Chapter 5 • Unit II

Exercise 5.3: Representation and Types of Relations

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

5.3 Representation and Types of Relations

In Applied Mathematics, operations research, computer science, and economic theory, understanding the structural behavior of relationships is vital. Whether modeling parity across financial instruments, establishing equivalence classes in risk categories, or verifying transitivity in corporate organizational hierarchies, the taxonomy of relations provides the necessary algebraic tools.

Core Structural Theorems & Classification

1. Modes of Representation

  • Roster (Tabular) Form: Explicitly listing all ordered pairs: \(R = \{(1, 2), (2, 4), (3, 6)\}\).
  • Set-Builder Form: Specifying the governing rule: \(R = \{(x, y) : y = 2x, \, x, y \in \mathbb{N}\}\).
  • Arrow Diagram: Directed arrows drawn from elements in domain set \(A\) to target elements in co-domain \(B\).
  • Matrix Representation: A binary matrix \(M_R\) where entry \(m_{ij} = 1\) if \((a_i, b_j) \in R\) and \(0\) otherwise.

2. Primary Types of Relations on a Set \(A\)

  • Empty (Void) Relation: \(R = \emptyset \subseteq A \times A\). No element of \(A\) is related to any element.
  • Universal Relation: \(R = A \times A\). Every element of \(A\) is related to every element of \(A\). (Empty and Universal relations are called trivial relations).
  • Identity Relation (\(I_A\)): \(I_A = \{ (a, a) : a \in A \}\). Every element of \(A\) is related to itself and only itself.
  • Reflexive Relation: A relation \(R\) is reflexive if: \[ \forall a \in A, \quad (a, a) \in R \] Key Distinction: Every identity relation is reflexive, but a reflexive relation may contain additional non-diagonal pairs \((a, b)\) with \(a \neq b\).
  • Symmetric Relation: A relation \(R\) is symmetric if: \[ (a, b) \in R \implies (b, a) \in R, \quad \forall a, b \in A \] Equivalently, \(R = R^{-1}\).
  • Transitive Relation: A relation \(R\) is transitive if: \[ (a, b) \in R \quad \text{and} \quad (b, c) \in R \implies (a, c) \in R, \quad \forall a, b, c \in A \]
  • Equivalence Relation: A relation \(R\) on set \(A\) is an equivalence relation if and only if it is simultaneously: \[ \textbf{Reflexive}, \quad \textbf{Symmetric}, \quad \text{and} \quad \textbf{Transitive.} \]
  • Equivalence Class: For an equivalence relation \(R\) on \(A\), the equivalence class containing \(a\) is: \[ [a] = \{ x \in A : (x, a) \in R \} \] Equivalence classes form a partition of \(A\): they are mutually disjoint and their union equals \(A\).

Systematic Verification Workflow for Relation Types

Property Mathematical Condition How to Prove How to Disprove (Counterexample)
Reflexive \((x, x) \in R, \, \forall x \in A\) Show the algebraic condition holds when \(y = x\) Find a single \(x_0 \in A\) such that \((x_0, x_0) \notin R\)
Symmetric \((x, y) \in R \implies (y, x) \in R\) Assume \((x, y) \in R\) and algebraically deduce \((y, x) \in R\) Find a pair with \((a, b) \in R\) but \((b, a) \notin R\)
Transitive \((x, y) \in R \land (y, z) \in R \implies (x, z) \in R\) Assume \((x, y) \in R\) and \((y, z) \in R\) and deduce \((x, z) \in R\) Find a chain where \((a, b) \in R\) and \((b, c) \in R\), but \((a, c) \notin R\)

Comprehensive Applied Example: Congruence Modulo \(m\)

Problem: Let \(\mathbb{Z}\) be the set of all integers. Define a relation \(R\) on \(\mathbb{Z}\) by: \[ R = \{ (a, b) : a - b \text{ is divisible by 5} \} \] Prove that \(R\) is an equivalence relation and determine all distinct equivalence classes.

Proof:

  1. Reflexive: For any \(a \in \mathbb{Z}\), \(a - a = 0 = 5 \times 0\), which is divisible by 5. Thus \((a, a) \in R\) for all \(a \in \mathbb{Z}\). \(R\) is reflexive.
  2. Symmetric: Let \((a, b) \in R\). Then \(a - b = 5k\) for some \(k \in \mathbb{Z}\). Then: \[ b - a = -(a - b) = -5k = 5(-k) \] Since \(-k \in \mathbb{Z}\), \(b - a\) is divisible by 5. Thus \((b, a) \in R\). \(R\) is symmetric.
  3. Transitive: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(a - b = 5k\) and \(b - c = 5m\) for integers \(k, m\). Adding these two equations: \[ (a - b) + (b - c) = 5k + 5m \implies a - c = 5(k + m) \] Since \(k + m \in \mathbb{Z}\), \(a - c\) is divisible by 5. Thus \((a, c) \in R\). \(R\) is transitive.

Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.

Equivalence Classes: By the Division Algorithm, when any integer is divided by 5, the remainder is \(0, 1, 2, 3,\) or \(4\). There are exactly 5 distinct equivalence classes: \[ [0] = \{ \dots, -10, -5, 0, 5, 10, \dots \} \] \[ [1] = \{ \dots, -9, -4, 1, 6, 11, \dots \} \] \[ [2] = \{ \dots, -8, -3, 2, 7, 12, \dots \} \] \[ [3] = \{ \dots, -7, -2, 3, 8, 13, \dots \} \] \[ [4] = \{ \dots, -6, -1, 4, 9, 14, \dots \} \]

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