- INTRODUCTION
- REPRESENTATION OF A SET
- SUBSETS AND SUPERSETS
- COMPLEMENT OF A SET
- UNION AND INTERSECTION OF SETS
- VENN DIAGRAMS
- PROPERTIES OF SETS
- CARTESIAN PRODUCT
Sets
We have all come across the word ‘set’ in our day-to-day lives. For example, “This is a set of the most creative students in class.” or “This is the most beautiful set of diamonds I have ever seen.” Hence, a set basically implies a group of certain similar objects.
Mathematically, a set can be defined as a group of certain similar and well defined objects. These objects are called the elements of the set.
For example, consider the set of vowels in the English alphabet. Let this set be represented by the letter V. This set contains 5 elements – a, e, i, o, u. This can also be written as:
V 
{a, e, i, o, u}
The following two conditions should be taken care of while writing a set:
A set must be denoted by a capital letter.
The elements of the set must be denoted in small letters.
Following are certain examples of sets:
A
{1, 2, 3, 4, 5}
B
{a, b, c, d, e}
In the above examples, A and B are sets while 1, 2, 3, 4, 5 and a, b, c, d, e are the elements belonging to these sets respectively.
The symbol ϵ is used to denote that an element belongs to a set. So, if we wish to write that 1 belongs to set A, we write it as follows:
1 ϵ A
ϵ is a Greek letter called epsilon.
Certain sets are used very frequently, like the set of all natural numbers. Following are certain standard letters used to denote commonly used sets:
N
Set of all natural numbers
{1, 2, 3, 4, …}
W
Set of all whole numbers
{0, 1, 2, 3, …}
I
Set of all integers
{…, 
3,
2,
1,
0, 1, 2, 3, …}
Q
Set of all rational numbers
R
Set of all real numbers
P
Set of all prime numbers
C
Set of all complex numbers
Cardinality of a set refers to the total number of elements that are present in the set.
For example, V
{a, e, i, o, u}
The cardinality of this set of vowels V is 5 as the total number of elements in the set is 5.
This can be represented as:
|V| = 5 or n(V) = 5
Based on the cardinality of sets, sets can be classified into the following types.
Null Set/ Void Set/ Empty Set
Consider the set B
{ }
This set has no elements. Hence, the cardinality of this set is zero.
This can be represented as:
|B| = 0 or n(B) = 0
Such a set is called a null set or an empty set. A null set can be represented as:
Infinite Set
Certain sets have infinite elements. Such sets are called infinite sets.
For example, the set of natural numbers N, set of integers I, etc.
Finite Set
Sets which have a finite number of elements are called finite sets.
For example, the set of vowels
Singleton Set/Singlet
A set which has one and only one element in it is called a singleton set.
For example,
1. A
{a}
2. B
{0}
In the above cases, A and B are both singleton sets.
REMEMBER:
- In the second example, B is not an empty set. It has the element 0 in it.
- B would have been an empty set if it did not
contain any elements, i.e. B
{ }.
Equivalent Sets
Two finite sets are said to be equivalent if their cardinality is the same,
i.e. n(A) = n(B)
For example, if A
{1, 2, 3} and B
{4, 5, 6}
Then, n(A) = n(B) = 3
Hence, A and B are equivalent sets.
Equal Sets
Two sets are said to be equal if and only if they contain the same elements.
This means that every element of set B is present in set A and every element of set A is present in set B.
For example, if A
{a, b, c} and B
{c, b, a}, then A and B are said to be equal sets.
Equality is denoted as A = B. Note that the order in which the elements of a set are written is not important.
There are two ways of representing a set. They are
- Roster notation
- Set Builder notation
Let us consider the set of vowels:
V
{a, e, i, o, u}
Here, each vowel is individually listed down in the set V.
This method in which each element of a set is specified within the set is called the Roster method.
REMEMBER:
- To denote an infinite set using the Roster method, like the set of natural numbers, we write the first few elements followed by ‘…’ to indicate continuation up to infinity.
- N
{1, 2, 3, …}
In the Set Builder notation, instead of writing down every single element of a set, we describe the common property of the elements of the set.
