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Complete Course of Mathematics
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Content On This Page
Introduction to Sets: Definition and Basic Concepts Representations of a Set (Roster Form, Set-Builder Form) Standard Sets of Numbers and Their Notations
Ordered Pair Cartesian Product of Two Sets


Sets: Fundamentals and Representation



Introduction to Sets: Definition and Basic Concepts

In mathematics, a set is one of the most fundamental and basic concepts. It provides a framework for grouping objects together. These objects, which can be anything from numbers, letters, symbols, points, or even other sets, are called the elements or members of the set.

Definition of a Set

A set is formally defined as a well-defined collection of distinct objects.

The objects that constitute a set are referred to as its elements or members.

Notation Used for Sets and Elements

Fundamental Characteristics of a Set

  1. Well-definedness:

    The membership rule must be clear and unambiguous for every potential object.
  2. Distinctness:

    Every element in a set is unique. Duplicates are not counted as separate elements.
  3. Order of Elements:

    The order in which the elements are listed within the set notation does not matter. For example, the set containing the numbers 1, 5, and 10 is the same whether written as $\{1, 5, 10\}$, $\{5, 10, 1\}$, or $\{10, 1, 5\}$. $\{1, 5, 10\} = \{5, 10, 1\}$.

Example 1. Let $C$ be the collection of colours in the Indian national flag.

Is $C$ a set? If yes, name its elements.

Answer:

Yes, $C$ is a set.

The colours in the Indian national flag are Saffron, White, and Green. This collection is well-defined because for any given colour, we can definitively say if it is one of these three colours or not. The elements are also distinct.

The elements of the set $C$ are Saffron, White, Green.

Using set notation, we can denote the set as:

$C = \{\text{Saffron, White, Green}\}$

We can write:

  • Saffron $\in C$
  • White $\in C$
  • Green $\in C$

For a colour like Blue:

  • Blue $\notin C$

Example 2. Consider the collection of "the 10 most beautiful cities in the world". Is this collection a set?

Answer:

No, this collection is not a set.

The term "beautiful" is subjective. What is considered beautiful varies greatly among individuals based on personal taste, cultural background, and other factors. There is no universal, objective criterion to definitively list "the 10 most beautiful cities". Different people would produce different lists.

Since the collection is not well-defined, it does not qualify as a set in mathematics.


Representations of a Set (Roster Form, Set-Builder Form)

After understanding what a set is, the next step is to learn how to represent or describe sets effectively. There are two primary methods used for representing sets: the Roster Form and the Set-Builder Form. Each method has its advantages and is suitable for different types of sets.


1. Roster Form (or Tabular Form)

In the Roster Form, a set is described by listing all its elements explicitly. The elements are written inside curly braces $\{\}$, with each element separated by a comma. This method is straightforward and intuitive, especially for sets with a small number of elements.

Key Characteristics and Conventions of Roster Form:


Example 1. Write the set of vowels in the English alphabet in Roster Form.

Answer:

The vowels in the English alphabet are a, e, i, o, u. Listing these inside curly braces gives the Roster Form representation.

$V = \{a, e, i, o, u\}$


Example 2. Write the set of all integers between -2 and 5 (inclusive) in Roster Form.

Answer:

Integers are whole numbers $(\mathbb{Z})$, which can be positive, negative, or zero. We need integers $x$ such that $-2 \leq x \leq 5$.

The integers starting from -2 up to 5 are -2, -1, 0, 1, 2, 3, 4, 5.

In Roster Form, the set is:

$I = \{-2, -1, 0, 1, 2, 3, 4, 5\}$


Example 3. Write the set of factors of 12 in Roster Form.

Answer:

The factors of 12 are the numbers that divide 12 evenly. These are 1, 2, 3, 4, 6, and 12.

In Roster Form, the set of factors of 12 is:

$F = \{1, 2, 3, 4, 6, 12\}$


2. Set-Builder Form (or Rule Method)

In the Set-Builder Form, a set is described by stating a common property that all its elements share. This property is not possessed by any object outside the set. This method is particularly useful for describing sets with a large or infinite number of elements where listing every element is impossible or impractical. It concisely defines the set based on the characteristics of its members.

