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Question
On Z, the set of all integers, a binary operation * is defined by a * b = a + 3b − 4. Prove that * is neither commutative nor associative on Z.
Solution
Commutativity:
\[\text{Let a}, b \in Z . \text{Then}, \]
\[a * b = a + 3b - 4\]
\[b * a = b + 3a - 4\]
\[a * b \neq b * a\]
\[\text{Let }a = 1, b = 2\]
\[1 * 2 = 1 + 6 - 4\]
\[ = 3\]
\[2 * 1 = 2 + 3 - 4\]
\[ = 1\]
\[\text{Therefore}, \exists \text{ a} = 1, b = 2 \in \text{Z such that a} * b \neq b * a\]
Thus, * is not commutative on Z.
Associativity:
\[\text{Let a}, b, c \in Z . \text{Then}, \]
\[a * \left( b * c \right) = a * \left( b + 3c - 4 \right)\]
\[ = a + 3\left( b + 3c - 4 \right) - 4\]
\[ = a + 3b + 9c - 12 - 4\]
\[ = a + 3b + 9c - 16\]
\[\left( a * b \right) * c = \left( a + 3b - 4 \right) * c\]
\[ = a + 3b - 4 + 3c - 4\]
\[ = a + 3b + 3c - 8\]
\[\text{Thus, a} * \left( b * c \right) \neq \left( a * b \right) * c\]
\[\text{ If a } = 1, b = 2, c = 3\]
\[1 * \left( 2 * 3 \right) = 1 * \left( 2 + 9 - 4 \right)\] \[ = 1 * 7 \]
\[ = 1 + 21 - 4\]
\[ = 18\]
\[\left( 1 * 2 \right) * 3 = \left( 1 + 6 - 4 \right) * 3\]
\[ = 3 * 3\]
\[ = 3 + 9 - 4\]
\[ = 8\]
\[\text{Therefore}, \exists \text{ a} = 1, b = 2, c = 3 \in \text{Z such that a } * \left( b * c \right) \neq \left( a * b \right) * c\]
Thus, * is not associative on Z.
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