Practical Problems on Union and Intersection of Two Sets: A Swiss mathematician, Euler, gave an idea to represent a set by the points in a closed curve. Hence, the diagrams drawn to represent sets are called Venn-Euler diagrams or simply Venn diagrams. In such diagrams, the universal set \( \cup \) is represented by points within a rectangle, and circles within the rectangle represent subsets.

Similar to constants and variables, various operations can be performed on sets too. To solve practical problems on union and intersection of sets, we need to define the union, intersection and difference of sets. In this article, let us discuss the practical problems on the union and intersection of two sets with the help of Venn diagrams.

Learn About Sets Here

Operations on Sets

This section will introduce some operations on sets to construct new sets from the given sets.

Union of Sets

Let \(A\) and \(B\) be any two sets. Then the union of \(A\) and \(B\) is the set of elements that belong to \(A\) or\(B\) or both \(A\) and \(B\).

We use the notation of \(A \cup \,B\) (read as “\(A\) union \(B\)”) to denote the union of \(A\) and \(B\). This can be represented as
\(A\, \cup \,B\, = \left\{ {x:x \in A\,{\text{or}}\,x \in B} \right\}\)

Note that \(x \in \,A\, \cup \,B \Leftrightarrow \,x \in A\) or \(x \in B\) and \(x \notin A\, \cup B \Leftrightarrow \,x\, \notin A\) and \(x \notin B\)

The shaded part in the Venn diagram given below represents the \(A \cup \,B\).

It is evident from the definition that,
\(A \subset \,A \cup B\)
\(B \subset \,A \cup B\)

If \(A\) and \(B\) are any two sets, such that \(A \subset B\) then \(A\, \cup \,B\, = B\). Also \(A\, \cup \,B\, = A\) if \(B \subset A\)

Note: If \({A_1},\,{A_2},\,…..,\,{A_n}\) are finite family of sets, then their union is represented by
\(\bigcup\limits_{i = 1}^n {{A_i}} \)
or
\({A_1}\, \cup {A_2}\, \cup \,{A_3}\,…..\, \cup \,{A_n}\)

Intersection of Sets

Let \(A\) and \(B\) be any two sets. The intersection of \(A\) and \(B\) is the set of elements that belong to both \(A\) and \(B\).

We use the notation of \(A \cap B\) (read as, “\(A\) intersection \(B\)”) to denote the intersection of \(A\) and \(B\). It is represented as,
\(A \cap B = \left\{ {x:x \in A\,{\text{and}}\,x \in B} \right\}\)
Hence \(x \in A \cap B \Leftrightarrow \,x \in A\,{\text{and}}\,x \in B\)

The shaded part in the Venn diagram given below represents \(A \cap B\).

Evidently,
\(A \cap B \subset A\)
\(A \cap B \subset B\)
Then \(A \cap B = A\), if \(A \subset B\) and \(A \cap B = B\) if \(B \subset A\) .

Note: If \({A_1},\,{A_2},\,…..,\,{A_n}\) are finite family of sets, then their intersection is represented by
\( \cap _{i = 1}^n\,{A_i}\)
or
\({A_1}\, \cap \,{A_{2\,}} \cap \,….. \cap \,{A_n}\)

Disjoint Sets

If two sets \(A\) and \(B\) are said to be disjoint, then \(A \cap B = \phi \)

Difference of Sets

Let \(A\) and \(B\) be two sets. The difference of \(A\) and \(B\) written as \(A – B\), is the set of all those elements of \(A\) which do not belong to \(B\).
Thus,
\(A – B = \left\{ {x:x \in A\,{\text{and}}\,x \notin B} \right\}\)
or
\(A – B = \left\{ {x \in A\,:\,x \notin B} \right\}\)
Here, \(x \in A – B \Leftrightarrow x \in A\) and \(x \notin B\)

The shaded portion in the Venn diagram of \(A – B\) is represented as shown.

Similarly, the difference \(B-A\) is the set of all those elements of \(B\) which do not belong to \(A\)
Thus,
\(B – A = \left\{ {x \in B\,:\,x \notin A} \right\}\)

The shaded portion in the Venn diagram represents \(B-A\) .

Symmetric Difference of Two Sets

Let \(A\) and \(B\) be two sets. The symmetric difference of sets \(A\) and \(B\) is the set \(\left( {A – B} \right)\, \cup \,\left( {B – A} \right)\) and it is denoted by \(A\,\vartriangle \,B\).
Thus,
\(A\,\vartriangle \,B\, = \,\left( {A – B} \right)\, \cup \,\left( {B – A} \right)\, = \,\left\{ {x:x \notin A\, \cap B} \right\}\)

The shaded part represents \(A\,\vartriangle \,B\) in the Venn diagram.

