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Applied Mathematics for Class 11th & 12th (Concepts and Questions)
11th Concepts Questions
12th Concepts Questions

Applied Maths Class 11th Chapters (Concepts)
1. Numbers and Quantification 2. Numbers Applications 3. Sets
4. Relations 5. Sequences and Series 6. Permutations and Combinations
7. Mathematical Reasoning 8. Calculus 9. Probability
10. Descriptive Statistics 11. Financial Mathematics 12. Coordinate Geometry

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Ordered Pairs Cartesian Product of Two Sets Relations


Chapter 4 Relations (Concepts)

Moving beyond the concept of sets as mere collections, this chapter introduces the fundamental mathematical structure of Relations. Relations provide the formal framework for describing specific connections, associations, or relationships that exist between elements, either within a single set or between elements of two different sets. Building directly upon set theory, understanding relations is a crucial stepping stone towards grasping more complex structures like functions and graphs, which are ubiquitous in applied mathematics, computer science, data analysis, and logical reasoning. We will explore how to precisely define, represent, and analyze these relationships using the rigorous language of mathematics.

The foundation for defining relations lies in the concept of an ordered pair, denoted as $(a, b)$, where the order of elements $a$ and $b$ is significant – $(a, b)$ is generally different from $(b, a)$. From ordered pairs, we construct the Cartesian Product of two sets, $A$ and $B$, denoted by $A \times B$. This product is the set containing all possible ordered pairs $(a, b)$ such that the first element $a$ comes from set $A$ ($a \in A$) and the second element $b$ comes from set $B$ ($b \in B$). If sets $A$ and $B$ are finite with cardinalities $n(A)$ and $n(B)$ respectively, then the cardinality of their Cartesian product is $n(A \times B) = n(A) \times n(B)$. With the Cartesian product established, a Relation $R$ from a set $A$ to a set $B$ is formally defined as simply any subset of $A \times B$, i.e., $R \subseteq A \times B$. If the relation connects elements within the same set $A$, it is called a relation on set A, meaning $R \subseteq A \times A$.

Representing relations clearly is essential. We explore several methods:

Associated with every relation $R$ from $A$ to $B$ are three important sets: the Domain of $R$ (the set of all first elements 'a' appearing in the ordered pairs of $R$), the Codomain (which is always the entire set $B$), and the Range of $R$ (the set of all second elements 'b' appearing in the ordered pairs of $R$). It's crucial to note that the Range is always a subset of the Codomain, but they are not necessarily equal.

A significant focus, particularly for relations defined on a single set $A$, is analyzing their fundamental properties. We will learn how to rigorously determine if a relation $R$ on $A$ is:

Verifying these properties often involves careful logical checking based on the relation's definition. Relations that exhibit all three of these properties – reflexive, symmetric, and transitive – are called Equivalence Relations. These are particularly important because they partition the underlying set $A$ into disjoint subsets known as Equivalence Classes, denoted $[a] = \{x \in A \mid (x, a) \in R\}$. This chapter equips you with the essential tools to formally define, represent, and classify relationships, paving the way for understanding functions and their properties.



Ordered Pairs

In mathematics, we often deal with collections of objects. When we consider a Set, the order in which the elements are listed is irrelevant. For instance, the set $\{a, b\}$ is exactly the same as the set $\{b, a\}$. However, there are numerous mathematical contexts and real-world applications where the order of elements is of paramount importance. For example, the coordinates $(2, 3)$ represent a specific point in a graph, which is distinct from the point represented by $(3, 2)$. To handle such situations where the position or order of elements matters, we use the concept of an Ordered Pair.

Definition of an Ordered Pair

An Ordered Pair is a collection of two elements, say 'a' and 'b', in which the order of the elements is specified. It is denoted by enclosing the two elements within parentheses and separating them by a comma, like $(a, b)$.

The term "ordered" is key here. Unlike a set, the position of each element in an ordered pair is significant.

