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Question
Let A = R0 × R, where R0 denote the set of all non-zero real numbers. A binary operation '⊙' is defined on A as follows (a, b) ⊙ (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R0 × R :
Show that '⊙' is commutative and associative on A ?
Solution
Commutativity:
\[\text{Let }X = \left( a, b \right) \text{and Y} = \left( c, d \right) \in A, \forall a, c \in R_0 \text{ & } b, d \in R . Then, \]
\[ X \odot Y = \left( ac, bc + d \right)\]
\[\text{ & }Y \odot X = \left( ca, da + b \right)\]
\[\text{Therefore}\]
\[X \odot Y = Y \odot X, \forall X, Y \in A\]
Thus, \[\odot\] is commutative on A.
Associativity :
\[\text{ Let } X = (a, b), Y = (c, d) \text{and} Z = ( e, f),\forall a, c, e \in R_0 \text{ & }b, d, f \in R\]
\[X \odot \left( Y \odot Z \right) = (a, b) \odot \left( ce, de + f \right)\]
\[ = \left( ace, bce + de + f \right)\]
\[\left( X \odot Y \right) \odot Z = \left( ac, bc + d \right) \odot \left( e, f \right)\]
\[ = \left( ace, \left( bc + d \right)e + f \right) \]
\[ = \left( ace, bce + de + f \right)\]
\[ \therefore X \odot \left( Y \odot Z \right) = \left( X \odot Y \right) \odot Z, \forall X, Y, Z \in A\]
\[\text{Thus}, \odot \text{is associative on A} . \]
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