Question

In the group $Q-\{-1\}$ under the binary operation $+$ defined by $a\ast b=a+b+ab$ the inverse of $10$ is

• A. $\frac{1}{10}$
• B. $\frac{11}{10}$
• C. $\frac{-11}{10}$
• D. $\frac{-10}{11}$

Answer 1

The correct option is D $\frac{-10}{11}$

Let the inverse element of $10$ be $e$, then
$a\ast e=0$.

Therefore, $10\ast e=10+e+10e=0$
$⟹10+11e=0$
$⟹e=\frac{-10}{11}$

Hence, option D is correct.