Question

For given binary operation $\ast$ defined below, determine whether $\ast$ is binary, commutative or associative.
(vi) On R-{-1},define $a\ast b=\frac{a}{b+1}$

Answer 1

On R-{-1},define $a\ast b=\frac{a}{b+1}$
$\frac{a}{b+1}\in R$ for $b\ne -1$ so that operation $\ast$ is binary.
It can be observed that $1\ast 2=\frac{1}{2+1}=\frac{1}{3}$ and $2\ast 1=\frac{2}{1+1}=1$
Therefore, $1\ast 2\ne 2\ast 1$ where $1,2\in R--1$
Therefore, the operation $\ast$ is not commutative.
It can also be obseved that $(1\ast 2)\ast 3=\frac{1}{3}\ast 3=\frac{\frac{1}{3}}{3+1}=\frac{1}{12}$
$1\ast (2\ast 3)=1\ast \frac{2}{3+1}=1\ast \frac{2}{4}=1\ast \frac{1}{2}=\frac{1}{\frac{1}{2}+1}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$
Therefore, $(1\ast 2\ast 3)\ne 1\ast (2\ast 3)$ where $1,2,3\in$ R -{-1}
Therefore, the operation $\ast$ is not associative.