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Question
Let S be the set of all rational numbers except 1 and * be defined on S by a * b = a + b \[-\] ab, for all a, b \[\in\] S:
Prove that * is commutative as well as associative ?
Solution
Commutativity :
\[\text{Let }a, b \in S . \text{Then}, \]
\[a * b = a + b - ab\]
\[ = b + a - ba\]
\[ = b * a\]
\[\text{Therefore}, \]
\[a * b = b * a, \forall a, b \in S\]
Thus, * is commutative on S.
Associativity:
\[\text{Let} a, b, c \in S . \text{Then}, \]
\[a * \left( b * c \right) = a * \left( b + c - bc \right)\]
\[ = a + b + c - bc - a\left( b + c - bc \right)\]
\[ = a + b + c - bc - ab - ac + abc\]
\[\left( a * b \right) * c = \left( a + b - ab \right) * c\]
\[ = a + b - ab + c - \left( a + b - ab \right)c\]
\[ = a + b + c - ab - ac - bc + abc\]
\[\text{Therefore}, \]
\[a * \left( b * c \right) = \left( a * b \right) * c, \forall a, b, c \in S\]
Thus , * is associative on S.
So, * is commutative as well as associative.
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