Question

Show that the set Q+ of all positive rational numbers forms an abelian group under the operation * defined by
a*b= 1/2(a.b); for all a,b ¢ Q+​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

The set Q+ of all positive rational numbers forms an abelian group under the operation * defined by

$\displaystyle \: \sf{ a*b = \frac{1}{2}ab \: \: \: \forall \: \: a, b \in \mathbb{Q}^{ + } }$

PROOF

1. CHECKING FOR CLOSURE PROPERTY

$\displaystyle \: \sf{ Let \: \: a, b \in \mathbb{Q}^{ + } }$

$\implies \: \displaystyle \: \sf{ \frac{1}{2} ab \: \in \mathbb{Q}^{ + } }$

$\implies \: \displaystyle \: \sf{ a* b \: \in \mathbb{Q}^{ + } }$

$\sf{So \: \: \mathbb{Q}^{ + } \: is \: closed \: under \: the \: operation \: * \: \: }$

2. CHECKING FOR ASSOCIATIVE PROPERTY

$\displaystyle \: \sf{ Let \: \: a, b ,c \: \in \mathbb{Q}^{ + } }$

Then

$\displaystyle \: \sf{ a*(b*c) = a* \big( \frac{1}{2} bc\big) \: } = \frac{1}{4} abc$

$\displaystyle \: \sf{ (a*b)*c = \big( \frac{1}{2} ab\big) *c\: } = \frac{1}{4} abc$

So

$\: \sf{ a*(b*c ) =\: \sf{ (a*b)*c \: }}$

$\sf{So \: \: \mathbb{Q}^{ + } \: is \: associative \: under \: the \: operation \: * \: \: }$

3. EXISTENCE OF IDENTITY ELEMENT

$\displaystyle \: \sf{ Let \: \: a \in \mathbb{Q}^{ + } }$

$\displaystyle \: \sf{ Let \: \: e \in \mathbb{Q}^{ + } } \: be \: the \: identity \: element \:$

$\sf{ \: Then \: \: e*a= a*e= a \: }$

$\sf{ \:Now \: \: e*a= a \: } \: \: implies$

$\: \displaystyle \: \sf{ \frac{1}{2} ea \: = a \:}$

$\implies \: \displaystyle \: \sf{ e = 2 }$

$\sf{So \: 2 \: \: is \: the \: identity \: element \: under \: the \: operation \: * \:}$

4. EXISTENCE OF INVERSE ELEMENT

$\displaystyle \: \sf{ Let \: \: a \in \mathbb{Q}^{ + } }$

$\displaystyle \: \sf{ Let \: \: b \in \mathbb{Q}^{ + } } \: be \: the \: inverse \: \: element \: of \: a\:$

$\sf{ \: Then \: \: a*b= b*a= e\: }$

$\sf{ \:Now \: \: a*b= e\: } \: \: implies$

$\: \displaystyle \: \sf{ \frac{1}{2} ab\: = 2 \:}$

$\implies \: \: \displaystyle \: \sf{ b\: = \frac{4}{a} \:}$

$\displaystyle \: \sf{ \: \: a \in \mathbb{Q}^{ + } \: \: implies \: \: \: \frac{4}{a} \in \mathbb{Q}^{ + } \: }$

$\displaystyle \: \sf{ \therefore \: \: \frac{4}{a} \: }\: is \: the \: inverse \: \: element \: of \: a\: \: under \: the \: operation \: *$

$\sf{So \: \: \mathbb{Q}^{ + } \: is \: a \: group \: under \: the \: operation \: * \: \: }$

CHECKING FOR COMMUTATIVE PROPERTY

$\displaystyle \: \sf{ Let \: \: a, b \in \mathbb{Q}^{ + } }$

$\displaystyle \: \sf{ a*b = \frac{1}{2}ab \: \: \: }$

$\displaystyle \: \sf{ b*a = \frac{1}{2}ba = \frac{1}{2}ab \: \: \: }$

$\sf{ \therefore \: \: \: a*b= b*a}$

$\sf{So \: \: \mathbb{Q}^{ + } \: is \: commutative \: \: under \: the \: operation \: * \: \: }$

$\sf{So \: \: \mathbb{Q}^{ + } \: is \: a \: commutative \: \: group \: under \: the \: operation \: * \: \: }$