Chapter 2 · Relations and Functions

2.1 Cartesian Products of Sets

An ordered pair $(a,b)$ remembers which entry comes first. The set $A\times B$ of all such pairs has $n(A)\cdot n(B)$ elements, and $\mathbb{R}\times\mathbb{R}$ is the coordinate plane itself.

A canteen offers three main dishes — idli, dosa and poha — and two drinks — tea and coffee. A "combo" is one dish together with one drink, so the canteen has six combos: (idli, tea), (idli, coffee), (dosa, tea), and so on. Each combo is a pair, and the order in which we write it carries meaning: the first entry is always the dish, the second always the drink. Writing (tea, idli) would break the convention and confuse the kitchen.

The same structure appears whenever two independent choices are made together: a square of a chessboard is a pair (file, rank) such as (e, 4); a point of the plane is a pair $(x,y)$ of numbers; the outcome of throwing a red die and a blue die is a pair such as $(3,5)$. In this section we make "pair" precise, collect all possible pairs into a new set called the Cartesian product, and learn to count and picture it. Relations and functions (Sections 2.2 and 2.3) will be chosen from inside such products.

Ordered Pairs

📘 Definition 2.1 — Ordered pair

An ordered pair $(a,b)$ is a pair of objects written in a definite order: $a$ is its first component (or first element) and $b$ its second component. Two ordered pairs are equal exactly when their corresponding components are equal: $$(a,b)=(c,d)\quad\iff\quad a=c\ \text{ and }\ b=d.$$

So $(2,5)\ne(5,2)$: the components are the same numbers, but in different positions. Contrast this with sets, where $\{2,5\}=\{5,2\}$ because a set has no order. The rule of equality turns one statement about pairs into two equations about numbers — which is how most problems on ordered pairs are solved.

✏️ Example 1 — Equal ordered pairs

Find $x$ and $y$ in each case. (a) $(x+3,\ 5)=(6,\ 2x+y)$ (b) $(2a-b,\ a+b)=(4,\ 5)$ (c) $(x^2,\ y+1)=(9,\ 4)$

💡 Show solution▶

(a) First components: $x+3=6$, so $x=3$. Second components: $5=2x+y=6+y$, so $y=-1$.

(b) $2a-b=4$ and $a+b=5$. Adding the two equations: $3a=9$, so $a=3$ and then $b=2$.

Check (a) $(3+3,5)=(6,5)$ and $(6,2\cdot3-1)=(6,5)$ ✓. (b) $(6-2,3+2)=(4,5)$ ✓.

The Cartesian Product of Two Sets

📘 Definition 2.2 — Cartesian product

Let $A$ and $B$ be non-empty sets. The Cartesian product $A\times B$ (read "$A$ cross $B$") is the set of all ordered pairs whose first component is in $A$ and whose second component is in $B$: $$A\times B=\{(a,b) : a\in A,\ b\in B\}.$$ If $A=\phi$ or $B=\phi$, we put $A\times B=\phi$. The product $A\times A$ is also written $A^2$.

✏️ Example 2 — Writing out a product

Let $A=\{1,2,3\}$ and $B=\{x,y\}$. Write $A\times B$ and $B\times A$ in roster form. Are they equal? How many elements does each have?

💡 Show solution▶

Method Take the elements of the first set one at a time and pair each with every element of the second set — row by row, so nothing is missed.

$A\times B$ $\{(1,x),(1,y),(2,x),(2,y),(3,x),(3,y)\}$ — six pairs.

$B\times A$ $\{(x,1),(x,2),(x,3),(y,1),(y,2),(y,3)\}$ — also six pairs.

Compare $(1,x)\in A\times B$ but $(1,x)\notin B\times A$ (its first component would have to be in $B$). So $A\times B\ne B\times A$, although both have $6$ elements.

📐 Theorem 2.1 — Number of elements in a Cartesian product

If $A$ and $B$ are finite sets with $n(A)=p$ and $n(B)=q$, then $$n(A\times B)=p\,q=n(A)\cdot n(B).$$ If one of $A$, $B$ is infinite and the other is non-empty, then $A\times B$ is infinite.

