diagonal Pairs: Exactly $n^2 - n = n(n-1)$ pairs of the form $(x, y)$ where $x \neq y$.

1. Total Number of Reflexive Relations: All $n$ diagonal pairs must be present ($1^n = 1$ choice). Each of the $n(n-1)$ non-diagonal pairs has 2 choices (in or out): $$N_{\text{reflexive}} = 1^n \times 2^{n(n-1)} = 2^{n(n-1)} = 2^{n^2 - n}$$
2. Total Number of Irreflexive Relations: All $n$ diagonal pairs must be absent ($1^n = 1$ choice). Each non-diagonal pair has 2 choices: $$N_{\text{irreflexive}} = 1^n \times 2^{n(n-1)} = 2^{n(n-1)} = 2^{n^2 - n}$$
3. Neither Reflexive Nor Irreflexive Relations: Relations containing at least one diagonal pair, but not all $n$ diagonal pairs: $$N_{\text{neither}} = 2^{n^2} - 2 \cdot 2^{n^2 - n} \quad (\text{for } n \ge 2)$$

Foundational Invariants of Reflexivity

• Extremal Relations:
  • The smallest reflexive relation is the identity relation $\Delta_A = \{(x, x) \mid x \in A\}$ with cardinality $n$.
  • The largest reflexive relation is the universal relation $A \times A$ with cardinality $n^2$.
  • The smallest irreflexive relation is the empty relation $\emptyset$ with cardinality $0$.
  • The largest irreflexive relation is $(A \times A) - \Delta_A$ with cardinality $n^2 - n$.
• Intersection and Union:
  • If $R_1$ and $R_2$ are reflexive, then $R_1 \cup R_2$ and $R_1 \cap R_2$ are both reflexive.
  • If $R_1$ and $R_2$ are irreflexive, then $R_1 \cup R_2$ and $R_1 \cap R_2$ are both irreflexive.
• Complementation Duality:
  • $R$ is reflexive if and only if its complement $R^c$ is irreflexive.
• Supersets and Subsets:
  • Any superset of a reflexive relation is reflexive.
  • Any subset of an irreflexive relation is irreflexive.

Diagnostic Evaluation: Reflexive vs Irreflexive

The following table evaluates classic candidate relations over set $A = \{1, 2, 3\}$ to demonstrate why Reflexivity and Irreflexivity are not logical opposites (relations can be neither, or for $A = \emptyset$, both!):

Relation on $A = \{1, 2, 3\}$	Reflexive?	Irreflexive?	Structural Verification Rule
$A \times A$	Yes	No	Contains all 3 diagonal pairs: $(1, 1), (2, 2), (3, 3)$.
$\emptyset$	No	Yes	Contains zero diagonal elements.
$\{(1, 1), (2, 2), (3, 3)\}$	Yes	No	Identity relation $\Delta_A$: contains all diagonals and nothing else.
$\{(1, 2), (2, 3), (1, 3)\}$	No	Yes	Strictly off-diagonal pairs; zero self-loops.
$\{(1, 1), (1, 2), (2, 1), (2, 2)\}$	No	No	Has $(1, 1), (2, 2)$ but misses $(3, 3)$ (Neither!).
$\{(1, 1), (2, 2), (3, 3), (1, 3), (2, 1)\}$	Yes	No	Contains all 3 diagonals plus additional non-diagonal edges.
$\{(1, 3), (2, 1), (2, 3), (3, 2)\}$	No	Yes	Zero diagonal elements present.

---

⚖️ Symmetric, Antisymmetric & Asymmetric Relations

How a relation handles directional reversals $(a, b) \to (b, a)$ defines its symmetry classification.

Symmetric vs Anti-Symmetric vs Asymmetric Relations

Comparing directional pairing requirements across discrete relations

Symmetric

Symmetric Relation

🔄

Dominant Architecture / DomainUndirected Graph Edges (a ⇄ b)

• •If (a, b) ∈ R, then (b, a) must also be in R.
• •Algebraic matrix invariant: M_R = (M_R)^T (symmetric matrix) and R = R⁻¹.
• •Zero constraints on diagonal elements: self-loops may be present or absent freely.
• •Total count: 2^(n(n+1)/2).

"Symmetry demands mutual reciprocity: whenever a forward path exists, the reverse path must exist."