Complement of a Set

Let \( \cup \) be the universal set and \(A\) be any set, such that \(A \subseteq  \cup \), then the complement of \(A\) with respect to \( \cup \) is denoted by \(A’\) or \({A^c}\) or \( \cup  – A\) and is defined as the set of all those elements of \( \cup \) which are not in \(A\).

Laws of Algebra of Sets and Proofs

Let us state and prove some fundamental laws of the algebra of sets.

Idempotent Law

For any set \(A\), \(A \cup A = A\) and \(A \cap A = A\).
Proof: 
\(A \cup A = \left\{ {x:\,x \in A\,{\text{or}}\,x\, \in A} \right\}\, = \left\{ {x:\,x \in A} \right\}\, = A\)
\(A \cap A = \,\left\{ {x:\,x \in A\,{\text{and}}\,x\, \in A} \right\}\, = \left\{ {x:\,x \in A} \right\}\, = A\)

Identity Law

For any set \(A\), \(A \cup \phi \, = \,A\) and \(A \cap \, \cup \, = \,A\) i.e., \(\phi \) and \( \cup \) are identity elements for union and intersection, respectively.
Proof: 
\(A \cup \phi \, = \,\left\{ {x:\,x \in A\,{\text{or}}\,x\, \in \phi } \right\}\, = \left\{ {x:\,x \in A} \right\}\, = A\)
\(A \cap \, \cup \, = \,\left\{ {x:\,x \in A\,{\text{and}}\,x\, \in \, \cup } \right\}\, = \left\{ {x:\,x \in A} \right\}\, = A\)

Commutative Law

For any two sets \(A\) and \(B\), \(A \cup B = \,B \cup A\) and \(A \cap B = \,B \cap A\) i.e., union and intersection are commutative.
Proof: Let \(x\)be an arbitrary element of \(A \cup \,B\). Then
\(x \in A \cup B \Rightarrow \,x \in A\) or \(x \in B\)
\( \Rightarrow x \in B\) or \(x \in A \Rightarrow \,x \in B \cup A\)
\(\therefore \,A \cup B \subseteq \,B \cup A\)
Similarly, \(B \cup A \subseteq A \cup B\)
Hence, \(A \cup B = \,B \cup A\)

De-Morgan’s Law

If \(A\) and \(B\) are two sets then \(\left( {A \cup B} \right)’\, = \,A’\, \cap \,B’\) and \(\left( {A \cap B} \right)’ = \,A’\, \cup \,B’\).
Proof:
Let \(x\) be an arbitrary element of \(\left( {A \cup B} \right)’\). Then \(x \in \left( {A \cup B} \right)’\)
\( \Rightarrow x \notin \left( {A \cup B} \right)\)
\( \Rightarrow x \notin \,A\) and \(x \notin \,B\)
\( \Rightarrow x \notin \,A’\) and \(x \notin \,B’\)
\( \Rightarrow x \in \,A’\, \cap \,B’\)
\(\therefore \,\left( {A \cap B} \right)’\, \subseteq \,A’\, \cap \,B’\)
Let \(y\) be an arbitrary element of \(A’\, \cap \,B’\). Then \(y \in A’\, \cap \,B’\)
\( \Rightarrow y \in A’\) and \(y \in B’\)
\( \Rightarrow y \notin A\) and \(y \notin B\)
\( \Rightarrow y \notin A\, \cup B\)
\( \Rightarrow y \in \left( {A\, \cup B} \right)’\)
\(\therefore \,A’\, \cap \,B’\, \subseteq \,\left( {A\, \cup B} \right)’\)
Hence, \(\left( {A\, \cup B} \right)’ = A’\, \cap \,B’\)
Similarly, we can prove another result also in the same way.

Cardinal of a Set

The number of distinct elements or members in a finite set is known as the cardinal number of a set. The cardinal number of a set \(P\) is denoted as \(n\left( P \right)\), where \(P\) is any set and \(n\left( P \right)\) is the number of members in set \(P\).

Cardinality of Union, Intersection, and Difference of Sets

Let us now learn to calculate the cardinality of the union, difference and intersection of two or three sets.

Union of Disjoint Sets

If \(A\) and \(B\) are two finite sets and if \(A \cap B = \phi \), then
\(n\left( {A\, \cup B} \right) = n\left( A \right) + n\left( B \right)\)

Union of Two Sets

If \(A\) and \(B\) are two finite sets, is represented as a Venn diagram as shown below.

