Relations and Functions: Class 12 Maths · Chapter 1

Mathematics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

105 min
1 simulation
41 practice questions
17 MCQs

What you will learn in Relations and Functions

Describe a relation as a set of ordered pairs and recognise empty and universal relations.
Test a relation for the reflexive, symmetric and transitive properties, giving a counterexample whenever a property fails.
Prove that a relation is an equivalence relation and find its equivalence classes as a partition of the set.
Decide whether a function is one-one, onto or bijective, using algebra, graphs or counting.
Explain why a one-one function on a finite set is onto, and why this can fail on an infinite set.
Form the composition of two functions and show that the order of composition matters.
Decide whether a function is invertible, find its inverse and check it through both compositions.
Count functions and relations with given properties on small finite sets.

Original companion text aligned to NCERT Mathematics Class 12, Chapter 1, pp. 1–17. Not affiliated with or endorsed by NCERT.

1.1 Introduction

A class register pairs each student with a roll number, a railway chart pairs each passenger with a berth, and a scorecard pairs each batter with a score. Mathematics keeps only the bare bones of such pairings, a collection of ordered pairs, and studies the shapes that collection can take. You met relations, functions, domain, codomain and range in Class XI. In this chapter we sort relations by three properties, sort functions by how they cover their codomain, and then chain functions together and run them backwards.

Recall that for sets $A$ and $B$ , the Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ with $a \in A$ and $b \in B$ . A relation (संबंध) from $A$ to $B$ is nothing more than a subset of $A \times B$ . The pairs may be chosen by a meaningful rule, such as is a factor of, or by no rule at all; once they are chosen, the relation simply is that set of pairs. When $(a, b) \in R$ we write $a R b$ and say that $a$ is related to $b$ .

Definition

Relation

A relation $R$ from $A$ to $B$ is any subset of $A \times B$ . When $A = B$ , we call $R$ a relation in $A$ ; it is then a subset of $A \times A$ .

ExampleWith $A = {2, 3, 5}$ and $B = {4, 6, 9, 10}$ , the rule ' $a$ divides $b$ ' selects $R = {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (5, 10)}$ .

2/3The rule $a$ divides $b$ keeps 6 of them. That set of pairs is the relation $R$ .

Fig. 1.1A relation is a choice of cells in the grid of all pairs.

A grid has rows 2, 3 and 5 for the set A and columns 4, 6, 9 and 10 for the set B, with a small dot in each of its 12 cells. In the first step every cell is plain: these are all the pairs of A × B, and the label says there are 3 × 4 = 12 of them. In the second step six cells are ringed, the pairs where a divides b: (2, 4), (2, 6), (2, 10), (3, 6), (3, 9) and (5, 10); the other dots fade and the label names the relation R. In the third step the grid slides to the left and the same six pairs appear on the right as arrows from the circles 2, 3 and 5 to the circles 4, 6, 9 and 10.

A function is a relation of a special kind: every element of the first set has exactly one arrow leaving it. In Fig. 1.1 the number 2 has three arrows, so this relation is not a function from $A$ to $B$ . Keep both pictures in mind, the grid and the arrows, as we classify relations in Section 1.2 and functions in Section 1.3.

Quick check

1 of 3

A set

A

has 3 elements and a set

B

has 4. How many different relations from

A

B

are there? Count every subset of

A \times B

, the empty one included.

Show answer

Answer 4096

A \times B

has

3 \times 4 = 12

pairs, and each pair is either in

R

or out of it, so there are

2^{12} = 4096

relations.

1.2 Types of Relations

From now on most of our relations live inside a single set: a relation in $A$ is a subset of $A \times A$ . Two subsets are always available, however strange $A$ is. One is the empty set $\emptyset$ , which relates nothing to anything. The other is the whole of $A \times A$ , which relates everything to everything.

Definition

Empty relation

The relation $R = \emptyset$ : the subset of $A \times A$ that contains no pairs at all, so nothing in $A$ is linked to anything.

ExampleIn $A = {1, 2, \dots, 9}$ the rule $a + b = 20$ is never met, because the largest possible sum is $9 + 9 = 18$ .

Definition

Universal relation

The relation $R = A \times A$ , which contains every ordered pair from $A$ ; each element is linked to all elements, itself included.

ExampleIn the same $A$ , the rule $∣ a - b ∣ < 10$ holds for all $9 \times 9 = 81$ pairs.

These two extremes, the empty (रिक्त) and the universal (सार्वत्रिक) relation, are called the trivial relations. On their own they are dull, but they make excellent test cases for the properties that follow.

Example 1NCERT Example 1 · p. 2

Months of the year

Let $M$ be the set of the 12 months of a non-leap year. In $M$ , let $a R_{1} b$ mean that months $a$ and $b$ together have more than 62 days, and let $a R_{2} b$ mean that the numbers of days in $a$ and $b$ differ by at most 3. Show that $R_{1}$ is the empty relation and $R_{2}$ is the universal relation.

Try it yourself first. Then reveal the solution one step at a time.

Three ways to write one relation. Take $R$ in ${1, 2, 3, 4}$ where $b$ is twice $a$ . In roster form, $R = {(1, 2), (2, 4)}$ . In set-builder form, $R = {(a, b) : b = 2 a}$ . In a third, very common style, we just say that $a R b$ if and only if $b = 2 a$ . All three describe the same set of pairs, so use whichever is clearest at the moment.

Reflexive, symmetric and transitive relations

The relations that matter most in mathematics, such as equality, congruence of triangles and 'leaves the same remainder', share three habits. Draw a relation in $A$ as arrows between the elements of $A$ , with a small loop when an element is related to itself. Each habit then becomes a rule about the arrows (Fig. 1.2).

Definition

Reflexive relation

A relation $R$ in $A$ with $(a, a) \in R$ for every $a \in A$ : each element is related to itself.

Example'was born in the same year as', in a set of people.

Definition

Symmetric relation

A relation $R$ in $A$ such that $(a, b) \in R$ implies $(b, a) \in R$ , for all $a, b \in A$ .

Example $∣ a - b ∣ = 5$ in the set $𝐙$ of integers.

Definition

Transitive relation

A relation $R$ in $A$ such that $(a, b) \in R$ and $(b, c) \in R$ together imply $(a, c) \in R$ , for all $a, b, c \in A$ .

Example'is older than', in a set of people.

4/4Side by side. A relation with all three habits is an equivalence relation.

Fig. 1.2Loops, return arrows and shortcuts: the three properties as pictures.

Three panels each show the points 1, 2 and 3 as circles. In the first step the left panel is highlighted: every point has a loop arrow back to itself, and one ordinary arrow runs from 1 to 2, so the relation is reflexive. In the second step the middle panel is highlighted: arrows join 1 and 2 in both directions, and 2 and 3 in both directions, so every arrow has a partner coming back and the relation is symmetric. In the third step the right panel is highlighted: arrows go from 1 to 2 and from 2 to 3, and a dashed shortcut arrow goes straight from 1 to 3, as transitivity demands. In the last step all three panels are shown at full strength side by side.

Two warnings save many marks. First, symmetry and transitivity are if–then statements. If a relation has no two-step paths at all, transitivity demands nothing, and the relation is transitive automatically (we say vacuously). Second, reflexivity is a claim about every element, so one missing loop spoils it: in ${1, 2, 3}$ the relation ${(1, 1), (2, 2)}$ is not reflexive, because $(3, 3)$ is absent.

To prove that a property holds, argue with arbitrary elements. To show that it fails, one specific counterexample is enough, and you should name it.

Definition

Equivalence relation

A relation that is reflexive, symmetric and transitive.

ExampleEquality, $a = b$ , in any set; 'leaves the same remainder on division by 7 as' in $𝐙$ .

InteractiveRelation lab

A 4 by 4 grid of switches, one for each ordered pair (a, b) with a and b from 1 to 4, next to an arrow diagram of the same relation on four points. Tapping a switch adds or removes that pair, and you can shrink the set to three elements. A panel says whether the relation is reflexive, symmetric and transitive; when a property fails it names a missing pair that proves it, and when all three hold it lists the equivalence classes. Buttons load ready-made relations, clear the grid, or add the pairs that each property needs, and a challenge menu asks you to build relations with given properties, such as symmetric and transitive but not reflexive.

