Question

If $R\subseteq A\times B$ and $S\subseteq B\times C$ be two relations, then $(\text{ SoR }{)}^{-1}$ is equal to

• A. $S^{-1}o{R}^{-1}$
• B. $RoS$
• C. $R^{-1}o{S}^{-1}$
• D. None of these

Answer 1

Answer: (C)

$SoR$ is a relation from $A$ to $C$.
$\therefore (SoR{)}^{-1}$ is a relation from $C$ to $A$.
$R^{-1}$ is a relation from $B$ to $A$.
$S^{-1}$ is a relation from $C$ to $B$.
$\therefore R^{-1}o{S}^{-1}$ is a relation from $C$ to $A$.
Let $(c,a)\in (SoR{)}^{-1}.$
$\therefore (a,c)\in SoR$
$\therefore \exists b\in B:(a,b)\in R$ and $(b,c)\in S$
$\therefore (b,a)\in R^{-1}$ and $(c,b)\in S^{-1}$
$\therefore (c,a)\in R^{-1}o{S}^{-1}$
$\therefore (SoR{)}^{-1}\subseteq R^{-1}o{S}^{-1}$
Conversely, let $(c,a)\in R^{-1}o{S}^{-1}$
$\therefore \exists b\in B:(c,b)\in S^{-1}$ and $(b,a)\in R^{-1}$
$\therefore (b,c)\in S$ and $(a,b)\in R$
$\Rightarrow (a,c)\in SoR$
$\Rightarrow (c,a)\in (SoR{)}^{-1}$
$\therefore R^{-1}o{S}^{-1}\subseteq (SoR{)}^{-1}$
Combining, we get $(SoR{)}^{-1}=R^{-1}o{S}^{-1}$