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Show that the binary operation * on A = R – { – 1} defined as a*b = a + b + ab for all a, b ∈ A is commutative and associative on A. Also find the identity element of * in A and prove that every element of A is invertible. - Mathematics

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Show that the binary operation * on A = R – { – 1} defined as a*b = a + b + ab for all a, b ∈ A is commutative and associative on A. Also find the identity element of * in A and prove that every element of A is invertible.

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Solution

We have ​a*b = a + b + ab for all a, b ∈ A, where A = R – { – 1}
Commutativity: For any a, b ∈ R – { – 1}
To prove: a*b = b*a  
Now, a*b = a + b + ab               .....(1)
b*a = b + a + ab                        .....(2)
From (1) and (2), we get
a*b = b*a 
Hence, * is commutative.

Associative: For any a, b, c ∈ R – { – 1}
To prove: a*(b*c) = (a*b)*c
a*(b*c) = a*(b + c +bc)
             = a + (b + c + bc) + a(b + c + bc)
             = a + b + c + ab + ac + bc + abc           .....(3)
(a*b)*c = (a + b + ab)*c
             = a + b + ab + c + (a + b + ab)c
             = a + b + c + ab + bc + ca + abc            .....(4)
From (3) and (4), we have
a*(b*c) = (a*b)*c
Hence, a*b is associative.

Identity element:
Let e be the identity element. Then,
a*e = e*a = a
a*e = a + e + ae = a
e(1 + a) = 0
Therefore, e = 0            [∵ a ≠ – 1]
Hence, the identity element for* is e = 0.

Existence of inverse: Let a = R – { – 1} and b be the inverse of a. 
Then, a * b = e = b * a
a*b = ea + b + ab = 0b= `−a/(a+1)`


Since,

aR(1)

a1

a+10

`⇒b=−a/(a+1)∈R`

Also, `−a/(a+1)` =1a = a11=0, which is not possible.

Hence, ` −a/(a+1)∈R−(−1)`

So, every element of R – { – 1} is invertible and the inverse of an element a is `−a/(a+1).`

Concept: Concept of Binary Operations
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