Question

Let * be a binary operation on Q₀ (set of non-zero rational numbers) defined by
$a*b=\frac{ab}{5}\mathrm{for}\mathrm{all}a,b\in Q_0$.
Show that * is commutative as well as associative. Also, find its identity element if it exists.

Answer 1

Commutativity:
$$\begin{gathered} \text{Let}a,b\in Q_0 \\ a*b=\frac{ab}{5} \\ =\frac{ba}{5} \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in Q_0 \end{gathered}$$
Thus, * is commutative on Q_o.

Associativity:
$$\begin{gathered} \text{Let}a,b,c\in Q_0 \\ a*\left(b*c\right)=a*\left(\frac{bc}{5}\right) \\ =\frac{a\left({\displaystyle \frac{bc}{5}}\right)}{5} \\ =\frac{abc}{25} \\ \left(a*b\right)*c=\left(\frac{ab}{5}\right)*c \\ =\frac{\left({\displaystyle \frac{ab}{5}}\right)c}{5} \\ =\frac{abc}{25} \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in Q_0 \end{gathered}$$
Thus, * is associative on Q_o.

Finding identity element:

Let e be the identity element in Z with respect to * such that
$$\begin{gathered} a*e=a=e*a,\forall a\in Q_0 \\ a*e=a\text{and}e*a=a,\forall a\in Q_0 \\ \Rightarrow \frac{ae}{5}=a\text{and}\frac{ea}{5}=a,\forall a\in Q_0 \\ \Rightarrow e=5,\forall a\in Q_0\left[\text{∵}a\ne 0\right] \end{gathered}$$

Thus, 5 is the identity element in Q_o with respect to *.