Table of Contents
Last modified on July 12th, 2024
A set represents a collection of distinct objects considered as an entity.
Venn Diagrams are visual representations of distinct mathematical sets and their relationships. They are also widely used in other fields to illustrate the logical relationships between sets.
A Venn diagram consists of overlapping and non-overlapping circles (or sometimes other closed shapes like squares) representing the relation between sets. The overlapping parts denote the elements common to the sets, and the non-overlapping part contains elements distinct to a particular set.
If A = {odd natural numbers from 1 to 20} and B = {multiples of 3 between 1 and 20}
Representing them on a Venn diagram, we get
In the above Venn diagram, the rectangular box represents the universal set, which has two overlapping circles representing sets A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} and B = {3, 6, 9, 12, 15, 18}
The overlapping part of the circles contains elements common to both sets, A ∩ B = {3, 9, 15}
In a Venn diagram, a subset is depicted by drawing a circle or another shape inside the larger circle or shape that represents the larger set.
If A = {a, c, d, g, h} and B = {a, b, c, d, e, f, g, h}, A is a subset of B, and B is the bigger set.
Constructing a Venn diagram with the above sets, we get:
Here is the formula used to solve problems involving two sets:
n(A ∪ B) = n(A) + n(B) – n (A ∩ B)
Similar to representing 2 sets, a Venn diagram can also be constructed involving three sets.
Here,
Here is the formula used to solve problems involving three sets:
n (A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
Venn diagrams can have unlimited circles.
If A = {1, 2, 3, 4, 8, 12, 13}, B = {2, 3, 4, 5, 6, 7, 14}, C = {4, 7, 8, 9, 11}, and D = {2, 6, 10, 12}, the Venn diagram can be formed by involving these four sets, as shown.
To draw a Venn diagram, we start with the universal set U, which contains all other sets as subsets, and then draw all the circles, where each circle represents a different set. Thus, every set will be within the universal set. If some elements are common to both sets, we place them within the intersecting circles.
Let us construct a Venn diagram to represent sets A = {1, 2, 3, 4, 5}, B = {5, 6, 7, 8, 9}, C = {2, 4, 6, 8}, and U = {set of natural numbers}.
First, the rectangle is formed with the universal set, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then three overlapping circles are drawn within it representing sets A, B, and C.
Here,
A ∩ B = {5}
B ∩ C = {6, 8}
A ∩ C = {2, 4}
Constructing a Venn diagram with the above sets, we get:
The various set operations are represented in the Venn diagram as follows:
If A and B are two sets, A union B, denoted by A ∪ B, represents all the combined elements of both sets.
A ∪ B = {x | x ∈ A or x ∈ B}
In the Venn diagram, A ∪ B is shown by the shaded region.
[Upload the diagram with the filename: Union of Sets]
If A and B are two sets, A intersection B, denoted by A ∩ B, represents the common elements of both sets.
A ∩ B = {x | x ∈ A and x ∈ B}
In the Venn diagram, A ∩ B is shown by the shaded region.
[Upload the diagram with the filename: Intersection of Sets]
The complement of a set contains all elements that are in the original set. If A is a set, its complement is denoted by A’ or Ac.
A’ = {x | x ∈ U and x ∉ A}
In the Venn diagram, A’ or Ac is shown by the shaded region.
[Upload the diagram with the filename: Complement of a Set]
Complement of Union (A ∪ B)’
The complement of A union B is a set representing elements neither in set A nor in set B.
Complement of Intersection (A ∩ B)’
The complement of A intersection B is a set representing elements of the universal set that are not common between sets A and B.
The difference between sets A and B is a set representing elements of set A but not in set B.
[Upload the diagram with the filename: Difference in Sets]
The symmetric difference of sets A and B is a set consisting of elements in either sets A or B but not in both.
Here is a summary of the commonly used set operations with their Venn diagrams:
In a class of 70 students, 15 play football, 25 play cricket, and 8 play both. Find the number of students who don’t play either football or cricket.
Here, n(F) = 15
n(C) = 25
n(F ∩ C) = 8
Now, n(F ∪ C) = n(F) + n(C) – n(F ∩ C) = 15 + 25 – 8 = 32
Thus, the number of students who don’t play either football or cricket = 70 – n(F ∪ C) = 70 – 32 = 38.
Out of 150 students in a school, 63% passed in Mathematics, 52% in Science, and 47% in both. What is the percentage of students who only passed in Science?
Here, n(M) = 63
n(S) = 52
n(M ∩ S) = 47
Thus, the percentage of students who only pass in Science = n(S) – n(M ∩ S) = 52 – 47 = 5%
Last modified on July 12th, 2024