General Representation in Set-Builder Form:

A set $A$ in Set-Builder Form is typically written as:

$A = \{x : P(x)\}$ or $A = \{x \mid P(x)\}$

Understanding the Notation:

The entire expression $\{x : P(x)\}$ is read as "the set of all $x$ such that $x$ has the property $P(x)$".

Often, the Set-Builder Form includes the domain from which the elements are taken to make the definition more precise. For example, $\{x \in \mathbb{Z} \mid x^2 = 9\}$ means "the set of all elements $x$ such that $x$ is an integer ($\mathbb{Z}$) AND the square of $x$ is 9 ($x^2=9$)".

Common sets of numbers used as domains include:


Example 4. Write the set of all prime numbers in Set-Builder Form.

Answer:

A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. Let $x$ represent an element of the set.

The property $P(x)$ is "$x$ is a prime number". Since prime numbers are natural numbers, we can specify the domain as $\mathbb{N}$.

In Set-Builder Form, the set is:

$P = \{x \in \mathbb{N} \mid x \text{ is a prime number}\}$

If we were to list the first few elements in Roster Form, it would be $P = \{2, 3, 5, 7, 11, 13, ...\}$.


Example 5. Write the set $A = \{1, 8, 27, 64, 125\}$ in Set-Builder Form.

Answer:

Let's observe the pattern in the elements:

  • $1 = 1^3$
  • $8 = 2^3$
  • $27 = 3^3$
  • $64 = 4^3$
  • $125 = 5^3$

The elements are the cubes of the first five natural numbers. A generic element can be written as $n^3$, where $n$ is a natural number ($\mathbb{N}$) such that $1 \leq n \leq 5$.

In Set-Builder Form, the set is:

$A = \{n^3 \mid n \in \mathbb{N} \text{ and } 1 \leq n \leq 5\}$

Alternatively, one could write:

$A = \{x \mid x = n^3, \text{ where } n \in \{1, 2, 3, 4, 5\}\}$


Example 6. Write the set $\{x \in \mathbb{R} \mid 0 < x < 1\}$ in Roster Form, if possible.

Answer:

The set is described in Set-Builder Form. It contains all real numbers ($x \in \mathbb{R}$) that are strictly greater than 0 and strictly less than 1. These are numbers like 0.1, 0.01, 0.99, 0.5, $1/\sqrt{2}$, etc.

It is not possible to write this set in Roster Form. There are infinitely many real numbers between 0 and 1, and they do not follow a listable pattern (there is no "next" real number after 0). Any attempt to list them would be incomplete and incorrect.


Comparing Roster Form and Set-Builder Form

Feature Roster Form (Tabular Form) Set-Builder Form (Rule Method)
Method Lists all elements. Describes elements by a common property.
Notation Elements separated by commas inside $\{\}$. Eg., $\{1, 2, 3\}$. $\{x : P(x)\}$ or $\{x \mid P(x)\}$. Eg., $\{x \mid x \text{ is a natural number less than 4}\}$.
Suitability Best for finite sets with few elements, or infinite sets with very clear listing patterns. Essential for infinite sets without obvious listing patterns (like intervals of real numbers); useful for finite sets defined by a specific property.
Element Duplication Elements are listed only once. Property defines distinct elements automatically.
Order of Elements Does not matter. Not relevant, as membership is determined by the property.


Standard Sets of Numbers and Their Notations

In mathematics, certain sets of numbers are used so frequently that they have been given standard names and symbols. Understanding these standard sets and their notations is fundamental to working with sets and other mathematical concepts.

Key Standard Sets of Numbers:

Notations for Subsets of Standard Sets:

Superscripts are often used to denote specific subsets of these standard number sets:

Similarly, $\mathbb{Q}_{\geq 0}$ denotes the set of non-negative rational numbers, and $\mathbb{R}_{\geq 0}$ denotes the set of non-negative real numbers.


Example 1. Classify the following numbers into the standard sets they belong to: $5, -7, 0, \frac{2}{3}, \sqrt{4}, \sqrt{7}, -2.5, 3.\overline{14}$.

