Question

Let $\ast$ be binary operation defined on $R$ by $a\ast b=1+ab,\forall a,b\in R.$ Then the opertion $\ast$ is

• A. associative but not commutative
• B. both commutative and associative
• C. commutative but not associative
• D. neither commutative nor associative

Answer 1

The correct option is C commutative but not associative

Checking commutativity

Given : $\ast$ be binary operation defined on $R$ by

$a\ast b=1+ab,\forall a,b\in R$

$\Rightarrow b\ast a=1+ba,\forall a,b\in R$

Since, for any $a,b\in R$

$ab=ba$

$(\because \text{Multiplication is commutative for real numbers)}$

$\Rightarrow 1+ab=1+ba$

$\Rightarrow (a\ast b)=(b\ast a)\forall a,b\in R$

Therefore, $\ast$ is commutative binary operation

Checking associativity

$a\ast (b\ast c)=1+a(b\ast c)$

$=1+a(1+bc)$

$=1+a+abc...\left(1\right)$

and

$(a\ast b)\ast c=1+(a\ast b)\ast c$

$=1+(1+ab)c$

$=1+c+abc...\left(2\right)$

Clearly, from equation (1) and (2)

$1+a+abc\ne 1+c+abc$

$\therefore a\ast (b\ast c)\ne (a\ast b)\ast c$

Therefore, $\ast$ is not associative binary operation.

Hence, $\ast$ is commutative but not associative.

$\therefore$ Option (A) is correct.