Set theory

Question

Let A = set of all real numbers. '*' be a binary operation

a*b  = a + b + a. b

(A, *)  is a group or not.

 

Doubt : Till finding identity element (=0) every thing is okay.

While finding inverse:

a * $a^{-1}$ = e

a +  $a^{-1}$ + a$a^{-1}$ = 0

a + $a^{-1}$ + 1  = 0

 

Or

a + $a^{-1}$(1+a) = 0

 

Both will lead to different answer.

What is correct.

Comments

Write the complete question.

Identity element does not exist because of  -1

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$a+a^{-1}+a.a^{-1}=0$

here $a.a^{-1}$ is multiplication not group operator (*)  so this multiplication is not zero but if it is written as $(a*a^{-1})$ then it will be zero .

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@raja11sep @Kabir5454 

note : * does not means multiplication

it is defined as 

a*b = a + b+ ab

e.g.

1*0 = 1+0+1.0

         = 1

 

Semigroup:

LHS

(a*b) *c

=( a+b +ab) *c

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

RHS

a*(b*c) 

a*(b+c+bc) 

a+b+c+bc+ ab+ac + ABC

LHS = RHS

 

let e be identity element

a*e = a

a + e + ae = a

e  + ae = 0

e(1+a) = 0

e = 0

Group:

number* inverse = e

number*inverse = 0

number+ inverse + number. inverse= 0

 

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The identity element does not exist because a can be -1

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$a+a^{-1}(1+a) =0$ is correct because if you write $aa^{-1}=1,$ it means you already know the value of $a^{-1}$ which is $\frac{1}{a}$ then what is point of finding $a^{-1}$ again. And, if you assume $a^{-1}=\frac{1}{a}$ then you will not get the real 'a' because then your equation becomes $a^2+a+1=0$ which has the complex roots but 'a' should be real.

So, if you follow $a+ a^{-1}(1+a)=0$ then $a^{-1}=\frac{-a}{1+a}$ and for this $a \neq -1$

So, $(\mathbb{A} \setminus  \{-1\},*)$ is a group.

Here, identity element is zero because for e=0, a+e+ae=a is true for all real a.

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@ankitgupta.1729

thanks you, your comment contain the answer which I was searching for.

Answer 1

A set of real numbers with the binary operation defined as a*b = a + b + a.b is not a group, because the inverse element does not exist for all elements in the set.

To show this, we need to find an inverse element a^-1 for a given element a such that a*a^-1 = e, where e is the identity element.

In this case, we have aa^-1 = a + a^-1 + aa^-1 = 0.

This equation does not have a real solution for a^-1 for all a, thus the set of real numbers with this binary operation does not form a group.