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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.

πŸ”‘Concepts

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A Relation RR from a set AA to a set BB is a subset of the Cartesian product AΓ—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 (Ο•\phi) and the Universal Relation (AΓ—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).

πŸ“Formulae

Total number of relations from set AA to set BB: 2n(A)Γ—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: βˆ€a∈A,(a,a)∈R\forall a \in A, (a, a) \in R

Symmetric Relation Condition: (a,b)∈Rβ€…β€ŠβŸΉβ€…β€Š(b,a)∈R(a, b) \in R \implies (b, a) \in R

Transitive Relation Condition: (a,b)∈RΒ andΒ (b,c)∈Rβ€…β€ŠβŸΉβ€…β€Š(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: 2n2βˆ’n2^{n^2 - n}

πŸ’‘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):T1β‰…T2}R = \{(T_1, T_2) : T_1 \cong T_2\} (where β‰…\cong denotes congruence). Show that RR is an equivalence relation.

Solution:

  1. Reflexive: For any triangle T1∈TT_1 \in T, T1β‰…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 T1β‰…T2T_1 \cong T_2. Since congruence is symmetric, T2β‰…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 T1β‰…T2T_1 \cong T_2 and T2β‰…T3T_2 \cong T_3. By the property of congruence, T1β‰…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):2Β dividesΒ (aβˆ’b)}R = \{(a, b) : 2 \text{ divides } (a - b)\}. Prove RR is an equivalence relation.

Solution:

  1. Reflexive: For any a∈Za \in Z, aβˆ’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 aβˆ’b=2ka - b = 2k for some integer kk. Multiplying by βˆ’1-1, bβˆ’a=βˆ’2k=2(βˆ’k)b - a = -2k = 2(-k). Since βˆ’k-k is an integer, 22 divides bβˆ’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 aβˆ’b=2ka - b = 2k and bβˆ’c=2mb - c = 2m for integers k,mk, m. Adding these: (aβˆ’b)+(bβˆ’c)=2k+2mβ€…β€ŠβŸΉβ€…β€Šaβˆ’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 aβˆ’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.