Question

Let U = {3, 6, 9, 12, 5, 18, 21, 24} be the universal set and Let A = {6, 12, 18, 24} be its subset.

Verify that:
(i) A Ո A’ = ϕ
(ii) A U A’ = U​

Answer 1

Given :-

• Let U = {3, 6, 9, 12, 5, 18, 21, 24} be the universal set and Let A = {6, 12, 18, 24} be its subset.

To find :-

• A Ո A’ = ϕ
• A U A’ = U

Solution :-

• Empty set

→ A set which does not contain any element is known as empty set

→ It is also called null or void set

→ It is denoted by ϕ or { }

• Complement set

→ The complement of a subset of universal set U is the set of all elements of its which are not the element of A

• Union of sets

→ The union of A and B is the set of all those element which belong either A or to B or to both A and B is known as union of sets

→ It is denoted by " A U B " (read as A union B)

• Intersection of sets

→ The intersections of A and B is a set of all those elements that belongs to both A and B is known as intersection of sets

→ It is denoted by " A Ո B " (read as A intersection B)

It is given that

• U = {3, 6, 9, 12, 5, 18, 21, 24}

• A = {6, 12, 18, 24}

According to first question

→ A Ո A’ = ϕ

• LHS = A Ո A’

→ A' = {3, 5, 9, 21}

→ A = {6, 12, 18, 24}

• A Ո A’ = ϕ = RHS Verified
• Note : It has no any common element

According to second question

→ A U A’ = U

• LHS = A U A’

→ A' = {3, 5, 9, 21}

→ A = {6, 12, 18, 24}

• A U A’ = {3, 5, 6, 9, 12, 18, 21, 24}
• A U A’ = U = RHS Verified

• Note : {3, 5, 6, 9, 18, 21, 24} is equal to universal set i.e U