Basics of Set Theory with Venn Diagrams

Friday, April 5th, 2019


Basics of Set Theory with Venn Diagrams

Set Theory is a very important branch of mathematics that has wide ranging applications. It is a very basic and easy concept but has significant use in Algebra, Logic, Combinatorics, probability etc. Thus, for the same reason understanding of this topic is of grave importance to solve various questions in competitive exams such as Banking and MBA entrance exams.

A set is nothing but a collection of distinct objects. The objects could be anything numbers, letters, vectors, etc.  For ex, all the students in a class can be grouped into a set or integers such as 2, 3, 4 can form a set etc.

SET NOTATION

A set is generally denoted with a capital alphabet S or A or and its elements are generally enclosed in curly brackets like this

S = {2, 3 ,4}

Here, 2, 3 and 4 are elements of set S and an element is generally denoted using the symbol Є (belongs to) like this: 2 Є S.

When a set is very large and cannot be contained it can also be denoted in the following way:

A = {x: x is an integer} or,

B = {x: 0<x<10, x Є R}, where R is a set of real numbers

CARDINALITY OF A SET

Cardinality of a set means number of elements in any set. For ex, Consider the set above S = {2, 3 ,4}. Now the set contains 3 elements in it and thus its cardinality is 3. It is denoted by n(S) = 3 or |S| = 3.

TYPES OF SETS

EMPTY SET or NULL SET: A empty set is the one which does not contain any element in it.  The set is denoted by the symbol

 and n() = 0, as the set is empty.

  • SINGLETON SET: A set is known as singleton set contains only single element in it, For ex, A = {5}. It is also sometimes called as unit set. Since, it contains single element in it its cardinality is 1.
  • FINITE SET: A set is called finite set when if the elements in that set are finite. For ex, S = {1, 2, 3, 4, 5}. Here in this case the elements in set S are fixed and finite i.e. |S| = 5.
  • INFINITE SET: A set is called infinite set when its elements are not finite whether countable or Uncountable. For ex, A = {x: x Є N} is a countable infinite set whereas, B = {x: x Є R} is an uncountable infinite set.
  • SUBSET OF A SET: A set B is subset of set S, if the set B is contained in the set S i.e. all the elements of set B are present in set S. For ex, consider

B = {1,2} and S = {1, 2, 3, 4, 5}

Then. Both 1 and 2 are present in the set S. Hence, B is subset of S.

Note: A null set is subset of every set.

Every set is subset of itself.

It is denoted by symbol ’. Thus, if B is contained in S then it can be written as B ⊂ S.

  • EQUAL SETS: Two sets are called to be equal if they contain same elements. For ex,

A = {x: x<5, x Є Z+} and B = {1, 2, 3, 4}

Notice that both the sets are equal as A = {1, 2, 3, 4}

Hence, A = B

Also, notice that n(A) = n(B) in equal sets.

 

  • UNIVERSAL SET: A universal set is the set containing all the all objects including itself of which all other sets are subsets. It is generally denoted by letter ‘U’. A perfect example for universal set can be set of Real Numbers and the set of Natural numbers, Integers, Rational Numbers etc., are the subsets of it.
  • POWER SET: A power set is the set of all the subsets of a set say S, it contains both null set and S itself. It is denoted by the letter P in this way ‘P(S)’. For ex,

S = {1, 2, 3}, then

P(S) = {,{1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}

If a set is a finite set, then the number of subsets of that sets of that set is |P(S)| = 2n.

  • UNION OF SET: In set theory, union of collection of sets is a set containing every element from those sets i.e. if there are two sets A and B then their Union contain elements which are in A, in B, or both A and B. It is denoted by ‘’. If we have two set A and B then, their union is denoted as A B and is called A union B. Mathematically, A  B = {x: x Є A or x Є B}.  For ex,

A = {1, 2} and B = {2, 3, 4}, then

A  B = {1, 2, 3, 4}

 

  • INTERSECTION OF SETS: Intersection of collection of sets containing elements that are common to all the set i.e. if A and B are two sets then their intersection is the set containing elements that are commonly present on both of them. It is denoted by symbol ‘∩’. Mathematically, A ∩ B = {x: x Є A and x Є B}. For ex,

A = {1, 2} and B = {2, 3, 4}, then

A ∩ B = {2}

  • DIFFERENCE OF SETS: If A and B are two sets then their difference (A – B) is a set containing elements that are only present in A and not in B. Mathematically,

 A – B = {x: x Є A and x   B} and B – A = {x: x Є B and x   A}. For ex,

A = {1, 2} and B = {2, 3, 4}, then

A – B = {1}, and B – A = {3,4}

  • COMPLEMENT OF A SET: Let A be any set then Ac is the set containing all the elements except that in A i.e. AC = U – A.
  • DISJOINT SETS: A collections of sets are called Disjoint sets if their intersection is empty or  e. if A and B are two sets then A ∩ B = .