Here, differently coloured regions depict the different disjoint sets, i.e., \(A – B,\,B – A\) and \(A \cap B\). Note that the sum of these represents \(A \cup B\).
Hence,
\(n\,\left( {A \cup B} \right) = n\left( {A – B} \right) + n\left( {B – A} \right) + n\left( {A \cap B} \right)\)

Difference of Sets

If \(A\) and \(B\) are two finite sets, then the cardinality of their difference is given by

\(n\left( {A\, – B} \right) = n\left( A \right) – n\left( {A \cap B} \right)\)

\(n\left( {B\, – A} \right) = n\left( B \right) – n\left( {A \cap B} \right)\)

Solved Examples

Q.1. Suppose \(X\) and \(Y\) are two sets, such that \(n\left( X \right) = 17,\,n\left( Y \right) = 23\) and \(n\left( {X \cup Y} \right) = \,38\,\), then find \(n\left( {X \cap Y} \right)\).
Sol:
Given:\(n\left( X \right) = 17,\,n\left( Y \right) = 23\) and \(n\left( {X \cup Y} \right) = \,38\,\),
We know, \(n\left( {X \cup Y} \right) = \,n\left( X \right) + n\left( Y \right) – n\left( {X \cap Y} \right)\)
Substituting the known values, we get
\(38 = 17 + 23 – n\left( {X \cap Y} \right)\)
\( \Rightarrow n\left( {X \cap Y} \right) = 40 – 38\)
\(\therefore n\left( {X \cap Y} \right) = 2\)

Q.2. Suppose \(A\) and \(B\) are two sets, such that \(n\left( A \right) = 17,\,n\left( B \right) = 23\) and \(n\left( {A \cup B} \right) = \,38\,\), then find \(n\left( {A \cap B} \right),\,n\left( {A – B} \right),\,n\left( {B – A} \right)\).
Sol:
Given: \(n\left( A \right) = 17,\,n\left( B \right) = 23\) and \(n\left( {A \cup B} \right) = \,38\,\),
We know, \(n\left( {A \cup B} \right) = \,n\left( A \right) + n\left( B \right) – n\left( {A \cup B} \right)\)
\( \Rightarrow n\left( {A \cap B} \right) = 17 + 23 – 38\)
\(\therefore n\left( {A \cap B} \right) = 2\)
Now, \(n\left( {A – B} \right) = \,n\left( A \right) – n\left( {A \cap B} \right)\)
\( \Rightarrow n\left( {A – B} \right) = 17 – 2\)
\(\therefore n\left( {A – B} \right) = 15\)
Similarly, \(n\left( {B – A} \right) = \,n\left( B \right) – n\left( {A \cap B} \right)\)
\( \Rightarrow n\left( {B – A} \right) = 23 – 2\)
\(\therefore n\left( {B – A} \right) = 21\)

Q.3. In a group of \(800\) people, \(550\) people can speak Hindi and \(450\) can speak English. How many speak both Hindi and English?
Sol:
Let \(H\) denote the set of people speaking Hindi
\(E\) denote the set of people speaking English
Therefore, \(n\left( H \right) = 550,\,n\left( E \right) = 450\) and \(n\left( {H \cup E} \right) = 800\)
Now, we have to find \(n\left( {H \cap E} \right)\).
\(n\left( {H \cup E} \right) = n\left( H \right) + n\left( E \right) – n\left( {H \cap E} \right)\)
\( \Rightarrow n\left( {H \cap E} \right) = n\left( H \right) + n\left( E \right) – n\left( {H \cup E} \right) = 550 + 450 – 800\)
\(\therefore n\left( {H \cap E} \right) = 200\)
Hence, \(200\) people can speak both Hindi and English.

Q.4. In a survey of \(700\) students in a college, \(180\) were listed as drinking Limca, \(275\) as drinking Miranda, and \(95\) were listed as both drinking Limca and Miranda. Find how many students were drinking, neither Limca nor Miranda.
Sol:
Let \( \cup \) be the set of all students
\(A\) denote the set of students drinking Limca
\(B\) denote the set of students drinking Miranda
Therefore, we have \(n\left(  \cup  \right) = 700,\,n\left( A \right) = 180,\,n\left( B \right) = 275\) and \(n\left( {A \cap B} \right) = 95\)
We have to find \(n\left( {A’ \cap B’} \right)\)
We know, \(n\left( {A’ \cap B’} \right) = n\left( {A \cup B} \right)’ = n\left(  \cup  \right) – n\left( {A \cup B} \right)\)
\( = n\left(  \cup  \right) – \left\{ {n\left( A \right) + n\left( B \right) – n\left( {A \cap B} \right)} \right\}\)
\( = 700 – \left( {180 + 275 – 95} \right)\)
\( = 700 – 360\)
\(\therefore n\left( {A’ \cap B’} \right) = 340\)
Hence, \(340\) students drink neither Limca nor Miranda.

