Question

On the set Q⁺ of all positive rational numbers a binary operation * is defined by $a*b=\frac{ab}{2}\mathrm{for}\mathrm{all},a,b\in Q^{+}$. The inverse of 8 is
(a) $\frac{1}{8}$

(b) $\frac{1}{2}$

(c) 2

(d) 4

Answer 1

(b) $\frac{1}{2}$

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

$$\begin{gathered} a*e=a=e*a,\forall a\in Q^{+} \\ a*e=a\text{and}e*a=a,\forall a\in Q^{+} \\ \mathrm{Then}, \\ \frac{ae}{2}=a\text{and}\frac{ea}{2}=a,\forall a\in Q^{+} \\ e=2,\forall a\in Q^{+} \end{gathered}$$

Thus, 2 is the identity element in Q⁺ with respect to *.

$$\begin{gathered} \text{Let}b\in Q^{+}\text{be the inverse of 8}.\text{Then,} \\ \text{8}*b=e=b*8 \\ 8*b=e\text{and}b*8=e \\ \frac{8b}{2}=2\text{and}\frac{b\left(8\right)}{2}\text{=2}\left[\because e=2\right] \\ b=\frac{1}{2} \\ \text{Thus,}\frac{1}{2}\text{is the inverse of}8. \end{gathered}$$