Fig. 1.3Switch pairs on and off in a relation on

{1, 2, 3, 4}

and watch the three properties update, with a reason for every failure.

Example 2NCERT Example 2 · p. 3

Rectangles with equal area

Let $S$ be the set of all rectangles drawn on a sheet of paper, and let let $P R Q$ mean that $P$ and $Q$ have the same area. Show that $R$ is an equivalence relation in $S$ .

Try it yourself first. Then reveal the solution one step at a time.

Notice the pattern in Example 2: the relation was 'has the same area as', an equality of some attribute. Every relation of the form 'has the same ___ as' inherits all three properties directly from equality. Example 21 proves this once and for all.

Example 3NCERT Example 3 · p. 3

Opposite signs

In the set $𝐑^{*}$ of non-zero real numbers, let $R = {(a, b) : a b < 0}$ , so that $a R b$ when $a$ and $b$ have opposite signs. Prove that $R$ is symmetric, and that it fails to be reflexive and fails to be transitive.

Try it yourself first. Then reveal the solution one step at a time.

There is a simple reason behind Example 3. Two 'opposite sign' steps bring you back to your original sign, so the shortcut that transitivity asks for would always join two numbers of the same sign, and $R$ never contains such a pair.

Example 4NCERT Example 4 · p. 3

Reading properties off a list

Let $A = {4, 5, 6}$ and $R = {(4, 4), (5, 5), (6, 6), (5, 6), (6, 4)}$ . Decide which of the three properties $R$ has.

Try it yourself first. Then reveal the solution one step at a time.

For a small relation given as a list, check transitivity systematically: for every pair $(a, b)$ , look at each pair $(b, c)$ that starts where it ends, and confirm that $(a, c)$ is present. Loops never cause trouble, since $(a, a)$ followed by $(a, c)$ only asks for $(a, c)$ itself.

Example 5NCERT Example 5 · p. 3

Same remainder on division by 4

Show that $R = {(a, b) : 4 divides a - b}$ is an equivalence relation in the set $𝐙$ of integers.

Try it yourself first. Then reveal the solution one step at a time.

Equivalence classes

Run the relation of Example 5 on the numbers 0 to 11 and watch what happens (Fig. 1.4). The number 1 is related to 5 and 9, which differ from it by multiples of 4, and to nothing else in the list. The number 2 is related to 6 and 10. The twelve numbers fall into four clusters, and every number lands in exactly one of them.

3/3Every number belongs to exactly one of $[0], [1], [2], [3]$ , and the four classes together make up the whole set.

Fig. 1.4An equivalence relation sorts a set into classes that do not overlap.

Twelve circles numbered 0 to 11. In the first step they are scattered with no order, and the rule a R b if 4 divides a minus b is written below. In the second step only 1, 5 and 9 stay bright and are ringed, forming the class of 1, while the other numbers fade. In the third step all the circles slide into four dashed boxes labelled [0], [1], [2] and [3]: the box [0] holds 0, 4 and 8; [1] holds 1, 5 and 9; [2] holds 2, 6 and 10; and [3] holds 3, 7 and 11. The boxes do not overlap and together they contain every number, as the label below them states.

Why does the remainder decide everything? Write $a = 4 q + r$ and $b = 4 q^{'} + r^{'}$ , with remainders $r, r^{'} \in {0, 1, 2, 3}$ . Then $a - b = 4 (q - q^{'}) + (r - r^{'})$ , so 4 divides $a - b$ exactly when 4 divides $r - r^{'}$ . Since $- 3 \leq r - r^{'} \leq 3$ , the only multiple of 4 it can equal is 0. So $a R b$ if and only if $r = r^{'}$ .

Definition

Equivalence class

For an equivalence relation $R$ in $A$ and an element $a \in A$ , the class of $a$ is $[a] = {x \in A : x R a}$ , the set of everything related to $a$ .

ExampleUnder 'same remainder on division by 4' in $𝐙$ , $[1] = {\dots, - 7, - 3, 1, 5, 9, \dots}$ , and $[1] = [5] = [- 3]$ .

PropertyEquivalence relations and partitions

Let $R$ be an equivalence relation in a set $A$ . Then (i) every element lies in its own class, $a \in [a]$ ; (ii) any two classes are either identical or disjoint; (iii) the union of all the classes is $A$ . Conversely, if $A$ is split into non-empty, pairwise disjoint pieces whose union is $A$ (a partition of $A$ ), then 'lies in the same piece as' is an equivalence relation whose classes are exactly those pieces.

(i) holds because $R$ is reflexive: $a R a$ . It also gives (iii), since each $a \in A$ lies in at least one class, its own.

(ii) Suppose $[a]$ and $[b]$ share an element $c$ , so $c R a$ and $c R b$ . By symmetry $a R c$ , and then transitivity gives $a R b$ . Now take any $x \in [a]$ : from $x R a$ and $a R b$ , transitivity gives $x R b$ , so $x \in [b]$ . Hence $[a] \subseteq [b]$ . Swapping the roles of $a$ and $b$ gives $[b] \subseteq [a]$ , so overlapping classes are equal.

For the converse, 'in the same piece as' is reflexive (each element lies in some piece), symmetric (the wording ignores order) and transitive: if $x, y$ share a piece and $y, z$ share a piece, both pieces contain $y$ , so they are the same piece because distinct pieces are disjoint.

So the same information can be given in two ways: as an equivalence relation, or as a partition of the set into classes. The integers split into 2 classes under 'same parity', into 4 classes under 'same remainder on division by 4', and into $n$ classes $[0], [1], \dots, [n - 1]$ under 'same remainder on division by $n$ '. In the other direction, any way of sorting a set into non-overlapping groups defines an equivalence relation: 'is in the same group as'.

Example 6NCERT Example 6 · p. 4

Sorting by number of digits

Let $A = {1, 2, \dots, 20}$ , and let $a R b$ mean that $a$ and $b$ have the same number of digits. Prove that $R$ is an equivalence relation, list its classes, and confirm that $R$ never links two numbers from different classes.

Try it yourself first. Then reveal the solution one step at a time.

Classes need not be finite lists. In the plane, call two points related when they are at the same distance from the origin $O$ . This is again 'has the same ___ as', so it is an equivalence relation. Use the simulation to explore the class of the point $P (3, 4)$ .

SimulationPoints related to P

A coordinate grid from minus 6 to 6 on both axes, with the origin O and a fixed point P at (3, 4), whose distance 5 from O is drawn and measured. A second point Q can be dragged to any grid point; its distance from O is drawn and updated as it moves, so you can hunt for the points that are related to P under the relation 'same distance from O'.

Fig. 1.5Drag

Q

to grid points whose distance from

O

equals

O P = 5

Try this

Drag $Q$ to $(4, 3)$ , then to $(0, - 5)$ . Is each one related to $P$ ?
Find all 12 grid points at distance 5 from $O$ . What shape do they lie on?
Drag $Q$ to $(1, 1)$ . Which grid points are in the same class as $(1, 1)$ ?

The class of $P (3, 4)$ is the whole circle of radius 5 centred at $O$ . It contains infinitely many points, only twelve of which have whole-number coordinates: $(\pm 5, 0)$ , $(0, \pm 5)$ , $(\pm 3, \pm 4)$ and $(\pm 4, \pm 3)$ . The classes of this relation are the circles centred at $O$ , together with the one-point class ${O}$ , and they partition the plane.