Proof. Arrange the pairs in a table with one row for each $a\in A$ and one column for each $b\in B$, writing $(a,b)$ where row $a$ meets column $b$. Every pair of $A\times B$ appears in exactly one cell — its first component fixes the row and its second fixes the column — and every cell holds a pair of $A\times B$. The table has $p$ rows of $q$ cells each, so $p\,q$ cells in all. If $B$ is infinite and $a\in A$, the pairs $(a,b)$ with $b\in B$ are already infinitely many (and similarly if $A$ is infinite). $\blacksquare$

✏️ Example 3 — Recovering the sets from the product

(a) $A\times B$ has $6$ elements, and three of them are $(1,a)$, $(2,b)$ and $(3,a)$. Find $A$, $B$ and the remaining elements of $A\times B$.
(b) If $n(A)=4$ and $n(A\times B)=20$, find $n(B)$ and $n(B\times B)$.

💡 Show solution▶

(a) Read off First components come from $A$: so $1,2,3\in A$. Second components come from $B$: so $a,b\in B$. Hence $n(A)\ge3$ and $n(B)\ge2$.

(a) Count $n(A)\cdot n(B)=6$ with $n(A)\ge3$, $n(B)\ge2$ forces $n(A)=3$, $n(B)=2$. So $A=\{1,2,3\}$ and $B=\{a,b\}$.

(a) The rest $A\times B=\{(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)\}$; the three remaining elements are $(1,b)$, $(2,a)$, $(3,b)$.

(b) $4\cdot n(B)=20$ gives $n(B)=5$, and then $n(B\times B)=5\cdot5=25$.

📘 Definition 2.3 — Ordered triple; product of three sets

An ordered triple $(a,b,c)$ has three components in a definite order, and $(a,b,c)=(d,e,f)$ exactly when $a=d$, $b=e$ and $c=f$. For sets $A$, $B$, $C$, $$A\times B\times C=\{(a,b,c) : a\in A,\ b\in B,\ c\in C\},$$ and, by the same table argument applied twice, $n(A\times B\times C)=n(A)\cdot n(B)\cdot n(C)$. In particular $\mathbb{R}\times\mathbb{R}\times\mathbb{R}$ is the set of points of space (Chapter 11).

✏️ Example 4 — A product of three sets

Let $A=\{-1,1\}$. Write $A\times A\times A$ in roster form and state its number of elements.

💡 Show solution▶

Count first $n(A\times A\times A)=2\cdot2\cdot2=8$, so we must find eight triples.

List Fix the first component, then the second, then vary the third: $(-1,-1,-1),\ (-1,-1,1),\ (-1,1,-1),\ (-1,1,1),\ (1,-1,-1),\ (1,-1,1),\ (1,1,-1),\ (1,1,1)$.

Picture In space these are the eight corners of a cube of side $2$ centred at the origin.

Products of Sets of Real Numbers: the Plane

The most important Cartesian product is $\mathbb{R}\times\mathbb{R}=\{(x,y) : x,y\in\mathbb{R}\}$. Its elements are exactly the coordinates of points in the plane, which is why the plane of Class 10 coordinate geometry is called the Cartesian plane. Products of subsets of $\mathbb{R}$ therefore have pictures: if $A$ and $B$ are intervals, $A\times B$ is a rectangle; if one is a finite set, the picture is made of points or line segments.

📐 Theorem 2.2 — Products distribute over union, intersection and difference

For any sets $A$, $B$, $C$: $$A\times(B\cup C)=(A\times B)\cup(A\times C),$$ $$A\times(B\cap C)=(A\times B)\cap(A\times C),$$ $$A\times(B-C)=(A\times B)-(A\times C).$$

Proof of the second law. For any ordered pair $(x,y)$: $$\begin{aligned}(x,y)\in A\times(B\cap C)&\iff x\in A \text{ and } y\in B \text{ and } y\in C\\&\iff (x\in A \text{ and } y\in B) \text{ and } (x\in A \text{ and } y\in C)\\&\iff (x,y)\in(A\times B)\cap(A\times C).\end{aligned}$$ The other two are proved in the same way (Exercise III). $\blacksquare$

✏️ Example 5 — Checking a product law

Let $A=\{1,2\}$, $B=\{1,2,3,4\}$, $C=\{3,4,5\}$. Verify that $A\times(B\cap C)=(A\times B)\cap(A\times C)$.

