Question

For the binary operation * defined on R − {1} by the rule a * b = a + b + ab for all a, b ∈ R − {1}, the inverse of a is
(a) $-a$

(b) $-\frac{a}{a+1}$

(c) $\frac{1}{a}$

(d) $a^2$

Answer 1

(b) $-\frac{a}{a+1}$

Let e be the identity element in R $-$ {1} with respect to * such that

$$\begin{gathered} a*e=a=e*a,\forall a\in R-\left\{1\right\} \\ a*e=a\text{and}e*a=a,\forall a\in R-\left\{1\right\} \\ \mathrm{Then}, \\ a+e+ae=a\text{and}e+a+ea=a,\forall a\in R-\left\{1\right\} \\ e\left(1+a\right)=0,\forall a\in R-\left\{1\right\} \\ e=0\in R-\left\{1\right\} \end{gathered}$$

Thus, 0 is the identity element in R $-$ {1}with respect to *.

$$\begin{gathered} \text{Let}a\in R-\left\{1\right\}\text{and}b\in R-\left\{1\right\}\text{be the inverse of}a.\text{Then,} \\ a*b=e=b*a \\ a*b=e\text{and}b*a=e \\ \Rightarrow a+b+ab=0\text{and}b+a+ba=0 \\ \Rightarrow b\left(1+a\right)=-a\in R-\left\{1\right\} \\ \Rightarrow b=\frac{-a}{1+a}\in R-\left\{1\right\} \\ \text{Thus,}\frac{-a}{1+a}\text{is the inverse of}a\in R-\left\{1\right\}. \end{gathered}$$