Mathematical-Reasoning · Unit 26 · Video 2 · Interactive Practice

Relations Defined: Any Subset of A × B, Domain and Range

IKey Formulas

Term	Definition
Relation from $A$ to $B$	any subset $R \subseteq A \times B$ , including $\emptyset$ and $A \times B$ itself
Relation on $A$	a relation from $A$ to $A$ , so $R \subseteq A \times A$
Domain	$\operatorname{dom}(R) = \{ u \in A \mid (u, y) \in R \text{ for at least one } y \in B \}$
Range	$\operatorname{range}(R) = \{ v \in B \mid (x, v) \in R \text{ for at least one } x \in A \}$

Key Insight: No rule is needed behind the pairs, and one element may have many partners: congruence modulo $3$ relates $5$ to $5$ , $2$ and $-1$ at once, which no function could do.

IIEvery Subset of $A \times B$ Is a Relation

Each cell of $A \times B$ is one ordered pair; the domain and range are the rows and columns that contain a chosen pair.

IIIOne Integer, Many Partners

Under congruence modulo $3$ , an integer $x$ is related to every $y$ with $3 \mid (x - y)$ , not to just one.

x 5

IVQuiz Questions

Problem 1 · Domain and Range of a Finite Relation

Given: $A = \{1, 2, 3\}$ , $B = \{p, q, r\}$ and the relation $R = \{ (1, p), (1, q), (3, q) \}$ from $A$ to $B$ . Find $\operatorname{dom}(R)$ and $\operatorname{range}(R)$ .

$\operatorname{dom}(R)$

$\operatorname{range}(R)$

✅ Correct! The first coordinates are

1, 1, 3

and the second coordinates are

p, q, q

; listed once each,

\operatorname{dom}(R) = \{1, 3\}

and

\operatorname{range}(R) = \{p, q\}

❌ Not quite.

2

is not the first coordinate of any pair in

R

, so

2 \notin \operatorname{dom}(R)

. The domain is a subset of

A

, not always all of

A

❌ Not quite. That is a set of second coordinates. The domain collects the first coordinates.

❌ Not quite.

1

has two partners,

p

and

q

. Several partners do not keep it out of the domain; one is enough to put it in.

❌ Not quite. List the first coordinate of each pair in

R

❌ Not quite. No pair in

R

has second coordinate

r

, so

r \notin \operatorname{range}(R)

❌ Not quite.

(1, p) \in R

, so

p

is a second coordinate too.

❌ Not quite. That is the domain. The range is a subset of

B

❌ Not quite. List the second coordinate of each pair in

R

Show solution

$\operatorname{dom}(R)$ is the set of $u \in A$ that occur as the first coordinate of at least one pair in $R$ . The pairs $(1, p)$ , $(1, q)$ and $(3, q)$ have first coordinates $1$ and $3$ , so $\operatorname{dom}(R) = \{1, 3\}$ ; no pair starts with $2$ .

$\operatorname{range}(R)$ is the set of $v \in B$ that occur as a second coordinate: $p$ and $q$ . So $\operatorname{range}(R) = \{p, q\}$ ; no pair ends with $r$ .

Problem 2 · Fixing the Second Coordinate on a Circle

Given: $T = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x^2 + y^2 = 64 \}$ . Find $\{ x \in \mathbb{R} \mid (x, 4) \in T \}$ .

✅ Correct!

x^2 + 16 = 64

gives

x^2 = 48

, so

x = \pm\sqrt{48} = \pm 4\sqrt{3}

❌ Close, but…

(-4\sqrt{3})^2 = 48

as well, so

(-4\sqrt{3}, 4) \in T

too. The number

4

is the second coordinate of two pairs.

❌ Not quite.

4\sqrt{5}

squares to

80 = 64 + 16

. Substituting

y = 4

gives

x^2 = 64 - 16

, not

64 + 16

❌ Not quite.

