Let R1 = {(a, b) $\in$ N $\times$ N : |a $-$ b| $\le$ 13} and
R2 = {(a, b) $\in$ N $\times$ N : |a $-$ b| $\ne$ 13}. Then on N :
Solution
$R_{1}=\{(a, b) \in N \times N:|a-b| \leq 13\}$ and
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$R_{2}=\{(a, b) \in N \times N:|a-b| \neq 13\}$
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In $R_{1}: \because|2-11|=9 \leq 13$
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$\therefore \quad(2,11) \in R_{1}$ and $(11,19) \in R_{1}$ but $(2,19) \notin R_{1}$
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$\therefore \quad R_{1}$ is not transitive
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Hence $R_{1}$ is not equivalence
<br/><br/>In $R_{2}:(13,3) \in R_{2}$ and $(3,26) \in R_{2}$ but $(13,26) \notin R_{2}$
$(\because|13-26|=13)$
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$\therefore R_{2}$ is not transitive
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Hence $R_{2}$ is not equivalence.
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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