Q.5. In a group of \(100\) people, \(72\) speak Hindi, and \(43\) speak English. How many speak Hindi only? How many speak English only? How many speak both Hindi and English?
Sol:
Let \(H\) and \(E\) be the sets of people who speak Hindi and English, respectively.
Given:
Number of people who speak Hindi, \(n\left( H \right) = 72\)
Number of people who speak English, \(n\left( E \right) = 43\)
Total number of people, \(n\left( {H \cup E} \right) = 100\)
(i) Number of people who speak both Hindi and English
\(n\left( {H \cap E} \right) = n\left( H \right) + n\left( E \right) – n\left( {H \cup E} \right)\)
\(n\left( {H \cap E} \right) = 72 + 43 – 100\)
\(n\left( {H \cap E} \right) = 115 – 100\)
\(\therefore n\left( {H \cap E} \right) = 15\)
(ii) Number of people who speak only Hindi
\(n\left( {H – E} \right) = \,n\left( H \right) – n\left( {H \cap E} \right)\)
\(n\left( {H – E} \right) = 72 – 15\)
\(\therefore n\left( {H – E} \right) = 57\)
(iii) Similarly, number of people who speak only English
\(n\left( {E – H} \right) = \,n\left( E \right) – n\left( {H \cap E} \right)\)
\(n\left( {E – H} \right) = 43 – 15\)
\(\therefore n\left( {E – H} \right) = 28\)

Summary

The practical problems related to the intersection and union of sets can be solved using the definition of union and intersection of sets. Let \(A\) and \(B\) be any two sets. Then the union of \(A\) and \(B\) is the set of elements that belong to either \(A\) or \(B\) or both \(A\) and \(B\). Similarly, let \(A\) and \(B\) be any two sets. Then, the intersection of \(A\) and \(B\) is the set of elements that belong to both \(A\) and \(B\). If the intersection of two sets is a null set, then they are disjoint sets. The cardinal number of a set is the number of distinct elements that are present in the given set.

Frequently ASked Questions (FAQs)

Q.1. How do you find the intersection and union of two sets?
Ans: Let \(A\) and \(B\) be finite two sets. Then, the union of \(A\) and \(B\) is the set of all the elements that belong to either \(A\) or \(B\) or both \(A\) and \(B\).
Then, the intersection of \(A\) and \(B\) is the set of all the elements that belong to both \(A\) and \(B\).

Q.2. What is the formula for the cardinality of the intersection of two sets?
Ans: Let \(A\) and \(B\) be finite sets, and \( \cup \) be a universal set. Then the cardinality of the intersection of \(A\) and \(B\) is given by:
\(n\,\left( {A \cap B} \right) = n\left( A \right) + n\left( B \right) – n\left( {A \cup B} \right)\)

Q.3. How do you solve practical problems on union and intersection of two sets?
Ans: To solve the practical problems on union and intersection of sets, we need to apply some of formulas which are listed below:
1. \(n\,\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) – n\left( {A \cap B} \right)\)
2. \(n\,\left( {A \cup B} \right) = n\left( {A – B} \right) + n\left( {A \cap B} \right) + n\left( {B – A} \right)\)
3. \(n\left( {A – B} \right) = \,n\left( A \right) – n\left( {A \cap B} \right)\)
4. \(n\left( {B – A} \right) = \,n\left( B \right) – n\left( {A \cap B} \right)\)

Q.4. What are the examples of union and intersection of sets?
Ans: If \(X = \left\{ {1,2} \right\}\) and \(Y = \left\{ {2,3} \right\}\), then \(X \cap Y = \left\{ 2 \right\}\)
If \(A = \left\{ {2,\,3,\,6} \right\}\) and \(B = \left\{ {2,\,3,\,4,\,5} \right\},\) then \(A \cup B = \left\{ {2,\,3,\,4,\,5,\,6} \right\}.\)

Q.5. How do you find the cardinality of the union of two sets?
Ans: Let \(A\) and \(B\) be any finite sets and \( \cup \) be a universal set. Then, the cardinality of the union of \(A\) and \(B\) is given by:
\(n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) – n\left( {A \cap B} \right)\)

Learn About Algebraic Expressions Here

We hope this detailed article on the Practical Problems on Union and Intersection of Two Sets will make you familiar with the topic. If you have any inquiries, feel to post them in the comment box. Stay tuned to embibe.com for more information.