Last updated at May 29, 2018 by Teachoo

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Ex 1.1, 9 (Introduction) Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} , given by (i) R = { (a, b):|a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 in each case. Modulus function 1 = 1 2 = 2 0 = 0 −1 = 1 −3 = 3 Ex 1.1, 9 Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} , given by (i) R = { (a, b):|a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 in each case. A = {x ∈ Z: 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} R = {(a,b) : |a – b| is a multiple of 4} Note: Multiple of 4 are 0, 4, 8, 12 i.e. |a – b| can be 0, 4, 8, 12 only Check reflexive Since |a – a| = |0| = 0 & 0 is a multiple of 4 , eg: 4 × 0 = 0 ⇒ |a – a| is a multiple of 4 ∴ (a, a) ∈ R, ∴ R is reflexive. Check symmetric We know that |a – b| = |b – a| Hence, if |a – b| is a multiple of 4, then |b – a| is also a multiple of 4 Hence, If (a, b) ∈ R, then (b, a) ∈ R ∴ R is symmetric Check transitive If |a – b| is a multiple of 4, then (a – b) is a multiple of 4 Similarly, if |b – c| is a multiple of 4 , then (b – c) is a multiple of 4 Now, Sum of multiple of 4 is also a multiple of 4 a – b + b – c is a multiple of 4 ⇒ a – c is a multiple of 4 Hence, |a – c| is a multiple of 4 ∴ If |a – b| &|b – c| is a multiple of 4 , then |a – c| is a multiple of 4 i.e. If (a, b) ∈ R & (b, c) ∈ R , then (a, c) ∈ R ∴ R is transitive Since R is reflexive, symmetric and transitive, it is equivalence relation We need to find set of elements related to 1 R = {(a, b) : |a – b| is a multiple of 4} & A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Here a = 1, The set of elements related to 1 are {1, 5, 9} Ex 1.1, 9 Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} , given by (ii) R = {(a, b): a = b} is an equivalence relation. Find the set of all elements related to 1 in each case. A = {x ∈ Z: 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} R = {(a, b) : a = b} Check reflexive Since a = a is always true, So, (a, a) ∈ R, ∴ R is reflexive. Check symmetric We know that if a = b, then b = a Hence, if (a, b) ∈ R, then (b, a) ∈ R Hence, R is symmetric Check transitive If a = b & b = c, then a = c If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R Hence, R is transitive Hence, R is reflexive, symmetric and transitive, ∴ it is equivalence relation We need to find set of elements related to 1 R = {(a, b) : a = b} All elements related to 1 means a = 1, If a = 1, then b = a = 1 ∴ Only (1, 1) satisfies the relation. Hence, the set of elements related to 1 is {1}.

Chapter 1 Class 12 Relation and Functions

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.