Antisymmetric

Anti-Symmetric Relation

🛡️

Dominant Architecture / DomainPosets & Hierarchies (a ≤ b)

• •If (a, b) ∈ R and (b, a) ∈ R, then a must equal b.
• •For distinct elements a ≠ b, at most one of (a, b) or (b, a) can be present.
• •Diagonal elements (a, a) are fully permitted without penalty: (R ∩ R⁻¹) ⊆ Δ_A.
• •Total count: 2^n · 3^(n(n-1)/2).

"Antisymmetry forbids 2-cycles between distinct nodes, but freely allows self-loops."

Asymmetric

Asymmetric Relation

⚡

Dominant Architecture / DomainStrict Partial Orders (a < b)

• •If (a, b) ∈ R, then (b, a) is strictly forbidden: (b, a) ∉ R.
• •Every asymmetric relation is both Anti-Symmetric AND Irreflexive.
• •Zero diagonal elements are permitted: R ∩ R⁻¹ = ∅.
• •Total count: 3^(n(n-1)/2).

"Asymmetry forbids all reciprocity and strictly outlaws self-loops."

📖Definition 2.7: Symmetry Formal Predicatesdefinition

1. Symmetric Relation: $$\forall a, b \in A, \quad (a, b) \in R \implies (b, a) \in R \iff R = R^{-1}$$
2. Anti-Symmetric Relation: $$\forall a, b \in A, \quad \Big( (a, b) \in R \land (b, a) \in R \Big) \implies a = b \iff (R \cap R^{-1}) \subseteq \Delta_A$$
3. Asymmetric Relation: $$\forall a, b \in A, \quad (a, b) \in R \implies (b, a) \notin R \iff R \cap R^{-1} = \emptyset$$

📐Formula 2.4: Combinatorial Counting for Symmetry Variantsformula

For a set $A$ of cardinality $n$, the $n^2$ pairs partition into $n$ diagonal pairs and $\frac{n(n-1)}{2}$ unordered non-diagonal pairs $\{\,(a, b), (b, a)\,\}$:

1. Total Number of Symmetric Relations: Each of the $n$ diagonal pairs has 2 choices ($2^n$). Each of the $\frac{n(n-1)}{2}$ symmetric pairs has 2 choices (both in or both out): $$N_{\text{symmetric}} = 2^n \times 2^{\frac{n(n-1)}{2}} = 2^{n + \frac{n(n-1)}{2}} = 2^{\frac{n(n+1)}{2}}$$
2. Total Number of Anti-Symmetric Relations: Each diagonal pair has 2 choices ($2^n$). Each unordered pair has 3 mutually exclusive choices: (i) only $(a, b) \in R$, (ii) only $(b, a) \in R$, or (iii) neither in $R$: $$N_{\text{antisymmetric}} = 2^n \times 3^{\frac{n(n-1)}{2}}$$
3. Total Number of Asymmetric Relations: All $n$ diagonal pairs must be absent ($1^n = 1$). Each unordered pair has 3 choices: $$N_{\text{asymmetric}} = 1^n \times 3^{\frac{n(n-1)}{2}} = 3^{\frac{n(n-1)}{2}}$$
4. Total Number of Reflexive and Symmetric Relations: All $n$ diagonals fixed in ($1$), each unordered pair has 2 choices: $$N_{\text{reflexive \& symmetric}} = 2^{\frac{n(n-1)}{2}}$$

Diagnostic Decision Matrix on a Three-Element Set

The following table demonstrates the precise boundaries between symmetry classifications:

Relation on $A = \{1, 2, 3\}$	Symmetric	Anti-Symmetric	Asymmetric	Reason / Structural Witness
$A \times A$	Yes	No	No	Has $(1, 2)$ and $(2, 1)$ with $1 \neq 2$
$\emptyset$	Yes	Yes	Yes	Vacuously satisfies all three definitions
$\{(1, 1), (2, 2), (3, 3)\}$	Yes	Yes	No	Contains diagonals, which breaks asymmetry
$\{(1, 2), (2, 3), (1, 3)\}$	No	Yes	Yes	Zero reciprocal pairs and zero diagonals
$\{(1, 1), (1, 2), (2, 1)\}$	Yes	No	No	Reciprocal pair $(1, 2)$ and $(2, 1)$ with $1 \neq 2$
$\{(1, 1), (1, 2)\}$	No	Yes	No	Anti-symmetric, but $(1, 1)$ breaks asymmetry

---

⛓️ Transitive Relations: Multi-Hop Reachability Invariants

A relation is transitive if indirect reachability through an intermediate hop implies direct connectivity.