For example, in the Roster notation the set of the first five natural numbers is written as follows:
D
{1, 2, 3, 4, 5}
Now, if we were to describe this set in words, we would say, “Set D is a set of all elements x such that x is a natural number and x lies between 1 and 5 (inclusive).”
This is written in Set Builder notation as follows:
D
{x | x ( N and 1 
x
5}
REMEMBER:
- The symbol | is read as “such that”.
- Let us take another example,
- V
{a, e, i, o, u}- Here, we can say, “the set V is a set of
elements x, such that x is a vowel from the English
alphabet.”
- This can be written in Set Builder notation
as:
- V
{x | x is a vowel from the English alphabet}
- Hence, in general, if all the elements of the set
A have a common property B, then it can be expressed in the set
builder form in the following way:
- A
{x | x has property B}
|
Convert the following sets from the Set Builder notation to the Roster notation:
i) P ii) M
Solution: i) Here, P is a set of all numbers y, such that y is a whole number less than 3. The whole numbers less than 3 are 0, 1 and 2. Hence, set P can be written in Roster notation as:
P
ii) Here, M is a set of all such numbers x where x2 – 5x + 6 = 0. To find the values of x, we need to solve this equation.
x2 – 5x + 6 = 0 x2 – 3x – 2x + 6 = 0 x(x – 3) – 2(x – 3) = 0 (x – 3)(x – 2) = 0 x = 3 or x = 2
Hence, we can write set M in the Roster form as: M |
Let us consider the set of all students in a college. Now, the set of first year students is a part of this bigger set of all students. Such a set, all of whose elements are present in a bigger set, is called a subset of the bigger set. The bigger set is called the superset of the smaller set.
Let us consider the following example:
A
{1, 2, 3, 4, 5, 6} and
B
{1, 2, 3}
We can see that every element of the set B is an element of the set A.
Hence, we can say that B is a subset of A. We can also say that B is contained in A.
A is said to be the superset of B.
REMEMBER:
- A Null Set is a subset of every set.
- Every set is a subset of itself.
Definition of a Subset
We say that A is a subset of B if and only if every element of A is an element of B.
This is denoted as follows:
A 
B
For example,
A
{1, 2}, B
{1, 2}
A
{1, 2, 3}, B
{1, 2, 3, 4}
In both these examples, A
B.
In the first example, B is also a subset of A. Hence,
A B
and B
A.
Set B = Set A
Such a subset is called an improper subset.
Subset A is an improper subset of B if A = B.
In the second example, B is not a
subset of A, which can be denoted by B 
A.
Such a subset, where the superset also contains other elements, is called a proper subset.
Set A is said to be a proper subset
of B if A
B but B
A.
A proper subset is denoted by the symbol
.
Thus, in the second example, A
B.
A power set is the set of all subsets of a set.
Consider a set A
{1, 2, 3}
Now, consider a set {1}. This set is a subset of A. Similarly, {1, 2} is also a subset of A. So, let us try and list down all the subsets of A. They would be:
{1}
{2}
{3}
{1, 2}
{1, 3}
{2, 3}
{1, 2, 3} … (Every set is a subset of itself)
{ } … (Null set is a subset of every set)
Total number of subsets = 8
Hence, a set of these subsets is called the power set of A.
P
{{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, { }}
If the number of elements (cardinality) in a set is n, then the cardinality of the power set is given by 2n.
Explanation:
To form a subset B of a set A containing n elements, each element of A can either be chosen or not chosen to be a part of B. Thus, the first, second, third…, nth elements of A can either be chosen or not chosen in 2 ways each.
The total number of ways in which a subset can be formed = 2 
2
2 … n times = 2n
The power set of a set containing n elements, has 2n
elements.
For example, in the above case, the cardinality of set A, n(A) = 3
So, the cardinality of power set P, n(P) = 23 = 8
Let us consider another example,
If B
{a, b}, then n(B) = 2
( Cardinality of the power set = 22 = 4
Verifying the above statement, power set
P
{{ }, {a}, {b}, {a, b}}
Let us consider a college, consisting of students from three streams—Arts (set A), Commerce (set C) and Science (set S). All these students together form the set of all the students in the entire college. Such a set is called a universal set (set U).
