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Let a relation $R$ be defined by $R = \{(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)\}$ then ${R^{ - 1}}oR$ is
$\{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)\}$
$\{(1, 1), (4, 4), (7, 7), (3, 3)\}$
$\{(1, 5), (1, 6), (3, 6)\}$
None of these
Solution
(a) We first find ${R^{ – 1}},$ we have
${R^{ – 1}} = \{ (5,\,4);\,(4,\,1);\,(6,\,4);\,(6,\,7);\,(7,\,3)\} $.
We now obtain the elements of ${R^{ – 1}}oR$ we first pick the element of
$R$ and then of ${R^{ – 1}}$. Since $(4,\,5) \in R$ and $(5,\,4) \in {R^{ – 1}}$, we have $(4,\,4) \in {R^{ – 1}}oR$
Similarly, $(1,\,4) \in R,\,(4,\,1) \in {R^{ – 1}} \Rightarrow \,(1,\,1) \in {R^{ – 1}}oR$
$(4,\,6) \in R,\,(6,\,4) \in {R^{ – 1}} \Rightarrow \,(4,\,4) \in {R^{ – 1}}oR,$
$(4,\,6) \in R,\,(6,\,7) \in {R^{ – 1}} \Rightarrow \,(4,\,7) \in {R^{ – 1}}oR$
$(7,\,6) \in R,\,(6,\,4) \in {R^{ – 1}} \Rightarrow \,(7,\,4) \in {R^{ – 1}}oR,$
$(7,\,6) \in R,\,(6,\,7) \in {R^{ – 1}} \Rightarrow \,(7,\,7) \in {R^{ – 1}}oR$
$(3,\,7) \in R,\,(7,\,3) \in {R^{ – 1}} \Rightarrow \,(3,\,3) \in {R^{ – 1}}oR,$
Hence,${R^{ – 1}}oR = \{(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)\}.$
