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.

Answer 1

$$\begin{gathered} \text{Let}a,b,c\in Q.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(\frac{b+c}{2}\right) \\ =\frac{a+\left({\displaystyle \frac{b+c}{2}}\right)}{2} \\ =\frac{2a+b+c}{4} \\ \left(a*b\right)*c=\left(\frac{a+b}{2}\right)*c \\ =\frac{\left({\displaystyle \frac{a+b}{2}}\right)+c}{2} \\ =\frac{a+b+2c}{4} \\ \text{Thus,}a*\left(b*c\right)\ne \left(a*b\right)*c \\ \text{If}a=1,b=2,c=3 \\ 1*\left(2*3\right)=1*\left(\frac{2+3}{2}\right) \\ =1*\frac{5}{2} \\ =\frac{1+{\displaystyle \frac{5}{2}}}{2} \\ =\frac{7}{4} \\ \left(1*2\right)*3=\left(\frac{1+2}{2}\right)*3 \\ =\frac{3}{2}*3 \\ =\frac{{\displaystyle \frac{3}{2}}+3}{2} \\ =\frac{9}{4} \\ \text{Therefore, ∃}a=1,b=2,c=3\in Q\text{such that}a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on Q.