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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 on set AA is Reflexive if every element maps to itself. Mathematically, ∀a∈A,(a,a)∈R\forall a \in A, (a, a) \in R. In a directed graph representation, this means every node has a self-loop.

Reflexive relation showing self-loops on elements a1 and a2.
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A relation RR is Symmetric if for every pair (a,b)∈R(a, b) \in R, the reverse pair (b,a)(b, a) is also in RR. If there is an arrow from aa to bb, there must be one from bb back to aa.

Symmetric relation showing bidirectional connections between two points.
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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. Visually, if you can go from aa to cc via bb, a direct path from aa to cc must exist.

Transitive relation showing that paths through an intermediate node imply a direct path.
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An Equivalence Relation is a relation that is simultaneously reflexive, symmetric, and transitive. It partitions the set into disjoint subsets called Equivalence Classes.

Partition of a set into disjoint equivalence classes.

📐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.

Problem 3:

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

Three parallel lines illustrating the transitive property.

Solution:

  1. Reflexive: Every line L1L_1 is parallel to itself (L1∥L1L_1 \parallel L_1). Thus, (L1,L1)∈R(L_1, L_1) \in R.
  2. Symmetric: If L1∥L2L_1 \parallel L_2, then L2∥L1L_2 \parallel L_1. Thus, (L1,L2)∈R  ⟹  (L2,L1)∈R(L_1, L_2) \in R \implies (L_2, L_1) \in R.
  3. Transitive: If L1∥L2L_1 \parallel L_2 and L2∥L3L_2 \parallel L_3, then L1∥L3L_1 \parallel L_3. Thus, (L1,L2)∈R(L_1, L_2) \in R and (L2,L3)∈R  ⟹  (L1,L3)∈R(L_2, L_3) \in R \implies (L_1, L_3) \in R. Since RR is reflexive, symmetric, and transitive, it is an equivalence relation.

Explanation:

The relation of being parallel satisfies all three criteria. Note that for lines, we consider a line parallel to itself to satisfy reflexivity.

Problem 4:

Check if the relation RR on the set A={1,2,3}A = \{1, 2, 3\} defined as R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3)}R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3)\} is an equivalence relation.

A directed graph showing a non-symmetric relation where 2 points to 3, but 3 does not point to 2.

Solution:

  1. Reflexive: (1,1),(2,2),(3,3)∈R(1, 1), (2, 2), (3, 3) \in R. So, it is reflexive.
  2. Symmetric: (1,2)∈R(1, 2) \in R and (2,1)∈R(2, 1) \in R. However, (2,3)∈R(2, 3) \in R but (3,2)∉R(3, 2) \notin R. Therefore, it is not symmetric.
  3. Transitive: (1,2)∈R(1, 2) \in R and (2,3)∈R(2, 3) \in R, but (1,3)∉R(1, 3) \notin R. Therefore, it is not transitive. Since it is not symmetric and not transitive, it is not an equivalence relation.

Explanation:

For a relation to be an equivalence relation, it must satisfy all three properties. Failing even one (like symmetry here) disqualifies it.