Mathematical-Reasoning · Unit 24 · Video 1 · Interactive Practice

A Function as a Set of Ordered Pairs: Each Input Once

IKey Formulas

Statement	Meaning
$A \times B = \{ (x, y) \mid x \in A \text{ and } y \in B \}$	The Cartesian product: every pair with first coordinate in $A$ and second coordinate in $B$
$f = \{ (x, f(x)) \mid x \in A \} = \{ (x, y) \in A \times B \mid y = f(x) \}$	Any function $f : A \to B$ is a subset of $A \times B$ : input first, output second
$F = \{ (1, a), (2, b), (3, b) \}$ , $\ g = \{ (x, x^2 - 2) \mid x \in \mathbb{R} \}$	A finite function by roster; an infinite one in set builder notation
Every element of $A$ is a first coordinate exactly once	When a subset of $A \times B$ is a function from $A$ to $B$ ; there is no condition on second coordinates

Key Insight: A set of ordered pairs fails to be a function from $A$ to $B$ in exactly two ways: some input appears in two pairs with different outputs, or some input appears in no pair at all.

IIWhich Subsets of A × B Are Functions?

A subset of $A \times B$ is a function from $A$ to $B$ when each row of the grid holds exactly one chosen pair.

Codomain

IIIInfinitely Many Pairs

The function $g(x) = x^2 - 2$ on $\mathbb{R}$ is the infinite set of its graph's points $(x, x^2 - 2)$ .

x 1

💡 The integer function $f(m) = 3m + 5$ can be written by roster, $\{ \ldots, (-1, 2), (0, 5), (1, 8), \ldots \}$ , because the integers can be listed in order; no such list exists for $\mathbb{R}$ , so $g$ needs set builder notation.

IVQuiz Questions

Problem 1 · From Arrows to Pairs

Given: $A = \{1, 2, 3\}$ , $B = \{p, q\}$ , and $h : A \to B$ with $h(1) = q$ , $h(2) = p$ , $h(3) = q$ . Which set of ordered pairs is $h$ ?

✅ Correct! Each input is paired with its output:

(x, h(x))

for

x = 1, 2, 3

. The repeated output

q

is allowed.

❌ Order matters. The input is the first coordinate. These pairs have the output first, so they lie in

B \times A

, not

A \times B

❌ An input is missing. Every element of

A

must appear as a first coordinate; this set gives

3

no output.

❌ Not quite. Read the outputs again:

h(1) = q

, not

p

Show solution

The function $h$ is the set $\{ (x, h(x)) \mid x \in A \}$ . Its pairs are $(1, h(1)) = (1, q)$ , $(2, h(2)) = (2, p)$ and $(3, h(3)) = (3, q)$ , so

h = \{ (1, q), (2, p), (3, q) \} \subseteq A \times B.

Two inputs share the output $q$ ; that is allowed, since the conditions only concern first coordinates.

Problem 2 · Which Set Is a Function?

Given: $A = \{1, 2, 3\}$ and $B = \{a, b\}$ . Which of these subsets of $A \times B$ is a function from $A$ to $B$ ?

✅ Correct! Each of

1

2

3

is a first coordinate exactly once. Every input sharing the output

a

is fine: outputs may repeat, and

b

may go unused.

❌ An input has no output. The element

3

is the first coordinate of no pair.

❌ An input has two outputs. One element of

A

appears in two pairs with different second coordinates.

❌ Not quite. Check each element of

A

: it must be a first coordinate exactly once.

Show solution

Test each set: every element of $A$ must be the first coordinate of exactly one pair.

(a) $3$ is a first coordinate of no pair: not a function.
(b) $1$ , $2$ , $3$ each appear once: a function, with every output equal to $a$ .
(c) $1$ appears in $(1, a)$ and $(1, b)$ : not a function.
(d) $3$ appears in $(3, a)$ and $(3, b)$ : not a function.

Problem 3 · An Integer Function as a Set

Given: $f : \mathbb{Z} \to \mathbb{Z}$ with $f(m) = 2m - 3$ .

Which ordered pair belongs to $f$ ?

Which set builder description is $f$ ?

✅ Correct!

f(-1) = 2(-1) - 3 = -5

, and

f = \{ (m, 2m - 3) \mid m \in \mathbb{Z} \}

is the same set as

\{ (m, n) \in \mathbb{Z} \times \mathbb{Z} \mid n = 2m - 3 \}

❌ Input first.

(-5, -1)

would mean

f(-5) = -1

, but

f(-5) = -13

❌ Check the output. A pair

(m, n)

is in

f

exactly when

n = 2m - 3

; compute

2m - 3

for the first coordinate.

❌ The roles are reversed. In this set the first coordinate is computed from the second, so it describes the swapped pairs, not

f

❌ Check the description. The second coordinate must be the output

2m - 3

of the first coordinate

m

Show solution

A pair $(m, n) \in \mathbb{Z} \times \mathbb{Z}$ belongs to $f$ exactly when $n = f(m) = 2m - 3$ .

$(-1, -5)$ : $2(-1) - 3 = -5$ ✓.
$(-5, -1)$ : $2(-5) - 3 = -13 \ne -1$ ; $(0, 3)$ : $2(0) - 3 = -3 \ne 3$ ; $(1, -5)$ : $2(1) - 3 = -1 \ne -5$ .

f = \{ (m, 2m - 3) \mid m \in \mathbb{Z} \} = \{ (m, n) \in \mathbb{Z} \times \mathbb{Z} \mid n = 2m - 3 \}.

The set $\{ (2m - 3, m) \}$ lists the same pairs with the coordinates swapped, so it is not $f$ .

Problem 4 · Counting the Functions Inside A × B

Given: $A = \{1, 2, 3, 4\}$ and $B = \{x, y\}$ , so $A \times B$ has $8$ elements and $2^8 = 256$ subsets. How many of those subsets are functions from $A$ to $B$ ?

✅ Correct! Each of the

4

inputs must be paired with exactly one of the

2

outputs, independently:

2^4 = 16

❌ Not every subset qualifies. Most subsets give some input no pair or two pairs.

❌ That counts pairs, not functions. A function is a set of

4

pairs, one for each input.

❌ Not quite. Count the choices: each input independently picks one output.

Show solution

A subset of $A \times B$ is a function from $A$ to $B$ exactly when each element of $A$ is a first coordinate exactly once. So the subset is built by choosing, for each input $1, 2, 3, 4$ , one of the two pairs $(k, x)$ or $(k, y)$ .

2 \cdot 2 \cdot 2 \cdot 2 = 2^4 = 16.

The other $256 - 16 = 240$ subsets either miss an input or give an input two outputs.

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