Significance of Order vs. Sets

The fundamental difference between a set $\{a, b\}$ and an ordered pair $(a, b)$ lies solely in the importance given to the order of elements:

Example: The set $\{1, 5\}$ is the same as $\{5, 1\}$. But the ordered pair $(1, 5)$ is different from the ordered pair $(5, 1)$. In a coordinate system, $(1, 5)$ represents a point 1 unit along the x-axis and 5 units along the y-axis, while $(5, 1)$ represents a point 5 units along the x-axis and 1 unit along the y-axis.

Equality of Ordered Pairs

Two ordered pairs are considered equal if and only if their corresponding components are equal. This means the first element of the first ordered pair must be equal to the first element of the second ordered pair, AND the second element of the first ordered pair must be equal to the second element of the second ordered pair.

$(a, b) = (c, d) \iff a = c \text{ and } b = d$

This property is fundamental when solving problems involving unknown values within ordered pairs that are stated to be equal.

Example 1. Find the values of $x$ and $y$ if $(x+1, y-2) = (3, 5)$.

Answer:

We are given that the ordered pair $(x+1, y-2)$ is equal to the ordered pair $(3, 5)$.

According to the definition of equality of ordered pairs, the first components must be equal, and the second components must be equal.

Equating the first components:

$x+1 = 3$

[Equating first components]

Now, we solve this simple linear equation for $x$. Subtract 1 from both sides:

$x = 3 - 1$

$x = 2$

Equating the second components:

$y-2 = 5$

[Equating second components]

Now, we solve this simple linear equation for $y$. Add 2 to both sides:

$y = 5 + 2$

$y = 7$

Thus, the values that satisfy the equality of the ordered pairs are $x = 2$ and $y = 7$.

We can verify this by substituting the values back into the original ordered pairs: $(2+1, 7-2) = (3, 5)$. This matches the given second ordered pair.

Example 2. Find the values of $a$ and $b$ if the ordered pair $(2a - b, 3a + b)$ is equal to $(8, 7)$.

Answer:

We are given the equality of ordered pairs $(2a - b, 3a + b) = (8, 7)$.

Using the definition of equality of ordered pairs, we equate the corresponding components.

Equating the first components gives us the equation:

$2a - b = 8$

Equating the second components gives us a second equation:

$3a + b = 7$

We now have a system of two linear equations in two variables, $a$ and $b$. We can solve this system using the elimination method.

Notice that the coefficients of $b$ in the two equations are $-1$ and $+1$. If we add the two equations, the $b$ terms will cancel out.

$(2a - b) + (3a + b) = 8 + 7$

Combine like terms:

$5a = 15$

Solve for $a$ by dividing both sides by 5:

$a = \frac{15}{5}$

$a = 3$

Now that we have the value of $a$, substitute $a=3$ into the second equation ($3a + b = 7$) to find the value of $b$.

$3(3) + b = 7$

Simplify and solve for $b$:

$9 + b = 7$

$b = 7 - 9$

$b = -2$

Thus, the values are $a = 3$ and $b = -2$.

Check: Substitute $a=3$ and $b=-2$ back into the original ordered pair $(2a - b, 3a + b)$.

First component: $2a - b = 2(3) - (-2) = 6 + 2 = 8$. (Matches the given 8)

Second component: $3a + b = 3(3) + (-2) = 9 - 2 = 7$. (Matches the given 7)

The ordered pairs are indeed equal with these values.

Significance and Uses of Ordered Pairs

Ordered pairs are not merely an abstract concept; they are fundamental building blocks used across many areas of mathematics:

Thus, the concept of ordered pairs provides the necessary structure to handle situations where the sequence or position of elements is important, paving the way for the study of relations, functions, and graphical representations.



Cartesian Product of Two Sets

Building upon the concept of ordered pairs, we can define an operation that combines elements from two sets in an ordered manner. This operation is known as the Cartesian Product, named after the French philosopher and mathematician René Descartes, whose work in coordinate geometry relies heavily on representing points using ordered pairs.

