NCERT Class 11 Mathematics — Relations and Functions, Exercise 2.1. All 10 questions solved.

The chapter opens with the Cartesian product, which pairs every element of one set with every element of another:

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

Two facts carry the whole exercise. First, ordered pairs are ordered: $(a, b) = (c, d)$ only when $a = c$ and $b = d$, so $(1, 2) \ne (2, 1)$ and $A \times B \ne B \times A$ in general. Second, the counting rule

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

Key insight. A Cartesian product can be read backwards. Given a set of ordered pairs, the first coordinates (with repeats dropped) rebuild $A$ and the second coordinates rebuild $B$. Questions 6, 9 and 10 are all this one idea, and question 10 adds the counting rule to pin down how many elements $A$ must have before the reading even starts.

Question 1

If $\left(\dfrac{x}{3} + 1,\ y – \dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right)$, find the values of $x$ and $y$.

Solution. Two ordered pairs are equal exactly when their corresponding entries are equal. That splits one equation between pairs into two ordinary equations:

$$\frac{x}{3} + 1 = \frac{5}{3} \qquad\text{and}\qquad y – \frac{2}{3} = \frac{1}{3}$$

From the first:

$$\frac{x}{3} = \frac{5}{3} – 1 = \frac{2}{3} \quad\Longrightarrow\quad x = 2$$

From the second:

$$y = \frac{1}{3} + \frac{2}{3} = 1$$

Question 2

If the set $A$ has $3$ elements and the set $B = \{3, 4, 5\}$, then find the number of elements in $(A \times B)$.

Solution. $B$ has $3$ elements. Each of $A$’s $3$ elements is paired with each of $B$’s $3$, so

$$n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9$$

Note that the identities of $A$’s elements are never needed — only how many there are.

Question 3

If $G = \{7, 8\}$ and $H = \{5, 4, 2\}$, find $G \times H$ and $H \times G$.

Solution. Take each element of the first set in turn and pair it with every element of the second, keeping the order.

$$G \times H = \{(7,5),\ (7,4),\ (7,2),\ (8,5),\ (8,4),\ (8,2)\}$$

$$H \times G = \{(5,7),\ (5,8),\ (4,7),\ (4,8),\ (2,7),\ (2,8)\}$$

Both have $2 \times 3 = 6$ elements, yet they share not a single ordered pair — $(7,5)$ and $(5,7)$ are different objects. This is the clearest demonstration that $G \times H \ne H \times G$.

Question 4

State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.

(i) If $P = \{m, n\}$ and $Q = \{n, m\}$, then $P \times Q = \{(m,n), (n,m)\}$. (ii) If $A$ and $B$ are non-empty sets, then $A \times B$ is a non-empty set of ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$. (iii) If $A = \{1, 2\}$, $B = \{3, 4\}$, then $A \times (B \cap \varphi) = \varphi$.

Solution.

(i) False. $P$ and $Q$ are in fact the same set, $\{m, n\}$, so the product has $2 \times 2 = 4$ pairs, not $2$. The statement lists only the pairs with unequal entries and drops $(m,m)$ and $(n,n)$. Corrected:

$$P \times Q = \{(m,m),\ (m,n),\ (n,m),\ (n,n)\}$$

(ii) True. This is the definition, together with the observation that if $A$ and $B$ each contain at least one element then at least one pair can be formed.

(iii) True. $B \cap \varphi = \varphi$, and pairing anything with the empty set produces no pairs at all:

$$A \times \varphi = \varphi$$

Question 5

If $A = \{-1, 1\}$, find $A \times A \times A$.

Solution. Build it in stages. First

$$A \times A = \{(-1,-1),\ (-1,1),\ (1,-1),\ (1,1)\}$$

Then attach each element of $A$ to each of these four pairs, producing ordered triples. The count is $2 \times 2 \times 2 = 8$, which is the check that none has been dropped:

$$A \times A \times A = \{(-1,-1,-1),\ (-1,-1,1),\ (-1,1,-1),\ (-1,1,1),$$ $$(1,-1,-1),\ (1,-1,1),\ (1,1,-1),\ (1,1,1)\}$$

Listing them in this order — varying the last coordinate fastest — is what makes the eight easy to produce without repetition or omission.

Question 6

If $A \times B = \{(a, x), (a, y), (b, x), (b, y)\}$, find $A$ and $B$.

Solution. Read the product backwards. $A$ is the set of all first entries and $B$ the set of all second entries, each written without repetition:

$$A = \{a, b\}, \qquad B = \{x, y\}$$

Check: $n(A) \times n(B) = 2 \times 2 = 4$, matching the four pairs given. That check matters — if the four pairs had been $\{(a,x), (a,y), (b,x)\}$, no pair of sets would produce exactly that list, because $A \times B$ must contain every combination.

Question 7

Let $A = \{1, 2\}$, $B = \{1, 2, 3, 4\}$, $C = \{5, 6\}$ and $D = \{5, 6, 7, 8\}$. Verify that

(i) $A \times (B \cap C) = (A \times B) \cap (A \times C)$ (ii) $A \times C$ is a subset of $B \times D$

Solution.

