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 ?

Answer 1

\[\text{ Commutativity }: \] 
\[ \text{ Let } \left( a, b \right) \text{ & }\left( c, d \right) \in A \forall a, b, c, d \in R_0 . \text{ Then }, \] 
\[\left( a, b \right) * \left( c, d \right) = \left( ac, bd \right)\] 
                      \[ = \left( ca, db \right)\] 
                      \[ = \left( c, d \right) * \left( a, b \right)\] 
\[ \therefore \left( a, b \right) * \left( c, d \right) = \left( c, d \right) * \left( a, b \right)\] 
\[\text{Thus, * is commutaive on A} . \] 

\[\text{ Associativity }: \] 

\[\text{ Let } \left( a, b \right), \left( c, d \right) \text{&}\left( e, f \right) \in A \forall a, b, c, d, e, f \in R_{0,} . \text{ Then }, \] 
\[\left( a, b \right) * \left( \left( c, d \right) * \left( e, f \right) \right) = \left( a, b \right) * \left( ce, df \right)\] 
                                       \[ = \left( ace, bdf \right)\] 
\[\left( \left( a, b \right) * \left( c, d \right) \right) * \left( e, f \right) = \left( ac, bd \right) * \left( e, f \right)\] 
                                        \[ = \left( ace, bdf \right)\] 
\[ \therefore \left( a, b \right) * \left( \left( c, d \right) * \left( e, f \right) \right) = \left( \left( a, b \right) * \left( c, d \right) \right) * \left( e, f \right)\] 
\[\text{ Thus, * is associative on A } . \]