What is a relation in mathematics?

A relation from set A to set B is any subset of the Cartesian product A × B. If (a, b) is in the relation R, we say a is related to b and write aRb. Relations are the foundation for functions in higher mathematics.

What is the domain and range of a relation?

The domain of a relation R is the set of all first coordinates of the ordered pairs in R. The range is the set of all second coordinates. If R ⊆ A × B, then Dom(R) ⊆ A and Range(R) ⊆ B.

How many relations are possible from set A to set B?

If |A| = m and |B| = n, then |A × B| = mn. A relation is any subset of A × B, so the total number of distinct relations from A to B is 2^(mn). For a relation on a single set A with |A| = n, the count is 2^(n²).

What is an equivalence relation?

An equivalence relation on set A is a relation that is reflexive (aRa for all a), symmetric (aRb implies bRa), and transitive (aRb and bRc imply aRc). An equivalence relation partitions the set into disjoint equivalence classes.

What is the difference between symmetric and anti-symmetric relations?

Symmetric means aRb implies bRa. Anti-symmetric means aRb and bRa together imply a = b. A relation can be symmetric, anti-symmetric, both (the identity relation), or neither.

What are equivalence classes?

For an equivalence relation R on A, the equivalence class of an element a, written [a], is the set of all elements of A related to a: [a] = {x ∈ A : xRa}. Two equivalence classes are either identical or disjoint, and their union covers A.

What is the inverse of a relation?

If R ⊆ A × B is a relation, its inverse R⁻¹ ⊆ B × A is obtained by swapping each ordered pair: R⁻¹ = {(b, a) : (a, b) ∈ R}. Also (R⁻¹)⁻¹ = R and Dom(R⁻¹) = Range(R).

What does congruence modulo n mean?

Two integers a and b are congruent modulo n, written a ≡ b (mod n), if n divides a − b. This is an equivalence relation on the integers and partitions Z into n equivalence classes: [0], [1], ..., [n − 1].

Relations

MathsSets, Relations And Number SystemFor JEE aspirants

A relation from a set $A$ to a set $B$ is any subset of the Cartesian product $A \times B$ . Relations extend the idea of pairing elements and lead directly to functions in JEE mathematics. This page covers the definition of a relation, its domain and range, the total number of relations, inverse and composition of relations, special relations in a set (identity, universal, void), the four defining properties (reflexive, symmetric, anti-symmetric, transitive), equivalence relations with equivalence classes, and congruence modulo $n$ - all with worked JEE-style examples.

Key Formulas - Quick Reference

If $∣ A ∣ = m$ and $∣ B ∣ = n$ , then $∣ A \times B ∣ = mn$ and total number of relations from $A$ to $B$ is $2^{mn}$
Number of relations on a set $A$ with $∣ A ∣ = n$ : $2^{n^{2}}$
Domain of $R$ : set of first coordinates; Range of $R$ : set of second coordinates
Inverse relation: $R^{- 1} = {(b, a) : (a, b) \in R}$ , and $(R^{- 1})^{- 1} = R$
Composition: $(S \circ R) (a) = {c : \exists b with (a, b) \in R, (b, c) \in S}$
Reversal rule: $(S \circ R)^{- 1} = R^{- 1} \circ S^{- 1}$
Reflexive: $(a, a) \in R \forall a \in A$ ; Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$ ; Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$
Equivalence relation: reflexive + symmetric + transitive; partitions $A$ into disjoint equivalence classes

1. What is a Relation?

Let $A$ and $B$ be two non-empty sets. Every subset of $A \times B$ defines a relation from $A$ to $B$ , and every relation from $A$ to $B$ is a subset of $A \times B$ .

If $R \subseteq A \times B$ and $(a, b) \in R$ , we say $a$ is related to $b$ under $R$ and write $a R b$ . If $(a, b) \in / R$ , we write $a \neq R b$ .

