Question

het u= {1,2,3,4,5,6,7, 8, 9, 107 A={2, 4, 6, 8} B = {2,3,5,6,7} Verify (AUB)' A' intersection B'​

Answer 1

Answer:

Step-by-step explanation:

Using set theory and given sets, find the values of A′,C′,(A∪B)′

and (A∩B)′

. Substitute these values in the expression to verify and prove that LHS = RHS.

Complete step-by-step answer:

We have been given the Universal set, i.e. it contains all the elements from the universal set.

i.e. U = {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Thus we have A = {2, 4, 6, 8} and B = {2, 3, 5, 7}.

We can find (A∪B)

which is A union B, and it will contain all elements of A and B.

(A∪B)

= {2, 3, 4, 5, 6, 7, 8}.

(A∩B)

, which is A intersection B, it will contain only the common elements of A and B.

(A∩B)

= {2}.

A’ represents the complement of A, which contains the elements of the universal set that is not in set A.

∴

A’ = {1, 3, 5, 7, 9}

Similarly, B’ = {1, 4, 6, 8, 9}

Now let us verify the first case.

(i) (A∪B)′=A′∩B′

We got (A∪B)

= {2, 3, 4, 5, 6, 7, 8}.

∴

(A∪B)′={1,9}

We got A’ = {1, 3, 5, 7, 9} and B’ = {1, 4, 6, 8, 9}.

∴A′∩B′={1,9}

Thus we got, (A∪B)′=A′∩B′

.

Thus LHS = RHS.

Now let us take the second case where we need to prove that, (A∩B)′=A′∪B′

.

We got, A∩B={2}

.

∴(A∩B)′={1,3,4,5,6,7,8,9}

.

We got A’ = {1, 3, 5, 7, 9} and B’ = {1, 4, 6, 8, 9}.

∴A′∪B′={1,3,4,5,6,7,8,9}

Thus we got that, (A∩B)′=A′∪B′

.

i.e. LHS = RHS.

∴

We had verified both the cases.

Note: A’ is the complement of A, which can also be denoted as AC

. Be careful when you put symbols of union (∪

) and intersection (∩

). Don’t mix up symbols and you may get wrong answers, while writing the set be careful to mention all the elements.

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