📖Definition 2.8: Transitive Relationdefinition

A relation $R$ on set $A$ is transitive if and only if:

$$\forall a, b, c \in A, \quad \Big( (a, b) \in R \land (b, c) \in R \Big) \implies (a, c) \in R$$

In relational composition terminology:

$$R \text{ is transitive} \iff (R \circ R) \subseteq R$$

The Vacuous Truth Trap in Transitivity

Students often make mistakes by assuming that a relation must contain three pairs to be transitive. If the condition $(a, b) \in R \land (b, c) \in R$ is never met, the premise is False, and transitivity holds vacuously!

• $R_1 = \{(1, 2)\}$ on $\{1, 2, 3\}$:
  • We have $(1, 2) \in R_1$. Is there any pair starting with $2$? No.
  • The premise $(a, b) \in R \land (b, c) \in R$ is never satisfied.
  • Conclusion: $R_1$ is Transitive (vacuously true!).
• $R_2 = \{(1, 2), (2, 3)\}$:
  • Here $(1, 2) \in R_2$ and $(2, 3) \in R_2$.
  • For transitivity, $(1, 3)$ must be in $R_2$.
  • Since $(1, 3) \notin R_2$, $R_2$ is NOT Transitive.
• $R_3 = \{(1, 2), (2, 1)\}$:
  • $(1, 2) \in R_3$ and $(2, 1) \in R_3 \implies (1, 1)$ is required!
  • $(2, 1) \in R_3$ and $(1, 2) \in R_3 \implies (2, 2)$ is required!
  • Since neither $(1, 1)$ nor $(2, 2)$ is present, $R_3$ is NOT Transitive.

Number of Transitive Relations

Unlike reflexivity and symmetry, there is no simple closed-form formula for the total number of transitive relations on an $n$-element set. The sequence grows rapidly (OEIS sequence A006905):

Set Size $\vert A \vert = n$	Total Relations ($2^{n^2}$)	Total Transitive Relations
$n = 0$	$2^0 = 1$	$1$
$n = 1$	$2^1 = 2$	$2$
$n = 2$	$2^4 = 16$	$13$
$n = 3$	$2^9 = 512$	$171$
$n = 4$	$2^{16} = 65{,}536$	$3{,}994$
$n = 5$	$2^{25} = 33{,}554{,}432$	$154{,}303$

---

🏛️ Master Taxonomy & Counting Reference Table

The following master reference table synthesizes the algebraic signatures, matrix conditions, and closed-form counting formulas for all primary relational properties on a set of cardinality $n$:

Property	Defining Mathematical Predicate	Matrix Invariant ($M_R$)	Smallest	Largest	Total Relations on Set of Size $n$
Reflexive	$\forall x \in A, \; (x, x) \in R$	Diagonal entries all 1	$\Delta_A$ ($n$)	$A \times A$ ($n^2$)	$2^{n^2 - n}$
Irreflexive	$\forall x \in A, \; (x, x) \notin R$	Diagonal entries all 0	$\emptyset$ ($0$)	$(A \times A) - \Delta_A$ ($n^2 - n$)	$2^{n^2 - n}$
Symmetric	$(a, b) \in R \implies (b, a) \in R$	$M_R = (M_R)^T$ (Symmetric)	$\emptyset$ ($0$)	$A \times A$ ($n^2$)	$2^{\frac{n(n+1)}{2}}$
Anti-Symmetric	$(a, b), (b, a) \in R \implies a = b$	$(R \cap R^{-1}) \subseteq \Delta_A$	$\emptyset$ ($0$)	Cardinality $\frac{n(n+1)}{2}$	$2^n \cdot 3^{\frac{n(n-1)}{2}}$
Asymmetric	$(a, b) \in R \implies (b, a) \notin R$	$R \cap R^{-1} = \emptyset$ (No self-loops)	$\emptyset$ ($0$)	Cardinality $\frac{n(n-1)}{2}$	$3^{\frac{n(n-1)}{2}}$
Refl & Symm	Reflexive $\land$ Symmetric	$M_R = (M_R)^T$ with diag 1	$\Delta_A$ ($n$)	$A \times A$ ($n^2$)	$2^{\frac{n(n-1)}{2}}$
Transitive	$(a, b), (b, c) \in R \implies (a, c) \in R$	$M_R \odot M_R \le M_R$	$\emptyset$ ($0$)	$A \times A$ ($n^2$)	No closed formula (OEIS A006905)

