Solution to Let RR be a relation from the set AA to the set BB. Taking into … - Sikademy
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Archangel Macsika

Let RR be a relation from the set AA to the set BB. Taking into account that R^{-1}=\{(b,a)\ |\ (a,b)\in R\}\subset B\times AR −1 ={(b,a) ∣ (a,b)∈R}⊂B×A, we conclude that Ran (R)=\{b\ |\ (a,b)\in R\}=\{b\ |\ (b,a)\in R^{-1}\}=Dom (R^{-1} ).Ran(R)={b ∣ (a,b)∈R}={b ∣ (b,a)∈R −1 }=Dom(R −1 ).

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As R is a relation on set A.


It implies that The elemnts of A must belong to the R in cartesian product form.


Taking LHS

IA*R=A*R=R


Since The product of the set on which relation is defined is equal to that relation.


Now taking LHS-

R*IA=R*A=R


This implies that-

IA*R=R=R*IA , Hence proved.

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