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Question
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
Show that '*' is both commutative and associative on A ?
Solution
\[\text{ Commutativity }: \]
\[ \text{ Let } \left( a, b \right) \text{ & }\left( c, d \right) \in A \forall a, b, c, d \in R_0 . \text{ Then }, \]
\[\left( a, b \right) * \left( c, d \right) = \left( ac, bd \right)\]
\[ = \left( ca, db \right)\]
\[ = \left( c, d \right) * \left( a, b \right)\]
\[ \therefore \left( a, b \right) * \left( c, d \right) = \left( c, d \right) * \left( a, b \right)\]
\[\text{Thus, * is commutaive on A} . \]
\[\text{ Associativity }: \]
\[\text{ Let } \left( a, b \right), \left( c, d \right) \text{&}\left( e, f \right) \in A \forall a, b, c, d, e, f \in R_{0,} . \text{ Then }, \]
\[\left( a, b \right) * \left( \left( c, d \right) * \left( e, f \right) \right) = \left( a, b \right) * \left( ce, df \right)\]
\[ = \left( ace, bdf \right)\]
\[\left( \left( a, b \right) * \left( c, d \right) \right) * \left( e, f \right) = \left( ac, bd \right) * \left( e, f \right)\]
\[ = \left( ace, bdf \right)\]
\[ \therefore \left( a, b \right) * \left( \left( c, d \right) * \left( e, f \right) \right) = \left( \left( a, b \right) * \left( c, d \right) \right) * \left( e, f \right)\]
\[\text{ Thus, * is associative on A } . \]
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