Let R₀ denote the set of all non-zero real numbers and let A = R₀ × R₀. If '' is a binary operation on A defined by
(a, b) (c, d) = (ac, bd) for all (a, b), (c, d) ∈ A
(i) Show that '*' is both commutative and associative on A
(ii) Find the identity element in A
(iii) Find the invertible element in A.

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Solution

$(i) Commutativity : Let (a, b) & (c, d) \in A \forall a, b, c, d \in R_{0} . Then, (a, b) * (c, d) = (a c, b d) = (c a, d b) = (c, d) * (a, b) ∴ (a, b) * (c, d) = (c, d) * (a, b) Thus, * is commutaive on A . Associativity : Let (a, b), (c, d) & (e, f) \in A \forall a, b, c, d, e, f \in R_{0,} . Then, (a, b) * ((c, d) * (e, f)) = (a, b) * (c e, d f) = (a c e, b d f) ((a, b) * (c, d)) * (e, f) = (a c, b d) * (e, f) = (a c e, b d f) ∴ (a, b) * ((c, d) * (e, f)) = ((a, b) * (c, d)) * (e, f) Thus, * is associative on A .$

$(ii) Let (x, y) be the identity element in A \forall (x, y) \in A . Then, (a, b) * (x, y) = (a, b) = (x, y) * (a, b) \Rightarrow (a, b) * (x, y) = (a, b) and (x, y) * (a, b) = (a, b) \Rightarrow (a x, b y) = (a, b) and (x a, y b) = (a, b) \Rightarrow x = 1 and y = 1 Thus, (1, 1) is the identity element of A .$

$(iii) Let (m, n) be the inverse of (a, b) \forall (a, b) \in A . Then, (a, b) * (m, n) = (1, 1) \Rightarrow (a m, b n) = (1, 1) \Rightarrow a m = 1 & b n = 1 \Rightarrow m = \frac{1}{a} & n = \frac{1}{b} Thus, (\frac{1}{a}, \frac{1}{b}) is the inverse of (a, b) \forall (a, b) \in A .$

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