State True or False: The sets \1, 2, 3, 4\ and \3, 4, 5, 6\ are equal.

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\{1, 2, 3, 4\}

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Concept understanding — Cartesian Product

Cartesian Product: From Intuition to Definition

Imagine you're ordering a pizza. You have two choices to make: the size (Small, Medium, Large) and the topping (Cheese, Pepperoni, Veggie). How many different pizzas can you order?

You can pair each size with each topping:

Small + Cheese, Small + Pepperoni, Small + Veggie
Medium + Cheese, Medium + Pepperoni, Medium + Veggie
Large + Cheese, Large + Pepperoni, Large + Veggie

That's $3 \times 3 = 9$ possible pizzas. What you just did — systematically pairing every element of one set with every element of another — is the Cartesian product in action.

The Intuition

The Cartesian product is a way to combine two sets to create a new set of ordered pairs. The order matters: (Small, Cheese) is different from (Cheese, Small) — one is a pizza order, the other is nonsense.

Think of it like a multiplication table for sets. If set $A$ has $m$ items and set $B$ has $n$ items, their Cartesian product has $m \times n$ items.

Note

The name comes from René Descartes, who used this idea to create the coordinate plane — every point $(x, y)$ on a graph is an element of the Cartesian product of the x-axis and y-axis.

The Precise Definition

Let $A$ and $B$ be two sets. The Cartesian product of $A$ and $B$ , written $A \times B$ , is the set of all ordered pairs $(a, b)$ where $a$ is from $A$ and $b$ is from $B$ .

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

The vertical bar means "such that." So read it as: "The set of all ordered pairs (a, b) such that a belongs to A and b belongs to B."

Key Properties to Remember

Order matters: $A \times B$ is generally not the same as $B \times A$ . For example, if $A = \{1, 2\}$ and $B = \{x, y\}$ :
- $A \times B = \{(1, x), (1, y), (2, x), (2, y)\}$
- $B \times A = \{(x, 1), (x, 2), (y, 1), (y, 2)\}$
These are different sets because the pairs are ordered differently.
Size formula: If $|A| = m$ and $|B| = n$ , then $|A \times B| = m \times n$ . This holds even if one set is empty — then the product is empty.
Empty set: $A \times \emptyset = \emptyset$ and $\emptyset \times B = \emptyset$ . You can't form any pairs if one set has nothing to contribute.

Watch out

A common mistake: thinking $A \times B$ contains all possible combinations of elements from $A$ and $B$ without caring about order. But $(a, b)$ and $(b, a)$ are different pairs unless $a = b$ . Always treat ordered pairs as distinct based on position.

Examples to Cement the Idea

Example 1: $A = \{1, 2\}$ , $B = \{3, 4\}$

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

Four pairs, as expected ( $2 \times 2 = 4$ ).

Example 2: $A = \{a\}$ , $B = \{1, 2, 3\}$

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

Three pairs — every element of $B$ gets paired with the single element of $A$ .

Example 3: $A = \{0, 1\}$ , $B = \{0, 1\}$

A \times B = \{(0, 0), (0, 1), (1, 0), (1, 1)\}

This is the set of all possible 2-bit binary strings — a foundation for computer science.

Why This Matters

The Cartesian product is the mathematical backbone of: …

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Two sets are equal only if they contain exactly the same elements. Since the first set has $1$ and $2$ which the second set lacks, they are not equal — the statement is False.

The idea of set equality is simple but often rushed past. Two sets are equal precisely when every element of one is also an element of the other — and vice versa. There is no notion of order, no notion of repetition; it’s purely about membership.

Let’s check the given sets:

List the elements of each set.

Set $A = \{1, 2, 3, 4\}$ .

Set $B = \{3, 4, 5, 6\}$ .
Check if every element of $A$ is in $B$ .
- $1 \in A$ but $1 \notin B$ — that alone breaks equality.
- $2 \in A$ but $2 \notin B$ — another missing element.
Check the reverse: every element of $B$ in $A$ ?
- $5 \in B$ but $5 \notin A$ .
- $6 \in B$ but $6 \notin A$ .

Since neither set is a subset of the other, they cannot be equal. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 27 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$A = \{1, 2\}$
(a) 6
(b) 36
(c) 32
(d) 64
›Reveal solutionSolution

A relation from $A$ to $B$ is any subset of $A \times B$ ; with $|A \times B| = 6$ , there are $2^6 = 64$ subsets, hence 64 relations.

A relation from $A$ to $B$ is defined as any subset of the Cartesian product $A \times B$ . Here $|A| = 2$ and $|B| = 3$ , so $|A \times B| = 2 \times 3 = 6$ . The total number of subsets of a set with 6 elements is $2^6 = 64$ . Since every subset of …

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CBSE 2026Set ANNUAL1 markMCQ
$A = \{1, 2\}$
(a) 4
(b) $2^4$
(c) 2
(d) 1
›Reveal solutionSolution

The number of relations from a set A to a set B is $2^{|A|\cdot|B|}$ , since every relation is a subset of $A \times B$ .

Given $A=\{1,2\}$ , $B=\{3,4\}$ , so $|A|=2$ , $|B|=2$ .

$A \times B$ has $|A|\times|B| = 2 \times 2 = 4$ ordered pairs: $(1,3),(1,4),(2,3),(2,4)$ .

