Question

Let $\ast$ be a binary operation defined on $R$ by $a\ast b=\frac{a+b}{4}\forall a,b\in R,$ then the operation $\ast$ is

• A. Commutative and Associative
• B. Commutative but not Associative
• C. Associative but not Commutative
• D. Neither Associative nor Commutative

Answer 1

The correct option is B Commutative but not Associative

Let $a,b,c\in R$
Then $a\ast b=\frac{a+b}{4}=\frac{b+a}{4}=b\ast a\forall a,b\in R$
Thus, $\ast$ is commutative on $R.$

Now, $(a\ast b)\ast c=\frac{a+b}{4}\ast c$
$=\frac{1}{4}(\frac{a+b}{4}+c)$
$=\frac{a+b+4c}{16}$
$a\ast (b\ast c)=a\ast \frac{b+c}{4}$
$=\frac{1}{4}(a+\frac{b+c}{4})$
$=\frac{4a+b+c}{16}$
$\text{Thus,}(a\ast b)\ast c\ne a\ast (b\ast c)\forall a,b,c\in R$
Thus, $\ast$ is not associative on $R$.