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Relations and Functions - Types of relations: reflexive, symmetric, transitive and equivalence relations

Grade 12CBSE

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

🔑Concepts

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A Relation RR from a set AA to a set BB is a subset of the Cartesian product A×BA \times B. Visually, this is represented by an arrow diagram where elements in the domain AA are connected to elements in the codomain BB by directed arrows.

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A relation RR on a set AA is called Reflexive if every element of AA is related to itself. Symbolically, (a,a)∈R(a, a) \in R for every a∈Aa \in A. In a coordinate graph, this means all points of the form (x,x)(x, x) lie on the identity line y=xy = x.

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A relation RR on a set AA is Symmetric if whenever (a,b)∈R(a, b) \in R, then (b,a)∈R(b, a) \in R for all a,b∈Aa, b \in A. Visually, if there is an arrow pointing from aa to bb, there must be a corresponding return arrow pointing from bb back to aa.

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A relation RR on a set AA 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 for all a,b,c∈Aa, b, c \in A. Visually, if you can travel from node aa to bb and then from bb to cc, there must be a direct 'shortcut' arrow from aa to cc to complete the path.

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A relation RR in a set AA is an Equivalence Relation if it is simultaneously Reflexive, Symmetric, and Transitive. Visually, an equivalence relation partitions the set into disjoint 'clusters' or subsets where every element within a cluster is related to every other element in that same cluster.

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An Equivalence Class [a][a] of an element a∈Aa \in A is the set of all elements x∈Ax \in A such that (x,a)∈R(x, a) \in R. Visually, if the set is partitioned into disjoint islands, the equivalence class of aa is the specific island that contains the element aa.

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The Empty Relation (ϕ\phi) and the Universal Relation (A×AA \times A) are called trivial relations. In an empty relation, no element is related to any other (no arrows in the diagram), while in a universal relation, every element is related to every element including itself (every possible arrow exists).

📐Formulae

Total number of relations from set AA to set BB: 2n(A)×n(B)2^{n(A) \times n(B)}

Total number of relations on a set AA with nn elements: 2n22^{n^2}

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

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

Transitive Relation 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

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

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

💡Examples

Problem 1:

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

Solution:

  1. Reflexive: For any triangle T1∈TT_1 \in T, T1≅T1T_1 \cong T_1 (every triangle is congruent to itself). Thus, (T1,T1)∈R(T_1, T_1) \in R for all T1∈TT_1 \in T. RR is reflexive.
  2. Symmetric: Let (T1,T2)∈R(T_1, T_2) \in R. This implies T1≅T2T_1 \cong T_2. Since congruence is symmetric, T2≅T1T_2 \cong T_1. Therefore, (T2,T1)∈R(T_2, T_1) \in R. RR is symmetric.
  3. Transitive: Let (T1,T2)∈R(T_1, T_2) \in R and (T2,T3)∈R(T_2, T_3) \in R. This means T1≅T2T_1 \cong T_2 and T2≅T3T_2 \cong T_3. By the property of congruence, 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 an equivalence relation, we must independently verify the three properties (Reflexive, Symmetric, and Transitive) using the definition of the given relation (congruence of triangles).

Problem 2:

Let ZZ be the set of integers and RR be a relation defined by R={(a,b):2 divides (a−b)}R = \{(a, b) : 2 \text{ divides } (a - b)\}. Prove RR is an equivalence relation.

Solution:

  1. Reflexive: For any a∈Za \in Z, a−a=0a - a = 0. Since 22 divides 00, (a,a)∈R(a, a) \in R. Hence, RR is reflexive.
  2. Symmetric: Let (a,b)∈R(a, b) \in R. Then a−b=2ka - b = 2k for some integer kk. Multiplying by −1-1, b−a=−2k=2(−k)b - a = -2k = 2(-k). Since −k-k is an integer, 22 divides b−ab - a. Thus (b,a)∈R(b, a) \in R. RR is symmetric.
  3. Transitive: Let (a,b)∈R(a, b) \in R and (b,c)∈R(b, c) \in R. Then a−b=2ka - b = 2k and b−c=2mb - c = 2m for integers k,mk, m. Adding these: (a−b)+(b−c)=2k+2m  ⟹  a−c=2(k+m)(a - b) + (b - c) = 2k + 2m \implies a - c = 2(k + m). Since k+mk + m is an integer, 22 divides a−ca - c. Thus (a,c)∈R(a, c) \in R. RR is transitive.

Explanation:

This demonstrates the properties using algebraic manipulation. The relation effectively partitions integers into two equivalence classes: even and odd numbers.