Let R be a relation such that `R = {(1,4), (3,7), (4,5), (4,6), (7,6)}`, check R is a function or not ?

Explanation

(i) We know that, `(R O R)^(-1) = R^(-1) O R^(-1)`
Dom (R ) = {1, 3, 4, 7}
Range (R ) = {4, 5, 6, 7}

We see that,
`1rarr4rarr5implies(1,5)inROR`
`1rarr4rarr6implies(1,6)inROR`
`3rarr7rarr6implies(3,6)inROR`
`therefore ROR={(1,5),(1,6),(3,6)}`
Then, `R^(-1)O R^(-1)=(ROR)^(-1)`
`={(5,1),(6,1),(6,3)}`
(ii) We know that, `(R^(-1)OR)^(-1)=R^(-1)O(R^(-1))^(-1)=R^(-1)OR`
Since,
`R={(1,4),(3,7),(4,5),(4,6),(7,6)}`
`therefore R^(-1)={(4,1),(7,3),(5,4),(6,4),(6,7)}`
`therefore` Dom (R ) = {1, 3, 4, 7}, Range (R ) = {4, 5, 6, 7}
Dom `(R^(-1))={4,5,6,7}`, Range `(R^(-1))` = {1,3,4,7}

We see that,
`1oversetRrarr4overset(R^(-1))rarr1implies(1,1)inR^(-1)OR`
`3oversetRrarr7overset(R^(-1))rarr3implies(3,3)inR^(-1)OR`
`4oversetRrarr5overset(R^(-1))rarr4implies(4,4)inR^(-1)OR`
`4oversetRrarr6overset(R^(-1))rarr4implies(4,4)inR^(-1)OR`
`4oversetRrarr6overset(R^(-1))rarr7implies(4,7)inR^(-1)OR`
`7oversetRrarr6overset(R^(-1))rarr4implies(7,4)inR^(-1)OR`
`7oversetRrarr6overset(R^(-1))rarr7implies(7,7)inR^(-1)OR`
`therefore R^(-1)OR={(1,1),(3,3),(4,4),(7,7),(4,7),(7,4)}`
Hence, `(R^(-1)OR)^(-1)=R^(-1)OR={(1,1),(3,3),(4,4),(7,7),(4,7),(7,4)}`