NCERT Class 11 Mathematics — Relations and Functions, Exercise 2.2. All 9 questions solved.

A relation $R$ from $A$ to $B$ is nothing more exotic than a subset of $A \times B$ — a selection of ordered pairs, chosen by some rule. Three sets go with every relation, and the exercise checks all three repeatedly:

• Domain — the set of first entries actually appearing in $R$.
• Range — the set of second entries actually appearing in $R$.
• Codomain — the whole of $B$, whether or not every element of it gets used.

So $\text{range} \subset \text{codomain}$, and they need not be equal.

Counting relations follows from counting subsets: if $n(A) = p$ and $n(B) = q$, then $n(A \times B) = pq$ and the number of relations from $A$ to $B$ is

$$2^{pq}$$

Key insight. The domain of a relation is what the rule achieves, not what it was offered. In question 1 the rule $y = 3x$ is applied to $A = \{1, \ldots, 14\}$, but $x = 5$ produces $y = 15$, which is outside $A$, so the pair is never formed and $5$ never enters the domain. Codomain is what you were given; range and domain are what survives.

Question 1

Let $A = \{1, 2, 3, \ldots, 14\}$. Define a relation $R$ from $A$ to $A$ by $R = \{(x, y) : 3x – y = 0, \text{ where } x, y \in A\}$. Write down its domain, codomain and range.

Solution. The condition $3x – y = 0$ rearranges to $y = 3x$. Both entries must lie in $A$, so run $x$ through $A$ and keep only those whose image $3x$ is still at most $14$:

$$x = 1 \to y = 3, \quad x = 2 \to y = 6, \quad x = 3 \to y = 9, \quad x = 4 \to y = 12$$

At $x = 5$ we would need $y = 15$, which is outside $A$, so that pair is not admitted — and neither is any later one. Hence

$$R = \{(1,3),\ (2,6),\ (3,9),\ (4,12)\}$$

The domain collects the first entries, the range the second, and the codomain is the whole of $A$ because that is the set $R$ maps into.

Question 2

Define a relation $R$ on the set $\mathbf{N}$ of natural numbers by $R = \{(x, y) : y = x + 5,\ x \text{ is a natural number less than } 4;\ x, y \in \mathbf{N}\}$. Depict this relationship using roster form. Write down the domain and the range.

Solution. The restriction $x < 4$ limits $x$ to $1$, $2$ and $3$, and $y = x + 5$ then determines each partner:

$$x = 1 \to y = 6, \qquad x = 2 \to y = 7, \qquad x = 3 \to y = 8$$

$$R = \{(1,6),\ (2,7),\ (3,8)\}$$

Note that the codomain here is all of $\mathbf{N}$, an infinite set, while the range is just three numbers — a clean illustration of why the two words are not interchangeable.

Question 3

$A = \{1, 2, 3, 5\}$ and $B = \{4, 6, 9\}$. Define a relation $R$ from $A$ to $B$ by $R = \{(x, y) : \text{the difference between } x \text{ and } y \text{ is odd};\ x \in A,\ y \in B\}$. Write $R$ in roster form.

Solution. A difference is odd exactly when one number is even and the other odd, so instead of computing twelve subtractions, sort by parity:

• Odd elements of $A$: $1, 3, 5$. Even elements of $A$: $2$.
• Even elements of $B$: $4, 6$. Odd elements of $B$: $9$.

An odd $x$ pairs with an even $y$, and the even $x$ pairs with the odd $y$:

• $x = 1$: $(1,4)$, $(1,6)$
• $x = 3$: $(3,4)$, $(3,6)$
• $x = 5$: $(5,4)$, $(5,6)$
• $x = 2$: $(2,9)$

Question 4

The figure below shows a relationship between the sets $P$ and $Q$. Write this relation (i) in set-builder form, (ii) in roster form. What is its domain and range?

Solution. Read the arrows: $5 \mapsto 3$, $6 \mapsto 4$, $7 \mapsto 5$. In every case the image is $2$ less than the source, and that is the pattern the set-builder form should capture.

(i) Set-builder form:

$$R = \{(x, y) : y = x – 2,\ x \in P\} \quad\text{where } P = \{5, 6, 7\}$$

(ii) Roster form:

$$R = \{(5,3),\ (6,4),\ (7,5)\}$$

Question 5

Let $A = \{1, 2, 3, 4, 6\}$. Let $R$ be the relation on $A$ defined by $\{(a, b) : a, b \in A,\ b \text{ is exactly divisible by } a\}$.

(i) Write $R$ in roster form    (ii) Find the domain of $R$    (iii) Find the range of $R$

Solution. Take each possible $a$ in turn and list the multiples of $a$ that lie in $A$. Working in this order guarantees nothing is missed.

