Question

Show that the set z of all integers form a group with respect to binary operation defined by a*b-a+b+1 is an abelian group

Answer 1

Answer:

Z = Set of all integers

Given:

* is defined on Z by a*b=a+b+1

Closure axiom:

Let a, b ∈ Z

Then a and b are integers.

since sum of any integers is also an integer, a+b+1 is an integer

That is, a*b is an integer

Therefore, a*b ∈ Z

Hence closure axiom is true

Associative axiom:

Let a, b, c ∈ Z

$a*b=a+b+1$

$(a*b)*c = (a+b+1)*c$

$(a*b)*c = (a+b+1)+c+1$

$(a*b)*c = a+b+c+2$...........(1)

$b*c=b+c+1$

$a*(b*c)=a*(b+c+1)$

$a*(b*c)=a+(b+c+1)+1$

$a*(b*c)=a+b+c+2$..............(2)

From (1) and (2)

$a*(b*c)=(a*b)*c$

Hence associattive axiom is true.

Identity axiom:

Let e be the identity element of Z

Then, by definition

$a*e=e*a=a$

$a*e=a$

This implies

$a+e+1=a$

$e+1=0$

$e=-1$ ∈ Z

Hence identity axiom is true

Inverse axiom:

Let $a^{-1}$ be the inverse of a

Then, by definition

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

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

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

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

$a^{-1}=-2-a$

$a^{-1}=-(2+a)$ ∈ Z ( since a and 2 are integers)

Hence inverse axiom is true

Commutative axiom:

Let a, b ∈ Z

$a*b=a+b+1$

$a*b=b+a+1$

$a*b=b*a$

Hence commutative axiom is true

Therefore, Z is an abelian group under *

Answer 2

Answer:

Clearly z is a non-empty set

Closure:-Let a,b∈z

=>a+b+1 ∈ z

=>a*b ∈z

Closure law holds

Associative:-Let a,b,c∈ z

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

L.H.S:- a*(b*c)=a*(b+c+1)

=a+b+c+1+1

=a+b+c+2

R.H.S:- (a*b)*c= (a+b+1)*c

=a+b+1+c+1

=a+b+c+2

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

Associative law holds

Identity:-Let a ∈ z

e is the identity in z

a*e=e*a=a

a*e=a

a+e+1=a

e+1=0

e=-1 ∈ z

-1 is the identity

Inverse:-

Let a ∈ z

b is the inverse of a

a*b=b*a=e

a*b=e

a+b+1=e

a+b+1=-1

a+b=-2

b=-2-a

b=-(2+a)∈Z

b is the inverse of a in z

(z,*) is a group

Commutative:-Let a,b∈Z

a*b=a+b+1

=b+a+1

=b*a

a*b =b*a, ∀ a,b∈z

∴ (z, *) is an abelian group.

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Step-by-step explanation: