Question

On Q, the set of all rational numbers a binary operation * is defined by \[a * b = \frac{a + b}{2}\] Show that * is not associative on Q.

Answer 1

\[\text{Let } a, b, c \in Q . \text{Then}, \] 
\[a * \left( b * c \right) = a * \left( \frac{b + c}{2} \right)\] 
                       \[ = \frac{a + \left( \frac{b + c}{2} \right)}{2}\] 
                      \[ = \frac{2a + b + c}{4}\] 
\[\left( a * b \right) * c = \left( \frac{a + b}{2} \right) * c\] 
              \[ = \frac{\left( \frac{a + b}{2} \right) + c}{2}\] 
                  \[ = \frac{a + b + 2c}{4}\] 
\[\text{Thus,a} * \left( b * c \right) \neq \left( a * b \right) * c\] 
\[\text{ If a } = 1, b = 2, c = 3 \] 
\[1 * \left( 2 * 3 \right) = 1 * \left( \frac{2 + 3}{2} \right)\] 
        \[ = 1 * \frac{5}{2}\] 
       \[ = \frac{1 + \frac{5}{2}}{2}\] \[ = \frac{7}{4}\] 
\[\left( 1 * 2 \right) * 3 = \left( \frac{1 + 2}{2} \right) * 3\] 
        \[ = \frac{3}{2} * 3\] 
         \[ = \frac{\frac{3}{2} + 3}{2}\] 
        \[ = \frac{9}{4}\] 
\[\text{Therefore}, \exists \text{ a} = 1, b = 2, c = 3 \in \text{Q such that a } * \left( b * c \right) \neq \left( a * b \right) * c\]

Thus, * is not associative on Q.