Question

For binary operation $\ast$ defined on $R-\{1\}$ such that $a\ast b=\frac{a}{b+1}$ is

• A. not associative
• B. not commutative
• C. commutative
• D. both $(a)$ and $(b)$

Answer 1

Answer: (D)

Commutative : $a\ast b$ $=\frac{a}{b+1}$ and $b\ast a=\frac{b}{a+1}$
$a\ast b\ne b\ast a$
$\Rightarrow \ast$ is not commutative.
Associative:
$(a\ast b)\ast c=\left(\frac{a}{b+1}\right)\ast c=\frac{\left(\frac{a}{b+1}\right)}{c+1}$
$=\frac{a}{\left(b+1\right)\left(c+1\right)}$
and $a\ast \left(b\ast c\right)=a\ast \left(\frac{b}{c+1}\right)$
$=\frac{a}{\left(\frac{b}{c+1}\right)+1}$
$=\frac{a\left(c+1\right)}{b+c+1}$
So, $(a\ast b)\ast c\ne a\ast (b\ast c)$.
Hence, $\ast$ is not associative.