💡 Show solution▶

LHS $B\cap C=\{3,4\}$, so $A\times(B\cap C)=\{(1,3),(1,4),(2,3),(2,4)\}$.

RHS $A\times B$ has the $8$ pairs $(a,b)$ with $a\in\{1,2\}$, $b\in\{1,2,3,4\}$; $A\times C$ has the $6$ pairs with $b\in\{3,4,5\}$. The pairs in both are those with $b\in\{3,4\}$: $\{(1,3),(1,4),(2,3),(2,4)\}$.

Check Both sides are the same four pairs ✓, and $n=2\cdot n(B\cap C)=2\cdot2=4$ agrees with Theorem 2.1.

✏️ Example 6 — Pictures of products in the plane

Describe each set as a picture in the Cartesian plane: (a) $\{1,2,3\}\times\{1,2\}$ (b) $[1,3]\times\{2\}$ (c) $\{2\}\times\mathbb{R}$ (d) $[1,3]\times[1,2]$.

💡 Show solution▶

(a) Six separate points: $(1,1),(2,1),(3,1),(1,2),(2,2),(3,2)$ — two rows of three dots.

(b) All points $(x,2)$ with $1\le x\le3$: a horizontal line segment at height $2$, from $(1,2)$ to $(3,2)$, both ends included.

(d) All $(x,y)$ with $1\le x\le3$ and $1\le y\le2$: the filled rectangle with corners $(1,1)$, $(3,1)$, $(3,2)$, $(1,2)$, including its edges.

$A$ is drawn on the $x$-axis in red and $B$ on the $y$-axis in blue; $A\times B$ is shaded in purple. The test point $P$ hops around the plane. It belongs to $A\times B$ only when both tests pass — its $x$ in $A$ and its $y$ in $B$. Combine a finite set with an interval and see dots become segments, and two intervals become a rectangle.

$A$:$B$:

test point $P$—

$x\in A$?—

$y\in B$?—

$P\in A\times B$?—

picture—

⚠️ Common mistakes with ordered pairs and products

1. Order matters. $(1,2)\ne(2,1)$, and $A\times B\ne B\times A$ unless $A=B$ (or one of them is empty).

2. Multiply, do not add. $n(A\times B)=n(A)\cdot n(B)$, not $n(A)+n(B)$. With $3$ dishes and $2$ drinks there are $6$ combos, not $5$.

3. A pair is one element. $(1,2)$ is a single element of $A\times B$, while $\{1,2\}$ is a set with two elements. Do not write $A\times B=\{1,2,x,y\}$.

4. The empty set kills the product. $A\times\phi=\phi$: with no second component available, no pair can be formed.

5. Same brackets, different meanings. $(1,3)$ may be an ordered pair (a point) or an open interval. Context decides; in this chapter, a pair inside $A\times B$ is always an ordered pair.

Exercises

Ordered pairs. Find $x$ and $y$.

1 $(x+1,\ y-2)=(3,\ 1)$
2 $(2x,\ x+y)=(6,\ 2)$
3 $(x-y,\ x+y)=(6,\ 10)$
4 $(x^2,\ y)=(4,\ x)$ — find all solutions.
5 Is $(1,2)=(2,1)$? Is $\{1,2\}=\{2,1\}$? Explain the difference.

💡 Show solutions▶

1 $x+1=3$, $y-2=1$: $x=2$, $y=3$.

2 $2x=6$ gives $x=3$; $3+y=2$ gives $y=-1$.

3 Adding: $2x=16$, $x=8$; then $y=2$.

4 $x^2=4$ gives $x=\pm2$, and $y=x$. Solutions: $(x,y)=(2,2)$ or $(-2,-2)$.

5 $(1,2)\ne(2,1)$, because the first components $1$ and $2$ differ. But $\{1,2\}=\{2,1\}$, because a set has no order — only membership matters.

Writing and counting products.

