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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 ________________ .
Options
\[- a\]
\[- \frac{a}{a + 1}\]
\[\frac{1}{a}\]
\[a^2\]
Solution
\[- \frac{a}{a + 1}\]
Let e be the identity element in R \[-\] {1} with respect to * such that
\[a * e = a = e * a, \forall a \in R - \left\{ 1 \right\}\]
\[a * e = a \text{ and }e * a = a, \forall a \in R - \left\{ 1 \right\}\]
\[\text{ Then }, \]
\[a + e + ae = a \text{ and }e + a + ea = a, \forall a \in R - \left\{ 1 \right\}\]
\[e\left( 1 + a \right) = 0 , \forall a \in R - \left\{ 1 \right\}\]
\[e = 0 \in R - \left\{ 1 \right\}\]
Thus, 0 is the identity element in R \[-\] {1}with respect to *.
\[\text{ Let }a \in R - \left\{ 1 \right\} \text{ and }b \in R - \left\{ 1 \right\} \text{ be the inverse of a . Then },\]
\[a * b = e = b * a\]
\[a * b = e \text{ and }b * a = e\]
\[ \Rightarrow a + b + ab = 0 \text{ and }b + a + ba = 0\]
\[ \Rightarrow b\left( 1 + a \right) = - a \in R - \left\{ 1 \right\}\]
\[ \Rightarrow b = \frac{- a}{1 + a} \in R - \left\{ 1 \right\}\]
\[\text{Thus},\frac{- a}{1 + a}\text{ is the inverse of a } \in R - \left\{ 1 \right\} . \]
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