Question

On Q, the set of all rational numbers, * is defined by \[a * b = \frac{a - b}{2}\] , shown that * is no associative ?

Answer 1

\[\text{Let }a, b, c \in Q . \text{Then}, \] 
\[a * \left( b * c \right) = a * \left( \frac{b - c}{2} \right)\] 
                   \[ = \frac{a - \left( \frac{b - c}{2} \right)}{2}\] 
                   \[ = \frac{2a - b + c}{4}\] 
\[\left( a * b \right) * c = \left( \frac{a - b}{2} \right) * c\] 
 \[ = \frac{\left( \frac{a - b}{2} \right) - c}{2}\] 

\[ = \frac{a - b - 2c}{4}\] 
\[\text{Thus, a} * \left( b * c \right) \neq \left( a * b \right) * c\] 
\[\text{If a} = 1, b = 2, c = 3 \] 
\[1 * \left( 2 * 3 \right) = 1 * \left( \frac{2 - 3}{2} \right)\] 
                  \[ = 1 * \frac{- 1}{2}\] 
                 \[ = \frac{1 + \frac{1}{2}}{2}\]

                  \[ = \frac{3}{4}\] 
\[\left( 1 * 2 \right) * 3 = \left( \frac{1 - 2}{2} \right) * 3\] 
                  \[ = \frac{- 1}{2} * 3\] 
                 \[ = \frac{\frac{- 1}{2} - 3}{2}\] 
                 \[ = \frac{- 7}{4}\] 
\[\text{Therefore}, \exists \text{ a} = 1, b = 2, c = 3 \in \text{R such that a} * \left( b * c \right) \neq \left( a * b \right) * c\]

Thus, * is not associative on Q.