A universal set is a set containing all the elements under consideration. It is represented by the capital letter U.
The universal set for the set of all numbers is as follows:
U
{∞
… ∞}. It includes all natural numbers, integers, rational numbers, irrational
numbers, etc.
Once we understand the idea of a universal set, it becomes easy for us to understand what the complement of a set is.
The complement of a set A, which is a subset of a universal set U, includes all the elements present in U which are not present in A. It is denoted by A’.
Suppose,
U
{a, b, c, d, e, f, g, h, i} and
A
{a, b, c, d, e},
Then A’, or complement of the set
A
{f, g, h, i}
Let us consider the following example. In a room, there are 5 people a, b, c, d, e. Out of them, a, b and c are men while d and e are women. Also, a and e study science while b, c and d study commerce.
The set of males is:
M
{a, b, c}
The set of females is:
F
{d, e}
The set of science students is
S
{a, e}
The set of commerce students is
C
{b, c, d}
If we wish to find out all female students
who have taken science, we need to find out what is common in set F and
set S. This is called an intersection of set F and set
S and is denoted by F 
S.
Here, F
S
{e}
Thus, an intersection of two sets is formed by the elements which are common to both the sets.
Similarly, if we consider sets M and F, there is no common element between them.
Hence, M
F = ∅
Such sets which have no element in common are called disjoint sets.
Now, let us find out those females who have not taken science. Here, we have to check the set F and remove all elements of set S present in this set. This is called the difference between two sets.
F
S
{d}
Thus, difference of set A and set B is defined as the elements present in A but not present in B.
A
B
{x | x ( A and x ∉ B}
Now, suppose you want to represent a set
containing “either males or commerce students or both”. This would mean taking
all the elements from set M and set C together into one set. This
is called the union of set M and set C and is denoted by
M 
C.
Thus, M
C
{a, b, c, d}
IMPORTANT:
Though b and c exist in both sets, they are written only once while writing the union. This is because no element is ever written twice while writing a set.
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Consider the set S = {2, 3, 4, ..., 2n + 1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y? [CAT 2007]
(1) 0(2) 1 (3) n/2(4) n + 1/2n (5) 2008
Solution: Y = (2 + 4 + 6 + 8 + … + 2n)/n
X = (3 + 5 + 7 + 9 + … + (2n + 1))/n
= [(2 + 1) + (4 + 1) + (6 + 1) + (8 + 1) + … + (2n + 1)]/n
= (2 + 4 + 6 + 8 + … + 2n)/n + (1 + 1 + 1 + 1 + … n times)/n
= Y + 1
Hence, option 2.
Note: The information that 'n is a positive integer larger than 2007' does not affect the answer in any way. |
|
Let the set consisting of the squares of the positive integers be called u; thus u is the set 1, 4, 9, ..... If a certain operation on one or more members of the set always yields a member of the set, we say that the set is closed under that operation. Then u is closed under : [FMS 2010]
(1) addition (2) multiplication (3) division (4) none of these
Solution: Let x and y be any two elements of set u.
x = a2 and y = b2 for some positive integers a and b
Hence, option 2. |
Every element of set u is a perfect square,
Sets can be represented using diagrams known as Venn Diagrams. Sets can be represented in the form of circles, triangles, rectangles, etc.
Let us take an example to understand Venn diagrams.
Let U be the universal set containing all the natural numbers between (and not including) 0 and 11.
Hence, U
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Let P be the set containing all the prime numbers between 0 and 11.
Thus, P
{2, 3, 5, 7}
Let E be the set containing all the even numbers between 0 and 11.
Hence, E
{2, 4, 6, 8, 10}
Hence, P
E
{2, 3, 4, 5, 6, 7, 8, 10}
Also, P
E
{2}
This can be represented using a Venn diagram in the following manner.
In the above diagram, the universal set is represented by the rectangle while all other sets are represented by circles. The common portion between the two sets P and E is the intersection of these two sets.