Definition of Cartesian Product

Let $A$ and $B$ be two non-empty sets. The Cartesian Product of set $A$ and set $B$ is a set that consists of all possible ordered pairs $(a, b)$ such that the first element, $a$, is taken from set $A$, and the second element, $b$, is taken from set $B$.

The Cartesian Product of $A$ and $B$ is denoted by $A \times B$, read as "A cross B".

Using Set-builder notation, the Cartesian Product is formally defined as:

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

This means that to form the set $A \times B$, you systematically pair every element from set $A$ with every element from set $B$, creating ordered pairs where the first component is always from $A$ and the second component is always from $B$.

Similarly, the Cartesian product of set $B$ and set $A$, denoted by $B \times A$, is the set of all ordered pairs $(b, a)$ where $b \in B$ and $a \in A$.

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

In general, $A \times B \neq B \times A$, because the order of elements within an ordered pair matters, and the ordered pairs in $A \times B$ have the first component from A and the second from B, while in $B \times A$ it is the other way around.

Example 1. Let $A = \{1, 2\}$ and $B = \{a, b\}$. Find $A \times B$ and $B \times A$.

Answer:

Given sets $A = \{1, 2\}$ and $B = \{a, b\}$.

To find $A \times B$, we form all ordered pairs $(p, q)$ where $p \in A$ and $q \in B$. We take each element from A (1, then 2) and pair it with each element from B (a, then b).

Starting with element 1 from A, pair it with 'a' from B to get $(1, a)$. Pair it with 'b' from B to get $(1, b)$.

Starting with element 2 from A, pair it with 'a' from B to get $(2, a)$. Pair it with 'b' from B to get $(2, b)$.

Listing all these ordered pairs within curly braces gives $A \times B$:

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

To find $B \times A$, we form all ordered pairs $(q, p)$ where $q \in B$ and $p \in A$. We take each element from B (a, then b) and pair it with each element from A (1, then 2).

Starting with element 'a' from B, pair it with 1 from A to get $(a, 1)$. Pair it with 2 from A to get $(a, 2)$.

Starting with element 'b' from B, pair it with 1 from A to get $(b, 1)$. Pair it with 2 from A to get $(b, 2)$.

Listing all these ordered pairs gives $B \times A$:

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

Comparing the elements of $A \times B$ and $B \times A$, we see that the ordered pairs are different (e.g., $(1, a) \neq (a, 1)$). Therefore, $A \times B \neq B \times A$, which is generally true for Cartesian products unless the sets are equal or at least one is empty.

Number of Elements in the Cartesian Product (Cardinality)

If set $A$ is a finite set with $m$ elements, and set $B$ is a finite set with $n$ elements, then the number of ordered pairs in the Cartesian product $A \times B$ is the product of the number of elements in $A$ and the number of elements in $B$.

$n(A \times B) = n(A) \times n(B)$

If $n(A) = m$ and $n(B) = n$, then $n(A \times B) = m \times n$.

Similarly, $n(B \times A) = n(B) \times n(A) = n \times m$. Since $m \times n = n \times m$, the number of elements in $A \times B$ is always equal to the number of elements in $B \times A$, even if the sets themselves are not equal.

Example 2. If $n(A) = 5$ and $n(B) = 3$, find $n(A \times B)$ and $n(B \times A)$.

Answer:

Given the number of elements in set A is $n(A) = 5$.

Given the number of elements in set B is $n(B) = 3$.

Using the formula for the number of elements in the Cartesian product (Equation 35):

$n(A \times B) = n(A) \times n(B)$

Substitute the given values:

$n(A \times B) = 5 \times 3 = 15$

For $n(B \times A)$:

$n(B \times A) = n(B) \times n(A)$

Substitute the given values:

$n(B \times A) = 3 \times 5 = 15$

Both $A \times B$ and $B \times A$ have 15 elements.

Cartesian Product Involving the Empty Set

If either set $A$ or set $B$ (or both) is the empty set ($\emptyset$), then their Cartesian product is also the empty set.