Practice · 16 questions

Practice 1.1: questions with answers

1
Short answerMediumNCERT Exercise 1.1, Q1
Decide whether each relation is reflexive, symmetric and transitive. Give a counterexample for every property that fails.
1. $R = {(x, y) : x + y = 10}$ in $A = {1, 2, \dots, 10}$
  Show solution
  AnswerSymmetric only.
  Not reflexive: $1 + 1 \neq 10$ , so $(1, 1) \notin R$ (the only loop is $(5, 5)$ ). Symmetric: $x + y = 10$ gives $y + x = 10$ . Not transitive: $(1, 9)$ and $(9, 1)$ are in $R$ , but $(1, 1)$ is not.
2. $R = {(x, y) : x < 0 and y > 0}$ in $𝐙$
  Show solution
  AnswerTransitive only.
  Not reflexive: $(1, 1) \notin R$ . Not symmetric: $(- 1, 1) \in R$ but $(1, - 1) \notin R$ . Transitive: $(x, y), (y, z) \in R$ would need $y > 0$ and $y < 0$ at once, which never happens, so there are no two-step paths and the condition holds vacuously.
3. In $A = {2, 3, \dots, 10}$ , $x R y$ when $x$ and $y$ have a common factor greater than 1
  Show solution
  AnswerReflexive and symmetric, not transitive.
  Reflexive: each $x \geq 2$ shares the factor $x$ with itself. Symmetric: a common factor of $x$ and $y$ is a common factor of $y$ and $x$ . Not transitive: $(2, 6)$ and $(6, 3)$ are in $R$ , but 2 and 3 have no common factor greater than 1.
4. $R = {(x, y) : ⌊ x ⌋ = ⌊ y ⌋}$ in $𝐑$ , where $⌊ x ⌋$ is the greatest integer not exceeding $x$
  Show solution
  AnswerAn equivalence relation.
  This is 'has the same greatest integer as', an equality of an attribute, so it is reflexive, symmetric and transitive. Its classes are the intervals $[n, n + 1)$ for $n \in 𝐙$ .
5. In the set of students of a school: ' $x$ and $y$ belong to the same house'
  Show solution
  AnswerAn equivalence relation.
  Equality of an attribute (the house), so all three properties hold.
6. In the same set: ' $x$ was born exactly one day before $y$ '
  Show solution
  AnswerNone of the three.
  Not reflexive: nobody is born a day before themselves. Not symmetric: if $x$ was born a day before $y$ , then $y$ was born a day after $x$ . Not transitive: if $x$ is a day older than $y$ and $y$ a day older than $z$ , then $x$ is two days older than $z$ .
7. In the same set: ' $x$ is younger than $y$ '
  Show solution
  AnswerTransitive only.
  Not reflexive and not symmetric (if $x$ is younger than $y$ , then $y$ is not younger than $x$ ). Transitive: younger than someone who is younger than $z$ means younger than $z$ .
ProofMediumNCERT Exercise 1.1, Q2
Show that $R = {(a, b) : a \geq 2 b}$ in $𝐑$ is neither reflexive, nor symmetric, nor transitive.
Show solution
Not reflexive: $1 \geq 2 \times 1$ is false, so $(1, 1) \notin R$ .
Not symmetric: $(4, 1) \in R$ since $4 \geq 2$ , but $(1, 4) \notin R$ since $1 \geq 8$ is false.
Not transitive: $(- 8, - 4) \in R$ since $- 8 \geq - 8$ , and $(- 4, - 2) \in R$ since $- 4 \geq - 4$ . But $(- 8, - 2) \notin R$ , since $- 8 \geq - 4$ is false.
Short answerEasyNCERT Exercise 1.1, Q3
In $A = {1, 2, \dots, 10}$ , list the relation $R = {(a, b) : b = 2 a}$ and check whether it is reflexive, symmetric or transitive.
Show solution
Answer $R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}$ ; none of the three properties holds.
$R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}$ . Not reflexive: $(1, 1) \notin R$ . Not symmetric: $(1, 2) \in R$ but $(2, 1) \notin R$ . Not transitive: $(1, 2), (2, 4) \in R$ but $(1, 4) \notin R$ .
ProofEasyNCERT Exercise 1.1, Q4
In a group of people, let $x R y$ mean ' $x$ is not taller than $y$ '. Prove that $R$ has the reflexive and transitive properties but lacks symmetry (assume the group has two people of different heights).
Show solution
$x R y$ means $h (x) \leq h (y)$ , where $h$ is height. Reflexive: $h (x) \leq h (x)$ . Transitive: $h (x) \leq h (y)$ and $h (y) \leq h (z)$ give $h (x) \leq h (z)$ .
Not symmetric: if Asha is 150 cm and Ravi is 165 cm tall, then Asha $R$ Ravi, but Ravi is taller than Asha, so Ravi is not related to Asha.
Short answerMediumNCERT Exercise 1.1, Q5
Check whether $R = {(a, b) : a \leq b + 1}$ in $𝐑$ is reflexive, symmetric or transitive.
Show solution
AnswerReflexive only.
Reflexive: $a \leq a + 1$ always. Not symmetric: $(0, 5) \in R$ since $0 \leq 6$ , but $(5, 0) \notin R$ since $5 \leq 1$ is false.
Not transitive: $(2, 1) \in R$ since $2 \leq 2$ , and $(1, 0) \in R$ since $1 \leq 1$ , but $(2, 0) \notin R$ since $2 \leq 1$ is false.
ProofEasyNCERT Exercise 1.1, Q6
Show that $R = {(4, 6), (6, 4), (5, 5)}$ in $A = {4, 5, 6}$ is symmetric but neither reflexive nor transitive.
Show solution
Symmetric: $(4, 6)$ and $(6, 4)$ come as a pair, and the loop $(5, 5)$ is its own reverse.
Not reflexive: $(4, 4) \notin R$ . Not transitive: the path $(4, 6), (6, 4)$ demands $(4, 4)$ , which is missing.
ProofEasyNCERT Exercise 1.1, Q7
In the set of mobile phones on sale in a shop, show that ' $x$ and $y$ have the same price' is an equivalence relation.
Show solution
Let $p (x)$ be the price of phone $x$ . Reflexive: $p (x) = p (x)$ . Symmetric: $p (x) = p (y) \Rightarrow p (y) = p (x)$ . Transitive: $p (x) = p (y)$ and $p (y) = p (z)$ give $p (x) = p (z)$ .
ProofMediumNCERT Exercise 1.1, Q8
Let $A = {10, 11, \dots, 17}$ and $R = {(a, b) : a + b is even}$ . Prove that $R$ is reflexive, symmetric and transitive. Then check that any two numbers from ${10, 12, 14, 16}$ are linked by $R$ , that the same holds inside ${11, 13, 15, 17}$ , and that $R$ never links a number of the first set with a number of the second.
Show solution
Reflexive: $a + a = 2 a$ is even. Symmetric: $a + b = b + a$ . Transitive: if $a + b$ and $b + c$ are even, then $a + c = (a + b) + (b + c) - 2 b$ is even.
Two even numbers have an even sum, and so do two odd numbers, so each of the two sets is related within itself. An even number plus an odd number is odd, so no element of ${10, 12, 14, 16}$ is related to any element of ${11, 13, 15, 17}$ . These two sets are the equivalence classes.
9
Short answerMediumNCERT Exercise 1.1, Q9
Let $A = {x \in 𝐙 : - 6 \leq x \leq 6}$ . Show that each relation is an equivalence relation in $A$ , and find the class of 1 in each case.
1. $R_{1} = {(a, b) : 5 divides a - b}$
  Show solution
  Answer $[1] = {- 4, 1, 6}$
  Reflexive: $a - a = 0 = 5 \times 0$ . Symmetric: if $a - b = 5 k$ then $b - a = 5 (- k)$ . Transitive: if $a - b = 5 k$ and $b - c = 5 m$ , then $a - c = 5 (k + m)$ . The elements of $A$ that differ from 1 by a multiple of 5 are $1 - 5 = - 4$ , $1$ and $1 + 5 = 6$ , so $[1] = {- 4, 1, 6}$ .
2. $R_{2} = {(a, b) : a^{2} = b^{2}}$
  Show solution
  Answer $[1] = {- 1, 1}$
  This is 'has the same square as', an equality of an attribute, so it is an equivalence relation. $b^{2} = 1$ means $b = \pm 1$ , so $[1] = {- 1, 1}$ .
10
Short answerHardNCERT Exercise 1.1, Q10
Using $A = {1, 2, 3}$ , give an example of a relation in $A$ that is:
1. reflexive, but neither symmetric nor transitive;
  Show solution
  AnswerFor example, the pairs $(2, 1)$ and $(1, 3)$ together with the three loops $(1, 1)$ , $(2, 2)$ , $(3, 3)$ .
  All loops are present. $(2, 1) \in R$ but $(1, 2) \notin R$ , so it is not symmetric; $(2, 1), (1, 3)$ demands $(2, 3)$ , which is missing, so it is not transitive.
2. symmetric and transitive, but not reflexive;
  Show solution
  AnswerFor example, ${(1, 2), (2, 1), (1, 1), (2, 2)}$ .
  Every pair among 1 and 2 is present, so it is symmetric and transitive, but $(3, 3)$ is missing.
3. transitive, but neither reflexive nor symmetric;
  Show solution
  AnswerFor example ${(1, 2)}$ .
  There is no two-step path, so it is transitive vacuously; there are no loops, and $(2, 1)$ is missing.
4. reflexive and symmetric, but not transitive;
  Show solution
  AnswerFor example, the pairs $(1, 2)$ , $(2, 1)$ , $(2, 3)$ , $(3, 2)$ together with the three loops.
  All loops are present and every pair has its reverse, but $(1, 2), (2, 3)$ demands $(1, 3)$ , which is missing.
5. an equivalence relation other than the identity and universal relations.
  Show solution
  AnswerFor example, the pairs $(1, 3)$ and $(3, 1)$ together with the three loops.
  Its classes are ${1, 3}$ and ${2}$ , a partition of $A$ , so by the partition theorem it is an equivalence relation.
ProofHardNCERT Exercise 1.1, Q11
In the set of points of the coordinate plane, let $P R Q$ when $∣ x_{P} ∣ + ∣ y_{P} ∣ = ∣ x_{Q} ∣ + ∣ y_{Q} ∣$ . Show that $R$ is an equivalence relation, and describe the class of the point $(2, 1)$ .
Show solution
AnswerThe square with vertices $(3, 0)$ , $(0, 3)$ , $(- 3, 0)$ and $(0, - 3)$ .
$R$ is 'has the same value of $∣ x ∣ + ∣ y ∣$ as', an equality of an attribute, so it is an equivalence relation.
For $(2, 1)$ the value is 3, so its class is ${(x, y) : ∣ x ∣ + ∣ y ∣ = 3}$ . In the first quadrant this is the segment $x + y = 3$ from $(3, 0)$ to $(0, 3)$ ; the other quadrants give its mirror images. Together they form a square with vertices $(\pm 3, 0)$ and $(0, \pm 3)$ , standing on one corner.
Short answerMediumNCERT Exercise 1.1, Q12
Two rectangles are similar when the ratio of the longer side to the shorter side is the same. Show that similarity is an equivalence relation in the set of all rectangles. Which of these are related: $R_{1}$ is $2 cm \times 3 cm$ , $R_{2}$ is $4 cm \times 5 cm$ , $R_{3}$ is $6 cm \times 10 cm$ , $R_{4}$ is $4 cm \times 6 cm$ and $R_{5}$ is $3 cm \times 5 cm$ ?
Show solution
Answer $R_{1}$ and $R_{4}$ are related; $R_{3}$ and $R_{5}$ are related; $R_{2}$ is related only to itself.
Similarity is 'has the same side ratio as', so it is an equivalence relation.
The ratios are $R_{1} : \frac{3}{2}$ , $R_{2} : \frac{5}{4}$ , $R_{3} : \frac{10}{6} = \frac{5}{3}$ , $R_{4} : \frac{6}{4} = \frac{3}{2}$ and $R_{5} : \frac{5}{3}$ . So $R_{1} R R_{4}$ and $R_{3} R R_{5}$ , while $R_{2}$ is in a class of its own among these five.
Short answerMediumNCERT Exercise 1.1, Q13
In the set of all polyhedra (solids with flat faces), let $P R Q$ when $P$ and $Q$ have the same number of faces. Show that $R$ is an equivalence relation. Is a square pyramid related to a cube? Is a pentagonal pyramid?
Show solution
AnswerA square pyramid (5 faces) is not related to a cube (6 faces); a pentagonal pyramid (6 faces) is.
$R$ is 'has the same number of faces as', an equality of an attribute, so it is an equivalence relation.
A cube has 6 faces. A square pyramid has $1 + 4 = 5$ faces, so it is not related to the cube (it is related to a triangular prism, which also has 5). A pentagonal pyramid has $1 + 5 = 6$ faces, so it is in the cube's class, together with every cuboid.
Short answerMediumNCERT Exercise 1.1, Q14
In the set of all circles in the $x y$ -plane, let $C_{1} R C_{2}$ when $C_{1}$ and $C_{2}$ have the same centre. Show that $R$ is an equivalence relation, and describe the class of the circle $x^{2} + y^{2} - 4 x + 6 y - 3 = 0$ .
Show solution
AnswerAll circles $(x - 2)^{2} + (y + 3)^{2} = r^{2}$ with $r > 0$ .
'Has the same centre as' is an equality of an attribute, so $R$ is an equivalence relation.
$x^{2} - 4 x + y^{2} + 6 y = 3$ gives $(x - 2)^{2} + (y + 3)^{2} = 3 + 4 + 9 = 16$ , a circle with centre $(2, - 3)$ and radius 4. Its class is the family of all circles centred at $(2, - 3)$ : $(x - 2)^{2} + (y + 3)^{2} = r^{2}$ , $r > 0$ .
Quick answerMediumNCERT Exercise 1.1, Q15
In ${1, 2, 3, 4}$ , let $R$ consist of the four loops together with $(2, 3), (3, 2), (1, 2), (2, 1)$ . Which statement is correct?
$R$ has all three properties. $R$ has only the symmetric and transitive properties. $R$ has only the reflexive and transitive properties. $R$ has only the reflexive and symmetric properties.
Show answer
Answer (D) $R$ has only the reflexive and symmetric properties.
All four loops are present and every pair comes with its reverse, but $(1, 2), (2, 3) \in R$ while $(1, 3) \notin R$ .
Quick answerEasyNCERT Exercise 1.1, Q16
Let $R = {(a, b) : a^{2} + b = 10, b < 0}$ in $𝐙$ . Which pair belongs to $R$ ?
$(2, 6)$ $(- 3, 1)$ $(4, - 6)$ $(3, - 1)$
Show answer
Answer (C) $(4, - 6)$
$4^{2} + (- 6) = 10$ and $- 6 < 0$ . The pairs $(2, 6)$ and $(- 3, 1)$ satisfy the equation but have $b > 0$ , and $(3, - 1)$ gives $9 - 1 = 8$ .

