Question

Let 'o' be a binary operation on the set Q₀ of all non-zero rational numbers defined by

$aob=\frac{ab}{2},\mathrm{for}\mathrm{all}a,b\in Q_0$.

(i) Show that 'o' is both commutative and associate.
(ii) Find the identity element in Q₀.
(iii) Find the invertible elements of Q₀.

Answer 1

(i) Commutativity:
$$\begin{gathered} \text{Let}a,b\in Q_0.\mathrm{Then}, \\ aob=\frac{ab}{2} \\ =\frac{ba}{2} \\ =boa \\ \text{Therefore,} \\ aob=boa,\forall a,b\in Q_0 \end{gathered}$$
Thus, o is commutative on Q_o.

Associativity:
$$\begin{gathered} \text{Let}a,b,c\in Q_0.\mathrm{Then}, \\ ao\left(boc\right)=ao\left(\frac{bc}{2}\right) \\ =\frac{a\left({\displaystyle \frac{bc}{2}}\right)}{2} \\ =\frac{abc}{4} \\ \left(aob\right)oc=\left(\frac{ab}{2}\right)oc \\ =\frac{\left({\displaystyle \frac{ab}{2}}\right)c}{2} \\ =\frac{abc}{4} \\ \text{Therefore,} \\ ao\left(boc\right)=\left(aob\right)oc,\forall a,b,c\in Q_0 \end{gathered}$$
Thus, o is associative on Q_o.

(ii) Let e be the identity element in Q_o with respect to * such that
$$\begin{gathered} aoe=a=eoa,\forall a\in Q_0 \\ aoe=a\text{and}eoa=a,\forall a\in Q_0 \\ \Rightarrow \frac{ae}{2}=a\text{and}\frac{ea}{2}=a,\forall a\in Q_0 \\ e=2\in Q_0,\forall a\in Q_0 \end{gathered}$$
Thus, 2 is the identity element in Q_o with respect to o.

$$\begin{gathered} \left(\mathrm{iii}\right)\text{Let}a\in Q_0\text{and}b\in Q_0\text{be the inverse of}a.\text{Then,} \\ aob=e=boa \\ \Rightarrow aob=e\text{and}boa=e \\ \Rightarrow \frac{ab}{2}=2\text{and}\frac{ba}{2}=2 \\ \Rightarrow b=\frac{4}{a}\in Q_0 \\ \text{Thus,}\frac{4}{a}\text{is the inverse of}a\in Q_0. \end{gathered}$$