1 $A=\{a,b\}$, $B=\{1,2,3\}$. Write $A\times B$ and $B\times A$.
2 $A=\{0,1\}$. Write $A\times A$ and find $n(A\times A\times A)$.
3 $n(A)=4$, $n(B)=5$. Find $n(A\times B)$, $n(B\times A)$ and $n(A\times A)$.
4 $A\times A$ has $9$ elements, two of which are $(-1,0)$ and $(0,1)$. Find $A$ and the remaining elements of $A\times A$.
5 $A\times B=\{(p,x),(p,y),(q,x),(q,y)\}$. Find $A$ and $B$.
6 $G=\{7,8\}$, $H=\{5,4,2\}$. Write $G\times H$ and $H\times G$.
7 If $A\times B=\phi$, what can you say about $A$ and $B$?
8 $n(A)=2$, $n(B)=3$, $n(C)=4$. Find $n(A\times B\times C)$.
9 If $n(A\times B)=15$ and $n(A)\gt n(B)\gt1$, find $n(A)$ and $n(B)$.

Check: see each product of parts 1, 2, 4 and 6 as a table of pairs

Product:

product—

number of pairs—

💡 Show solutions▶

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

2 $A\times A=\{(0,0),(0,1),(1,0),(1,1)\}$; $n(A\times A\times A)=2^3=8$.

3 $n(A\times B)=n(B\times A)=20$; $n(A\times A)=16$.

4 $n(A)^2=9$ gives $n(A)=3$, and $-1,0,1\in A$ (they appear as components). So $A=\{-1,0,1\}$. The other seven pairs: $(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),(1,1)$.

5 First components: $A=\{p,q\}$; second components: $B=\{x,y\}$.

6 $G\times H=\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\}$; $H\times G=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}$.

7 At least one of $A$, $B$ is empty. (If both were non-empty, picking $a\in A$, $b\in B$ would give $(a,b)\in A\times B$.)

8 $2\cdot3\cdot4=24$.

9 $15=5\times3$ is the only factorisation with $n(A)\gt n(B)\gt1$, so $n(A)=5$, $n(B)=3$.

III

Laws and pictures.

1 $A=\{1,2\}$, $B=\{3,4\}$, $C=\{4,5\}$. Find $A\times(B\cap C)$ and $(A\times B)\cap(A\times C)$.
2 With the sets of part 1, find $A\times(B\cup C)$ and check its number of elements with Theorem 2.1.
3 Prove that $A\times(B\cup C)=(A\times B)\cup(A\times C)$.
4 If $A\subset B$, show that $A\times C\subset B\times C$.
5 Describe $\{2\}\times\mathbb{R}$ and $\mathbb{R}\times\{-1\}$ in the plane.
6 Describe $[0,1]\times[0,1]$ in the plane and find its area.
7 $A=\{-1,1\}$. Plot $A\times A$ and show that its points are the corners of a square. Find the area of that square.
8 Is $(2,3)\in[1,2)\times[3,5]$? Is $(1,5)$?

💡 Show solutions▶

1 $B\cap C=\{4\}$, so $A\times(B\cap C)=\{(1,4),(2,4)\}$. $A\times B=\{(1,3),(1,4),(2,3),(2,4)\}$, $A\times C=\{(1,4),(1,5),(2,4),(2,5)\}$, intersection $\{(1,4),(2,4)\}$. Equal ✓

2 $B\cup C=\{3,4,5\}$: $A\times(B\cup C)=\{(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)\}$, with $2\cdot3=6$ elements ✓.

3 $(x,y)\in A\times(B\cup C)\iff x\in A$ and ($y\in B$ or $y\in C$) $\iff$ ($x\in A$ and $y\in B$) or ($x\in A$ and $y\in C$) $\iff(x,y)\in(A\times B)\cup(A\times C)$.

4 Let $(x,y)\in A\times C$. Then $x\in A\subset B$, so $x\in B$, and $y\in C$. Hence $(x,y)\in B\times C$.

5 $\{2\}\times\mathbb{R}$ is the vertical line $x=2$; $\mathbb{R}\times\{-1\}$ is the horizontal line $y=-1$.