1.3 Types of Functions

In Class XI you met the identity, constant, polynomial, rational, modulus and signum functions, drew their graphs and combined functions by adding and multiplying them. Here we ask a new question about a function $f : X \to Y$ : how does it cover the codomain $Y$ ? Does it ever send two inputs to the same output? Does it reach every element of $Y$ ?

Definition

Function

A relation $f$ from $X$ to $Y$ in which every $x \in X$ is related to exactly one element of $Y$ , written $f (x)$ . $X$ is the domain, $Y$ the codomain, and ${f (x) : x \in X}$ the range.

ExampleThe PIN code of a post office: every post office has exactly one PIN code, though one PIN code may serve several post offices.

Fig. 1.6 shows the four possibilities with four inputs. Look at two things in each picture: whether two arrows ever land on the same square, and whether some square is left without an arrow.

1/4Two inputs share the output $a$ , and $b$ is never reached.

Fig. 1.6Two questions about every function: are outputs ever shared, and is anything in the codomain missed?NCERT Fig. 1.2 · p. 8

Four circles labelled 1 to 4, the set X, sit in a row above a row of squares, the set Y, and each circle sends one arrow down to a square. Step one: 1 and 2 both point to a, 3 points to c and 4 to d, while b receives no arrow and is drawn dashed; the labels read one-one: no and onto: no. Step two: there are five squares a to e, and the arrows go to a, b, d and e, all different, but c is missed and dashed: one-one yes, onto no. Step three: there are three squares; 1 goes to a, 2 and 3 both go to b, and 4 goes to c, so every square is hit but b is hit twice: one-one no, onto yes. Step four: four squares, with arrows 1 to b, 2 to d, 3 to a and 4 to c, so every square is hit exactly once: one-one yes, onto yes.

