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.

Answer 1

Commutativity:
$$\begin{gathered} \text{Let}a,b\in R-\left\{1\right\}.\mathrm{Then}, \\ a*b=a+b-ab \\ =b+a-ba \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in R-\left\{1\right\} \end{gathered}$$
Thus, * is commutative on R $-$ {1}.

Associativity:
$$\begin{gathered} \text{Let}a,b,c\in R-\left\{1\right\}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b+c-bc\right) \\ =a+b+c-bc-a\left(b+c-bc\right) \\ =a+b+c-bc-ab-ac+abc \\ \left(a*b\right)*c=\left(a+b-ab\right)*c \\ =a+b-ab+c-\left(a+b-ab\right)c \\ =a+b+c-ab-ac-bc+abc \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in R-\left\{1\right\} \end{gathered}$$
Thus, * is associative on R $-$ {1}.

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

$$\begin{gathered} a*e=a=e*a,\forall a\in R-\left\{1\right\} \\ a*e=a\text{and}e*a=a,\forall a\in R-\left\{1\right\} \\ \Rightarrow a+e-ae=a\text{and}e+a-ea=a,\forall a\in R-\left\{1\right\} \\ e\left(1-a\right)=0,\forall a\in R-\left\{1\right\} \\ e=0\in \forall a\in R-\left\{1\right\},\forall a\in R-\left\{1\right\}\left[\text{∵}a\ne \text{1}\right] \end{gathered}$$

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

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