JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 1)
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 :
Both R1 and R2 are equivalence relations
Neither R1 nor R2 is an equivalence relation
R1 is an equivalence relation but R2 is not
R2 is an equivalence relation but R1 is not
Explanation
$R_{1}=\{(a, b) \in N \times N:|a-b| \leq 13\}$ and
$R_{2}=\{(a, b) \in N \times N:|a-b| \neq 13\}$
In $R_{1}: \because|2-11|=9 \leq 13$
$\therefore \quad(2,11) \in R_{1}$ and $(11,19) \in R_{1}$ but $(2,19) \notin R_{1}$
$\therefore \quad R_{1}$ is not transitive
Hence $R_{1}$ is not equivalence
In $R_{2}:(13,3) \in R_{2}$ and $(3,26) \in R_{2}$ but $(13,26) \notin R_{2}$ $(\because|13-26|=13)$
$\therefore R_{2}$ is not transitive
Hence $R_{2}$ is not equivalence.
$R_{2}=\{(a, b) \in N \times N:|a-b| \neq 13\}$
In $R_{1}: \because|2-11|=9 \leq 13$
$\therefore \quad(2,11) \in R_{1}$ and $(11,19) \in R_{1}$ but $(2,19) \notin R_{1}$
$\therefore \quad R_{1}$ is not transitive
Hence $R_{1}$ is not equivalence
In $R_{2}:(13,3) \in R_{2}$ and $(3,26) \in R_{2}$ but $(13,26) \notin R_{2}$ $(\because|13-26|=13)$
$\therefore R_{2}$ is not transitive
Hence $R_{2}$ is not equivalence.
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