Question #133610
On Q/{-1} define * by a*b=a+b+an for all a,b element of Q/{-1} show that (Q/{-1},*) is an abelian group
1
Expert's answer
2020-09-17T10:27:26-0400

aQ/{1}a\in Q/\{-1\}

bQ/{1}b∈Q/\{−1\}

ab=a+b+abQ/{1}a*b=a+b+ab\in Q/\{-1\}


,is\therefore *, is  a binary operation in(Q/{1},)(Q/\{-1\}, *)

Associativity: Suppose a,b,cQ/{1}a,b,c \in Q/\{-1\}

(ab)c=(a+b+ab)c(a*b)*c=(a+b+ab)*c

=(a+b+ab)+c+(a+b+ab)c=(a+b+ab)+c+(a+b+ab)c

=a+b+c+ab+bc+ca+abc=a+b+c+ab+bc+ca+abc


So,

a(bc)=a(b+c+bc)a*(b*c)=a*(b+c+bc)

=a+(b+c+bc)+a(b+c+bc)=a+(b+c+bc)+a(b+c+bc)


=a+(b+c+bc)+a(b+c+bc)=a+(b+c+bc)+a(b+c+bc)

=a+b+c+ab+bc+ca+abc=a+b+c+ab+bc+ca+abc

(ab)c=a(bc)(a*b)*c=a*(b*c)

* is associative.


ExistenceExistence ofof Identity:Identity:

0Q/{1}0\in Q/\{-1\}

isis the identity element.

Because aQ/{1}a\in Q/\{-1\}

0a=0+a+0a=a0*a=0+a+0a=a

a0=a+0+a0=aa*0=a+0+a0=a

ExistenceExistence of InverseInverse :

Every aa\in Q/{1}Q/\{-1\}

For -a/a+1 \in Q/{-1}

a1,a\not=-1, which is inverse of a

Because

(a/(a+1))a\bigg(-a/(a+1) \bigg) *a ==

=(a/(a+1))+a+(a/(a+1))a=0=\bigg(-a/(a+1)\bigg) +a+\bigg(-a/(a+1)\bigg) *a=0


a(a/(a+1))=a*\bigg(-a/(a+1)\bigg) =

=a+(a)/(a+1)+a(a/(a+1))=0=a+(-a)/(a+1)+a\bigg(-a/(a+1)\bigg) =0


Commutativity:Commutativity:- a,bQ/{1}a, b \in Q/\{-1\}

ab=a+b+aba*b=a+b+ab

ba=b+a+bab*a=b+a+ba


So,

ab=baa*b=b*a

Operation * is commutative.


The above discussion proves that the group is abelian. 






Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
18.09.20, 20:25

Dear PRATHIBHA ROSE C S, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

PRATHIBHA ROSE C S
18.09.20, 07:37

Thank you so much for your response

LATEST TUTORIALS
APPROVED BY CLIENTS