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Set theoretic operations using Venn diagrams

To represent set theoretic operations using Venn diagrams

A set is a collection of well defined objects or groups of objects.

These objects are often called elements or members of a set. For example, a group of players in a cricket team is a set.

__The symbols used while representing the operations of sets__

- Union of sets symbol: ∪
- Intersection of sets symbol: ∩
- Complement of set: A’ or A
^{c} - Subset of set: ⊂

- Union of sets : A∪B
- Intersection of Sets : A∩B
- Complement of Sets : A’
- Difference of Sets : A-B

Most of the relationships between sets can be represented by means of diagrams which are known as Venn diagrams.

These diagrams consist of rectangles and closed curves usually circles. The universal set is represented by rectangle and its subsets by circles.

A Venn diagram is a diagram that helps us visualize the logical relationship between sets and their elements and helps us solve examples based on these sets.

__ A ∪ B __: The union of two sets A and B is the set C which consists of all those elements which are either in A or in B (including those which are in both).

Symbolically we write A ∪ B = {x : x ∈A or x ∈B}

The union of two sets can be represented by a Venn diagram as shown in following figure.

The shaded portion in figure represents A ∪ B.

Let A and B be any two sets. The union of A and B is the set which consists of all the elements of A and all the elements of B, the common elements being taken only once.

The symbol ‘∪’ is used to denote the union.

Symbolically, we write A ∪ B and usually read as ‘A union B’.

__Example __: Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A ∪ B.

__Solution__ : We have A ∪ B = {2, 4, 6, 8, 10, 12}

**A ∩ B **The intersection of two sets A and B is the set of all those elements which belong to both A and B.

Symbolically, we write A ∩ B = {x : x ∈ A and x ∈ B}.

The shaded portion in Figure indicates the intersection of A and B.

__Example:__ Consider the sets A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A ∩ B.

__Solution: __We see that 6, 8 are the only elements which are common to both A and B.

A ∩ B = { 6, 8 }

The required formula can be written in any of the following forms: A Union B Complement is equal to the intersection of the complements of the two sets A and B.

__Example:__ Determine the elements of A union B complement if U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, and B = {1, 3, 5}

__Solution:__ We have A = {2, 4, 6}, and B = {1, 3, 5}, then A U B is given by, A U B = {1, 2, 3, 4, 5, 6}, then A union B complement is given by, (A ∪ B)' = U - (A ∪ B) = {1, 2, 3, 4, 5, 6, 7} - {1, 2, 3, 4, 5, 6} = {7}

Answer: (A ∪ B)' = {7}

Mathematically, it is written as (A ∩ B)' = A' U B'.

__Example__: Consider U = {1, 2, 3, 4, 5, 6, 7, 8, 9} and A' = {2, 3, 5} and B' = {1, 2, 3, 4, 5}. Find the elements in A Intersection B Complement.

__Solution:__ We know that (A ∩ B)' = A' U B' Now, A' U B' = {2, 3, 5} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5} ⇒ (A ∩ B)' = {1, 2, 3, 4, 5}

Answer:The elements in A Intersection B Complement are {1, 2, 3, 4, 5}.

Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A.

Symbolically, we write A′ to denote the complement of A with respect to U.

Thus, A′ = {x : x ∈ U and x ∉ A}.

Obviously A′ = U – A. We note that the complement of a set A can be looked upon, alternatively, as the difference between a universal set U and the set A.

Let U be the universal set which consists of all prime numbers and A be the subset of U which consists of all those prime numbers that are not divisors of 42.

Thus, A = {x : x ∈ U and x is not a divisor of 42}. We see that 2 ∈ U but 2 ∉ A, because 2 is divisor of 42.

Similarly, 3 ∈ U but 3 ∉ A, and 7 ∈ U but 7 ∉ A. Now 2, 3 and 7 are the only elements of U which do not belong to A.

The set of these three prime numbers, i.e., the set {2, 3, 7} is called the Complement of A with respect to U, and is denoted by A′. So we have A′ = {2, 3, 7}.

__Example__: Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.

__Solution:__ We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A. Answer : A′ = {2, 4, 6, 8,10}