Answer:

  • $5$: Is a natural number, whole number, integer, rational number (since $5 = \frac{5}{1}$), and real number.

    $5 \in \mathbb{N}, 5 \in \mathbb{W}, 5 \in \mathbb{Z}, 5 \in \mathbb{Q}, 5 \in \mathbb{R}$

  • $-7$: Is an integer, rational number (since $-7 = \frac{-7}{1}$), and real number. It is a negative integer ($\mathbb{Z}^-$).

    $-7 \in \mathbb{Z}, -7 \in \mathbb{Q}, -7 \in \mathbb{R}, -7 \in \mathbb{Z}^-$

  • $0$: Is a whole number, integer, rational number (since $0 = \frac{0}{1}$), and real number.

    $0 \in \mathbb{W}, 0 \in \mathbb{Z}, 0 \in \mathbb{Q}, 0 \in \mathbb{R}$

  • $\frac{2}{3}$: Is a rational number (by definition) and a real number. It is not a natural number, whole number, or integer.

    $\frac{2}{3} \in \mathbb{Q}, \frac{2}{3} \in \mathbb{R}$

  • $\sqrt{4}$: First, simplify the expression: $\sqrt{4} = 2$. This is a natural number, whole number, integer, rational number, and real number.

    $\sqrt{4} = 2 \implies 2 \in \mathbb{N}, 2 \in \mathbb{W}, 2 \in \mathbb{Z}, 2 \in \mathbb{Q}, 2 \in \mathbb{R}$

  • $\sqrt{7}$: 7 is not a perfect square. $\sqrt{7}$ has a non-terminating, non-repeating decimal expansion ($2.64575...$). Therefore, it is an irrational number and a real number.

    $\sqrt{7} \in \mathbb{T}, \sqrt{7} \in \mathbb{R}$

  • $-2.5$: This decimal terminates. It can be written as a fraction: $-2.5 = -\frac{25}{10} = -\frac{5}{2}$. Therefore, it is a rational number and a real number. It is a negative rational number ($\mathbb{Q}^-$).

    $-2.5 \in \mathbb{Q}, -2.5 \in \mathbb{R}, -2.5 \in \mathbb{Q}^-$

  • $3.\overline{14}$: This decimal is non-terminating and repeating (the block '14' repeats). Therefore, it is a rational number and a real number. It can be written as a fraction (e.g., $3.\overline{14} = \frac{314-3}{99} = \frac{311}{99}$).

    $3.\overline{14} \in \mathbb{Q}, 3.\overline{14} \in \mathbb{R}$


Ordered Pair

In set theory, the order in which elements are listed does not matter. For instance, $\{a, b\}$ and $\{b, a\}$ represent the exact same set. However, in many areas of mathematics, the order of elements is crucial. This is where the concept of an ordered pair becomes essential. An ordered pair is used to group two objects in a specific sequence.

Definition of an Ordered Pair

An ordered pair is a collection of two objects in a specific order. It is denoted by $(a, b)$, where $a$ is called the first component (or first element) and $b$ is called the second component (or second element).

The key difference from a set $\{a, b\}$ is the significance of the position of the elements.

Key Properties of Ordered Pairs


Example 1. If the ordered pair $(2x + 1, 3y - 5)$ is equal to the ordered pair $(x + 3, y - 1)$, find the values of $x$ and $y$.

Answer:

Given: $(2x + 1, 3y - 5) = (x + 3, y - 1)$.

To Find: The values of $x$ and $y$.

Solution:

By the definition of equality of ordered pairs, the corresponding components must be equal.

Equating the first components:

$2x + 1 = x + 3$

Subtracting $x$ from both sides:

$2x - x + 1 = x - x + 3$

$x + 1 = 3$

Subtracting 1 from both sides:

$x = 3 - 1$

$x = 2$

Equating the second components:

$3y - 5 = y - 1$

Subtracting $y$ from both sides:

$3y - y - 5 = y - y - 1$

$2y - 5 = -1$

Adding 5 to both sides:

$2y = -1 + 5$

$2y = 4$

Dividing both sides by 2:

$y = \frac{4}{2}$

$y = 2$

Thus, the values are $x = 2$ and $y = 2$.