(i) Left side first. $B$ holds $\{1,2,3,4\}$ and $C$ holds $\{5,6\}$, which share nothing:

$$B \cap C = \varphi \quad\Longrightarrow\quad A \times (B \cap C) = A \times \varphi = \varphi$$

Right side:

$$A \times B = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4)\}$$ $$A \times C = \{(1,5), (1,6), (2,5), (2,6)\}$$

Every pair in $A \times B$ has second entry at most $4$, and every pair in $A \times C$ has second entry at least $5$, so no pair is common:

$$(A \times B) \cap (A \times C) = \varphi$$

Both sides equal $\varphi$, so the identity holds here. ✓

(ii) From above, $A \times C = \{(1,5), (1,6), (2,5), (2,6)\}$. Now

$$B \times D = \{(b, d) : b \in \{1,2,3,4\},\ d \in \{5,6,7,8\}\}$$

which has $4 \times 4 = 16$ pairs. Each of the four pairs of $A \times C$ has first entry in $\{1,2\} \subset B$ and second entry in $\{5,6\} \subset D$, so each lies in $B \times D$. Hence

$$A \times C \subset B \times D \qquad ✓$$

The general principle behind (ii) is worth stating: $A \subset B$ and $C \subset D$ together give $A \times C \subset B \times D$.

Question 8

Let $A = \{1, 2\}$ and $B = \{3, 4\}$. Write $A \times B$. How many subsets will $A \times B$ have? List them.

Solution.

$$A \times B = \{(1,3),\ (1,4),\ (2,3),\ (2,4)\}$$

This has $4$ elements, and a set with $n$ elements has $2^n$ subsets, so $A \times B$ has

$$2^4 = 16 \text{ subsets}$$

Listing them by size keeps the count honest — $1 + 4 + 6 + 4 + 1 = 16$. Writing $p = (1,3)$, $q = (1,4)$, $r = (2,3)$, $s = (2,4)$ for brevity:

• None: $\varphi$
• One: $\{p\}$, $\{q\}$, $\{r\}$, $\{s\}$
• Two: $\{p,q\}$, $\{p,r\}$, $\{p,s\}$, $\{q,r\}$, $\{q,s\}$, $\{r,s\}$
• Three: $\{p,q,r\}$, $\{p,q,s\}$, $\{p,r,s\}$, $\{q,r,s\}$
• Four: $\{p,q,r,s\}$

Question 9

Let $A$ and $B$ be two sets such that $n(A) = 3$ and $n(B) = 2$. If $(x, 1)$, $(y, 2)$, $(z, 1)$ are in $A \times B$, find $A$ and $B$, where $x$, $y$ and $z$ are distinct elements.

Solution. Every ordered pair in $A \times B$ has its first entry in $A$ and its second in $B$. Reading off the three given pairs:

• First entries $x$, $y$, $z$ all lie in $A$. They are distinct, so that is already three elements — and $n(A) = 3$, so there is no room for any more.
• Second entries $1$ and $2$ lie in $B$, which is two elements, and $n(B) = 2$.

$$A = \{x, y, z\}, \qquad B = \{1, 2\}$$

The counting hypotheses are what make the answer unique: without $n(A) = 3$, the set $A$ could have contained further elements not appearing among the three pairs listed.

Question 10

The Cartesian product $A \times A$ has $9$ elements among which are found $(-1, 0)$ and $(0, 1)$. Find the set $A$ and the remaining elements of $A \times A$.

Solution. Start with the count. Since $n(A \times A) = n(A)^2$,

$$n(A)^2 = 9 \quad\Longrightarrow\quad n(A) = 3$$

Now read the two given pairs. From $(-1, 0)$ we get $-1 \in A$ and $0 \in A$; from $(0, 1)$ we get $0 \in A$ and $1 \in A$. That is three distinct elements $-1$, $0$, $1$ — exactly the three $A$ is allowed:

$$A = \{-1, 0, 1\}$$

The full product has all nine pairs from these three elements:

$$A \times A = \{(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)\}$$

Removing the two already named leaves seven.

Common mistakes

• Question 3, thinking $G \times H = H \times G$. They have the same number of elements and none in common. Cartesian product is not commutative, and this is the question that proves it.
• Question 4(i), not noticing $P$ and $Q$ are the same set. $\{m,n\}$ and $\{n,m\}$ are equal, so the product includes $(m,m)$ and $(n,n)$. Writing sets in a different order does not make them different sets.
• Question 5, producing pairs instead of triples. $A \times A \times A$ consists of ordered triples; there should be eight of them, not four.
• Question 6, writing $A$ with a repeated element. The first entries read $a, a, b, b$, but $A = \{a, b\}$ — a set never lists an element twice.
• Question 7(i), computing $B \cap C$ carelessly. $\{1,2,3,4\}$ and $\{5,6\}$ are disjoint, so the left side collapses to $\varphi$ immediately. Expanding $A \times B$ and $A \times C$ in full is still worth doing on the right, because “verify” means showing both sides, not asserting the identity.
• Question 9, adding elements to $A$ that were never given. The count $n(A) = 3$ is what closes the set. With three distinct first entries already found, nothing further can belong.
• Question 10, taking $n(A) = 9$. The product has nine elements, so $A$ has three. Squaring is the step that is easy to skip and fatal to the rest of the question.

Practise next

• Exercise 2.2 — relations, which are simply subsets of a Cartesian product. Question 8’s list of the $16$ subsets of $A \times B$ is, in the next exercise’s language, the list of all relations from $A$ to $B$.
• Miscellaneous Exercise on Chapter 1 — worth revisiting for the subset counting $2^n$ used in question 8.