…

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CBSE 2026Set ANNUAL1 mark
$A \times B = \{(p, q), (p, r)\}$
›Reveal solutionSolution

In $A\times B=\{(p,q),(p,r)\}$ , A is the set of first coordinates and B is the set of second coordinates; here $A=\{p\}$ , not $\{p,q,r\}$ .

Every ordered pair $(x,y)\in A\times B$ has $x\in A$ and $y\in B$ .

Here both ordered pairs $(p,q)$ and $(p,r)$ have first component $p$ , so the set of first components — which is exactly A — is $A=\{p\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{1, 2\}, B = \{3\} \Rightarrow A \times B =$
(a) $\{1, 2, 3\}$
(b) $\{(1,3), (2,3), (1,2)\}$
(c) $\{(1,3), (2,3)\}$
(d) $\{(1,2), (3,1)\}$
›Reveal solutionSolution

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

For sets $A$ and $B$ , $A \times B = \{(a, b) : a \in A, b \in B\}$ — every element of $A$ is paired, in order, with every element of $B$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{a, b\}, B = \{c, d\}$
(a) 8
(b) 16
(c) 32
(d) 64
›Reveal solutionSolution

The number of relations from $A$ to $B$ is $2^{|A \times B|} = 2^4 = 16$ .

A relation from $A$ to $B$ is defined as any subset of the Cartesian product $A \times B$ . If $A$ has $m$ elements and $B$ has $n$ elements, $A \times B$ has $mn$ elements, and a set with $mn$ elements has $2^{mn}$ subsets.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If A = {1, 2} and B = {3, 4}, then A \times B is:$
(a) {3, 4, 6, 8}
(b) {3, 8}
(c) {(1, 3), (1, 4), (2, 3), (2, 4)}
(d) None of these
›Reveal solutionSolution

$A \times B$ is the set of all ordered pairs $(a,b)$ with $a \in A$ , $b \in B$ .

Given $A=\{1,2\}$ and $B=\{3,4\}$ .

Pair each element of $A$ with every element of $B$ :

$1 \to (1,3),(1,4)$

$2 \to (2,3),(2,4)$

…

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CBSE 2025Set ANNUAL1 mark
$A = \{x, y, z\}$
›Reveal solutionSolution

A relation from $A$ to $B$ is any subset of $A \times B$ ; with $|A|=3$ and $|B|=2$ , $|A \times B| = 6$ , so there are $2^6 = 64$ relations.

$A = \{x, y, z\}$ has 3 elements, $B = \{1, 2\}$ has 2 elements.

…

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CBSE 2024Set ANNUAL1 markMCQ
$A$
(a) 8
(b) 9
(c) 10
(d) 6
›Reveal solutionSolution

If $A$ has $m$ elements and $B$ has $n$ elements, then $A\times B$ has $mn$ elements.

The Cartesian product $A\times B$ consists of all ordered pairs $(a,b)$ with $a\in A$ and $b\in B$ . Since $A$ has 3 elements and $B=\{3,4,5\}$ has 3 elements, each of the 3 choices for the first coordinate can be paire …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Let A = {x, y, z} and B = {1, 2}, then number of relations from A into B will be:$
(a) 6
(b) 9
(c) 24
(d) 64
›Reveal solutionSolution

A relation from A to B is any subset of $A\times B$ ; with $|A|=3,|B|=2$ , there are $2^{3\times2}=64$ subsets.

A relation from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$ .

$|A| = 3$ (elements $x,y,z$ ), $|B| = 2$ (elements $1,2$ ).

So $|A \times B| = 3 \times 2 = 6$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$n((A\times B)\cap(A\times C))=8$
(a) $8$
(b) $6$
(c) $16$
(d) $4$
›Reveal solutionSolution

Using $(A\times B)\cap(A\times C)=A\times(B\cap C)$ , we get $n(A)=4$ .

For sets, $(A\times B)\cap(A\times C)=A\times(B\cap C)$ , so

$n\big((A\times B)\cap(A\times C)\big)=n(A)\times n(B\cap C).$ …

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CBSE 2024Set ANNUAL1 markMCQ
$n$
(a) $2^n$
(b) $2^{n^2}$
(c) $n^2$
(d) $n^n$ .
›Reveal solutionSolution

The number of relations on a set of $n$ elements is $2^{n^2}$ .

A relation $R$ on set $A$ is defined as any subset of the Cartesian product $A\times A$ . Since $A$ has $n$ elements, $A\times A$ has $n\times n = n^2$ ordered pairs. The number of subsets of a set with $n^2$ elements is $2^{n^2}$ (by the subset-counting rule). Each such subset is a valid relation on $A$ (including the …

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CBSE 2023Set ANNUAL1 markMCQ
$A = \{a, b, c\}$
(a) 8
(b) 5
(c) $2^9$
(d) 9
›Reveal solutionSolution

The Cartesian product's size is the product of the two sets' sizes: $3 \times 3 = 9$ .

The Cartesian product $A \times B$ consists of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$ . For each of the $n(A)$ choices of $a$ , there are $n(B)$ choices of $b$ , giving:

$n(A \times B) = n(A) \cdot n(B)$ …

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Q.State True or False: The sets {1,2,3,4}\{1, 2, 3, 4\}{1,2,3,4} and {3,4,5,6}\{3, 4, 5, 6\}{3,4,5,6} are equal.

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Q.State True or False: The sets $\{1, 2, 3, 4\}$ and $\{3, 4, 5, 6\}$ are equal.

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