The universal set contains two numbers, 1 and 9, which do not belong to the set of prime or even numbers. Hence, they are written outside both circles but within the universal set.
Now, let us look at the cardinality of these sets.
n(P) = 4, n(E) =
5 and n(P
E) = 8
Thus if we observe, n(P
E) ≠ n(P) + n(E)
This is because the element 2 is included in both P and E but while taking the union of the two sets, we need to count it only once.
Therefore, if we subtract n(P
E) from n(P) + n(E), then we will get
n(P
E).
n(P
E) = n(P) + n(E) – n(P
E)
i.e. n(P
E) = 4 + 5 – 1 = 8
To generalise this, for any two sets A and B,
n(A
B) = n(A) + n(B)
n(A
B)
|
In a class of 100 students, 60 play football, 40 play hockey while 10 do not play any sport. Find the number of students who play both hockey and football.
Solution: n(U) = 100
Since 10 do not play anything, n(F
Hence, n(F
|
Similarly, now let us consider a universal set as follows:
U
{1, 2, 3, 4, 5, 6, 7, 8, 9, 15}
Let A be the set of all even numbers.
Hence, A
{2, 4, 6, 8}
Let B be the set of all perfect squares.
Hence, B
{1, 4, 9}
Let C be the set of all composite numbers.
Hence, C
{4, 6, 8, 9, 15}
This can be represented using a Venn diagram as follows:
For a three level diagram, the formula is
n(A
B
C) = n(A) + n(B) + n(C)
n(A
B)
n(A
C) – n(B
C) + n(A
B
C)
So in this case,
n(A
B
C) = 4 + 3 + 5
1
3
2 + 1 = 7
|
Draw Venn diagrams for: A A A
Solution: i) A
ii) A
iii) A
|
|
Express the following as Venn diagrams:
Solution: Let A = Set of Animals, B = Set of bulls, S = Set of things in the stock market
Here, A and B are disjoint sets. |
Let us now look at some properties of sets:
- Union of any set with a null set is equal to that set.
- Intersection of any set with a null set is equal to a null set.
- The intersection of any set with its complement is a null set.
- The union of any set with its complement is the universal set.
A
A’ = U
- (A
B)’ = A’
B’
- (A
B)’ = A’
B’
The last two properties are called De Morgan’s laws.
EXPLANATION:
We can prove De Morgan’s laws using Venn Diagrams.
Law 1: (A
B)’ = A’
B’
In the above diagram,
(A
B) = a + b + c
Hence,
(A
B)’ = U
(A
B) = a + b + c + d
(a + b + c) = d
(A
B)’ = d
A’ = c + d
B’ = a + d
A’
B’ = d
(A
B)’ = A’
B’
Law 2: (A
B)’ = A’
B’
(A
B) = b
(A
B)’ = a + c + d
A’ = c + d
B’ = a + d
A’
B’ = a + c + d
(A
B)’ = A’
B’
- Associative Property
(A
B)
C = A
(B
C)
(A
B)
C = A
(B
C)
- Distributive Property
A
(B
C) = (A
B)
(A
C)
A
(B
C) = (A
B)
(A
C)
An ordered pair (x, y) represents two objects x and y taken in that order. Thus (x, y) is different from (y, x).
The Cartesian product of any two sets A and B is given by
A
B
{(x, y) | x ( A and y ( B}
For example, consider two sets A and B as follows:
A
{1, 2, 3}
B
{a, b, c}
Then the Cartesian product of these sets is given by
A
B
{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3,
c)}
Note that A
B is different from B
A
|
Find the Cartesian product A A
Solution: A |
|
A teacher tested 80 students of class V for their abilities in four areas - sports, academics, music and art. She found that all her students had at least one of these four abilities strongly developed. 15, 24 and 45 students respectively were good at music, art and sports.
How many students were good only in academics?
What was the difference between the number of students who were not good at music but good at the other three activities and the number of students who were good only at sports and music?
Solution: Here we need a four level Venn diagram. From statements (i) and (ii), there was 1 student who showed only musical and artistic abilities. From statements (iii), (iv) and (v), (30 – 9 – 3 – 8) = 10 students were good only in academics.