$A \times \emptyset = \emptyset$

$\emptyset \times B = \emptyset$

$\emptyset \times \emptyset = \emptyset$

Explanation: The definition of $A \times B$ requires forming ordered pairs $(a, b)$ where $a \in A$ and $b \in B$. If set $A$ is empty, there are no elements $a$ to choose from. Thus, no ordered pairs can be formed, and the resulting set is empty. The same logic applies if set $B$ is empty.

Using the cardinality formula, if $n(A)=m$ and $n(\emptyset)=0$, then $n(A \times \emptyset) = n(A) \times n(\emptyset) = m \times 0 = 0$. A set with 0 elements is the empty set.

Cartesian Product of a Set with Itself

The Cartesian product of a set $A$ with itself is denoted by $A \times A$ or $A^2$. It is the set of all ordered pairs $(a_1, a_2)$ where both the first and second elements are taken from the same set $A$.

$A \times A = \{(a_1, a_2) \mid a_1 \in A \text{ and } a_2 \in A\}$

If $n(A) = m$, then the number of elements in $A \times A$ is $n(A \times A) = n(A) \times n(A) = m \times m = m^2$.

The graphical representation of $A \times A$ (when A is a set of numbers) forms points in a 2-dimensional plane, which is why coordinate systems are linked to Cartesian products.

Example 3. Let $A = \{p, q\}$. Find $A \times A$. What is the number of elements in $A \times A$?

Answer:

The given set is $A = \{p, q\}$. We need to form ordered pairs where both elements come from $A$.

Take the first element from A, which can be $p$ or $q$. Take the second element from A, which can also be $p$ or $q$. Form all combinations:

  • First element is $p$, second element is $p$: $(p, p)$
  • First element is $p$, second element is $q$: $(p, q)$
  • First element is $q$, second element is $p$: $(q, p)$
  • First element is $q$, second element is $q$: $(q, q)$

Listing these ordered pairs within curly braces gives $A \times A$:

$A \times A = \{(p, p), (p, q), (q, p), (q, q)\}$

The number of elements in set A is $n(A) = 2$.

The number of elements in $A \times A$ is $n(A \times A) = 4$.

This matches the formula $n(A \times A) = n(A)^2 = 2^2 = 4$.

Cartesian Product of Three or More Sets

The concept of the Cartesian product can be extended to three or more sets. The Cartesian product of three sets $A$, $B$, and $C$, denoted by $A \times B \times C$, is the set of all possible ordered triplets $(a, b, c)$ where the first element $a \in A$, the second element $b \in B$, and the third element $c \in C$.

$A \times B \times C = \{(a, b, c) \mid a \in A, b \in B, c \in C\}$

The number of elements in the Cartesian product of three finite sets is $n(A \times B \times C) = n(A) \times n(B) \times n(C)$.

This concept generalizes to the Cartesian product of $n$ sets, which results in ordered $n$-tuples.



Relations

In mathematics, a Relation is a concept used to describe how elements from one set are connected or associated with elements from another set (or even the same set). Intuitively, a relation establishes a link or connection between pairs of elements from the sets involved. For example, "is less than", "is a factor of", "is taller than", "lives in" are all examples of relations that link elements of different sets (or the same set). Formally, relations are defined using the concept of the Cartesian product of sets.

Definition of a Relation

A Relation $R$ from a non-empty set $A$ to a non-empty set $B$ is defined as any subset of the Cartesian product $A \times B$.

Since $A \times B$ is the set of all possible ordered pairs $(a, b)$ where $a \in A$ and $b \in B$, a relation $R$ from $A$ to $B$ is simply a collection of some (or all, or none) of these ordered pairs.

Mathematically, this is written as:

$R \subseteq A \times B$

If an ordered pair $(a, b) \in A \times B$ is included in the relation $R$, it means that element $a$ from set $A$ is related to element $b$ from set $B$ according to the rule or property that defines the relation $R$. When $(a, b) \in R$, we often write this as $a R b$, which is read as "$a$ is related to $b$ by the relation $R$".