Example: Let

A = {1, 2, 3, 4, 5}

B = {1, 3}

. Define

R

a R b

iff

a \leq b

, with

a \in A

b \in B

. Then

R = {(1, 1), (1, 3), (2, 3), (3, 3)} \subset A \times B .

2. Domain and Range of a Relation

Let $R \subseteq A \times B$ . Then:

Domain of $R$ = ${a : a \in A, (a, b) \in R for some b \in B}$ . It is the set of all first coordinates of the ordered pairs in $R$ .
Range of $R$ = ${b : b \in B, (a, b) \in R for some a \in A}$ . It is the set of all second coordinates.

Clearly $Dom (R) \subseteq A$ and $Range (R) \subseteq B$ .

3. Total Number of Distinct Relations from A to B

If $∣ A ∣ = m$ and $∣ B ∣ = n$ , then $∣ A \times B ∣ = mn$ . A relation is any subset of $A \times B$ , so the total number of relations is the number of subsets of $A \times B$ :

$Number of relations from A to B = 2^{mn} .$

In particular, if $∣ A ∣ = n$ , the number of relations on $A$ is $2^{n^{2}}$ .

4. Inverse Relation

Let $R \subseteq A \times B$ . The inverse relation $R^{- 1} \subseteq B \times A$ is defined by

$R^{- 1} = {(b, a) : (a, b) \in R, a \in A, b \in B} .$

Key properties:

$a R b ⟺ b R^{- 1} a$
$Dom (R^{- 1}) = Range (R)$ and $Range (R^{- 1}) = Dom (R)$
$(R^{- 1})^{- 1} = R$

Example: Let

A = {1, 2, 3, 4}

B = {a, b, c}

R = {(1, a), (1, c), (2, a)}

. Then

Dom (R) = {1, 2}

Range (R) = {a, c}

, and

R^{- 1} = {(a, 1), (c, 1), (a, 2)}

5. Composition of Relations

Let $R \subseteq A \times B$ and $S \subseteq B \times C$ . The composition $S \circ R \subseteq A \times C$ is defined by:

$(a, c) \in S \circ R ⟺ \exists b \in B such that (a, b) \in R and (b, c) \in S .$

Example:

A = {1, 2, 3}

B = {a, b, c, d}

C = {α, β, γ}

R = {(1, a), (1, c), (2, d)}

S = {(a, α), (a, γ), (c, β)}

.
Then

S \circ R = {(1, α), (1, γ), (1, β)}

Be careful: $S \circ R$ starts with $R$ ; $R \circ S$ starts with $S$ . In general $S \circ R \neq = R \circ S$ . The reversal rule holds: $(S \circ R)^{- 1} = R^{- 1} \circ S^{- 1}$ .

6. Relations in a Set

A relation in a set $A$ is a relation from $A$ to $A$ , i.e. a subset of $A \times A$ .

Identity Relation

$R$ is an identity relation if $(a, b) \in R$ iff $a = b$ . Every element is related only to itself: $I_{A} = {(a, a) : a \in A}$ .

Universal Relation

If $R = A \times A$ , then $R$ is called the universal relation on $A$ . Every element is related to every element.

Void (Empty) Relation

The empty subset $ϕ$ of $A \times A$ is the void relation. No element is related to any element.

7. Properties of Relations

Reflexive Relation

$R$ is reflexive if $(a, a) \in R$ for every $a \in A$ . If there is even one $a \in A$ with $(a, a) \in / R$ , then $R$ is not reflexive.

Example: On

A = {1, 2, 3, 4, 5}

R_{1} = {(1, 1), (3, 2), (4, 2), (4, 4), (5, 2), (5, 5)}

is not reflexive since

(3, 3) \in / R_{1}

R_{2} = {(1, 1), (2, 2), (3, 2), (3, 3), (4, 1), (4, 4), (5, 5)}

is reflexive.

