Is the set \mathbb{z} a group under the following operations:
a.) a*b = a + b - 1
b.) a*b = a - b + ab

a) yes

b) no

Step-by-step explanation:

is a gruop if satisfies the following conditions:

1. If a and b are two elements in , then the product a*b is also in .

2. The defined multiplication is associative, i.e., for all a,b,c in , (a*b)*c=a*(b*c).

3. There is an identity element e such that e*a=a*e=a for every element a in .

4. There must be an inverse of each element. Therefore, for each element a of , the set contains an element b=a^(-1) such that a*a^(-1)=a^(-1)*a=e.

Let's see if the conditions are satisfied:

a)

1. if x and y are integers then x+y-1=a*y is an integer

2. If x,y and z are integers then

   (x*y)*z= (x+y-1)*z= (x+y-1) + z - 1= x +y+z-2,

   x*(y*z)= x*(y+z-1)= x + (y+z-1) -1 = x+ y + z -2

  Then (x*y)*z=x*(y*z), i.e, * is associative.

3. Let e=1 and b an integer. Observe that

    1*b=1+b-1=b and b*1= b + 1 -1= b.

  Then e is an identity element.

4. a and integer and b= 2- a. Observe that

    b*a= 2-a+a-1= 1 and a*b= a+2-a-1=1,

the b= a^(-1) is the inverse of a.

We conclude that is a group.

b)

1. If x,y and z are integers then

   (x*y)*z= (x-y+xy)*z= (x-y+xy) - z + (x-y+xy)z= x -y-z+xy+xz-yz+xyz

   x*(y*z)= x*(y-z+yz)= x - (y-z+yz) +x(y-z+yz) = x-y +z + xy -xz -yz+xyz

  Then (x*y)*z≠x*(y*z), i.e, * isn't associative.

We conclude that isn't a group.

answer is 144

step-by-step explanation:

christopher is 272,195 year old

step-by-step explanation: