Question

For any two sets A and B, prove that $A\cap \left( A\cup B \right)’ = \phi $

Answer 1

Hint:Using De Morgan’s law find $\left( A\cup B \right)$ . Now using associative property and rearrange them using basic properties, find that it is a null set.

Complete step-by-step answer:
In set theory, De Morgan’s Law relates the intersection and union of sets through complements. In propositional logic, De Morgan’s Law relates conjunctions and disjunctions of propositions through negation. De Morgan’s laws states that the complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of 2 sets is equal to the union of their complements.
i.e $\left( A\cup B \right)'=A'\cap B'$
$\left( A\cap B \right)'=A'\cup B'$
These are called De Morgan’s law.
Now we have been given $A\cap \left( A\cup B \right)'$ , which we need to prove that the given expression proves out to be a null set. In a null set there are no elements. Its size or cardinality, which is the count of elements in the set is zero,
Now under De Morgan’s law,
$\left( A\cup B \right)'=A'\cap B'$
Thus, we can write,
$A\cap \left( A\cup B \right)'=A\cap \left( A'\cap B' \right)$
Now by associative property of set theory we can write,
$A\cap \left( A'\cap B' \right)=\left( A\cap A' \right)\cap B'$
In set theory, the complement of a set A refers to elements not in A. Here A’ is the complement of set A.
Thus $A\cap A'=\phi $ , which is a null set.
Now putting $A\cap A'=\phi $ in the above expression, we get
$\left( A\cap A' \right)\cap B'=\phi \cap B'$
The intersection of a null set with any set, results in a null set.
$\therefore \phi \cap B'=\phi $
Hence, we got, $A\cap \left( A\cup B \right)'=\phi $
Thus, we proved for any set A and B that $A\cap \left( A\cup B \right)'=\phi $

Note: We know the null set has no elements and consider $B'=\{2,3,5\}$ . Now the intersection of $\phi $ and B’ , will be \[\{\}\cap \{2,3,5\}.\] Both the set have no elements in common. Thus,
$\{\}\cap \{2,3,5\}=\{\}.$ Hence $\phi \cap B'=\phi ,$ a null set.