---

🌍 Real-World Engineering Context

1. Git Commit History as a Strict Partial Order: In Git, commits form a Directed Acyclic Graph (DAG). The relation "is an ancestor of" is irreflexive (a commit is not its own parent), asymmetric (no two distinct commits can be ancestors of each other), and transitive (parent of parent is ancestor). This makes Git history a strict partial order.
2. Database Transaction Schedules (Conflict Serializability): In database engines, conflict serializability is verified by constructing a precedence graph. Transactions $T_i \to T_j$ represent conflict dependencies. The schedule is serializable if and only if the transitive closure of this relation contains no self-loops (i.e. is acyclic).
3. Role-Based Access Control (RBAC) Hierarchies: Role inheritance in enterprise software is modeled as a partial order (reflexive, anti-symmetric, transitive). If Admin subsumes Editor and Editor subsumes Viewer, transitivity guarantees that Admin automatically inherits Viewer permissions.

---

📝 Step-by-Step Worked Examples

Example 1: Property Evaluation of Concrete Relations

Problem Statement:
Let $A = \{1, 2, 3\}$. Evaluate Reflexivity, Irreflexivity, Symmetry, Antisymmetry, Asymmetry, and Transitivity for:

$$R = \{(1, 1), \; (2, 2), \; (3, 3), \; (1, 2), \; (2, 3), \; (1, 3)\}$$

Step-by-Step Solution:

1. Reflexivity:
   • Check all diagonal pairs: $(1, 1) \in R$, $(2, 2) \in R$, and $(3, 3) \in R$.
   • All three are present. Reflexive: Yes.
2. Irreflexivity:
   • Requires zero diagonal pairs. Since all diagonals exist: Irreflexive: No.
3. Symmetry:
   • We have $(1, 2) \in R$. For symmetry, $(2, 1)$ must be in $R$.
   • Since $(2, 1) \notin R$: Symmetric: No.
4. Antisymmetry:
   • For all distinct pairs $a \neq b$, check if both $(a, b)$ and $(b, a)$ exist.
   • $(1, 2) \in R$ but $(2, 1) \notin R$.
   • $(2, 3) \in R$ but $(3, 2) \notin R$.
   • $(1, 3) \in R$ but $(3, 1) \notin R$.
   • No distinct pair exhibits two-way reciprocity. Anti-Symmetric: Yes.
5. Asymmetry:
   • Requires anti-symmetry AND irreflexivity (no diagonals).
   • Because $(1, 1) \in R$: Asymmetric: No.
6. Transitivity:
   • $(1, 2) \in R$ and $(2, 3) \in R \implies (1, 3) \in R$ (present!).
   • All other combinations trivially satisfy the condition. Transitive: Yes.

Key Takeaway:
Because this relation is Reflexive, Anti-Symmetric, and Transitive, it is a Partial Order Relation (POSET) (specifically, the standard $\le$ relation on $\{1, 2, 3\}$!).

---

Example 2: Combinatorial Counting of Relations with Mixed Properties

Problem Statement:
Let $A$ be a set with $|A| = 4$ elements. Determine the exact number of relations on $A$ that are:

1. Reflexive.
2. Symmetric.
3. Reflexive and Symmetric simultaneously.
4. Anti-Symmetric.

Step-by-Step Solution:

Given $n = 4$:

1. Reflexive Relations: $$N_{\text{reflexive}} = 2^{n(n-1)} = 2^{4 \times 3} = 2^{12} = 4{,}096$$
2. Symmetric Relations: $$N_{\text{symmetric}} = 2^{\frac{n(n+1)}{2}} = 2^{\frac{4 \times 5}{2}} = 2^{10} = 1{,}024$$
3. Reflexive and Symmetric Simultaneously: All 4 diagonals must be present ($1$). The $\frac{4 \times 3}{2} = 6$ symmetric non-diagonal pairs each have 2 choices (both in or both out): $$N_{\text{refl \& symm}} = 2^{\frac{n(n-1)}{2}} = 2^6 = 64$$
4. Anti-Symmetric Relations: Each of the 4 diagonals has 2 choices ($2^4 = 16$). Each of the 6 unordered pairs has 3 choices ($3^6 = 729$): $$N_{\text{antisymmetric}} = 2^4 \times 3^6 = 16 \times 729 = 11{,}664$$

Key Takeaway:
Decomposing the $n^2$ elements into $n$ diagonal slots and $\frac{n(n-1)}{2}$ symmetric partner slots simplifies all relational counting derivations.