Symmetric Relation

$R$ is symmetric if $(x, y) \in R \Rightarrow (y, x) \in R$ . Equivalently, $R$ is symmetric iff $R^{- 1} = R$ .

Example: On

A = {1, 2, 3}

R = {(1, 1), (1, 3), (3, 1)}

is symmetric.

Anti-symmetric Relation

$R$ is anti-symmetric if $(a, b) \in R$ and $(b, a) \in R \Rightarrow a = b$ . So if $a \neq = b$ , at most one of $a R b$ and $b R a$ can hold.

Example: On

N

, the divides relation

x R y

iff

x ∣ y

is anti-symmetric, because

x ∣ y

and

y ∣ x

imply

x = y

(for positive integers).

Transitive Relation

$R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \Rightarrow (a, c) \in R$ . Transitivity fails only when there exist $a, b, c$ with $a R b$ , $b R c$ but $(a, c) \in / R$ .

Example: On

A = {1, 2, 3}

R_{1} = {(1, 2), (1, 3)}

R_{2} = {(1, 2)}

R_{3} = {(1, 1)}

are transitive. But

R_{4} = {(1, 2), (2, 1), (1, 1)}

is not, since

(2, 1) \in R_{4}

and

(1, 2) \in R_{4}

, but

(2, 2) \in / R_{4}

Note: Every identity relation is reflexive, symmetric, and transitive, but not every reflexive relation is an identity relation.

8. Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation if it is:

Reflexive: $(a, a) \in R$ for all $a \in A$
Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$
Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$

The equivalence relation is usually denoted by $\sim$ .

Equivalence Classes

Let $R$ be an equivalence relation on $A$ (with $A \neq = ϕ$ ) and let $a \in A$ . The equivalence class of $a$ , denoted $[a]$ or $\overset{a}{ˉ}$ , is the set of all elements of $A$ related to $a$ :

$[a] = {x \in A : x R a} .$

Basic facts about equivalence classes:

$b \in [a] \Rightarrow a \in [b]$
$b \in [a] \Rightarrow [a] = [b]$
Any two equivalence classes are either disjoint or identical

Thus an equivalence relation partitions the set $A$ into disjoint equivalence classes.

A Standard Example - Congruence Modulo n

Fix a positive integer $n$ . Define $x \equiv y (mod n)$ iff $n ∣ (x - y)$ . This is an equivalence relation on $Z$ . Taking $n = 5$ :

$[0] = {x : x \equiv 0 (mod 5)} = {5 p : p \in Z} = {0, \pm 5, \pm 10, \pm 15, \dots} .$

$[1] = {x : x \equiv 1 (mod 5)} = {5 k + 1 : k \in Z} = {\dots, - 9, - 4, 1, 6, 11, \dots} .$

There are exactly 5 distinct equivalence classes: $[0], [1], [2], [3], [4]$ .

9. Congruences

Let $m$ be a positive integer. Two integers $a$ and $b$ are said to be congruent modulo $m$ if $a - b$ is divisible by $m$ , i.e. $a - b = mλ$ for some integer $λ$ . We write $a \equiv b (mod m)$ .

The relation "congruent modulo $m$ " is an equivalence relation on $Z$ .

Solved Examples

Solved Example 1

N

is the set of natural numbers. A relation

R

is defined on

N \times N

(a, b) R (c, d) ⟺ a + d = b + c

. Prove that

R

is an equivalence relation.

Solution:

Reflexive: $(a, b) R (a, b) ⟺ a + b = b + a$ , which is true. So $R$ is reflexive.

Symmetric: If $(a, b) R (c, d)$ , then $a + d = b + c$ , i.e. $c + b = d + a$ , i.e. $(c, d) R (a, b)$ . So $R$ is symmetric.

Transitive: Suppose $(a, b) R (c, d)$ and $(c, d) R (e, f)$ . Then $a + d = b + c$ and $c + f = d + e$ . Adding: $a + d + c + f = b + c + d + e$ , so $a + f = b + e$ , i.e. $(a, b) R (e, f)$ . So $R$ is transitive.

