Question

Let S be the set of all real numbers except −1 and let '*' be an operation defined by

a * b = a + b + ab for all a, b ∈ S.

Determine whether '*' is a binary operation on S. If yes, check its commutativity and associativity. Also, solve the equation (2 * x) * 3 = 7.

Answer 1

Checking for binary operation:

$$\begin{gathered} \text{Let}a,b\in S.\mathrm{Then}, \\ a,b\in R\text{and}a\ne -1,b\ne -1 \\ a*b=a+b+ab \\ \text{We need to prove that}a+b+ab\in S.\left[\text{For this we have to prove that}a+b+ab\in R\text{and}a+b+ab\ne -1\right] \\ \text{Since}a,b\in R,a+b+ab\in R,\mathrm{l}\text{et us assume that}a+b+ab=-1. \\ a+b+ab+1=0 \\ a+ab+b+1=0 \\ a\left(1+b\right)+1\left(1+b\right)=0 \\ \left(a+1\right)\left(b+1\right)=0 \\ a=-1,b=-1\left[\text{which is false}\right] \\ \text{Hence,}a+b+ab\ne -1 \\ \text{Therefore,} \\ a+b+ab\in S \end{gathered}$$

Thus, * is a binary operation on S.

Commutativity:

$$\begin{gathered} \text{Let}a,b\in S.\mathrm{Then}, \\ a*b=a+b+ab \\ =b+a+ba \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in S \end{gathered}$$
Thus, * is commutative on N.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in S \\ a*\left(b*c\right)=a*\left(b+c+bc\right) \\ =a+b+c+bc+a\left(b+c+bc\right) \\ =a+b+c+bc+ab+ac+abc \\ \left(a*b\right)*c=\left(a+b+ab\right)*c \\ =a+b+ab+c+\left(a+b+ab\right)c \\ =a+b+ab+c+ac+bc+abc \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in S \end{gathered}$$

Thus, * is associative on S.

Now,
$$\begin{gathered} \text{Given:\hspace{0.17em}}\left(2*x\right)*3=7 \\ \Rightarrow \left(2+x+2x\right)*3=7 \\ \Rightarrow \left(2+3x\right)*3=7 \\ \Rightarrow 2+3x+3+\left(2+3x\right)3=7 \\ \Rightarrow 5+3x+6+9x=7 \\ \Rightarrow 12x+11=7 \\ \Rightarrow 12x=-4 \\ \Rightarrow x=\frac{-4}{12} \\ \Rightarrow x=\frac{-1}{3} \end{gathered}$$