R={(a,b)€z ,a+b =even } prove that R is in equivalence with z

1.	R={(a,b)€z ,a+b =even } prove that R is in equivalence with z
Answer» R is reflexive , as 2 divides a-a = 0Now (a,b){tex}\\in {/tex}R implies (a-b) is divided by 2{tex}\\Rightarrow{/tex}\xa0(b-a) is also divided by 2Hence, (b,a){tex}\\in {/tex}RHence,\xa0R is symmetric.Now ,Let a,b,c {tex}\\in{/tex}ZIf (a,b) {tex}\\in {/tex}RAnd (b,c) {tex}\\in {/tex}RThen a-b and b-c divided by 2Therefore a-b is even and b-c is even{tex}\\Rightarrow{/tex}a-b +b-c is even, as sum of two even numbers is even\xa0{tex}\\Rightarrow{/tex}(a-c) is evenSo a-c is divided by 2{tex}\\Rightarrow{/tex}(a,c) {tex}\\in {/tex}RHence it is transitive.Therefore R is reflexive,symmetric and transitiveSo R is an equivalence relation

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