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  • Show that the relation R in the set A = {1, 2, 3, 4, 5, 6} given by R = {(a, b) : |a – b| is divisible by 2} is an equivalence relation.  

Show that the relation R in the set A = {1, 2, 3, 4, 5, 6} given by R = {(a, b) : |a – b| is divisible by 2} is an equivalence relation.

 

Answers (1)

A = {1, 2, 3, 4, 5, 6}

R = {(a, b) : |a – b| is divisible by 2} 

\\ $ Check for Reflexive: $ \\ $ if $ (\mathrm{a}, \mathrm{a}) $ for every $ a \in \mathrm{R}$ $\Rightarrow \mathrm{R}$ is reflexive $ \\ $ Here $|a-a|$ divisible by 2 $ \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 2 $ \Rightarrow |a-b| = 2 k \\ |b-a| = |-(a-b)| = 2k \Rightarrow |b-a| $ is divisible by 2 $\\ $ R is symmetric

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

\\ |a-b|= 2k, |b-c| = 2l \\ a-c = (a-b)+(b-c) = \pm 2k \pm 2l \\ |a-c| =\pm 2(k+l) \Rightarrow |a-c|$ is divisible by 2 $ \\ $ R is transitive

 Hence R an equivalence relation.

Posted by

Safeer PP

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