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Let S = R - {-1} and define a binary operation on S by a*b = a+b+ab, what is true about (S,*)
A. (S,*) is a group but is not commutative
B. (S,*) is a group and is also commutative
C. (S,*) is not a group because inverse of 1 doesn't exist

"How to check if inverse and identity element exist?
asked in Set Theory & Algebra by Veteran (14.6k points)  
edited by | 47 views

1 Answer

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Best answer

Here '*' be defined as a*b=a+b+ab.

i)it is closed because all values belongs to R(real numbers)

ii)Check associativity

a*(b*c)

=a*(b+c+bc)

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

(a*b)*c

=(a+b+ab)*c

=a+b+ab+c+ac+bc+abc.

it satisfies Assocaitivity

iii) Finding identity element check this condition a*e=a

a+e+ae=a

e(1+a)=0

e=0 // it is identity element

iii)Inverse check this condition a*b=e  

 a+b+ab=0

b=-a/1+a.  only when a!=-1.

it is given that R!=(-1) so inverse exist for every element

iv)Commutative

a*b=b*a

a+b+ab=b+a+ba.

Hence is is abelian group

answered by Veteran (10.7k points)  
selected by
perfect!! Bhagwan tumhara Bhala kare!!
u r done correct bro


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