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Relations and Functions - Types of Relations: Reflexive, Symmetric, Transitive and Equivalence

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A relation RR on a set AA is Reflexive if every element of AA is related to itself. Formally, ∀a∈A,(a,a)∈R\forall a \in A, (a, a) \in R. Visually, in a directed graph representation, every node must have a self-loop.

Reflexive relation diagram showing nodes with self-loops.
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A relation RR is Symmetric if whenever (a,b)∈R(a, b) \in R, then (b,a)∈R(b, a) \in R must also be true. In a mapping diagram, this means any arrow from one set to another is matched by an arrow in the opposite direction.

Symmetric relation diagram showing two-way arrows between nodes.
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A relation RR is Transitive if (a,b)∈R(a, b) \in R and (b,c)∈R(b, c) \in R implies (a,c)∈R(a, c) \in R. It represents a 'shortcut' or a chain of relations.

Transitive relation diagram showing a triangle of connections.
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An Equivalence Relation is one that is simultaneously Reflexive, Symmetric, and Transitive. It partitions the set into disjoint subsets called Equivalence Classes.

Diagram showing a set partitioned into equivalence classes.

📐Formulae

Reflexive condition: ∀a∈A,(a,a)∈R\forall a \in A, (a, a) \in R

Symmetric condition: (a,b)∈R  ⟹  (b,a)∈R(a, b) \in R \implies (b, a) \in R

Transitive condition: (a,b)∈R and (b,c)∈R  ⟹  (a,c)∈R(a, b) \in R \text{ and } (b, c) \in R \implies (a, c) \in R

Number of relations on a set with nn elements: 2n22^{n^2}

Number of reflexive relations on a set with nn elements: 2n2−n2^{n^2 - n}

Equivalence class of aa: [a]={x∈A:(x,a)∈R}[a] = \{x \in A : (x, a) \in R\}

💡Examples

Problem 1:

Let TT be the set of all triangles in a plane. Let RR be a relation in TT defined as R={(T1,T2):T1≅T2}R = \{(T_1, T_2) : T_1 \cong T_2\} (where ≅\cong denotes congruence). Prove that RR is an equivalence relation.

Solution:

  1. Reflexivity: Every triangle T1T_1 is congruent to itself (T1≅T1T_1 \cong T_1). Therefore, (T1,T1)∈R(T_1, T_1) \in R for all T1∈TT_1 \in T. RR is reflexive.
  2. Symmetry: If (T1,T2)∈R(T_1, T_2) \in R, then T1≅T2T_1 \cong T_2. By the property of congruence, T2≅T1T_2 \cong T_1. Therefore, (T2,T1)∈R(T_2, T_1) \in R. RR is symmetric.
  3. Transitivity: If (T1,T2)∈R(T_1, T_2) \in R and (T2,T3)∈R(T_2, T_3) \in R, then T1≅T2T_1 \cong T_2 and T2≅T3T_2 \cong T_3. This implies T1≅T3T_1 \cong T_3. Therefore, (T1,T3)∈R(T_1, T_3) \in R. RR is transitive. Since RR is reflexive, symmetric, and transitive, it is an equivalence relation.

Explanation:

To prove a relation is an equivalence relation, we must verify all three fundamental properties (reflexive, symmetric, and transitive) using the given logical definition of the relation.

Problem 2:

Check if the relation RR on the set of integers Z\mathbb{Z} defined by R={(a,b):a−b is divisible by 3}R = \{(a, b) : a - b \text{ is divisible by } 3\} is an equivalence relation.

Solution:

  1. Reflexive: For any a∈Za \in \mathbb{Z}, a−a=0a - a = 0. Since 00 is divisible by 33, (a,a)∈R(a, a) \in R. Thus, RR is reflexive.
  2. Symmetric: Let (a,b)∈R(a, b) \in R. Then a−b=3ka - b = 3k for some integer kk. Then b−a=−(a−b)=−3k=3(−k)b - a = -(a - b) = -3k = 3(-k). Since −k-k is an integer, b−ab - a is divisible by 33, so (b,a)∈R(b, a) \in R. Thus, RR is symmetric.
  3. Transitive: Let (a,b)∈R(a, b) \in R and (b,c)∈R(b, c) \in R. Then a−b=3ka - b = 3k and b−c=3mb - c = 3m for integers k,mk, m. Adding these: (a−b)+(b−c)=3k+3m  ⟹  a−c=3(k+m)(a - b) + (b - c) = 3k + 3m \implies a - c = 3(k + m). Since k+mk + m is an integer, a−ca - c is divisible by 33, so (a,c)∈R(a, c) \in R. Thus, RR is transitive. Since all three hold, RR is an equivalence relation.

Explanation:

This demonstrates an equivalence relation over an infinite set (Integers). We use algebraic manipulation to show that if the condition holds for specific pairs, it must hold for the reflexive and transitive requirements.

Problem 3:

Show that the relation RR defined on the set LL of all lines in a plane by R={(L1,L2):L1∥L2}R = \{(L_1, L_2) : L_1 \parallel L_2\} (parallelism) is an equivalence relation.

Three parallel lines L1, L2, and L3 demonstrating transitivity.

Solution:

  1. Reflexive: Every line L1L_1 is parallel to itself (L1∥L1L_1 \parallel L_1). So, (L1,L1)∈R(L_1, L_1) \in R.
  2. Symmetric: If (L1,L2)∈R(L_1, L_2) \in R, then L1∥L2  ⟹  L2∥L1L_1 \parallel L_2 \implies L_2 \parallel L_1, thus (L2,L1)∈R(L_2, L_1) \in R.
  3. Transitive: If (L1,L2)∈R(L_1, L_2) \in R and (L2,L3)∈R(L_2, L_3) \in R, then L1∥L2L_1 \parallel L_2 and L2∥L3L_2 \parallel L_3. This implies L1∥L3L_1 \parallel L_3, so (L1,L3)∈R(L_1, L_3) \in R. Since RR is reflexive, symmetric, and transitive, it is an equivalence relation.

Explanation:

The relation of parallelism satisfies all three criteria. Note: we assume a line is parallel to itself in this context (reflexivity).

Problem 4:

Check if the relation RR on the set of real numbers R\mathbb{R} defined by R={(a,b):a≤b2}R = \{(a, b) : a \le b^2\} is transitive.

Graph of y = x squared used to visualize the condition a <= b^2.

Solution:

To check transitivity, we need to see if (a,b)∈R(a, b) \in R and (b,c)∈R(b, c) \in R implies (a,c)∈R(a, c) \in R. Consider a counter-example: Let a=3a = 3, b=2b = 2, c=1.5c = 1.5. (3,2)∈R(3, 2) \in R because 3≤223 \le 2^2 (3≤43 \le 4). (2,1.5)∈R(2, 1.5) \in R because 2≤(1.5)22 \le (1.5)^2 (2≤2.252 \le 2.25). However, a≤c2a \le c^2 becomes 3≤(1.5)23 \le (1.5)^2 (3≤2.253 \le 2.25), which is False. Since (a,b)∈R(a, b) \in R and (b,c)∈R(b, c) \in R but (a,c)∉R(a, c) \notin R, the relation is not transitive.

Explanation:

Transitivity fails because squaring numbers between 0 and 1 reduces their value, or squaring small numbers doesn't always maintain the chain of magnitude required by the relation.