---

Example 3: Verifying Matrix and Algebraic Invariants

Problem Statement:
Let $R$ be an anti-symmetric relation on set $A$. Prove algebraically that $R \cap R^{-1} \subseteq \Delta_A$.

Step-by-Step Solution:

1. Let $(x, y) \in R \cap R^{-1}$ be an arbitrary element.
2. By definition of set intersection: $$(x, y) \in R \quad \text{and} \quad (x, y) \in R^{-1}$$
3. By definition of inverse relation: $$(x, y) \in R^{-1} \iff (y, x) \in R$$
4. Substituting back: $$(x, y) \in R \quad \text{and} \quad (y, x) \in R$$
5. Because $R$ is given to be anti-symmetric, the definition states that: $$\forall a, b \in A, \; \Big( (a, b) \in R \land (b, a) \in R \Big) \implies a = b$$
6. Therefore, $x = y$.
7. Since $x = y$, the pair is of the form $(x, x)$, which belongs to the diagonal identity relation $\Delta_A$: $$(x, y) = (x, x) \in \Delta_A$$
8. Since every element of $R \cap R^{-1}$ belongs to $\Delta_A$, we conclude: $$R \cap R^{-1} \subseteq \Delta_A$$

Key Takeaway:
The intersection of any relation with its inverse isolates all two-way reciprocal paths. For anti-symmetric relations, those paths can only exist on the main diagonal.

---

🎯 Exam & Interview Pitfall Check

Core Conceptual Questions

Question 1: "Can a relation be neither reflexive nor irreflexive? Can it be both?"
Key Focus Points:

• Neither: Yes! A relation is neither if it contains some diagonal pairs, but not all of them. For example, on $A = \{1, 2\}$, $R = \{(1, 1)\}$ has $(1, 1)$ so it is not irreflexive, but lacks $(2, 2)$ so it is not reflexive.
• Both: A relation can be both reflexive and irreflexive if and only if the underlying set $A = \emptyset$. When $A$ is empty, $\Delta_A = \emptyset$. The empty relation $\emptyset$ trivially satisfies both definitions vacuously! For any non-empty set, reflexive and irreflexive are strictly mutually exclusive.

Question 2: "What is the difference between Anti-Symmetric and Asymmetric relations?"
Key Focus Points:

• Diagonal Pairs: Anti-symmetric relations permit diagonal pairs $(x, x)$ without penalty. Asymmetric relations strictly forbid all diagonal pairs.
• Algebraic Link: An asymmetric relation is identical to an anti-symmetric relation that is also irreflexive ($\text{Asymmetric} \equiv \text{Anti-Symmetric} \land \text{Irreflexive}$).
• Subset Inclusion: Every asymmetric relation is anti-symmetric, but not every anti-symmetric relation is asymmetric.

Common Interview Traps: Architectural Trade-Offs

Trap 1: If R₁ and R₂ are anti-symmetric, is their union R₁ ∪ R₂ always anti-symmetric?
Answer: No! Let $R_1 = \{(1, 2)\}$ and $R_2 = \{(2, 1)\}$ on $\{1, 2\}$. Both $R_1$ and $R_2$ are anti-symmetric individually. But their union is $R_1 \cup R_2 = \{(1, 2), (2, 1)\}$, which contains a 2-cycle between distinct elements and is therefore symmetric, not anti-symmetric. Their intersection $R_1 \cap R_2$, however, is always anti-symmetric.

Trap 2: Is the empty relation ∅ symmetric, anti-symmetric, and transitive?
Answer: Yes, to all three! Because $\emptyset$ contains zero elements, the antecedent for symmetry, anti-symmetry, and transitivity is always False. Under propositional logic, $\mathbf{F} \implies P$ is universally True. Hence, $\emptyset$ is vacuously symmetric, anti-symmetric, asymmetric, and transitive!

---

🗺️View Complete Discrete Mathematics Roadmap & Syllabus

← Previous Topic2.1 Binary Relations, Representations & Relational Matrix AlgebraNext Topic →2.3 Relational Closures (Warshall's Algorithm) & Equivalence Partitions