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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.
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Solution
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 Qo.
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 Qo with respect to *.
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