Question

Let * be a binary operation defined on Q⁺ by the rule $a*b=\frac{ab}{3}\mathrm{for}\mathrm{all}a,b\in Q^{+}$. The inverse of 4 * 6 is

(a) $\frac{9}{8}$

(b) $\frac{2}{3}$

(c) $\frac{3}{2}$

(d) none of these

Answer 1

(a) $\frac{9}{8}$

Let e be the identity element in Q⁺ with respect to * such that

$$\begin{gathered} a*e=a=e*a,\forall a\in Q^{+} \\ a*e=a\text{and}e*a=a,\forall a\in Q^{+} \\ \mathrm{Then}, \\ \frac{ae}{3}=a\text{and}\frac{ea}{3}=a,\forall a\in Q^{+} \\ \Rightarrow e=3,\forall a\in Q^{+} \end{gathered}$$

Thus, 3 is the identity element in Q⁺ with respect to *.

$$\begin{gathered} \text{Let}a\in Q^{+}\text{and}b\in Q^{+}\text{be the inverse of}a.\text{Then,} \\ a*b=e=b*a \\ a*b=e\text{and}b*a=e \\ \therefore \frac{ab}{3}=3\text{and}\frac{ba}{3}\text{=3} \\ b=\frac{9}{a}\in Q^{+} \\ \text{Thus,}\frac{9}{a}\text{is the inverse of}a\in Q^{+}. \end{gathered}$$

$$\begin{gathered} \mathrm{Given}:a*b=\frac{ab}{3} \\ 4*6=\frac{4\times 6}{3}=8 \\ \mathrm{Now}\text{,} \\ {a}^{-1}=\frac{9}{a} \\ {\left(4*6\right)}^{-1}=8^{-1} \\ =\frac{9}{8} \end{gathered}$$