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Check whether the relation R in the set Z of integers defined as R = {(π‘Ž, 𝑏) ∢ π‘Ž + 𝑏 is "divisible by 2"} is reflexive, symmetric or transitive. Write the equivalence class containing 0 i.e. [0].

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Check whether relation R in set Z of integers defined as R = {(a, b)

Question 29 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 29 - CBSE Class 12 Sample Paper for 2021 Boards - Part 3
Question 29 - CBSE Class 12 Sample Paper for 2021 Boards - Part 4

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Note : This is similar to Example 5 of NCERT – Chapter 1 Class 12 Relations and Functions

Check the answer here https:// www.teachoo.com /3959/673/Example-5---R---(a--b)---2-divides-a-b--is-equivalence-relation/category/Examples/

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Transcript

Question 29 Check whether the relation R in the set Z of integers defined as R = {(π‘Ž, 𝑏) ∢ π‘Ž + 𝑏 is "divisible by 2"} is reflexive, symmetric or transitive. Write the equivalence class containing 0 i.e. [0]. R = {(a, b) : π‘Ž + 𝑏 is "divisible by 2"} Check reflexive Since a + a = 2a & 2 divides 2a Therefore, 2 divides a + a ∴ (a, a) ∈ R, ∴ R is reflexive. Check symmetric If 2 divides a + b , then 2 divides b + a Hence, If (a, b) ∈ R, then (b, a) ∈ R ∴ R is symmetric Check transitive If 2 divides (a + b) , & 2 divides (b + c) , So, we can write a + b = 2k b + c = 2p Adding (1) & (2) (a + b) + (b + c) = 2k + 2p a + c + 2b = 2k + 2p a + c = 2k + 2p βˆ’ 2b a + c = 2(k + p βˆ’ b) So, 2 divides (a + c) ∴ If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R Therefore, R is transitive. Thus, R is an equivalence relation in Z. Now, Equivalence class containing 0 i.e. [0] will be all values of a where one element is 0 Now, R = {(a, b) : π‘Ž + 𝑏 is "divisible by 2"} Putting b = 0 R = {(a, 0) : π‘Ž is "divisible by 2"} So, [0] = All possible values of a = {…., βˆ’6, βˆ’4, βˆ’2, 0, 2, 4, 6, ….}

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