If an ordered pair $(a, b) \in A \times B$ does NOT belong to $R$ (i.e., $(a, b) \notin R$), it means that $a$ is not related to $b$ by the relation $R$, and we can write $a \not R b$.

Example 1. Let $A = \{1, 2\}$ and $B = \{1, 2, 3\}$.

Let $R$ be a relation from A to B defined by the rule "a is less than b", where $a \in A$ and $b \in B$.

Write the relation $R$ as a set of ordered pairs.

Answer:

Given sets $A = \{1, 2\}$ and $B = \{1, 2, 3\}$. The relation $R$ is from $A$ to $B$.

First, let's list all possible ordered pairs in the Cartesian product $A \times B$. These are all pairs $(a, b)$ where $a \in A$ and $b \in B$.

$A \times B = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)\}$

The relation $R$ is defined by the property "$a$ is less than $b$" ($a < b$). We need to identify which ordered pairs $(a, b)$ from $A \times B$ satisfy this property.

Let's check each ordered pair in $A \times B$ against the condition $a < b$:

  • For $(1, 1)$: Is $1 < 1$? No. So, $(1, 1) \notin R$.
  • For $(1, 2)$: Is $1 < 2$? Yes. So, $(1, 2) \in R$.
  • For $(1, 3)$: Is $1 < 3$? Yes. So, $(1, 3) \in R$.
  • For $(2, 1)$: Is $2 < 1$? No. So, $(2, 1) \notin R$.
  • For $(2, 2)$: Is $2 < 2$? No. So, $(2, 2) \notin R$.
  • For $(2, 3)$: Is $2 < 3$? Yes. So, $(2, 3) \in R$.

The ordered pairs that satisfy the condition "$a < b$" are $(1, 2)$, $(1, 3)$, and $(2, 3)$.

The relation $R$, as a set of ordered pairs, is the collection of these pairs:

$R = \{(1, 2), (1, 3), (2, 3)\}$

We can see that $R$ is indeed a subset of $A \times B$.

Domain and Range of a Relation

For a relation $R$ from set $A$ to set $B$ ($R \subseteq A \times B$):


Visual Representation

An arrow diagram clearly illustrates the difference between these concepts. Consider a relation R from set A to set B.

An arrow diagram showing Set A = {1,2,3,4} and Set B = {a,b,c,d,e}. The Domain is highlighted as {1,3,4} within A. The Range is highlighted as {a,b,c} within B. The Codomain is the entire Set B.

In the diagram above:

Example 2. For the relation $R = \{(1, 2), (1, 3), (2, 3)\}$ from Example 1, identify the Domain, Range, and Codomain of R.

Answer:

The relation is given as $R = \{(1, 2), (1, 3), (2, 3)\}$. This relation is from set $A = \{1, 2\}$ to set $B = \{1, 2, 3\}$.

Domain of R: This is the set of all first elements of the ordered pairs in $R$.

The first elements are 1 (from (1, 2)), 1 (from (1, 3)), and 2 (from (2, 3)).

Listing the distinct first elements: {1, 2}.

Domain$(R) = \{1, 2\}$.

Note that the Domain of R is equal to set A in this case.

Range of R: This is the set of all second elements of the ordered pairs in $R$.

The second elements are 2 (from (1, 2)), 3 (from (1, 3)), and 3 (from (2, 3)).

Listing the distinct second elements: {2, 3}.

Range$(R) = \{2, 3\}$.

Codomain of R: This is the set B, which is the set to which the relation maps elements from A.

Codomain$(R) = B = \{1, 2, 3\}$.

We can observe that Range$(R) = \{2, 3\}$ is indeed a subset of Codomain$(R) = \{1, 2, 3\}$.

Number of Possible Relations

The definition states that a relation from set $A$ to set $B$ is any subset of the Cartesian product $A \times B$.

If set $A$ has $m$ elements ($n(A) = m$) and set $B$ has $n$ elements ($n(B) = n$), then the number of elements in the Cartesian product $A \times B$ is $m \times n$.

