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=[tex]A\cap B[/tex] form a group.
Suppose, x={1,2}
P(X)={[tex]\phi [/tex],{1},{2},{1,2}}
1.Closure property:[tex]A\cap B\in P(X)[/tex]
{1}[tex]\cap[/tex] {2}=[tex]\phi \in P(X)[/tex]
It is satisfied for all [tex]A,B\in P(X)[/tex]
2.Associative property:[tex](A\cap B)\cap C=A\cap (B\cap C)[/tex]
If A={1},B={2},C={1,2}
[tex]A\cap(B\cap C)[/tex]={1}[tex]\cap[/tex]({2}[tex]\cap[/tex]{1,2})={1}[tex]\cap[/tex] {2}=[tex]\phi[/tex]
[tex](A\cap B)\cap C[/tex]=({1}[tex]\cap[/tex]{2})[tex]\cap[/tex]{1,2}=[tex]\phi\cap[/tex]{1,2}=[tex]\phi[/tex]
Hence, P(X) satisfied the associative property.
3.Identity :[tex]A\cap B=A[/tex] Where B is identity element of P(X)
[tex]A\cap X=A[/tex]
It is satisfied for every element A in P(X).
Hence, X is identity element in P(X)
4.Inverse :[tex]A\cap B=X[/tex] Where B is an inverse element of A in P(x)
It can not be possible for every element that satisfied [tex]A\cap B=X[/tex]
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=[tex]A\cup B[/tex] form a group
Similarly,
For set X={1,2}
P(X)={[tex]\phi [/tex],{1},{2},{1,2}}
1.Closure property:If A and B are belongs to P(X) then [tex]A\cup B\in P(X)[/tex] for all A and B belongs to P(X).
2.Associative property:[tex]A\cup (B\cup C)=(A\cup B)\cup C[/tex]
If A={1},B={2},C{1,2}
[tex]A\cup B[/tex]={1}[tex]\cup [/tex]{2}={1,2}
[tex](A\cup B)\cup C[/tex]={1,2}[tex]\cup [/tex]{1,2}={1,2}
[tex]B\cup C[/tex]={2}[tex]\cup [/tex]{1,2}={1,2}
[tex]A\cup (B\cup C)[/tex]={1}[tex]\cup [/tex]{1,2}={1,2}
Hence, P(X) satisfied the associative property.
3.Identity :[tex]A\cup B=A[/tex] Where B is identity element of P(X)
Only [tex]\phi[/tex] is that element for every A in P(X) that satisfied [tex]A\cup B=A[/tex]
Hence, [tex]\phi[/tex] is identity element of P(X) w.r.t union.
4.Inverse element :
[tex]A\cup B=\phi[/tex] 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.
