Question

Let $\ast$ be a binary operation defined on $R$ by $a\ast b=\frac{a+b}{4}$ for all $a,b\in R$ 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

A binary operation $\ast$ on $A$ is associative if $\forall a,b,c\in A$,

$(a\ast b)\ast c=a\ast (b\ast c)$ and

A binary operation $\ast$ on $A$ is commutative if $\forall a,b\in A$,

$a\ast b=b\ast a$

$a\ast b=\frac{a+b}{4}$ is commutative as:

$a\ast b=b\ast a\Rightarrow \frac{a+b}{4}=\frac{b+a}{4}\Rightarrow \frac{a+b}{4}=\frac{a+b}{4}$

which is true.

$a\ast b=\frac{a+b}{4}$ is not associative as:

$(a\ast b)\ast c=a\ast (b\ast c)\Rightarrow \left(\frac{a+b}{4}\right)\ast c=a\ast \left(\frac{b+c}{4}\right)\Rightarrow \frac{\frac{a+b}{4}+c}{4}=\frac{a+\frac{b+c}{4}}{4}\Rightarrow \frac{a+b+4c}{16}=\frac{4a+b+c}{16}$

which is not true.

Hence, the binary operation $\ast$ is commutative but not associative.