Definition

One-one (injective) function

A function $f : X \to Y$ in which different inputs give different outputs. Equivalently, whenever $f (x_{1}) = f (x_{2})$ we must have $x_{1} = x_{2}$ . A function that is not one-one is called many-one.

Example $f (x) = x + 3$ on $𝐍$ is one-one; $g (x) = x^{2}$ on $𝐙$ is many-one, since $g (- 2) = g (2)$ .

The form ' $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ ' is the one to use in proofs: assume two outputs are equal and deduce that the inputs were equal. To show that $f$ is not one-one, exhibit two different inputs with the same output.

Definition

Onto (surjective) function

A function $f : X \to Y$ in which every $y \in Y$ is the image of at least one $x \in X$ ; that is, the equation $f (x) = y$ has a solution in $X$ for every $y \in Y$ .

Example $f : 𝐑 \to 𝐑$ , $f (x) = x^{3}$ is onto, because $y$ is the image of $\sqrt[3]{y}$ .

Put differently, $f$ is onto exactly when its range equals its codomain. So 'onto' depends on the codomain we declare: $x \mapsto x^{2}$ is not onto as a map $𝐑 \to 𝐑$ , since nothing goes to $- 1$ , but it is onto as a map $𝐑 \to [0, \infty)$ . Always read the arrow $f : X \to Y$ before you judge.

Definition

Bijective function

A function that is both one-one and onto. Every element of the codomain is then the image of exactly one element of the domain.

Example $f : 𝐑 \to 𝐑$ , $f (x) = 2 x - 7$ .

InteractiveMapping lab

Two columns of points: the domain X with 2 to 5 elements numbered from 1, and the codomain Y with 2 to 5 elements lettered from a, with menus to set the two sizes. For each element of X a drop-down menu chooses its image in Y, and the arrow diagram updates. A status panel says whether the function is one-one, naming two inputs that share an output if not, and onto, naming a missed output if not. A Reverse arrows switch flips every arrow and says whether the reversed picture is a function from Y to X. Challenge buttons ask for a function with given properties and explain when one is impossible for the chosen sizes, and a counter shows how many functions and how many one-one functions exist for those sizes.

Fig. 1.7Choose where each element of

X

goes and see at once whether your function is one-one, onto or bijective, and whether reversing its arrows gives a function.

Example 7NCERT Example 7 · p. 8

Accession numbers in a library

A school library has 1850 books, and each book is stamped with an accession number from 1 to 1850, no two books sharing a number. Let $B$ be the set of these books and $f : B \to 𝐍$ , $f (b) =$ the accession number of $b$ . Prove that $f$ is one-one and that it is not onto.

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 8

Shifting the natural numbers

Show that $f : 𝐍 \to 𝐍$ , $f (x) = x + 3$ , is one-one but not onto.

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 9

A bijection of the real line

Prove that $f : 𝐑 \to 𝐑$ , $f (x) = \frac{x + 4}{3}$ , is one-one and onto.

Try it yourself first. Then reveal the solution one step at a time.

Graphs give a quick picture of both properties (Fig. 1.8). Draw horizontal lines $y = c$ across the graph of a function $f : 𝐑 \to 𝐑$ . If some line meets the graph twice, two inputs share the output $c$ , and $f$ is not one-one. If some line misses the graph, $c$ is never reached, and $f$ is not onto. A bijection of $𝐑$ is a graph that every horizontal line meets exactly once; the straight line of Example 9, with slope $\frac{1}{3}$ , passes this test.

1/3The line $y = 5$ meets $y = x^{2} - 4 x$ at $x = - 1$ and $x = 5$ : not one-one.

Fig. 1.8The horizontal-line test for a function from

𝐑

𝐑

.NCERT Fig. 1.4 · p. 9

Axes with x from minus 2 to 6 and y from minus 7 to 9. Step one: the U-shaped graph of y = x squared minus 4x, lowest at the point (2, minus 4), is cut by the dashed horizontal line y = 5 at two marked points, x = minus 1 and x = 5. Step two: the dashed line moves down to y = minus 6, below the lowest point, and meets the graph nowhere. Step three: the U-shaped graph fades and an S-shaped cubic, y equals one half of (x minus 2) cubed plus 1, appears; three dashed lines at y = minus 5, minus 1 and 5 each meet it at exactly one marked point.

Example 10NCERT Example 10 · p. 9

Counting digits

Let $f : 𝐍 \to 𝐍$ give the number of digits of $n$ , so that $f (7) = 1$ , $f (42) = 2$ and $f (2026) = 4$ . Show that $f$ is onto but not one-one.

Try it yourself first. Then reveal the solution one step at a time.

Example 11NCERT Example 11 · p. 9

A parabola that fails twice

Show that $f : 𝐑 \to 𝐑$ , $f (x) = x^{2} - 4 x$ , is neither one-one nor onto.

Try it yourself first. Then reveal the solution one step at a time.

Example 12NCERT Example 12 · p. 9

Swapping neighbours

Define $f : 𝐍 \to 𝐍$ by $f (n) = n + 1$ if $n$ leaves remainder 1 on division by 3, $f (n) = n - 1$ if it leaves remainder 2, and $f (n) = n$ if $n$ is a multiple of 3. So $f$ swaps 1 and 2, swaps 4 and 5, swaps 7 and 8, and fixes 3, 6, 9, …. Show that $f$ is one-one and onto.

Try it yourself first. Then reveal the solution one step at a time.

On a finite set the two properties are tied together. Picture four letters being posted into four letter-boxes, each letter going into exactly one box.

Example 13NCERT Example 13 · p. 10

Onto forces one-one on a finite set

Let $S = {a, b, c, d}$ . Show that every onto function $f : S \to S$ is one-one.

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 10

One-one forces onto on a finite set

With $S = {a, b, c, d}$ again, show that every one-one function $f : S \to S$ is onto.

Try it yourself first. Then reveal the solution one step at a time.

PropertyOne-one and onto on a finite set

Let $X$ be a finite set and $f$ a map from $X$ to itself. Then $f$ is one-one exactly when it is onto, so either property alone makes $f$ a bijection.

1.4 Composition of Functions and Invertible Function

A taxi fare depends on the distance travelled, and the distance depends on the destination you choose. Feed a destination into the first rule, feed the result into the second, and you get the fare directly. Chaining two functions in this way produces a new function, called their composition.

Definition

Composition of functions

For $f : A \to B$ and $g : B \to C$ , the composite $g \circ f : A \to C$ is defined by $(g \circ f) (x) = g (f (x))$ for every $x \in A$ . Read it as ' $g$ after $f$ '. NCERT writes it as $g o f$ .

ExampleIf $f (x) = x + 2$ and $g (x) = x^{2}$ on $𝐑$ , then $(g \circ f) (3) = g (5) = 25$ .

Two details matter. The function written on the right, $f$ , acts first. And the composite only makes sense when every output of $f$ is a valid input for $g$ , which is why the codomain of $f$ is the domain of $g$ in the definition (Fig. 1.9).

4/4Seen from outside, the two machines form one function, $(g \circ f) (x) = (x + 2)^{2}$ .

Fig. 1.9

g \circ f

runs

f

first, then

g

.NCERT Fig. 1.5 · p. 12

A dashed conveyor belt runs from left to right through two square machines, f on the left and g on the right, with f(x) = x + 2 written under f and g(x) = x squared written under g. Step one: a chip showing 3 waits at the input end. Step two: a chip showing 5 slides in between the machines, since f(3) = 5, and the input chip fades. Step three: a chip showing 25 slides in at the output end, since g(5) = 25. Step four: a dashed box encloses both machines and is labelled g composed with f, with (g composed with f)(x) = (x + 2) squared written below, so the pair acts as one machine that turns 3 into 25.

