abstract algebra
let x be a set and let p(x) be the power set of x.
a) does p(x) with the binary operation a *b=anb form a group? justify your answer.
b) does p(x) with the binary operation a *b=aub form a group? justify your answer.

1.No, because inverse does not exist.

2.No, because inverse does not exist.

Step-by-step explanation:

We are given that X be  a set and let P(X) be the power set of X.

a. We have to tell P(X) with binary operation

A*B= form  a group.

Suppose, x={1,2}

P(X)={,{1},{2},{1,2}}

1.Closure  property:

{1} {2}=

It is satisfied for all

2.Associative property:

If A={1},B={2},C={1,2}

={1}({2}{1,2})={1} {2}=

=({1}{2}){1,2}={1,2}=

Hence, P(X) satisfied the associative property.

3.Identity : Where B is identity element of P(X)

It is satisfied for every element A in P(X).

Hence, X is identity element in  P(X)

4.Inverse : Where B is an inverse element of A in P(x)

It can not be possible for every element that satisfied

Hence, inverse does not exist.

Therefore, P(X) is not a  group w.r.t to given binary operation.

2.We have to tell P(X) with the binary operation

A*B= form a group

Similarly,

For set X={1,2}

P(X)={,{1},{2},{1,2}}

1.Closure property:If A and B are belongs to P(X) then for all A and B belongs to P(X).

2.Associative property:

If A={1},B={2},C{1,2}

={1}{2}={1,2}

={1,2}{1,2}={1,2}

={2}{1,2}={1,2}

={1}{1,2}={1,2}

Hence, P(X) satisfied the associative property.

3.Identity : Where B is identity element of P(X)

Only is that element for every A in P(X) that satisfied

Hence, is identity element of P(X) w.r.t union.

4.Inverse element :

where B is  an inverse element of A in P(X)

It is not possible for every element that satisfied the property.

Hence, inverse does not exist for each element in P(X).

Therefore, P(X) is not  a group w.r.t binary operation.

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1

your answer would be 12