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 ?

Answer 1

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.