If $A = \{x : x \in \mathbf{W},\ x < 2\}$, $B = \{x : x \in \mathbf{N},\ 1 < x < 5\}$, $C = \{3, 5\}$ find (i) $A \times (B \cap C)$ (ii) $A \times (B \cup C)$

> [!TLDR] > We first define the sets $A$, $B$, and $C$ based on the given conditions. Then, we perform the set operations (intersection and union) within the parentheses, and finally, compute the Cartesian product with set $A$. > The results are $A \times (B \cap C) = \{(0, 3), (1, 3)\}$ and $A \times (B \cup C) = \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\}$. To find the Cartesian products, we must first clearly define the elements of each set $A$, $B$, and $C$. Once the sets are explicitly listed, we can perform the set operations (intersection and union) and then the Cartesian product. Remember that the Cartesian product of two sets $X$ and $Y$, denoted $X \times Y$, is the set of all possible ordered pairs $(x, y)$ where $x \in X$ and $y \in Y$. Let's break this down step-by-step. 1. **Define the sets $A$, $B$, and $C$ explicitly.** * **Set $A$**: $A = \{x : x \in \mathbf{W},\ x < 2\}$ Here, $\mathbf{W}$ represents the set of **Whole Numbers**, which includes $0, 1, 2, 3, \dots$. The condition $x < 2$ means $x$ can be $0$ or $1$. So, $A = \{0, 1\}$. * **Set $B$**: $B = \{x : x \in \mathbf{N},\ 1 < x < 5\}$ Here, $\mathbf{N}$ represents the set of **Natural Numbers**, which includes $1, 2, 3, 4, \dots$. The condition $1 < x < 5$ means $x$ must be greater than $1$ and less than $5$. The natural numbers satisfying this are $2, 3, 4$. So, $B = \{2, 3, 4\}$. * **Set $C$**: $C = \{3, 5\}$ This set is given directly. Now we have: $A = \{0, 1\}$ $B = \{2, 3, 4\}$ $C = \{3, 5\}$ 2. **Calculate $B \cap C$ for part (i).** The intersection of two sets, $B \cap C$, contains all elements that are common to both set $B$ and set $C$. $B = \{2, 3, 4\}$ $C = \{3, 5\}$ The only element common to both sets is $3$. So, $B \cap C = \{3\}$. 3. **Calculate $A \times (B \cap C)$ for part (i).** We need to find the Cartesian product of set

If A = \x : x ∈ mathbfW,x < 2\, B = \x : x ∈ mathb – Class 11 Mathematics

NCERT Exemplar · Q3

A = \{x : x \in \mathbf{W},\ x < 2\}

(i)

A \times (B \cap C)

(ii)

A \times (B \cup C)

Manipur CohsemShort· 2mImportance★★★★★est

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✓ Free question

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:

Coordinate geometry: Every point $(x, y)$ in the plane is from $\mathbb{R} \times \mathbb{R}$ .
Database tables: A table's rows are elements of the Cartesian product of its column domains.
Probability: All possible outcomes of two independent events form a Cartesian product.
Functions: A function from $A$ to $B$ is a subset of $A \times B$ with special properties.

Important

The Cartesian product is not commutative ( $A \times B \neq B \times A$ in general), but it is associative: $(A \times B) \times C$ can be thought of as $A \times B \times C$ , the set of ordered triples. This extends naturally to any number of sets.

Quick Check for Yourself

If $A = \{1, 2\}$ and $B = \{2, 3\}$ , what is $A \times B$ ? What is $B \times A$ ? Are they the same?

Answer: $A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3)\}$ ; $B \times A = \{(2, 1), (2, 2), (3, 1), (3, 2)\}$ . They share only $(2, 2)$ — the rest are different because the order of coordinates is swapped.