64 - 16 = 48 \ge 0

, so real solutions exist. The set is empty for

y = 9

, where

x^2 = -17

❌ Not quite. Substitute

y = 4

and solve for

x

Show solution

Substitute $y = 4$ : $x^2 + 16 = 64$ , so $x^2 = 48$ . Since $48 = 16 \cdot 3$ ,

x = \pm\sqrt{48} = \pm 4\sqrt{3}.

So $\{ x \in \mathbb{R} \mid (x, 4) \in T \} = \{-4\sqrt{3}, 4\sqrt{3}\}$ , and in particular $4 \in \operatorname{range}(T)$ .

Problem 3 · A Sideways Parabola

Given: the relation $P = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x = y^2 \}$ on $\mathbb{R}$ . Find $\operatorname{dom}(P)$ and $\operatorname{range}(P)$ .

$\operatorname{dom}(P)$

$\operatorname{range}(P)$

✅ Correct! Every real

y

is the second coordinate of

(y^2, y)

, so the range is

\mathbb{R}

; a real

y

with

y^2 = x

exists exactly when

x \ge 0

, so the domain is

[0, \infty)

❌ Not quite. For

x = -1

we would need

y^2 = -1

, so

-1 \notin \operatorname{dom}(P)

❌ Close, but…

(0, 0) \in P

, since

0 = 0^2

, so

0 \in \operatorname{dom}(P)

❌ Not quite. For which

x

does

y^2 = x

have a real solution?

❌ Not quite. Here

y

is not a square;

x

is.

(4, -2) \in P

, so

-2 \in \operatorname{range}(P)

❌ Not quite.

(4, -2) \in P

and

(0, 0) \in P

, so

-2

and

0

are both in the range.

❌ Not quite. For a given

y

, is there an

x

with

x = y^2

Show solution

Domain. A real $x$ is in $\operatorname{dom}(P)$ when some real $y$ has $y^2 = x$ . If $x \ge 0$ , then $y = \sqrt{x}$ works. If $x \lt 0$ , no real $y$ works, because $y^2 \ge 0$ . So $\operatorname{dom}(P) = [0, \infty)$ .

Range. For every real $y$ , the pair $(y^2, y)$ satisfies $x = y^2$ , so $y \in \operatorname{range}(P)$ . So $\operatorname{range}(P) = \mathbb{R}$ .

The roles are the reverse of the parabola $y = x^2$ : the squared coordinate is restricted, the other is free.

Problem 4 · Counting Relations

Given: $A = \{1, 2\}$ and $B = \{p, q, r\}$ . How many different relations from $A$ to $B$ are there?

✅ Correct! A relation from

A

B

is any subset of

A \times B

, which has

2 \cdot 3 = 6

elements, so there are

2^6 = 64

relations.

❌ Not quite.

6

is the number of ordered pairs in

A \times B

. A relation is a set of such pairs, so count the subsets of

A \times B

❌ Not quite.

3^2 = 9

counts the functions from

A

B

(each of the

2

inputs has

3

choices of output). A relation needs no such rule: any subset of

A \times B

counts.

❌ Not quite.

2^3 = 8

is the number of subsets of

B

. The relations are the subsets of

A \times B

❌ Not quite. Count the subsets of

A \times B

Show solution

By definition, a relation from $A$ to $B$ is a subset of $A \times B$ , and every subset qualifies, including $\emptyset$ and $A \times B$ itself. So the relations from $A$ to $B$ are exactly the elements of $\mathcal{P}(A \times B)$ .

$A \times B = \{ (1, p), (1, q), (1, r), (2, p), (2, q), (2, r) \}$ has $6$ elements, and a set with $6$ elements has $2^6 = 64$ subsets. So there are $64$ relations from $A$ to $B$ .

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Relations Defined: Any Subset of A × B, Domain and Range

IKey Formulas

IIEvery Subset of A×BA \times BA×B Is a Relation

IIIOne Integer, Many Partners

IVQuiz Questions

IIEvery Subset of $A \times B$ Is a Relation