6 The filled unit square with corners $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$, edges included; area $1\times1=1$ square unit.

7 $A\times A=\{(-1,-1),(-1,1),(1,-1),(1,1)\}$. Adjacent points differ by $2$ in exactly one coordinate, so the four points are the corners of a square of side $2$; area $4$ square units.

8 $(2,3)$: $2\notin[1,2)$, so no. $(1,5)$: $1\in[1,2)$ and $5\in[3,5]$, so yes.

Products in real life.

1 Canteen. A canteen has $4$ kinds of sandwich and $3$ drinks. Describe the set of "one sandwich + one drink" combos as a Cartesian product and count it.
2 Chess. The files of a chessboard are $F=\{a,b,\dots,h\}$ and the ranks are $R=\{1,2,\dots,8\}$. Which product is the set of squares, and how many squares are there? Which square is $(e,4)$?
3 Dice. A red and a blue die are thrown. Write the set of outcomes as a Cartesian product, find its number of elements, and list the outcomes with sum $4$.
4 Codes. A locker code is one digit followed by one capital English letter, e.g. $(7,K)$. How many codes are possible?
5 Seating. In an auditorium the rows are labelled A to P and each row has seats numbered $1$ to $24$. How many seats are there? Write the set of seats in row C as a product.

💡 Show solutions▶

1 If $S$ is the set of sandwiches and $D$ the set of drinks, the combos form $S\times D$, with $4\times3=12$ elements.

2 $F\times R$, with $8\times8=64$ squares. $(e,4)$ is the square e4 — fifth file, fourth rank.

3 $\{1,\dots,6\}\times\{1,\dots,6\}$, with $36$ outcomes. Sum $4$: $(1,3),(2,2),(3,1)$ — three outcomes (the order red, blue matters).

4 Digits $\times$ letters: $10\times26=260$ codes.

5 $16$ rows (A to P) $\times\ 24$ seats $=384$ seats. Row C: $\{C\}\times\{1,2,\dots,24\}$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksJEE MainJEE Advanced

How this is examined: Board papers ask for the values of unknowns from equal ordered pairs (1 mark), for $A\times B$ in roster form and $n(A\times B)$ (2 marks), and for a verification of a product law such as $A\times(B\cap C)=(A\times B)\cap(A\times C)$ (3 marks). JEE combines products with intersections and counting.

E1. Board · 1 mark If $(a+2,\ b-3)=(5,\ 1)$, then $a+b$ equals (A) $7$ (B) $3$ (C) $4$ (D) $1$

💡 Show solution▶

Idea $a+2=5$ and $b-3=1$, so $a=3$, $b=4$.

Trap (C) is just $b$; (B) is $a$.

Answer (A) $a+b=7$.

E2. Board · 2 marks Let $A=\{1,2\}$ and $B=\{2,3\}$. Write $A\times B$ and $B\times A$, and show that $A\times B\ne B\times A$.

💡 Show solution▶

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

Answer $(1,2)\in A\times B$ but $(1,2)\notin B\times A$, so the sets are unequal (they share only $(2,2)$). (1 mark)

E3. Board · 3 marks Let $A=\{1,2,3\}$, $B=\{3,4\}$ and $C=\{4,5,6\}$. Find (i) $A\times(B\cap C)$ (ii) $(A\times B)\cap(A\times C)$ (iii) $A\times(B\cup C)$, giving the number of elements in (iii).

💡 Show solution▶

(i) $B\cap C=\{4\}$, so $A\times(B\cap C)=\{(1,4),(2,4),(3,4)\}$. (1 mark)

(ii) The pairs common to $A\times B$ and $A\times C$ have second component in $\{3,4\}\cap\{4,5,6\}=\{4\}$: $\{(1,4),(2,4),(3,4)\}$ — the same as (i). (1 mark)

Answer (iii) $B\cup C=\{3,4,5,6\}$, so $A\times(B\cup C)$ has $3\times4=12$ elements: $(a,b)$ with $a\in\{1,2,3\}$, $b\in\{3,4,5,6\}$. (1 mark)

E4. JEE Main Sets $A$ and $B$ have $3$ and $4$ elements respectively, and $A\cap B$ has $2$ elements. The number of elements of $(A\times B)\cap(B\times A)$ is (A) $4$ (B) $6$ (C) $9$ (D) $12$

💡 Show solution▶

Idea $(x,y)$ lies in both products exactly when $x\in A$, $y\in B$ and $x\in B$, $y\in A$, i.e. $x\in A\cap B$ and $y\in A\cap B$. So $(A\times B)\cap(B\times A)=(A\cap B)\times(A\cap B)$.

