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:

$$\begin{gathered} \text{Let}a,b\in Z.\mathrm{Then}, \\ a*b=a+3b-4 \\ b*a=b+3a-4 \\ a*b\ne b*a \\ \text{Let}a=1,b=2 \\ 1*2=1+6-4 \\ =3 \\ 2*1=2+3-4 \\ =1 \\ \text{Therefore, ∃}a=1,b=2\in Z\text{such that}a*b\ne b*a \end{gathered}$$

Thus, * is not commutative on Z.

Associativity:
$$\begin{gathered} \text{Let}a,b,c\in Z.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b+3c-4\right) \\ =a+3\left(b+3c-4\right)-4 \\ =a+3b+9c-12-4 \\ =a+3b+9c-16 \\ \left(a*b\right)*c=\left(a+3b-4\right)*c \\ =a+3b-4+3c-4 \\ =a+3b+3c-8 \\ \text{Thus,}a*\left(b*c\right)\ne \left(a*b\right)*c \\ \text{If}a=1,b=2,c=3 \\ 1*\left(2*3\right)=1*\left(2+9-4\right) \\ =1*7 \\ =1+21-4 \\ =18 \\ \left(1*2\right)*3=\left(1+6-4\right)*3 \\ =3*3 \\ =3+9-4 \\ =8 \\ \text{Therefore, ∃}a=1,b=2,c=3\in Z\text{such that}a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$
Thus, * is not associative on Z.