Question

The value of n[(A-B)U(B-A)]+n(A∩B) equals​

Answer 1

Answer:

(A−B)∪(B−A)=A∪B (A−B)∪(B−A)=A∪B

and you want to prove that

A∩B=∅. A∩B=∅.

Say that there is an element

x∈A∩B x∈A∩B . You want to prove that no such exist, so assume that is does. Then Now then

x∈A x∈A and

x∈B x∈B . So, certainly,

a∈A∪B=(A−B)∪(B−A). a∈A∪B=(A−B)∪(B−A). If an element is in the union of two sets, then it is one of the sets (maybe in both). So

x∈A−B x∈A−B or

x∈B−A x∈B−A . But both of these options don't hold. Saying that, for example,

x∈A−B x∈A−B is saying that

x∉B x∉B which contradicts that

x∈A∩B x∈A∩B . Hence no such

Answer 2

Answer:

Step-by-step explanation:

(A−B)∪(B−A)=A∪B (A−B)∪(B−A)=A∪B

and you want to prove that

A∩B=∅. A∩B=∅.

Say that there is an element

x∈A∩B x∈A∩B . You want to prove that no such exist, so assume that is does. Then Now then

x∈A x∈A and

x∈B x∈B . So, certainly,

a∈A∪B=(A−B)∪(B−A). a∈A∪B=(A−B)∪(B−A). If an element is in the union of two sets, then it is one of the sets (maybe in both). So

x∈A−B x∈A−B or

x∈B−A x∈B−A . But both of these options don't hold. Saying that, for example,

x∈A−B x∈A−B is saying that

x∉B x∉B which contradicts that

x∈A∩B x∈A∩B . Hence no such