Learn with visual diagrams, step-by-step solutions with marking scheme, MCQs, Assertion-Reasoning questions, and previous year questions.
What is a relation? What is a function? Let's start from the beginning with visuals.
A relation connects elements of one set to elements of another set.
If A and B are two non-empty sets, then any subset of A × B is called a relation from A to B.
Let $A$ and $B$ be two non-empty sets. Then a relation $R$ from $A$ to $B$ is a subset of $A \times B$, i.e., $R \subseteq A \times B$.
If $(a, b) \in R$, we write $a \ R \ b$ and say "a is related to b under R".
For a relation $R$ from set $A$ to set $B$:
Domain → Codomain → Range
There are 5 main types of relations. Let's visualize each one.
| Type | Condition | Meaning |
|---|---|---|
| Empty Relation | $R = \phi$ | No element is related to any other |
| Universal Relation | $R = A \times A$ | Every element related to every element |
| Reflexive | $(a, a) \in R \ \forall a$ | Every element related to itself |
| Symmetric | $(a, b) \in R \Rightarrow (b, a) \in R$ | Bidirectional |
| Transitive | $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ | Chain property |
A relation $R$ is an equivalence relation if it is Reflexive + Symmetric + Transitive.
A function $f: A \to B$ is a special relation where every element of A has exactly one image in B.
Distinct elements of A have distinct images in B.
Every element of B is the image of some element of A.
Both one-one and onto. Only bijective functions have an inverse.
Understand every concept with clear diagrams
Speed techniques to solve problems faster in board exams
To check reflexivity, just scan for (1,1), (2,2), (3,3)... in the relation.
For every pair $(a, b)$, check if $(b, a)$ is also present.
Look for chains: If $a \to b$ and $b \to c$ both exist, then $a \to c$ must exist.
All three must hold: Reflexive + Symmetric + Transitive.
On a graph, apply the Horizontal Line Test:
For set $A$ with $n$ elements and set $B$ with $m$ elements:
For finite sets with $n(A) = n(B)$:
"Right first, then left" — $gof$ means apply $f$ first, then $g$.
To find $f^{-1}$: Swap $x$ and $y$, solve for $y$, replace with $f^{-1}(x)$.
For a function to be onto, its range must equal its codomain. If any codomain element is "missed", it's not onto.
Click "Show Answer" to reveal step-by-step solutions with marks for each step
Q1. Show that the relation $R$ in the set $\mathbb{Z}$ of integers given by $R = \{(a, b) : 2 \text{ divides } a - b\}$ is an equivalence relation.
For any $a \in \mathbb{Z}$, we have $a - a = 0$, and $2$ divides $0$.
So, $(a, a) \in R$ for all $a \in \mathbb{Z}$.
Therefore, $R$ is reflexive. ✓
Suppose $(a, b) \in R$. Then $2$ divides $(a - b)$.
Since $b - a = -(a - b)$, we have $2$ divides $(b - a)$ also.
So $(b, a) \in R$. Therefore, $R$ is symmetric. ✓
Suppose $(a, b) \in R$ and $(b, c) \in R$. Then $2 | (a - b)$ and $2 | (b - c)$.
Now, $a - c = (a - b) + (b - c)$, which is divisible by $2$.
So $(a, c) \in R$. Therefore, $R$ is transitive. ✓
Q2. Show that $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = 2x$ is one-one and onto.
Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$.
Then $2x_1 = 2x_2 \Rightarrow x_1 = x_2$.
Therefore, $f$ is one-one. ✓
Let $y \in \mathbb{R}$ be arbitrary.
Choose $x = \frac{y}{2} \in \mathbb{R}$.
Then $f(x) = f\left(\frac{y}{2}\right) = 2 \cdot \frac{y}{2} = y$.
So every $y \in \mathbb{R}$ has a pre-image. Therefore, $f$ is onto. ✓
Q3. Let $A = \{1, 2, 3\}$. Find the number of equivalence relations containing $(1, 2)$.
To be an equivalence relation containing $(1, 2)$, we need:
• $(1, 1), (2, 2), (3, 3)$ for reflexivity
• $(2, 1)$ for symmetry of $(1, 2)$
So the smallest such relation is:
$R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$
Adding any remaining pair $(1,3), (3,1), (2,3), (3,2)$ forces us to add all of them (to maintain transitivity and symmetry), resulting in the universal relation.
So total relations = $R_1$ + universal relation = 2.
Select your answer and click "Check Answers"
Each question has an Assertion (A) and a Reason (R). Choose the correct option.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is NOT the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Frequently asked questions in board exams
Q. Show that the relation R in the set $\{1, 2, 3\}$ given by $R = \{(1, 2), (2, 1)\}$ is symmetric but neither reflexive nor transitive.
$(1, 1) \notin R$ → Not reflexive ✗
$(1, 2) \in R$ and $(2, 1) \in R$ → Symmetric ✓
$(1, 2) \in R$ and $(2, 1) \in R$, but $(1, 1) \notin R$ → Not transitive ✗
Q. Show that $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = 2x$ is one-one and onto.
Let $f(x_1) = f(x_2) \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2$. So $f$ is one-one. ✓
For any $y \in \mathbb{R}$, take $x = y/2$. Then $f(x) = 2(y/2) = y$. So $f$ is onto. ✓
Q. Let $A = \{1, 2, 3\}$. Then the number of equivalence relations containing $(1, 2)$ is:
$R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$
Adding any remaining pair forces the universal relation.
Q. Show that the relation R in the set $\mathbb{Z}$ given by $R = \{(a, b) : 2 \mid (a - b)\}$ is an equivalence relation.
$2 \mid (a - a) = 0$ → $(a, a) \in R$ ✓
$2 \mid (a - b) \Rightarrow 2 \mid (b - a)$ → $(b, a) \in R$ ✓
$2 \mid (a-b)$ and $2 \mid (b-c) \Rightarrow 2 \mid (a-c)$ ✓
Q. Check whether the relation R defined on $\{1, 2, 3, 4, 5, 6\}$ by $R = \{(a, b) : b = a + 1\}$ is reflexive, symmetric or transitive.
$(1,1) \notin R$ → not reflexive ✗
$(1,2) \in R$, but $(2,1) \notin R$ → not symmetric ✗
$(1,2), (2,3) \in R$, but $(1,3) \notin R$ → not transitive ✗
Q. Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = x^4$. Choose the correct answer:
$f(-1) = f(1) = 1$, so not one-one.
Negative numbers have no pre-image → not onto.