First slide

Relations XII

Question

Let a relation R be defined by $R = \{(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)\}$ then $R^{- 1} oR$ is

Easy

A

$\{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)\}$
B

$\{(1, 1), (4, 4), (4, 7), (3, 3)\}$
C

$\{(1, 5), (1, 6), (3, 6)\}$
D

None of these

Solution

We first find $R^{- 1} oR$ , 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, $R^{- 1} oR = \{(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)\} .$

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