Relations and Functions • Topic 1 of 3

Cartesian Product of Sets

An ordered pair $(a, b)$ is two objects written in a fixed order — a first component $a$ and a second component $b$. Unlike a set, order is essential: $(a, b) = (c, d)$ only if $a = c$ and $b = d$. So $(2, 3) \ne (3, 2)$, whereas $\{2,3\} = \{3,2\}$.

The Cartesian product $A \times B$ is the set of all ordered pairs whose first component comes from $A$ and second from $B$:

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

Think of it as every possible pairing — one element of $A$ matched with one element of $B$, exhaustively. Because each of the $n(A)$ first components can be paired with each of the $n(B)$ second components, the size multiplies:

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

Key properties:

Property	Statement
Not commutative	$A \times B \ne B \times A$ in general (unless $A = B$ or one is empty)
Empty factor	If $A = \varnothing$ or $B = \varnothing$, then $A \times B = \varnothing$
Distributes over $\cup,\cap$	$A \times (B \cup C) = (A \times B) \cup (A \times C)$
Plane and space	$\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$ (the plane); $\mathbb{R}^3$ is the ordered triples $(x,y,z)$

The familiar $xy$-plane is exactly $\mathbb{R} \times \mathbb{R}$: every point is an ordered pair of real numbers. This is why the chapter matters — coordinate geometry, graphs of functions and the whole of calculus live inside a Cartesian product.

Deeper Insight — why "ordered" changes everything: The single word ordered is what separates this chapter from the previous one on sets. A set answers only "is this object in or out?"; an ordered pair additionally records "in which role?" — first or second, input or output, $x$ or $y$. That extra bit of structure is precisely what lets a Cartesian product model relationships between two collections rather than just a lump of objects. Once you grasp that $A \times B$ lists every conceivable pairing, the next two ideas fall out naturally: a relation is just a chosen subset of those pairings (the ones that actually satisfy some rule), and a function is a relation disciplined by one extra condition. Everything downstream — domain, range, graphs, even the coordinate plane you have used since Class 9 — is built on this one move from unordered membership to ordered pairing.

Solved Examples

Worked Example

If $A = \{1, 2\}$ and $B = \{a, b, c\}$, write $A \times B$ and state $n(A \times B)$.

Solution

Pair each element of $A$ with every element of $B$.
$A \times B = \{(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)\}$.
$n(A \times B) = 2 \times 3 = 6$.

Answer: $6$ ordered pairs as listed above.

Worked Example

If $(x + 1, y - 2) = (3, 1)$, find $x$ and $y$.

Solution

Equal ordered pairs ⇒ components equal: $x + 1 = 3$ and $y - 2 = 1$.
$x = 2$, $y = 3$.

Answer: $x = 2,\ y = 3$.

Worked Example

If $n(A) = 3$ and $n(B) = 4$, how many elements are in $A \times B$, and how many in $A \times A$?

Solution

$n(A \times B) = n(A) \cdot n(B) = 3 \times 4 = 12$.
$n(A \times A) = 3 \times 3 = 9$.

Answer: $12$ and $9$ respectively.

Worked Example

If $A \times B = \{(p, q), (p, r), (m, q), (m, r)\}$, find $A$ and $B$.

Solution

First components form $A$: $\{p, m\}$.
Second components form $B$: $\{q, r\}$.
Check: $2 \times 2 = 4$ pairs ✓.

Answer: $A = \{p, m\}$, $B = \{q, r\}$.

Worked Example

Let $A = \{1, 2\}$. Is $(1, 2) \in A \times A$? Is $(1, 3) \in A \times A$?

Solution

$A \times A = \{(1,1),(1,2),(2,1),(2,2)\}$.
$(1,2)$ appears in the list — yes.
$3 \notin A$, so $(1,3)$ cannot be in $A \times A$ — no.

Answer: $(1,2) \in A \times A$; $(1,3) \notin A \times A$.

Worked Example

If $A = \{-1, 1\}$, find $A \times A \times A$ and state how many ordered triples it contains.

Solution

$n(A \times A \times A) = 2^3 = 8$.
The triples are all $(\pm1, \pm1, \pm1)$: $(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)$.

Answer: $8$ ordered triples (the eight $(\pm1,\pm1,\pm1)$ combinations).

Key Points

An ordered pair $(a,b)$ fixes order: $(a,b) = (c,d) \iff a = c$ and $b = d$.
$A \times B = \{(a,b) : a \in A,\ b \in B\}$ — every possible pairing.
$n(A \times B) = n(A) \cdot n(B)$; the product is empty if either factor is empty.
$A \times B \ne B \times A$ in general; the $xy$-plane is $\mathbb{R} \times \mathbb{R}$.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.If $A=\{1,2\}$ and $B=\{3\}$, then $A\times B=$

Explanation: $A\times B$ pairs each element of $A$ with each of $B$.

Q2.If $n(A)=3$ and $n(B)=4$, then $n(A\times B)=$

Explanation: $n(A\times B)=n(A)\cdot n(B)=12$.

Q3.$(a,b)=(c,d)$ if and only if:

Explanation: Ordered pairs are equal iff matching components are equal.

Q4.$A\times B=B\times A$ holds when:

Explanation: Cartesian product is not commutative unless $A=B$ (or one is empty).

Practice more MCQs → 📝 Assignment · 35 marks