Question

Let * be a binary operation on Q₀ (set of non-zero rational numbers) defined by \[a * b = \frac{ab}{5} \text{for all a, b} \in Q_0\]

 Show that * is commutative as well as associative. Also, find its identity element if it exists.

Answer 1

Commutativity:

\[\text{ Let }a, b \in Q_0 \] 
\[a * b = \frac{ab}{5}\] 
          \[ = \frac{ba}{5}\] 
          \[ = b * a \] 
\[\text{Therefore},\] 
\[a * b = b * a, \forall a, b \in Q_0\]

Associativity:

\[\text{Let}a, b, c \in Q_0 \] 
\[a * \left( b * c \right) = a * \left( \frac{bc}{5} \right)\] 
                   \[ = \frac{a\left( \frac{bc}{5} \right)}{5}\] 
                   \[ = \frac{abc}{25}\] 
\[\left( a * b \right) * c = \left( \frac{ab}{5} \right) * c\] 
                     \[ = \frac{\left( \frac{ab}{5} \right)c}{5}\] 
                      \[ = \frac{abc}{25}\] 
\[\text{Therefore},\] 
\[a * \left( b * c \right) = \left( a * b \right) * c, \forall a, b, c \in Q_0 \] 

Thus, * is associative on Q_o.

Finding identity element :

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

\[a * e = a = e * a, \forall a \in Q_0 \] 
\[a * e = a \text{ and }e * a = a, \forall a \in Q_0 \] 
\[ \Rightarrow \frac{ae}{5} = a \text{ and }\frac{ea}{5} = a, \forall a \in Q_0 \] 
\[ \Rightarrow e = 5 , \forall a \in Q_0 \left[ \because a \neq 0 \right]\]

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