Cartesian Product of Two Sets

Building upon the concept of ordered pairs, we can define an operation between two sets that produces a new set consisting of all possible ordered pairs where the first element comes from the first set and the second element comes from the second set. This operation is called the Cartesian Product.

Definition of Cartesian Product

Let $A$ and $B$ be two non-empty sets. The Cartesian product of set $A$ and set $B$, denoted by $A \times B$ (read as "A cross B"), is the set of all possible ordered pairs $(a, b)$ such that the first component $a$ is an element of set $A$, and the second component $b$ is an element of set $B$.

In Set-Builder Form, the Cartesian product is defined as:

$A \times B = \{ (a, b) \mid a \in A \text{ and } b \in B \}$

If either set $A$ or set $B$ is the empty set ($\emptyset$), their Cartesian product is the empty set. This is because it's impossible to form an ordered pair $(a, b)$ if there are no elements in $A$ or no elements in $B$.

$A \times \emptyset = \emptyset$

$\emptyset \times B = \emptyset$

Key Properties and Characteristics of Cartesian Product


Example 1. Let $P = \{ \text{Red, Blue} \}$ and $Q = \{ \text{Car, Bike, Cycle} \}$. Find $P \times Q$ and $|P \times Q|$.

Answer:

Given: $P = \{ \text{Red, Blue} \}$ and $Q = \{ \text{Car, Bike, Cycle} \}$.

To Find: $P \times Q$ and $|P \times Q|$.

Solution:

We need to form all ordered pairs $(p, q)$ where $p \in P$ and $q \in Q$.

Taking the first element from $P$ (Red) and pairing it with each element in $Q$:

  • (Red, Car)
  • (Red, Bike)
  • (Red, Cycle)

Taking the second element from $P$ (Blue) and pairing it with each element in $Q$:

  • (Blue, Car)
  • (Blue, Bike)
  • (Blue, Cycle)

The Cartesian product $P \times Q$ is the set containing all these ordered pairs:

$P \times Q = \{ (\text{Red, Car}), (\text{Red, Bike}), (\text{Red, Cycle}), (\text{Blue, Car}), (\text{Blue, Bike}), (\text{Blue, Cycle}) \}$

To find the cardinality $|P \times Q|$, we use the formula $|P \times Q| = |P| \times |Q|$.

$|P| = 2$ (since P has 2 elements)

$|Q| = 3$ (since Q has 3 elements)

$|P \times Q| = 2 \times 3 = 6$

Counting the elements in the set $P \times Q$ confirms this result; there are 6 ordered pairs.


Example 2. Let $A = \{x \in \mathbb{Z} \mid 1 \leq x \leq 2\}$ and $B = \{y \in \mathbb{N} \mid y < 3\}$. Find $A \times B$ and $B \times A$.

Answer:

Given: $A = \{x \in \mathbb{Z} \mid 1 \leq x \leq 2\}$ and $B = \{y \in \mathbb{N} \mid y < 3\}$.

To Find: $A \times B$ and $B \times A$.

Solution:

First, let's determine the elements of sets $A$ and $B$ by converting their Set-Builder Forms to Roster Forms.

Set $A$: Integers $x$ such that $1 \leq x \leq 2$. The integers are 1 and 2.

$A = \{1, 2\}$

Set $B$: Natural numbers $y$ such that $y < 3$. Natural numbers are $\{1, 2, 3, ...\}$. The natural numbers less than 3 are 1 and 2.

$B = \{1, 2\}$

Now we find the Cartesian products.

Finding $A \times B$:

We form all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

Take 1 from A, pair with elements in B: $(1, 1), (1, 2)$

Take 2 from A, pair with elements in B: $(2, 1), (2, 2)$

$A \times B = \{ (1, 1), (1, 2), (2, 1), (2, 2) \}$

Finding $B \times A$:

We form all ordered pairs $(b, a)$ where $b \in B$ and $a \in A$.

