Let * be a binary operation on Q0 (set of non-zero rational numbers) defined by a ^* b= ab 5 for all a, b _0. Show that * is commutative as well as associative. Also, find its identity element if it exists.

Commutativity: Let a, b _0 a ^* b= ab 5 = ba 5 =b ^* a Therefore, a ^* b=b ^* a, a, b _0 Thus, * is commutative on Q0. Associativity: Let a, b, c _0 a ^* (b ^* c)=a ^* ( bc 5 ) = a ( bc 5 ) 5 = abc 25 (a ^* b) ^* c= ( ab 5 ) ^* c = ( ab 5 )c 5 = abc 25 Therefore, a * (b * c) = (a * b) * c, a, b, c _0 Thus, * is associative on Q0. Finding identity element: Let e be the identity element in Z with respect to * such that, a * e = a = e * a, _0 a * e = a and e * a = a, _0 Implies that ae 5 =a and ea 5 =a, a _0 Implies that e=5, a _0 [ a 0] Thus, 5 is the identity element in with respect to *.

Let * be a binary operation on Q0 (set of non-zero rational numbe…

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Let * be a binary operation on Q₀ (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 Q₀.
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, $\forall\ \text{a, b, c}\in\text{Q}_0$
Thus, * is associative on Q₀.
Finding identity element:
Let e be the identity element in Z with respect to * such that,
a * e = a = e * a, $\forall\text{a}\in\text{Q}_0$
a * e = a and e * a = a, $\forall\text{a}\in\text{Q}_0$
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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