Prove that the relation R on Z, defined by R {(x, y) : (x – y) is divisible by 5} is an equivalence relation.

 

Answers (1)
S safeer

R on Z and R = {(x, y) : (x– y) is divisible by 5} 

\\ $ Check for Reflexive: $ \\ $ if $ (\mathrm{a}, \mathrm{a}) $ for every $ a \in \mathrm{R}$ $\Rightarrow \mathrm{R}$ is reflexive $ \\ $ Here $(a-a)$ divisible by 5 $ \Rightarrow (\mathrm{a}, \mathrm{a}) \in \mathrm{R} \\ $ \mathrm{R} is reflexive $ \\

\\ $ Check for Symmetric: $ \\ $ if $ (\mathrm{a}, \mathrm{b}) \in \mathrm{R}$ $ then $ (\mathrm{b}, \mathrm{a}) \in \mathrm{R} $\Rightarrow \mathrm{R}$ is Symmetric $

\\ $ Here $(a-b) $ is divisible by 5 $ \Rightarrow (a-b) = 5 k \\ (b-a) = -5k \Rightarrow (b-a) $ is divisible by 5 $\\ $ R is symmetric

\\ $ Check For transitive : $ \\ $ If $(a, b),(b, c) \in R $ then $ (a, c) \in R \Rightarrow $ R is transitive $

\\ (a-b)= 5k, (b-c) = 5l \\ (a-c )= (a-b)+(b-c) = 5k +5l \\ (a-c) = 5(k+l) \Rightarrow (a-c)$ is divisible by 5 $ \\ $ R is transitive

 Hence R an equivalence relation.

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