All this information can be presented in the following Venn diagram.
From (iv), x + y + z + 3 = 9
Also, m + y + p = 3 and n + z = 6
Hence, 10 students were good only at academics.
z represents the number of students who were not good at music but good at the other three activities and m represents the number of students who were good only at sports and music. The difference between z and m is zero. |
|
ABC coaching classes conducted a study in a locality to find out the number of students of class eighth who took tuitions for English, Maths and Science. 36 students took Maths tuitions, out of which 20 did not take tuitions for any other subject. 27 students took English tuitions. The number of students who took only English tuitions and the number of students who took English as well as Science tuitions but not Maths tuitions was equal. Also, an equal number of students took tuitions for Maths as well as English and Maths as well as Science.
Find the maximum possible number of students who took only English tuitions, the maximum possible number of students who took tuitions for all the three subjects and the minimum possible number of students who took tuitions for Math and English but not Science.
Solution:
From the figure,
2y + z + 20 = 36
2x + y + z = 27 …(ii)
When y = 3, x = 7
When y = 5, x = 8
When y = 7, x = 9
When y = 9, x = 10
But from (ii), 2x + y
From (ii), z is maximum when x and y are minimum.
= 27 – 1 – 12 = 14
|
Number theory questions often appear disguised as Set Theory questions. Also, number theory questions can be solved easily using concepts in set theory.
|
Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements of S. With how many consecutive zeros will the product end? [CAT 2000]
(1) 1(2) 4 (3) 5(4) 10
Solution: There is only one even prime number and one prime number ending with 5 (i.e 5 itself) in set S.
When all elements of set S are multiplied, there will be only 1 zero at the end of the product. Hence, option 1. |
|
How many numbers lying in the range 5000 to 10000 are divisible by 7, 11 or 13?
Solution: Let A, B and C be the set of numbers divisible by 7, 11 and 13 respectively.
Lowest number in A = 5005. Highest number in A = 9996
Lowest number in B = 5005. Highest number in B = 9999
Lowest number in C = 5005. Highest number in C = 9997
Similarly,
Number of elements divisible by 7 and 11 = Number of elements in the set A
Number of elements in the set A
Number of elements in the set B
Number of elements in the set A
n(A
|
= 714
|
A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of the three popular options- air conditioning, radio and power windows- were already installed. The survey found
What is the number of cars that had none of the options? [CAT 2003 Re-Test]
(1) 4(2) 3 (3) 1(4) 2
Solution:
Here, AC – Air Conditioning, R – Radio and PW – Power Windows
From the given conditions, we have the above Venn diagram.
When we add up all the values in the Venn diagram, we get (15 + 5 + 1 + 2) = 23 cars.
Hence, option 4. |
|
Three of a popular actor’s films, A, B and C, were nominated for the best film award. 60 persons were asked which film deserved the award the most.
16 people thought only C deserved it.
The number of people who thought that all the three films equally deserved the award was twice the number of people who thought that B and C deserved the award but A did not. The number of people who thought that B deserved the award was equal to twice the number of people who thought all the three equally deserved it.
The number of people who thought that A and B, but not C deserved the award was equal to the number of people who thought that only A or A and C but not B deserved it.
Find the maximum number of people who thought that all the three films deserved the award?
Solution: Let x be the number of people who thought that B and C deserved the award but A did not.
Then the number of people who thought all the three equally deserved it is 2x.
Let m be the number of people who thought that A and B, but not C deserved the award.
The number of people who thought that B deserved the award = 4x
The number of people who thought that only B
deserved the award = 4x
From the given data, we can draw the following Venn diagram.
Thus, 2m + x – m + 2x + x + 16 = 60
|
REMEMBER:
- For any two sets A and B,
- n(A
B) = n(A) + n(B) – n(A
B)
- For any three sets A, B and C,
- n(A
B
C) = n(A) + n(B) + n(C)
n(A
B) – n(A
C)
n(B
C) + n(A
B
C)
- De Morgan’s Laws
- (A
B)’ = A’
B’- (A
B)’ = A’
B’
