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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.
Solution
\[\text{Let } a, b, c \in Q . \text{Then}, \]
\[a * \left( b * c \right) = a * \left( \frac{b + c}{2} \right)\]
\[ = \frac{a + \left( \frac{b + c}{2} \right)}{2}\]
\[ = \frac{2a + b + c}{4}\]
\[\left( a * b \right) * c = \left( \frac{a + b}{2} \right) * c\]
\[ = \frac{\left( \frac{a + b}{2} \right) + c}{2}\]
\[ = \frac{a + b + 2c}{4}\]
\[\text{Thus,a} * \left( b * c \right) \neq \left( a * b \right) * c\]
\[\text{ If a } = 1, b = 2, c = 3 \]
\[1 * \left( 2 * 3 \right) = 1 * \left( \frac{2 + 3}{2} \right)\]
\[ = 1 * \frac{5}{2}\]
\[ = \frac{1 + \frac{5}{2}}{2}\] \[ = \frac{7}{4}\]
\[\left( 1 * 2 \right) * 3 = \left( \frac{1 + 2}{2} \right) * 3\]
\[ = \frac{3}{2} * 3\]
\[ = \frac{\frac{3}{2} + 3}{2}\]
\[ = \frac{9}{4}\]
\[\text{Therefore}, \exists \text{ a} = 1, b = 2, c = 3 \in \text{Q such that a } * \left( b * c \right) \neq \left( a * b \right) * c\]
Thus, * is not associative on Q.
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