Let * be a binary operation on Q_0 (set of non - zero rational nu… - Vidyadip

Question

Let $*$ be a binary operation on $Q_0 $(set of non $-$ zero rational numbers$)$ defined by $\text{a}\ ^* \ \text{b}=\frac{\text{ab}}{5}$ for all $\text{a, b}\in\text{Q}_0.$ Show that $*$ is commutative as well as associative. Also, find its identity element if it exists.

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Answer

Commutativity: Let $\text{a, b}\in\text{Q}_0$
$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{5}$
$=\frac{\text{ba}}{5}$
$=\text{b}\ ^*\ \text{a}$
Therefore, $\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a},\forall\ \text{a, b}\in\text{Q}_0$
Thus $, *$ is commutative on $Q_0.$
Associativity: Let $\text{a, b, c}\in\text{Q}_0$
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{bc}}{5}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{5}\big)}{5}$
$=\frac{\text{abc}}{25}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{ab}}{5}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{5}\big)\text{c}}{5}$
$=\frac{\text{abc}}{25}$
Therefore,
$a * (b * c) = (a * b) * c, $ for all $ \text{a, b, c}\in\text{Q}_0$
Thus $,*$ is associative on $Q_0$.
Finding identity element:
Let e be the identity element in $Z$ with respect to $*$ such that,
$a * e = a = e * a,$ for all $\text{a}\in\text{Q}_0$
$a * e = a$ and $e * a = a,$ for all $\text{a}\in\text{Q}_0$
Implies that $\frac{\text{ae}}{5}=\text{a}$ and $\frac{\text{ea}}{5}=\text{a},\forall\ \text{a}\in\text{Q}_0$
Implies that $\text{e}=5,\forall\ \text{a}\in\text{Q}_0\ [\because\ \text{a}\neq0]$
Thus $, 5$ is the identity element in with respect to $*$.

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