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Question
On R − {1}, a binary operation * is defined by a * b = a + b − ab. Prove that * is commutative and associative. Find the identity element for * on R − {1}. Also, prove that every element of R − {1} is invertible.
Solution
Commutativity:
\[\text{ Let }a, b \in R - \left\{ 1 \right\} . \text{Then}, \]
\[a * b = a + b - ab\]
\[ = b + a - ba\]
\[ = b * a\]
\[\text{Therefore},\]
\[a * b = b * a, \forall a, b \in R - \left\{ 1 \right\}\]
Thus, * is commutative on R \[-\] { 1 }
Associativity :
\[\text{ Let }a, b, c \in R - \left\{ 1 \right\} . \text{ Then }, \]
\[a * \left( b * c \right) = a * \left( b + c - bc \right)\]
\[ = a + b + c - bc - a\left( b + c - bc \right)\]
\[ = a + b + c - bc - ab - ac + abc\]
\[\left( a * b \right) * c = \left( a + b - ab \right) * c\]
\[ = a + b - ab + c - \left( a + b - ab \right)c\]
\[ = a + b + c - ab - ac - bc + abc\]
\[\text{Therefore},\]
\[a * \left( b * c \right) = \left( a * b \right) * c, \forall a, b, c \in R - \left\{ 1 \right\}\]
Thus, * is associative on R \[-\] {1} .
Finding identity element:
Let e be the identity element in R \[-\] {1} with respect to * such that
\[a * e = a = e * a, \forall a \in R - \left\{ 1 \right\}\]
\[a * e = a \text{ and }e * a = a, \forall a \in R - \left\{ 1 \right\}\]
\[ \Rightarrow a + e - ae = a \text{ and }e + a - ea = a, \forall a \in R - \left\{ 1 \right\}\]
\[e\left( 1 - a \right) = 0, \forall a \in R - \left\{ 1 \right\}\]
\[e = 0 \in \forall a \in R - \left\{ 1 \right\}, \forall a \in R - \left\{ 1 \right\} \left[ \because a \neq 1 \right]\]
Thus, 0 is the identity element in R \[-\] {1} with respect to *
Finding inverse :
\[\text{ Let }a \in R - \left\{ 1 \right\} \text{ and }b \in R - \left\{ 1 \right\}\text{be the inverse of a . Then},\]
\[a * b = e = b * a\]
\[a * b = e \text{ and }b * a = e\]
\[ \Rightarrow a + b - ab = 0 \text{ and }b + a - ba = 0\]
\[ \Rightarrow a = ab - b\]
\[ \Rightarrow a = b\left( a - 1 \right) \]
\[ \Rightarrow b = \frac{a}{a - 1}\]
\[\text{Thus},\frac{a}{a - 1} \text{is the inverse of a} \in R - \left\{ 1 \right\} . \]
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