• $a = 1$: every element is divisible by $1$ — $(1,1), (1,2), (1,3), (1,4), (1,6)$
• $a = 2$: the multiples of $2$ in $A$ are $2, 4, 6$ — $(2,2), (2,4), (2,6)$
• $a = 3$: the multiples of $3$ in $A$ are $3, 6$ — $(3,3), (3,6)$
• $a = 4$: only $4$ itself ($8 \notin A$) — $(4,4)$
• $a = 6$: only $6$ itself — $(6,6)$

That is $5 + 3 + 2 + 1 + 1 = 12$ pairs.

$$R = \{(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)\}$$

Every element of $A$ appears as a first entry (since $a$ divides itself) and as a second entry (for the same reason), so domain and range are both the whole of $A$.

Question 6

Determine the domain and range of the relation $R$ defined by $R = \{(x,\ x + 5) : x \in \{0, 1, 2, 3, 4, 5\}\}$.

Solution. The relation is given directly in terms of a parameter $x$, so the domain is precisely the set that $x$ runs over, and each image is $5$ more:

$$R = \{(0,5),\ (1,6),\ (2,7),\ (3,8),\ (4,9),\ (5,10)\}$$

Question 7

Write the relation $R = \{(x, x^3) : x \text{ is a prime number less than } 10\}$ in roster form.

Solution. The primes below $10$ are $2$, $3$, $5$ and $7$ — note that $1$ is not prime and $9 = 3^2$ is not either. Cubing each:

$$2^3 = 8, \qquad 3^3 = 27, \qquad 5^3 = 125, \qquad 7^3 = 343$$

Question 8

Let $A = \{x, y, z\}$ and $B = \{1, 2\}$. Find the number of relations from $A$ to $B$.

Solution. A relation from $A$ to $B$ is a subset of $A \times B$, so counting relations means counting subsets.

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

A set of $6$ elements has $2^6$ subsets, so

$$\text{number of relations} = 2^6 = 64$$

The empty relation and the whole of $A \times B$ are both included in that count — they are legitimate relations, just uninteresting ones.

Question 9

Let $R$ be the relation on $\mathbf{Z}$ defined by $R = \{(a, b) : a, b \in \mathbf{Z},\ a – b \text{ is an integer}\}$. Find the domain and range of $R$.

Solution. The condition looks restrictive but is not. If $a$ and $b$ are both integers then $a – b$ is automatically an integer — the set $\mathbf{Z}$ is closed under subtraction. So the condition excludes nothing:

$$R = \mathbf{Z} \times \mathbf{Z}$$

Every integer appears as a first entry (pair it with anything) and as a second entry, so

$$\text{Domain} = \mathbf{Z}, \qquad \text{Range} = \mathbf{Z}$$

The question is a deliberate test of whether you read a condition or merely react to it: had the relation been defined on $\mathbf{Q}$ or $\mathbf{R}$ with the same wording, the answer would have been very different.

Common mistakes

• Question 1, giving the domain as all of $A$. The rule $y = 3x$ fails for $x \ge 5$ because the image leaves $A$, so the domain stops at $4$. Domain means the first entries that actually occur, not the set the relation was defined on.
• Question 1, confusing range with codomain. The codomain is $\{1, \ldots, 14\}$ because $R$ maps into $A$; the range is only $\{3, 6, 9, 12\}$. Both are asked for, and they are different sets.
• Question 3, computing $x – y$ rather than the difference. “The difference between $x$ and $y$” is $|x – y|$, but parity is unaffected by sign, so the shortcut of checking odd-versus-even parity is both faster and safer.
• Question 5, forgetting the pairs $(a, a)$. Every number divides itself, so $(1,1), (2,2), (3,3), (4,4), (6,6)$ all belong. Leaving them out drops five of the twelve pairs and changes both the domain and the range.
• Question 5, including $(4, 8)$ or $(6, 12)$. The second entry must also lie in $A$, and neither $8$ nor $12$ does.
• Question 7, treating $9$ or $1$ as prime. The primes below $10$ are exactly $2, 3, 5, 7$.
• Question 8, answering $6$ instead of $2^6$. $6$ is the number of ordered pairs; the number of relations is the number of ways to choose a subset of those pairs.
• Question 9, hunting for a restriction that is not there. $a – b$ is an integer for all $a, b \in \mathbf{Z}$, so the relation is the whole Cartesian product.

Practise next

• Exercise 2.3 — functions, which are the relations in which each domain element has exactly one image. Question 5’s relation, where $1$ is paired with five different values, is precisely the kind that fails that test.
• Exercise 2.1 — worth revising if the counting in question 8 felt unfamiliar, since $2^{pq}$ is the subset count applied to $A \times B$.