Hence $R$ is an equivalence relation on $N \times N$ .

Solved Example 2

A relation

R

on the set of complex numbers is defined by

z_{1} R z_{2} ⟺ \frac{z _{1} - z _{2}}{z _{1} + z _{2}}

is real. Show that

R

is an equivalence relation.

Solution:

Reflexive: $\frac{z _{1} - z _{1}}{z _{1} + z _{1}} = 0$ , which is real. So $z_{1} R z_{1}$ for all $z_{1} \in C$ (with $z_{1} \neq = 0$ ).

Symmetric: If $\frac{z _{1} - z _{2}}{z _{1} + z _{2}}$ is real, so is $- \frac{z _{1} - z _{2}}{z _{1} + z _{2}} = \frac{z _{2} - z _{1}}{z _{1} + z _{2}}$ , which equals $\frac{z _{2} - z _{1}}{z _{2} + z _{1}}$ . Hence $z_{2} R z_{1}$ .

Transitive: Let $z_{j} = a_{j} + i b_{j}$ for $j = 1, 2, 3$ . Writing $\frac{z _{1} - z _{2}}{z _{1} + z _{2}}$ in the form (real part) $+ i$ (imag part) and setting the imaginary part to zero gives

$(a_{1} + a_{2}) (b_{1} - b_{2}) - (a_{1} - a_{2}) (b_{1} + b_{2}) = 0 \Rightarrow 2 a_{2} b_{1} - 2 a_{1} b_{2} = 0 \Rightarrow \frac{a _{1}}{b _{1}} = \frac{a _{2}}{b _{2}} .$

Similarly $z_{2} R z_{3}$ gives $\frac{a _{2}}{b _{2}} = \frac{a _{3}}{b _{3}}$ . Combining, $\frac{a _{1}}{b _{1}} = \frac{a _{3}}{b _{3}}$ , so $z_{1} R z_{3}$ .

Hence $R$ is an equivalence relation.

Solved Example 3

Find all congruent solutions of

8 x \equiv 6 (mod 14)

Solution:

We want $8 x - 6 = 14 λ$ for some non-negative integer $λ$ , so $x = \frac{14 λ + 6}{8} = \frac{7 λ + 3}{4}$ .

Rewriting: $x = \frac{4 λ + 3 ( λ + 1 )}{4} = λ + \frac{3 ( λ + 1 )}{4}$ , valid when $4 ∣ 3 (λ + 1)$ , i.e. when $λ + 1 \equiv 0 (mod 4)$ , i.e. $λ \equiv 3 (mod 4)$ .

The greatest common divisor of $8$ and $14$ is $2$ , so modulo $14$ there are $g cd (8, 14) = 2$ distinct solutions. Taking $λ = 3$ gives $x = 6$ ; taking $λ = 7$ gives $x = 13$ .

Solutions: $x \equiv 6, 13 (mod 14)$ .

Common Mistakes to Avoid

Watch out

Confusing symmetric with anti-symmetric. Symmetric means $a R b \Rightarrow b R a$ ; anti-symmetric means $a R b$ and $b R a \Rightarrow a = b$ . The identity relation is both.
Assuming that a relation with $(a, a)$ for some $a$ is reflexive. Reflexivity requires $(a, a) \in R$ for every $a \in A$ .
Applying composition in the wrong order. $S \circ R$ starts with $R$ : first apply $R$ , then $S$ .
Forgetting to check all three properties for an equivalence relation. Missing even one (reflexive, symmetric, transitive) invalidates it.
Treating equivalence classes as ordered. $[a] = [b]$ whenever $a R b$ ; the label chosen doesn't matter.
Counting relations from $A$ to $B$ as $2^{m} + 2^{n}$ or $mn$ . The correct count is $2^{mn}$ - the number of subsets of $A \times B$ .

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