The Cartesian product of two sets is introduced at the very start of the NCERT Class 11 Mathematics chapter on Relations and Functions, and "Cartesian product of sets definition and examples" is a commonly searched foundational topic for CBSE board and JEE Main preparation. This concept also underlies coordinate geometry and the formal definition of a function, both of which are frequently tested in "relations and functions important questions".

We first define the sets $A$ , $B$ , and $C$ based on the given conditions. Then, we perform the set operations (intersection and union) within the parentheses, and finally, compute the Cartesian product with set $A$ .

The results are $A \times (B \cap C) = \{(0, 3), (1, 3)\}$ and $A \times (B \cup C) = \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\}$ .

To find the Cartesian products, we must first clearly define the elements of each set $A$ , $B$ , and $C$ . Once the sets are explicitly listed, we can perform the set operations (intersection and union) and then the Cartesian product. Remember that the Cartesian product of two sets $X$ and $Y$ , denoted $X \times Y$ , is the set of all possible ordered pairs $(x, y)$ where $x \in X$ and $y \in Y$ .

Let's break this down step-by-step.

Define the sets $A$ , $B$ , and $C$ explicitly.
- Set $A$ : $A = \{x : x \in \mathbf{W},\ x < 2\}$
  
  Here, $\mathbf{W}$ represents the set of Whole Numbers, which includes $0, 1, 2, 3, \dots$ .
  
  The condition $x < 2$ means $x$ can be $0$ or $1$ .
  
  So, $A = \{0, 1\}$ .
- Set $B$ : $B = \{x : x \in \mathbf{N},\ 1 < x < 5\}$
  
  Here, $\mathbf{N}$ represents the set of Natural Numbers, which includes $1, 2, 3, 4, \dots$ .
  
  The condition $1 < x < 5$ means $x$ must be greater than $1$ and less than $5$ . The natural numbers satisfying this are $2, 3, 4$ .
  
  So, $B = \{2, 3, 4\}$ .
- Set $C$ : $C = \{3, 5\}$
  
  This set is given directly.
Now we have:

$A = \{0, 1\}$

$B = \{2, 3, 4\}$

$C = \{3, 5\}$
Calculate $B \cap C$ for part (i).

The intersection of two sets, $B \cap C$ , contains all elements that are common to both set $B$ and set $C$ .

$B = \{2, 3, 4\}$

$C = \{3, 5\}$

The only element common to both sets is $3$ .

So, $B \cap C = \{3\}$ .
Calculate $A \times (B \cap C)$ for part (i).

We need to find the Cartesian product of set $A$ and the set $(B \cap C)$ .

$A = \{0, 1\}$

$B \cap C = \{3\}$
$X \times Y$
We form all possible ordered pairs where the first element comes from $A$ and the second element comes from $B \cap C$ .

For $x=0 \in A$ , we pair it with $y=3 \in (B \cap C)$ , giving $(0, 3)$ .

For $x=1 \in A$ , we pair it with $y=3 \in (B \cap C)$ , giving $(1, 3)$ .

Therefore, $A \times (B \cap C) = \{(0, 3), (1, 3)\}$ .
Calculate $B \cup C$ for part (ii).

The union of two sets, $B \cup C$ , contains all elements that are in set $B$ or in set $C$ (or both). We list all unique elements from both sets.

$B = \{2, 3, 4\}$

$C = \{3, 5\}$

Combining the elements and listing each unique element once: $2, 3, 4, 5$ .

So, $B \cup C = \{2, 3, 4, 5\}$ .
Calculate $A \times (B \cup C)$ for part (ii).

We need to find the Cartesian product of set $A$ and the set $(B \cup C)$ .

$A = \{0, 1\}$

$B \cup C = \{2, 3, 4, 5\}$

We form all possible ordered pairs where the first element comes from $A$ and the second element comes from $B \cup C$ .

For $x=0 \in A$ , we pair it with each element in $B \cup C$ :

$(0, 2), (0, 3), (0, 4), (0, 5)$

For $x=1 \in A$ , we pair it with each element in $B \cup C$ :

$(1, 2), (1, 3), (1, 4), (1, 5)$

Combining these, we get:

$A \times (B \cup C) = \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\}$ .