Example 15NCERT Example 15 · p. 12

Composing finite functions

Let $f : {0, 1, 2, 3} \to {1, 2, 5, 10}$ be $f (x) = x^{2} + 1$ , and let $g : {1, 2, 5, 10} \to {0, 1, 2}$ give the remainder when $y$ is divided by 3. Find $g \circ f$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 16NCERT Example 16 · p. 12

Order matters

Let $f, g : 𝐑 \to 𝐑$ be $f (x) = x + 2$ and $g (x) = x^{2}$ . Find $g \circ f$ and $f \circ g$ , and show that they are different functions.

Try it yourself first. Then reveal the solution one step at a time.

So composition is not commutative. It is, however, associative: for $f : A \to B$ , $g : B \to C$ and $h : C \to D$ , both $h \circ (g \circ f)$ and $(h \circ g) \circ f$ send $x$ to $h (g (f (x)))$ . That is why we may write $h \circ g \circ f$ without brackets.

Invertible functions

Some functions can be run backwards. If $f (x) = x + 2$ sends 3 to 5, then $y \mapsto y - 2$ sends 5 back to 3, and it undoes $f$ for every input. To say this precisely we use the identity function $I_{X} : X \to X$ , $I_{X} (x) = x$ , which leaves every element where it is.

Definition

Invertible function

A function $f : X \to Y$ is invertible if there is a function $g : Y \to X$ with $g \circ f = I_{X}$ and $f \circ g = I_{Y}$ . We then call $g$ the inverse of $f$ and write it as $f^{- 1}$ .

Example $f : 𝐑 \to 𝐑$ , $f (x) = x + 2$ , is invertible with $f^{- 1} (y) = y - 2$ .

Both conditions are needed. $g \circ f = I_{X}$ says that $g$ undoes $f$ : start in $X$ , go to $Y$ and come back to where you began. $f \circ g = I_{Y}$ says that $f$ undoes $g$ . Which functions have an inverse? Reverse the arrows of $f$ and ask whether the result is still a function (Fig. 1.10).

2/3Reversed, each element of $B$ has exactly one arrow out: this is $f^{- 1}$ .

Fig. 1.10Reversing the arrows gives a function exactly when every element of

B

was hit exactly once.

Two columns of circles: A with 1, 2 and 3 on the left and B with p, q and r on the right. Step one: the arrows of f go from 1 to q, from 2 to r and from 3 to p, and the line below lists these values. Step two: the arrows are reversed and highlighted, running from q to 1, from r to 2 and from p to 3; every element of B has exactly one arrow leaving it, so the reversed picture is the function f inverse. Step three shows a different function h with h(1) = h(2) = p and h(3) = q, reversed: p now has two arrows leaving it, labelled two arrows, and r, drawn dashed, has none, labelled no arrow, so the reversed picture is not a function.

TheoremInvertible means bijective

A function $f : X \to Y$ is invertible if and only if it is one-one and onto.

Invertible implies bijective. Let $g$ satisfy $g \circ f = I_{X}$ and $f \circ g = I_{Y}$ . If $f (x_{1}) = f (x_{2})$ , apply $g$ to both sides: $x_{1} = g (f (x_{1})) = g (f (x_{2})) = x_{2}$ , so $f$ is one-one. For any $y \in Y$ , the element $x = g (y)$ satisfies $f (x) = f (g (y)) = y$ , so $f$ is onto.

Bijective implies invertible. If $f$ is onto, every $y \in Y$ has at least one preimage; if $f$ is one-one, it has at most one. So we may define $g (y)$ to be the unique $x$ with $f (x) = y$ . Then $g (f (x)) = x$ , because $x$ is the unique preimage of $f (x)$ , and $f (g (y)) = y$ by the choice of $g (y)$ .

This gives two practical routes. To show that $f$ is invertible without finding $f^{- 1}$ , prove that it is one-one and onto. To find $f^{- 1}$ , solve $y = f (x)$ for $x$ : the expression you get is $f^{- 1} (y)$ , and you can confirm it by checking both compositions.

Example 17NCERT Example 17 · p. 12

Finding and checking an inverse

Let $Y = {8, 11, 14, 17, \dots} = {y \in 𝐍 : y = 3 x + 5 for some x \in 𝐍}$ and let $f : 𝐍 \to Y$ be $f (x) = 3 x + 5$ . Show that $f$ is invertible and find $f^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

Restricting the domain

The rule $x \mapsto x^{2} + 4$ is not invertible as a map from $𝐑$ to $𝐑$ . Show that $f : [0, \infty) \to [4, \infty)$ , $f (x) = x^{2} + 4$ , is invertible and find $f^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Miscellaneous Examples

These examples mix the ideas of the chapter. Several are proofs about relations in general; others count relations or functions on small sets, where listing the cases carefully is the whole skill.

Example 18NCERT Example 18 · p. 13

Combining two equivalence relations

In $𝐙$ , let $R_{1}$ be 'same parity' (2 divides $a - b$ ) and $R_{2}$ be 'same remainder on division by 3' (3 divides $a - b$ ). (a) Show that if $R_{1}$ and $R_{2}$ are any two equivalence relations in a set $A$ , then $R_{1} \cap R_{2}$ is an equivalence relation. (b) Describe $R_{1} \cap R_{2}$ for the two relations above. (c) Show that $R_{1} \cup R_{2}$ is not an equivalence relation here.

Try it yourself first. Then reveal the solution one step at a time.

Example 19NCERT Example 19 · p. 13

Pairs with equal differences

In the set $A = 𝐍 \times 𝐍$ of ordered pairs of natural numbers, define $(a, b) R (c, d)$ if and only if $a + d = b + c$ . Show that $R$ is an equivalence relation.

Try it yourself first. Then reveal the solution one step at a time.

Example 20NCERT Example 20 · p. 13

One relation, two descriptions

Let $X = {1, 2, \dots, 10}$ , $R_{1} = {(x, y) : 4 divides x - y}$ and let $x R_{2} y$ mean that $x$ and $y$ both lie in the same one of the sets ${1, 5, 9}$ , ${2, 6, 10}$ , ${3, 7}$ and ${4, 8}$ . Show that $R_{1} = R_{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 21NCERT Example 21 · p. 14

Same output, same class

In $𝐑$ , call two numbers related when they have the same square: $a \sim b$ if $a^{2} = b^{2}$ . (a) Show that $\sim$ is an equivalence relation and find its classes. (b) Explain why 'have the same output under $f$ ' is an equivalence relation in $X$ for every function $f : X \to Y$ , and what its classes tell you about $f$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 22NCERT Example 22 · p. 14

Counting one-one functions

How many one-one functions are there from $A = {1, 2, 3}$ to $B = {p, q, r, s, t}$ ?

Try it yourself first. Then reveal the solution one step at a time.

Example 23NCERT Example 23 · p. 14

Counting relations with given properties

Let $A = {1, 2, 3}$ . How many relations in $A$ contain $(2, 1)$ and $(3, 1)$ and are reflexive and transitive, but not symmetric?

Try it yourself first. Then reveal the solution one step at a time.

Counting equivalence relations is the same as counting the ways to sort the elements into groups, which is easy to picture (Fig. 1.11).

4/4 $1 + 3 + 1 = 5$ partitions, so 5 equivalence relations.

Fig. 1.11Each way of sorting

{a, b, c}

into classes is one equivalence relation.

Five panels each show the points a, b and c with shaded outlines around the classes. In the first step only the first panel is highlighted: one outline holds all three points, which gives the universal relation. In the second step the second panel is highlighted: each point has its own outline, three classes, which gives the identity relation. In the third step the last three panels are highlighted: in each, two points share an outline and the third stands alone, once for each choice of the lone point. In the last step all five panels are shown and the count 1 + 3 + 1 = 5 is written below.

Example 24NCERT Example 24 · p. 14

All equivalence relations on three elements

Find all the equivalence relations in $A = {a, b, c}$ and count them.

Try it yourself first. Then reveal the solution one step at a time.

Example 25NCERT Example 25 · p. 14

A sum of onto functions

Let $f, g : 𝐙 \to 𝐙$ be $f (n) = n$ and $g (n) = - n$ . Show that $f$ and $g$ are both onto, but $f + g$ is not.

Try it yourself first. Then reveal the solution one step at a time.

