CBSE Class 12 Sample Paper for 2021 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## Check whether the relation R in the set Z of integers defined as R = {(π, π) βΆ π + π is "divisible by 2"} is reflexive, symmetric or transitive. Write the equivalence class containing 0 i.e. [0].

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

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### Transcript

Question 29 Check whether the relation R in the set Z of integers defined as R = {(π, π) βΆ π + π is "divisible by 2"} is reflexive, symmetric or transitive. Write the equivalence class containing 0 i.e. [0]. R = {(a, b) : π + π is "divisible by 2"} Check reflexive Since a + a = 2a & 2 divides 2a Therefore, 2 divides a + a β΄ (a, a) β R, β΄ R is reflexive. Check symmetric If 2 divides a + b , then 2 divides b + a Hence, If (a, b) β R, then (b, a) β R β΄ 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 β 2b a + c = 2(k + p β b) So, 2 divides (a + c) β΄ If (a, b) β R and (b, c) β R, then (a, c) β 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 + π is "divisible by 2"} Putting b = 0 R = {(a, 0) : π is "divisible by 2"} So, [0] = All possible values of a = {β¦., β6, β4, β2, 0, 2, 4, 6, β¦.}