Question

Let A = R₀ × R, where R₀ denote the set of all non-zero real numbers. A binary operation '⊙' is defined on A as follows :
(a, b) ⊙ (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R₀ × R.
(i) Show that '⊙' is commutative and associative on A
(ii) Find the identity element in A
(iii) Find the invertible elements in A.

Answer 1

(i) Commutativity:

$$\begin{gathered} \text{Let}X=\left(a,b\right)\text{and}Y=\left(c,d\right)\in A,\forall a,c\in R_0\&b,d\in R.\mathrm{Then}, \\ X\odot Y=\left(ac,bc+d\right) \\ \&Y\odot X=\left(ca,da+b\right) \\ \text{Therefore,} \\ X\odot Y=Y\odot X,\forall X,Y\in A \end{gathered}$$
Thus, $\odot$ is commutative on A.

Associativity:

$$\begin{gathered} \mathrm{Let}X=(a,b),Y=(c,d)andZ=(e,f),\forall a,c,e\in R_0\&b,d,f\in R \\ X\odot \left(Y\odot Z\right)=(a,b)\odot \left(ce,de+f\right) \\ =\left(ace,bce+de+f\right) \\ \left(X\odot Y\right)\odot Z=\left(ac,bc+d\right)\odot \left(e,f\right) \\ =\left(ace,\left(bc+d\right)e+f\right) \\ =\left(ace,bce+de+f\right) \\ \therefore X\odot \left(Y\odot Z\right)=\left(X\odot Y\right)\odot Z,\forall X,Y,Z\in A \\ \mathrm{Thus},\odot \mathrm{is}\mathrm{associative}\mathrm{on}\mathrm{A}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let}E=(x,y)\mathrm{be}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{in}\mathrm{A}\mathrm{with}\mathrm{respect}\mathrm{to}\odot ,\forall x\in R_0\&y\in R\mathrm{such}\mathrm{that} \\ X\odot E=X=E\odot X,\forall X\in A \\ \Rightarrow X\odot E=X\mathrm{and}E\odot X=X \\ \Rightarrow \left(ax,bx+y\right)=\left(a,b\right)\mathrm{and}\left(xa,ya+b\right)=\left(a,b\right) \\ \mathrm{Considering}\left(ax,bx+y\right)=\left(a,b\right) \\ \Rightarrow ax=a \\ \Rightarrow x=1 \\ \&bx+y=b \\ \Rightarrow y=0\left[\because x=1\right] \\ \mathrm{Considering}\left(xa,ya+b\right)=\left(a,b\right) \\ \Rightarrow xa=a \\ \Rightarrow x=1 \\ \&ya+b=b \\ \Rightarrow y=0\left[\because x=1\right] \\ \therefore \left(1,0\right)\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{in}\mathrm{A}\mathrm{with}\mathrm{respect}\mathrm{to}\odot . \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let}F=(m,n)\mathrm{be}\mathrm{the}\mathrm{inverse}\mathrm{in}A\forall m\in R_0\&n\in R \\ X\odot F=E\mathrm{and}F\odot X\mathit{=}E \\ \mathit{\Rightarrow }\left(am\mathit{,}\mathit{}bm\mathit{+}n\right)\mathit{=}\left(1,0\right)\mathit{}\mathrm{and}\mathit{}\left(ma\mathit{,}\mathit{}na\mathit{+}b\right)\mathit{=}\left(\mathit{1}\mathit{,}\mathit{}\mathit{0}\right) \\ \mathrm{Considering}\mathit{}\left(am\mathit{,}\mathit{}bm\mathit{+}n\right)\mathit{=}\left(1,0\right) \\ \Rightarrow am\mathit{=}1 \\ \Rightarrow m\mathit{=}\frac{\mathit{1}}{a} \\ \&\mathit{}bm+n=0 \\ \Rightarrow n\mathit{=}\frac{\mathit{-}b}{a}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathit{\because }\mathit{}m\mathit{=}\frac{\mathit{1}}{a}\right] \\ \mathrm{Considering}\left(ma\mathit{,}\mathit{}na\mathit{+}b\right)\mathit{=}\left(1,0\right) \\ \Rightarrow ma\mathit{=}1 \\ \Rightarrow m\mathit{=}\frac{\mathit{1}}{a} \\ \&na+b\mathit{=}0 \\ \Rightarrow n\mathit{=}\frac{\mathit{-}b}{a} \\ \therefore \mathrm{The}\mathrm{inverse}\mathrm{of}\left(a,b\right)\in A\mathit{}\mathrm{with}\mathrm{respect}\mathrm{to}\odot \mathrm{is}\mathit{}\left(\frac{\mathit{1}}{a}\mathit{,}\frac{\mathit{-}b}{a}\mathit{}\right). \end{gathered}$$