Example 26NCERT Example 26 · p. 15

A sum of one-one functions

Prove that the maps $x \mapsto x^{3}$ and $x \mapsto - 3 x$ of $𝐑$ are both one-one, and that their sum $x \mapsto x^{3} - 3 x$ is not. Call the maps $f$ and $g$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

How many functions are there from

{1, 2, 3, 4}

{0, 1}

functions

Show answer

Answer 16 functions

Each of the 4 inputs independently chooses one of 2 outputs:

2^{4} = 16

Recap

Relations and Functions: key points and summary

A relation from $A$ to $B$ is any subset of $A \times B$ ; a relation in $A$ is a subset of $A \times A$ (Section 1.1).
The empty relation $\emptyset$ and the universal relation $A \times A$ are the two trivial relations (Section 1.2).
Reflexive: $(a, a) \in R$ for every $a$ . Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$ . Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ (Reflexive, symmetric and transitive relations).
An equivalence relation has all three properties; 'has the same ___ as' is the typical example (Example 21).
The class $[a]$ is the set of elements related to $a$ . The classes partition the set, and every partition comes from an equivalence relation (Property).
$f : X \to Y$ is one-one if $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ , onto if its range is $Y$ , and bijective if it is both (Section 1.3).
For $f : 𝐑 \to 𝐑$ : one-one means no horizontal line meets the graph twice; onto means every horizontal line meets it (Fig. 1.8).
For maps from a finite set to itself, one-one and onto are equivalent; for infinite sets they are not (Property).
$(g \circ f) (x) = g (f (x))$ , so $f$ acts first. Composition is associative, but in general $g \circ f \neq f \circ g$ (Section 1.4).
$f$ is invertible when some $g$ has $g \circ f = I_{X}$ and $f \circ g = I_{Y}$ ; this happens exactly when $f$ is bijective (Theorem).
To find $f^{- 1}$ , solve $y = f (x)$ for $x$ , then check both compositions (Example 17).
There are $n^{k}$ functions from a $k$ -element set to an $n$ -element set, and $n (n - 1) \dots (n - k + 1)$ of them are one-one (Example 22).

Timeline

How the idea of a function grew

1637
Curves become equations
René Descartes's La Géométrie describes curves by equations in two unknowns, so that one quantity depends on another.
1673
Leibniz uses the word 'function'
In his notes on curves, Gottfried Leibniz uses it for quantities such as the slope or the tangent at a point.
1718
A function of a variable
Johann Bernoulli calls a function a quantity built from a variable and constants, and writes it $φ x$ .
1734
Euler writes $f (x)$
Leonhard Euler introduces the notation $f (x)$ , and later makes functions the central objects of analysis.
1797
Lagrange's analytic functions
Joseph-Louis Lagrange builds calculus on functions in Théorie des fonctions analytiques and writes $f^{'} (x)$ for the derivative.
1837
Dirichlet: any rule will do
Peter Gustav Lejeune Dirichlet lets $y$ depend on $x$ by any rule at all, not necessarily a formula.
Early 1900s
Functions as sets of pairs
With Georg Cantor's set theory in place, a function becomes a set of ordered pairs in which no first entry repeats: the definition used in this chapter.

Practice · 13 questions

Practice questions with answers (NCERT “Miscellaneous exercises”)

ProofHardNCERT Miscellaneous Exercise, Q1
Show that $f : (- 1, 1) \to 𝐑$ , $f (x) = \frac{x}{1 - ∣ x ∣}$ , is a bijection.
Show solution
For $0 \leq x < 1$ , $f (x) = \frac{x}{1 - x} \geq 0$ ; for $- 1 < x < 0$ , $f (x) = \frac{x}{1 + x} < 0$ . So $f (x)$ has the same sign as $x$ , and $f (x) = 0$ only at $x = 0$ .
One-one: if $f (x_{1}) = f (x_{2})$ , then $x_{1}$ and $x_{2}$ lie on the same side of 0. If both are $\geq 0$ , $\frac{x_{1}}{1 - x_{1}} = \frac{x_{2}}{1 - x_{2}}$ gives $x_{1} - x_{1} x_{2} = x_{2} - x_{1} x_{2}$ , so $x_{1} = x_{2}$ . If both are negative, $\frac{x_{1}}{1 + x_{1}} = \frac{x_{2}}{1 + x_{2}}$ gives $x_{1} = x_{2}$ in the same way.
Onto: for $y \geq 0$ , take $x = \frac{y}{1 + y} \in [0, 1)$ ; then $1 - x = \frac{1}{1 + y}$ and $f (x) = y$ . For $y < 0$ , take $x = \frac{y}{1 - y} \in (- 1, 0)$ ; then $1 + x = \frac{1}{1 - y}$ and $f (x) = y$ . In both cases $x = \frac{y}{1 + ∣ y ∣}$ , which is in fact the formula for $f^{- 1}$ .
ProofMediumNCERT Miscellaneous Exercise, Q2
Show that $f : 𝐑 \to 𝐑$ , $f (x) = x^{3} + x$ , is one-one.
Show solution
If $x_{1}^{3} + x_{1} = x_{2}^{3} + x_{2}$ , then $(x_{1} - x_{2}) (x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} + 1) = 0$ .
The second factor equals ${(x_{1} + \frac{1}{2} x_{2})}^{2} + \frac{3}{4} x_{2}^{2} + 1 \geq 1$ , so it is never zero. Hence $x_{1} - x_{2} = 0$ , and $f$ is one-one.
3
Short answerMediumNCERT Miscellaneous Exercise, Q3
Let $X$ be a non-empty finite set and $P (X)$ the set of all its subsets. For $A, B \in P (X)$ , let $A R B$ when $A$ and $B$ have the same number of elements.
1. Is $R$ an equivalence relation in $P (X)$ ? Justify your answer.
  Show solution
  AnswerYes.
  $R$ is 'has the same size as', an equality of an attribute, so it is reflexive, symmetric and transitive.
2. For $X = {1, 2, 3}$ , list the equivalence classes. How many are there?
  Show solution
  Answer4 classes.
  The classes are ${\emptyset}$ ; the three one-element subsets; the three two-element subsets; and ${X}$ . So there are 4 classes, with 1, 3, 3 and 1 members (8 subsets in all).
4
NumericalMediumNCERT Miscellaneous Exercise, Q4
Let $S = {1, 2, 3, 4, 5}$ .
1. How many functions $S \to S$ are there?
  Show answer
  Answer 3125
  Each of 5 inputs has 5 choices: $5^{5} = 3125$ .
2. How many of them are one-one?
  Show answer
  Answer 120
  $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$ .
3. How many of them are onto?
  Show answer
  Answer 120
  On a finite set, onto is the same as one-one, so again $5! = 120$ .
4. How many are neither one-one nor onto?
  Show answer
  Answer 3005
  The 120 one-one functions are exactly the onto ones, so $3125 - 120 = 3005$ functions are neither.
5
Short answerMediumNCERT Miscellaneous Exercise, Q5
Let $A = {1, 2, 3}$ , $B = {1, 8, 27}$ , and let $f, g : A \to B$ be $f (x) = x^{3}$ and $g (x) = 6 x^{2} - 11 x + 6$ . Two functions with the same domain and codomain are equal when they agree at every point of the domain.
1. Are $f$ and $g$ equal?
  Show solution
  AnswerYes.
  $g (1) = 6 - 11 + 6 = 1$ , $g (2) = 24 - 22 + 6 = 8$ and $g (3) = 54 - 33 + 6 = 27$ , which match $f (1) = 1$ , $f (2) = 8$ and $f (3) = 27$ . So $f = g$ , even though the formulas look different.
2. Would $f$ and $g$ still be equal on the domain ${1, 2, 3, 4}$ (with a large enough codomain)?
  Show solution
  AnswerNo.
  $f (4) = 64$ but $g (4) = 96 - 44 + 6 = 58$ . Indeed $f (x) - g (x) = (x - 1) (x - 2) (x - 3)$ , which vanishes only at 1, 2 and 3.
Quick answerMediumNCERT Miscellaneous Exercise, Q6
Let $A = {1, 2, 3}$ . How many relations in $A$ contain $(1, 2)$ and are reflexive and symmetric, but not transitive?
Show answer
Answer 2
The link ${1, 2}$ must be present, so there are 4 candidates: add nothing, add ${1, 3}$ , add ${2, 3}$ , or add both. Adding nothing gives an equivalence relation, and adding both gives the universal relation, both transitive. Adding exactly one link breaks transitivity (for example $(3, 1), (1, 2)$ needs $(3, 2)$ ), so the answer is 2.
Quick answerMediumNCERT Miscellaneous Exercise, Q7
How many equivalence relations in ${1, 2, 3, 4}$ contain the pair $(1, 2)$ ?
Show answer
Answer 5
The partitions with 1 and 2 together are ${12} {3} {4}$ , ${123} {4}$ , ${124} {3}$ , ${12} {34}$ and ${1234}$ : 5 in all.
Quick answerEasy
How many relations in a set with 3 elements are reflexive?
Show answer
Answer 64
Of the 9 pairs, the 3 loops must be in, and each of the other 6 can be in or out: $2^{6} = 64$ .
Quick answerMedium
How many relations in ${1, 2, 3}$ are both reflexive and symmetric?
Show answer
Answer 8
The loops are forced, and each of the three two-way links ${1, 2}$ , ${1, 3}$ , ${2, 3}$ is in or out: $2^{3} = 8$ .
Quick answerMedium
If $f : A \to B$ and $g : B \to C$ are both one-one, then $g \circ f$ is
always one-onealways ontoone-one only when $A = C$ never one-one
Show answer
Answer (A) always one-one
$g (f (x_{1})) = g (f (x_{2}))$ gives $f (x_{1}) = f (x_{2})$ because $g$ is one-one, and then $x_{1} = x_{2}$ because $f$ is one-one.
11
Quick answerMedium
Match each function with its type.
Tap an item on the left, then its match on the right.
Show answer
Answer
$x \mapsto x^{3}$ on $𝐑$ : bijective
$x \mapsto x + 3$ on $𝐍$ : one-one, not onto
$x \mapsto ∣ x ∣$ on $𝐑$ : neither
number of digits, $𝐍 \to 𝐍$ : onto, not one-one
$x^{3}$ has the inverse $\sqrt[3]{y}$ ; $x + 3$ misses 1, 2, 3; $∣ x ∣$ has $∣ - 1 ∣ = ∣ 1 ∣$ and misses $- 1$ ; the digit count is onto with $f (1) = f (2)$ .
Quick answerEasy
A function $f : X \to Y$ is invertible if and only if it is and .
Show answer
Answer (1) one-one, (2) onto
Invertible means bijective: one-one and onto (Theorem).
13
Quick answerEasy
Put the steps for finding the inverse of a bijection $f$ in order.
1. Check $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (y)) = y$
2. Call the result $f^{- 1} (y)$
3. Solve for $x$ in terms of $y$
4. Write $y = f (x)$
Show answer
Answer
Write $y = f (x)$
Solve for $x$ in terms of $y$
Call the result $f^{- 1} (y)$
Check $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (y)) = y$
Solving $y = f (x)$ for $x$ gives the inverse formula, and the two compositions confirm it.

