Dear Student , (i) Since a−a=0 and 0 is an even integer (a,a)∈R ∴ R is reflexive (ii) If (a−b) is even, then (b−a) is also even then, if (a−b)∈R,(b,a)∈R ∴ The relation is symmetric (iii) If (a,b)∈R,(b,c)∈R, then (a−b) is even, (b−c) is even, then &&(a-b +b-c)=a-c&& is even ∴ If (a,b)∈R,(b,c)∈R implies (a,c)∈R ∴ R is transitive Since R is reflexive, symmetric and transitive, it is an equivalence relation Regards

Please solve Q. No.3

Dear Student ,

(i) S i n c e a - a = 0 a n d 0 i s a n e v e n i n t e g e r (a, a) \in R ∴ R i s r e f l e x i v e . (i i) I f (a - b) i s e v e n, t h e n (b - a) i s a l s o e v e n . t h e n, i f (a - b) \in R, (b, a) \in R ∴ T h e r e l a t i o n i s s y m m e t r i c . (i i i) I f (a, b) \in R, (b, c) \in R, t h e n (a - b) i s e v e n, (b - c) i s e v e n, t h e n & & (a - b + b - c) = a - c & & i s e v e n . ∴ I f (a, b) \in R, (b, c) \in R i m p l i e s (a, c) \in R ∴ R i s t r a n s i t i v e . S i n c e R i s r e f l e x i v e, s y m m e t r i c a n d t r a n s i t i v e, i t i s a n e q u i v a l e n c e r e l a t i o n .

Regards

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