A) Symmetric
B) Reflexive and Transitive
C) Transitive and symmetric
D) Equivalence
Correct Answer: B
Solution :
\[R=\left\{ \left( 1,\,1 \right),\,\left( 1,\,2 \right),\,\left( 1,3 \right),\,\left( 1,4 \right)\,\left( 1,\,5 \right),\,\left( 1,6 \right) \right.\] \[\left. \left( 2,\,2 \right),\,\left( 2,\,4 \right),\,\left( 2,\,6 \right),\,\left( 3,\,3 \right),\,\left( 3,\,6 \right),\,\left( 4,\,4 \right),\left( 5,5 \right),\,\left( 6,\,6 \right) \right\}\] \[R=\left\{ \left( 1,\,1 \right),\,\left( 1,\,2 \right),\,\left( 2,\,2 \right),\,\left( 3,\,3 \right),\,\left( 4,\,4 \right),\,\left( 5,\,5 \right),\,\left( 6,\,6 \right) \right\}\] here \[\left( a,\,\,a \right)\in \,\,R\,\,\forall \,a\] \[\Rightarrow R\] is reflexive Also \[\left( 1,\,\,2 \right)\in R\] but \[\left( 2,\,\,1 \right)\notin R\] \[\Rightarrow R\] is not symmetric clearly R is transitive.
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