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 1.1, Q1	Practice 1.1 · Q1
NCERT Exercise 1.1, Q2	Practice 1.1 · Q2
NCERT Exercise 1.1, Q3	Practice 1.1 · Q3
NCERT Exercise 1.1, Q4	Practice 1.1 · Q4
NCERT Exercise 1.1, Q5	Practice 1.1 · Q5
NCERT Exercise 1.1, Q6	Practice 1.1 · Q6
NCERT Exercise 1.1, Q7	Practice 1.1 · Q7
NCERT Exercise 1.1, Q8	Practice 1.1 · Q8
NCERT Exercise 1.1, Q9	Practice 1.1 · Q9
NCERT Exercise 1.1, Q10	Practice 1.1 · Q10
NCERT Exercise 1.1, Q11	Practice 1.1 · Q11
NCERT Exercise 1.1, Q12	Practice 1.1 · Q12
NCERT Exercise 1.1, Q13	Practice 1.1 · Q13
NCERT Exercise 1.1, Q14	Practice 1.1 · Q14
NCERT Exercise 1.1, Q15	Practice 1.1 · Q15
NCERT Exercise 1.1, Q16	Practice 1.1 · Q16
NCERT Exercise 1.2, Q1	Practice 1.2 · Q1
NCERT Exercise 1.2, Q2	Practice 1.2 · Q2
NCERT Exercise 1.2, Q3	Practice 1.2 · Q3
NCERT Exercise 1.2, Q4	Practice 1.2 · Q4
NCERT Exercise 1.2, Q5	Practice 1.2 · Q5
NCERT Exercise 1.2, Q6	Practice 1.2 · Q6
NCERT Exercise 1.2, Q7	Practice 1.2 · Q7
NCERT Exercise 1.2, Q8	Practice 1.2 · Q8
NCERT Exercise 1.2, Q9	Practice 1.2 · Q9
NCERT Exercise 1.2, Q10	Practice 1.2 · Q10
NCERT Exercise 1.2, Q11	Practice 1.2 · Q11
NCERT Exercise 1.2, Q12	Practice 1.2 · Q12
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercises · Q7

Quick check: MCQs on Relations and Functions

1.1 Introduction3 questions
Equivalence classes4 questions
1.3 Types of Functions4 questions
Invertible functions4 questions
Miscellaneous Examples2 questions

Glossary · 15

Key terms (English–Hindi)

bijective functionएकैकी आच्छादक फलनekaiki aachchhaadak phalan: A function that is both one-one and onto.

compositionफलनों का संयोजनphalanon ka sanyojan: For $f : A \to B$ and $g : B \to C$ , the function $g \circ f : A \to C$ , $(g \circ f) (x) = g (f (x))$ .First seen in the chapter →

empty relationरिक्त संबंधrikt sambandh: The relation $\emptyset$ , in which no element is related to any element.
equivalence classतुल्यता वर्गtulyata varg: For an equivalence relation, the set $[a]$ of all elements related to $a$ .
equivalence relationतुल्यता संबंधtulyata sambandh: A relation that is reflexive, symmetric and transitive.

functionफलनphalan: A relation from $X$ to $Y$ that relates each element of $X$ to exactly one element of $Y$ .

identity functionतत्समक फलनtatsamak phalan: The function $I_{X} : X \to X$ with $I_{X} (x) = x$ for every $x$ .First seen in the chapter →
invertible functionव्युत्क्रमणीय फलनvyutkramaniya phalan: A function $f : X \to Y$ with a function $f^{- 1} : Y \to X$ such that $f^{- 1} \circ f = I_{X}$ and $f \circ f^{- 1} = I_{Y}$ .

one-one functionएकैकी फलनekaiki phalan: A function with $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ : different inputs give different outputs.
onto functionआच्छादक फलनaachchhaadak phalan: A function whose range equals its codomain: every element of the codomain is an image.

reflexive relationस्वतुल्य संबंधsvatulya sambandh: A relation with $(a, a) \in R$ for every $a$ in the set.
relationसंबंधsambandh: A subset of $A \times B$ ; a relation in $A$ is a subset of $A \times A$ .First seen in the chapter →

symmetric relationसममित संबंधsammit sambandh: A relation in which $(a, b) \in R$ always implies $(b, a) \in R$ .

transitive relationसंक्रामक संबंधsankraamak sambandh: A relation in which $(a, b) \in R$ and $(b, c) \in R$ always imply $(a, c) \in R$ .

universal relationसार्वत्रिक संबंधsaarvatrik sambandh: The relation $A \times A$ , in which every element is related to every element.

Relations and Functions: Class 12 Maths · Chapter 1

What you will learn in Relations and Functions

1.1 Introduction

1.2 Types of Relations

Reflexive, symmetric and transitive relations

Equivalence classes

Practice 1.1: questions with answers

1.3 Types of Functions

Practice 1.2: questions with answers

1.4 Composition of Functions and Invertible Function

Invertible functions

Miscellaneous Examples

Relations and Functions: key points and summary

Practice questions with answers (NCERT “Miscellaneous exercises”)

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Quick check: MCQs on Relations and Functions

Key terms (English–Hindi)

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