✓Final answer

A \times (B \cap C) = \boxed{\{(0, 3), (1, 3)\}}

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

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 mark
$A \times B = \{(a, x), (a,y), (b,x), (b,y)\}$
›Reveal solutionSolution

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

By definition, $A\times B$ consists of all ordered pairs $(p,q)$ with $p\in A$ and $q\in B$ . Given $A\times B=\{(a,x),(a,y),(b,x),(b,y)\}$ , the set of distinct first coordinates appearing is $\{a,b\}$ , so $A=\{a,b\}$ ; the set of distinct second coordinates is $\{x,y\}$ , so $B=\{x,y\}$ . Checking: $\{a,b\}\times\{x,y\}$ indeed produces exactly the four pairs $(a,x),(a,y),(b,x),(b,y)$ .

✓Final answer
$A=\{a,b\}$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$n(A) = p$
›Reveal solutionSolution

A relation from A to B is a subset of $A\times B$ ; since $A\times B$ has $pq$ elements, it has $2^{pq}$ subsets, i.e. $2^{pq}$ possible relations.

Since $n(A)=p$ and $n(B)=q$ , the Cartesian product $A\times B$ contains exactly $n(A)\times n(B)=pq$ ordered pairs.

A relation from $A$ to $B$ is defined as any subset of $A\times B$ (including the empty relation and the full relation). A set with $pq$ elements has exactly $2^{pq}$ subsets.

✓Final answer
$2^{pq}$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$A \times B = \{(a,1), (a,5), (1,2), (b,2), (b,5), (b,1)\}$
›Reveal solutionSolution

Reading the printed set as $A\times B=\{(a,1),(a,2),(a,5),(b,1),(b,2),(b,5)\}$ (correcting a scan misprint of "(a,2)" as "(1,2)"), $A=\{a,b\}$ and $B=\{1,2,5\}$ , giving $B\times A=\{(1,a),(1,b),(2,a),(2,b),(5,a),(5,b)\}$ .

Noting a likely scan issue: the printed set $\{(a,1),(a,5),(1,2),(b,2),(b,5),(b,1)\}$ has one pair, $(1,2)$ , whose first coordinate breaks the pattern — in a genuine Cartesian product $A\times B$ every first coordinate must come from the same set $A$ , and here " $b$ " already appears paired with all three of $1,2,5$ while " $a$ " appears with only $1,5$ (missing $2$ ). This is consistent with "(1,2)" being a scanning misread of "(a,2)" (the two characters look similar in some fonts). With that one correction the six pairs read as a complete, consistent product:

$A\times B=\{(a,1),(a,2),(a,5),(b,1),(b,2),(b,5)\}.$

From this, $A=\{a,b\}$ (the first coordinates) and $B=\{1,2,5\}$ (the second coordinates), and indeed $n(A)\times n(B)=2\times3=6$ matches the 6 pairs given — confirming the correction.

To find $B\times A$ , pair every element of $B$ with every element of $A$ , in that order:

$B\times A=\{(1,a),(1,b),(2,a),(2,b),(5,a),(5,b)\}.$

✓Final answer
$B\times A=\{(1,a),(1,b),(2,a),(2,b),(5,a),(5,b)\}$

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Q.If A={x:x∈W, x<2}A = \{x : x \in \mathbf{W},\ x < 2\}A={x:x∈W, x<2}, B={x:x∈N, 1<x<5}B = \{x : x \in \mathbf{N},\ 1 < x < 5\}B={x:x∈N, 1<x<5}, C={3,5}C = \{3, 5\}C={3,5} find

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Q.If $A = \{x : x \in \mathbf{W},\ x < 2\}$ , $B = \{x : x \in \mathbf{N},\ 1 < x < 5\}$ , $C = \{3, 5\}$ find

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