For number 1. The pls give me the answer
1 answer:
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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.
