Question

Consider an algebraic system (G, *), where G is the set of all non-zero real numbers and * is a binary operation defined by

a*b=(a*b)/4

Show that (G, *) is an abelian group

Answer 1

TO PROVE

The set G is the set of all non-zero real numbers forms an abelian group under the operation * defined by

$\displaystyle \: \sf{ a*b = \frac{1}{4}ab \: \: \: \forall \: \: a, b \in G}$

PROOF

1. CHECKING FOR CLOSURE PROPERTY

$\displaystyle \: \sf{ Let \: \: a, b \in G}$

$\implies \: \displaystyle \:\sf{ \frac{1}{4} ab \: \in G}$

$\implies \: \displaystyle \: \sf{ a* b \: \in G}$

So * is closed

2. CHECKING FOR ASSOCIATIVE PROPERTY

$\displaystyle \: \sf{ Let \: \: a, b ,c \: \in G}$

Then

$\displaystyle \: \sf{ a*(b*c) = a* \bigg( \frac{1}{4} bc\bigg) \: } = \frac{1}{16} abc$

$\displaystyle \: \sf{ (a*b)*c = \bigg( \frac{1}{4} ab\bigg) *c\: } = \frac{1}{16} abc$

So a * ( b * c ) = ( a * b ) * c

So * is associative

3. EXISTENCE OF IDENTITY ELEMENT

Let a ∈ G

Let e be the identity element

Then e*a= a*e= a

$\: \displaystyle \: \sf{ \frac{1}{4} ea \: = a \:}$

$\implies \: \displaystyle \: \sf{ e = 4 }$

So 4 is the identity element

4. EXISTENCE OF INVERSE ELEMENT

Let a ∈ G

Let there exists b ∈ G such that

a*b= b*a= e

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

$\displaystyle \: \sf{ \frac{1}{4} ab\: = 4 \:}$

$\implies \: \: \displaystyle \: \sf{ b\: = \frac{16}{a} \:}$

$\displaystyle \: \sf{ \therefore \: \: \frac{16}{a} \: }\: is \: the \: inverse \: \: element \: of \: a\: \: under \: the \: operation \: *$

So G is a group

CHECKING FOR COMMUTATIVE PROPERTY

Let a, b ∈ G

Now

$\displaystyle \: \sf{ a*b = \frac{1}{4}ab \: \: \: }$

$\displaystyle \: \sf{ b*a = \frac{1}{2}ba = \frac{1}{4}ab \: \: \: }$

So a * b = b * a

So ( G , * ) is commutative group

Hence proved

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