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  • Let <img alt="\mathrm{R}_{1}=\{(\mathrm{a}, \mathrm{b}) \in \mathbf{N} \times \mathbf{N}:|\mathrm{a}-\mathrm{b}| \leqslant 13\}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BR%7

Let \mathrm{R}_{1}=\{(\mathrm{a}, \mathrm{b}) \in \mathbf{N} \times \mathbf{N}:|\mathrm{a}-\mathrm{b}| \leqslant 13\} and \mathrm{}\quad R_{2}=\{(a, b) \in \mathbf{N} \times \mathbf{N}:|a-b| \neq 13\}. Then on \mathbf{N} :

Option: 1

Both \mathrm{R}_{1} and\: \mathrm{R}_{2} are equivalence relations


Option: 2

Neither \mathrm{R}_{1}$ nor $\mathrm{R}_{2} is an equivalence relation


Option: 3

\mathrm{R}_{1} is an equivalence relation but \mathrm{R}_{2} is not


Option: 4

\mathrm{R}_{2} is an equivalence relation but \mathrm{R}_{1} is not


Answers (1)

best_answer

\mathrm{R_{1}} is not transitive as  \mathrm{(1,12)\in \: R_{1}}  and  \mathrm{(12,24)\in \: R_{1}}  but  \mathrm{(1,24)\notin \: R_{1}}

\mathrm{R_{2}} is not transitive as  \mathrm{(1,10)\in \: R_{2}}  and  \mathrm{(10,14)\in \: R_{2}}  but  \mathrm{(1,14)\notin \: R_{2}}

Hence the correct option is 2

Posted by

Gaurav

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