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.

Answer 1

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.