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

If A × B = \(a, x), (a, y), (b, x), (b, y)\, then A = \a, b\, B = \x, y\

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A \times B = \{(a, x), (a, y), (b, x), (b, y)\}

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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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The Cartesian product $A \times B$ lists every ordered pair where the first element comes from $A$ and the second from $B$ . By collecting all first coordinates we get $A = \{a, b\}$ , and all second coordinates give $B = \{x, y\}$ .

The Cartesian product $A \times B$ is defined as the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$ . This means that if you are given the full set of pairs, you can recover $A$ and $B$ by simply looking at which elements appear in the first and second positions respectively.

Here, the given product is $\{(a, x), (a, y), (b, x), (b, y)\}$ . Notice that every possible combination of the two first-position elements $\{a, b\}$ with the two second-position elements $\{x, y\}$ appears exactly once. That is the hallmark of a complete Cartesian product between two finite sets.

Let’s extract the sets step by step.

Find $A$ : Look at the first coordinate of every ordered pair. The pairs are $(a, x)$ , $(a, y)$ , $(b, x)$ , $(b, y)$ . The first coordinates are $a$ and $b$ . So $A$ must contain exactly these two elements: $A = \{a, b\}$ .
Find $B$ : Now look at the second coordinate of every ordered pair. The second coordinates are $x$ and $y$ . So $B$ must contain exactly these two elements: $B = \{x, y\}$ . …

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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 $B=\{x,y\}$ . Ch …

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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.

…

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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:

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

…

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Q.If A×B={(a,x),(a,y),(b,x),(b,y)}A \times B = \{(a, x), (a, y), (b, x), (b, y)\}A×B={(a,x),(a,y),(b,x),(b,y)}, then A={a,b}A = \{a, b\}A={a,b}, B={x,y}B = \{x, y\}B={x,y}

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

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