Count $2\times2=4$.

Trap (D) is $n(A\times B)$; (B) multiplies $3\cdot2$ as if only one component were restricted.

Answer (A). ⏱ Fast route: $n\big((A\times B)\cap(B\times A)\big)=n(A\cap B)^2$.

E5. JEE Main Numerical value. Let $A=\{1,2,3,4\}$ and $B=\{1,2,3\}$. Find the number of elements $(a,b)\in A\times B$ for which $a+b$ is even.

💡 Show solution▶

Idea $a+b$ is even when $a$, $b$ are both odd or both even.

Count Both odd: $\{1,3\}\times\{1,3\}$, $4$ pairs. Both even: $\{2,4\}\times\{2\}$, $2$ pairs.

Answer $6$ (out of $n(A\times B)=12$).

E6. JEE Advanced One or more options correct. For non-empty sets $A$, $B$, $C$, which statements are always true? (A) $A\times(B\cup C)=(A\times B)\cup(A\times C)$ (B) $(A\times B)\cap(B\times A)=(A\cap B)\times(A\cap B)$ (C) $A\times B=B\times A$ (D) if $A\times B=A\times C$ then $B=C$

💡 Show solution▶

(A), (B) (A) is Theorem 2.2. (B) was shown in E4. Both true.

(D) True because $A\ne\phi$: pick $a\in A$; then $y\in B\iff(a,y)\in A\times B=A\times C\iff y\in C$. (If $A$ were empty, both products would be $\phi$ whatever $B$ and $C$ are — that is why the options insist on non-empty sets.)

Answer (A), (B), (D).

🔬 Research–Teaching NexusMini project · about 40 minutes

Research question. How do everyday "grid reference" systems — map grids, stadium and cinema seats, chessboards, spreadsheet cells such as C7 — use Cartesian products, and how large are they?

Why it matters. Any system that names an object by two independent labels is a Cartesian product, and $n(A\times B)=n(A)\cdot n(B)$ tells its designers how many objects it can name. Spreadsheet software, for example, must label over a million rows and thousands of columns; railway coaches label berths by (coach, berth number).

What to do.

Choose two grid systems you can observe: a cinema seating plan, a city map with a lettered grid, a spreadsheet, a parking lot or a railway coach chart.
For each, identify the two sets $A$ and $B$ whose product labels the objects, and state which component comes first.
Compute $n(A\times B)$ and compare it with the actual number of objects. Explain any difference (missing seats, aisles, reserved cells).
Design a labelling system of your own for a school hall of $600$ seats as a product $A\times B$, and justify your choice of $n(A)$ and $n(B)$.

Tools: Phone camera for seating plans; a spreadsheet app

Hand in: A one-page table (system, $A$, $B$, $n(A\times B)$, actual count) and your hall design.

Going further. A spreadsheet column label runs A, …, Z, AA, AB, …, ZZ, AAA, … How many columns can be labelled with at most three letters? (Each length is itself a product: $26$, $26\times26$, $26\times26\times26$.)

Instructor note: the "going further" total is $26+26^2+26^3=18\,278$. Students often notice that real seating plans are products with some pairs removed — a good bridge to Section 2.2, where a relation is exactly a subset of a product.

🧪 Self-check quiz

0/5 correct

Q1. If $(x,3)=(2,y)$, then

Q2. If $n(A)=3$ and $n(B)=2$, then $n(A\times B)$ is

Q3. Which of these is an element of $\{1,2\}\times\{3\}$?

Q4. For any set $A$, $A\times\phi$ equals

Q5. In the Cartesian plane, $[0,2]\times[0,1]$ is

← Chapter 2 — Relations and Functions 2.2 Relations →