Take 1 from B, pair with elements in A: $(1, 1), (1, 2)$

Take 2 from B, pair with elements in A: $(2, 1), (2, 2)$

$B \times A = \{ (1, 1), (1, 2), (2, 1), (2, 2) \}$

In this specific case, since $A = B = \{1, 2\}$, we have $A \times B = B \times A$. This confirms the property that $A \times B = B \times A$ if and only if $A=B$ (assuming neither is empty).

Also, $|A| = 2$, $|B| = 2$. So, $|A \times B| = 2 \times 2 = 4$, which matches the number of elements we found.


Example 3. If $A = \{1, 2, 3\}$ and $B = \emptyset$, find $A \times B$ and $B \times A$.

Answer:

Given: $A = \{1, 2, 3\}$ and $B = \emptyset$ (the empty set).

To Find: $A \times B$ and $B \times A$.

Solution:

According to the definition, if either set is empty, their Cartesian product is the empty set.

For $A \times B = \{ (a, b) \mid a \in A \text{ and } b \in B \}$, since $B$ has no elements, it is impossible to pick an element $b \in B$. Thus, no ordered pairs can be formed.

$A \times B = \emptyset$

Similarly, for $B \times A = \{ (b, a) \mid b \in B \text{ and } a \in A \}$, since $B$ has no elements, it is impossible to pick an element $b \in B$. Thus, no ordered pairs can be formed.

$B \times A = \emptyset$

Alternatively, using cardinality: $|A|=3$, $|B|=0$.

$|A \times B| = |A| \times |B| = 3 \times 0 = 0$. A set with 0 elements is the empty set.

$|B \times A| = |B| \times |A| = 0 \times 3 = 0$. A set with 0 elements is the empty set.

Thus, $A \times B = \emptyset$ and $B \times A = \emptyset$. In this case, $A \times B = B \times A = \emptyset$, which aligns with the property when one of the sets is empty.


Example 4. If the Cartesian product $X \times Y = \{ (a, x), (a, y), (a, z), (b, x), (b, y), (b, z) \}$, find the sets $X$ and $Y$.

Answer:

Given: $X \times Y = \{ (a, x), (a, y), (a, z), (b, x), (b, y), (b, z) \}$.

To Find: Sets $X$ and $Y$.

Solution:

By the definition of the Cartesian product, the first component of every ordered pair in $X \times Y$ must be an element of $X$, and the second component must be an element of $Y$.

Set $X$ consists of all the distinct first components of the ordered pairs in $X \times Y$.

The first components are $a, a, a, b, b, b$. The distinct first components are $a$ and $b$.

Therefore, $X = \{a, b\}$.

Set $Y$ consists of all the distinct second components of the ordered pairs in $X \times Y$.

The second components are $x, y, z, x, y, z$. The distinct second components are $x, y, z$.

Therefore, $Y = \{x, y, z\}$.

We can verify this by calculating the Cartesian product of the sets we found:

$X \times Y = \{a, b\} \times \{x, y, z\}$

$X \times Y = \{ (a, x), (a, y), (a, z), (b, x), (b, y), (b, z) \}$

This matches the given Cartesian product, so our sets $X$ and $Y$ are correct.

We can also check the cardinality: $|X| = 2$, $|Y| = 3$. $|X \times Y| = 2 \times 3 = 6$. The given set has 6 elements, which matches.


Summary for Competitive Exams

Ordered Pair: $(a, b)$. An ordered collection of two objects. Order matters: $(a, b) = (c, d) \iff a=c \text{ and } b=d$. Components can be the same.

Cartesian Product of Two Sets (A and B): $A \times B = \{ (a, b) \mid a \in A, b \in B \}$.

  • Result is a set of ordered pairs.
  • The first component of each pair is from the first set (A), the second from the second set (B).
  • Generally, $A \times B \neq B \times A$, unless $A=B$ or one/both sets are empty.
  • Cardinality: $|A \times B| = |A| \times |B|$.
  • If $A$ or $B$ is empty, $A \times B = \emptyset$.
  • $A \times B \times C = \{ (a, b, c) \mid a \in A, b \in B, c \in C \}$. $|A \times B \times C| = |A||B||C|$.
  • Given $A \times B$, set $A$ is the set of all first components; set $B$ is the set of all second components.