SOME IMPORTANT LAWS OF SET ALGEBRA

  • COMMUTATIVE LAWS
  1. A B = B  A
  2. A ∩ B = B ∩ A
  • ASSOCIATIVE LAWS
  1. (A B)  C = A  (B  C)    
  2. (A ∩ B) ∩ C = A ∩ (B ∩ C)
  • DISTRIBUTIVE LAWS
  1. A (B ∩ C) = (A  B) ∩ (A  C)
  2. A ∩ (B C) = (A ∩ B)  (A ∩ C)
  • IDENTITY LAWS
  1. A  = A
  2. A ∩ U = A
  • COMPLEMENT LAWS
  1. A Ac = U
  2. A ∩ Ac =
  • DE MORGAN’S LAW
  1. (A B)C = AC ∩ BC
  2. (A ∩ B)C = AC BC

Also, if A  B then,

  1. A ∩ B = A
  2. A B = B
  • A – B =
  1. Bc Ac

 

Since we have now covered all the important formulas and concepts related to sets except one thing i.e. Venn Diagrams, they are pictorial representation of sets and help in easily solving questions

VENN DIAGRAMS

A Venn diagram is a diagram that shows all possible logical relations between a finite collection of sets. They can be used to depict two sets, three sets, 4 sets or more. But for our purpose and level we only need their understanding up till four sets.

A two set Venn diagram is given below

set theory venn

The above diagram is called Venn Diagram where there are two sets A and B and U is the universal set.

Various Relationship between sets using Venn Diagrams

UNION OF SETS

INTERSECTION OF SETS

DISJOINT SETS

DIFFERENCE OF SETS

Also, notice that through the Venn diagram that A -B = A ∩ B’

COMPLEMENT OF A SET

Some basic formulas for Venn Diagram

n(A B) = n(A) + n(B) – n(A ∩ B)

n(A  B  C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

VENN DIAGRAM FOR 3 SETS

Let A, B and C be three sets and U be the universal set then the Venn diagram for 3 sets can be constructed as following

Solved Examples:

Example 1: In a group of 100 persons, 72 people can speak English and 43 can speak French. How many can speak English only? How many can speak French only and how many can speak both English and French?

Let A denote people who can speak English

Let B denote people who can speak French

Given that: n(A  B) = 100

To find:

  1. People who can speak English only i.e. only A
  2. People who can speak French only i.e. only B
  3. People who can speak both English and French only i.e. A ∩ B

Sol: Since, n(A B) = n(A) + n(B) – n(A ∩ B)

n(A ∩ B) = 72 + 43 – 100

= 15

Now only A can be written as = n(A) – n(A ∩ B)

= 72 – 15 = 57

Similarly, only B = n(B) – n(A ∩ B) = 43 -15 = 28

Example 2: Of the 200 candidates who were interviewed for a position at a call center, 100 had a two-wheeler, 70 had a credit card and 140 had a mobile phone. 40 of them had both, a two-wheeler and a credit card, 30 had both, a credit card and a mobile phone and 60 had both, a two wheeler and mobile phone and 10 had all three. How many candidates had none of the three?

Sol: Let A denote candidates who have 2-wheelers.

Let B denote candidates with credit card.

Let C denote candidates with mobile phones.

Now we need to find n(Ac∩ Bc∩ Cc)

This can be easily be calculated using D’morgans Law, i.e.

n (A  B  C)c =  n(Ac∩ Bc∩ Cc)

Since, n(A  B  C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

= 100 +70 +140 – 40 – 30 – 60 + 10 = 190

Total Candidates are 200, therefore,

n(A  B  C) + n (A  B  C)c =200

n (A  B  C)c = 10

 

Example 3: Study the following figure and answer the questions given below.

Find out:

1.) How many doctors are neither artist nor players?

2.) How many doctors are both players and artists?

3.) How many artists are players?

4.) How many players are neither artist nor doctors?

 

Sol: To solve such questions. Let’s construct the table like we did earlier in 3 set Venn diagram case corresponding to the value of the region. But first,

Let A be Artists

Let P be players

Let D be doctors

Now let’s see what we need to find out

1.) Doctors that are neither Artist nor Players => Only Doctors i.e. Only D = 17

2.) Doctors that are both Artist and Players => D ∩ A ∩ P = 3

3.) Artist that are players => (A ∩ P) + (D ∩ A ∩ P) = 22 +3 = 25

4.) Players that are neither doctors nor artist => Only Players = 25

There can be many questions formed on these kinds of sets. But you can easily solve them using little manipulation and basic addition and subtraction. With a little practice, you can easily and quickly solve questions of sets either using Venn diagrams.

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CAT Questions related to Quantitative Aptitude – Modern Maths

All questions from CAT Exam Quantitative Aptitude – Modern Maths
Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
Quantitative Aptitude – Modern Maths – Progressions – Q4: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q5: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?
Quantitative Aptitude – Modern Maths – P&C – Q5

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