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Question
On Q, the set of all rational numbers, * is defined by \[a * b = \frac{a - b}{2}\] , shown that * is no associative ?
Solution
\[\text{Let }a, b, c \in Q . \text{Then}, \]
\[a * \left( b * c \right) = a * \left( \frac{b - c}{2} \right)\]
\[ = \frac{a - \left( \frac{b - c}{2} \right)}{2}\]
\[ = \frac{2a - b + c}{4}\]
\[\left( a * b \right) * c = \left( \frac{a - b}{2} \right) * c\]
\[ = \frac{\left( \frac{a - b}{2} \right) - c}{2}\]
\[ = \frac{a - b - 2c}{4}\]
\[\text{Thus, a} * \left( b * c \right) \neq \left( a * b \right) * c\]
\[\text{If a} = 1, b = 2, c = 3 \]
\[1 * \left( 2 * 3 \right) = 1 * \left( \frac{2 - 3}{2} \right)\]
\[ = 1 * \frac{- 1}{2}\]
\[ = \frac{1 + \frac{1}{2}}{2}\]
\[ = \frac{3}{4}\]
\[\left( 1 * 2 \right) * 3 = \left( \frac{1 - 2}{2} \right) * 3\]
\[ = \frac{- 1}{2} * 3\]
\[ = \frac{\frac{- 1}{2} - 3}{2}\]
\[ = \frac{- 7}{4}\]
\[\text{Therefore}, \exists \text{ a} = 1, b = 2, c = 3 \in \text{R such that a} * \left( b * c \right) \neq \left( a * b \right) * c\]
Thus, * is not associative on Q.
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