We know that the total number of possible subsets of a set with $k$ elements is $2^k$.

Since a relation $R$ from $A$ to $B$ is any subset of $A \times B$, and $A \times B$ has $m \times n$ elements, the total number of possible relations from $A$ to $B$ is the number of subsets of $A \times B$.

$\text{Number of relations from A to B} = 2^{n(A \times B)} = 2^{n(A) \times n(B)}$

Example 3. If $n(A) = 2$ and $n(B) = 3$, how many relations are possible from A to B?

Answer:

Given the number of elements in set A is $n(A) = 2$.

Given the number of elements in set B is $n(B) = 3$.

The number of elements in the Cartesian product $A \times B$ is:

$n(A \times B) = n(A) \times n(B) = 2 \times 3 = 6$

... (i)

A relation from A to B is any subset of $A \times B$. The total number of possible relations is the number of subsets of $A \times B$.

Using the formula for the number of relations (Equation 37):

$\text{Number of relations} = 2^{n(A \times B)}$

Substitute the number of elements in $A \times B$ from (i):

$= 2^6$

Calculate the value of $2^6$: $2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.

$= 64$

Therefore, there are 64 possible relations that can be defined from set A to set B.

Representing Relations

Relations can be represented in several ways, each useful in different contexts:

Example 4. Let $A = \{1, 2, 3\}$ and $B = \{2, 4, 6\}$. Let $R$ be the relation from A to B defined by "$a$ divides $b$", where $a \in A$ and $b \in B$. Represent R using an arrow diagram.

Answer:

Given sets $A = \{1, 2, 3\}$ and $B = \{2, 4, 6\}$. The relation $R$ is from A to B with the rule "$a$ divides $b$".

First, let's determine the ordered pairs $(a, b)$ in $A \times B$ that satisfy the condition "$a$ divides $b$". An integer $a$ divides an integer $b$ if $b = ka$ for some integer $k$. We are checking divisibility for positive integers here.

Possible ordered pairs $(a, b)$ from $A \times B$:

$(1, 2)$: Does 1 divide 2? Yes ($2 = 2 \times 1$). So, $(1, 2) \in R$.

$(1, 4)$: Does 1 divide 4? Yes ($4 = 4 \times 1$). So, $(1, 4) \in R$.

$(1, 6)$: Does 1 divide 6? Yes ($6 = 6 \times 1$). So, $(1, 6) \in R$.

$(2, 2)$: Does 2 divide 2? Yes ($2 = 1 \times 2$). So, $(2, 2) \in R$.

$(2, 4)$: Does 2 divide 4? Yes ($4 = 2 \times 2$). So, $(2, 4) \in R$.

$(2, 6)$: Does 2 divide 6? Yes ($6 = 3 \times 2$). So, $(2, 6) \in R$.

$(3, 2)$: Does 3 divide 2? No.

$(3, 4)$: Does 3 divide 4? No.

$(3, 6)$: Does 3 divide 6? Yes ($6 = 2 \times 3$). So, $(3, 6) \in R$.

The relation $R$ in roster form is $R = \{(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 6)\}$.

Now, we draw the arrow diagram:

Arrow diagram for relation 'divides' from A={1,2,3} to B={2,4,6}

Draw an oval representing set A with elements 1, 2, and 3 inside it.

Draw another oval representing set B with elements 2, 4, and 6 inside it.

Draw arrows for each ordered pair in R:

  • From 1 in A to 2 in B.
  • From 1 in A to 4 in B.
  • From 1 in A to 6 in B.
  • From 2 in A to 2 in B.
  • From 2 in A to 4 in B.
  • From 2 in A to 6 in B.
  • From 3 in A to 6 in B.

The arrow diagram visually shows which elements of A are related to which elements of B.

Domain and Range from Arrow Diagram:

Domain is the set of elements in A from which arrows originate: {1, 2, 3}. (In this case, Domain(R) = A)

Range is the set of elements in B to which arrows point: {2, 4, 6}. (In this case, Range(R) = B = Codomain)