Mathematical-Reasoning · Unit 26 · Video 4 · Interactive Practice

Which Relations Are Functions? The Circle vs the Parabola

IKey Formulas

Statement	As a condition on pairs
A function is a set of pairs	$f = \{ (x, f(x)) \mid x \in A \} = \{ (x, y) \in A \times B \mid y = f(x) \} \subseteq A \times B$
Condition (1): every input has an output	for every $x \in A$ , there exists $y \in B$ with $(x, y) \in f$
Condition (2): no input has two outputs	for all $x \in A$ and $y, z \in B$ , if $(x, y) \in f$ and $(x, z) \in f$ , then $y = z$
Function from $A$ to $B$	a relation from $A$ to $B$ that satisfies (1) and (2)

Key Insight: Condition (2) restricts only pairs that share a first coordinate. Two inputs may share an output; one input may not have two.

IIBuilding a Relation Arrow by Arrow

A relation from $A$ to $B$ is a function exactly when every element of $A$ sends out one arrow.

IIIOne Input, How Many Outputs?

At $x = a$ , the pairs of the relation with first coordinate $a$ sit on the vertical line through $a$ .

Relation a 0

IVQuiz Questions

Problem 1 · Which Set of Pairs Is a Function?

Given: $A = \{1, 2, 3\}$ and $B = \{4, 5\}$ . Which relation from $A$ to $B$ is a function from $A$ to $B$ ?

✅ Correct! Each of

1, 2, 3

is a first coordinate exactly once. All three share the output

4

, which condition (2) allows.

❌ Not quite.

3

is not the first coordinate of any pair, so condition (1) fails at

3

❌ Not quite.

(1, 4)

and

(1, 5)

share the first coordinate

1

with

4 \neq 5

, so condition (2) fails.

❌ Not quite. These pairs have first coordinates in

B

, so this set is not even a subset of

A \times B

; it is a relation from

B

A

❌ Not quite. Check conditions (1) and (2) for each set.

Show solution

$\{ (1, 4), (2, 5) \}$ : no pair starts with $3$ , so condition (1) fails.
$\{ (1, 4), (1, 5), (2, 4), (3, 5) \}$ : $(1, 4)$ and $(1, 5)$ with $4 \neq 5$ , so condition (2) fails.
$\{ (1, 4), (2, 4), (3, 4) \}$ : every element of $A$ occurs once as a first coordinate, so (1) and (2) hold. It is the constant function with value $4$ .
$\{ (4, 1), (5, 2), (4, 3) \}$ : a subset of $B \times A$ , not of $A \times B$ .

Problem 2 · A Counterexample on the Circle

Given: $T = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x^2 + y^2 = 64 \}$ . Which two pairs in $T$ show that $T$ fails condition (2) at the input $6$ ?

✅ Correct!

36 + 28 = 64

for both pairs, they share the first coordinate

6

, and

2\sqrt{7} \neq -2\sqrt{7}

❌ Not quite. Both pairs are in

T

, but their first coordinates differ. Condition (2) only restricts pairs with the same first coordinate; two inputs may share an output.

❌ Not quite.

6^2 + 8^2 = 100 \neq 64

, so neither pair is in

T

. At

x = 0

the outputs are

\pm 8

; at

x = 6

they are smaller.

❌ Not quite.

6^2 + 10^2 = 136

. Solve

36 + y^2 = 64

instead.

❌ Not quite. Substitute

x = 6

and solve for

y

Show solution

Substitute $x = 6$ : $36 + y^2 = 64$ , so $y^2 = 28$ and $y = \pm\sqrt{28} = \pm 2\sqrt{7}$ . Both $(6, 2\sqrt{7})$ and $(6, -2\sqrt{7})$ lie in $T$ .

They have the same first coordinate $6$ and different second coordinates, since $2\sqrt{7} \neq -2\sqrt{7}$ . That is a counterexample to condition (2), so $T$ is not a function from $\mathbb{R}$ to $\mathbb{R}$ .

Problem 3 · Which Conditions Fail?

Given: $H = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid |y| = x \}$ , viewed as a relation from $\mathbb{R}$ to $\mathbb{R}$ . Which of conditions (1) and (2) does $H$ fail?

✅ Correct! The input

-1

has no output, since

|y| = -1

has no solution; the input

3

has two,

(3, 3)

and

(3, -3)

❌ Close, but… Condition (2) fails at

x = 3

, but condition (1) fails too: no real

y

has

|y| = -1

, so the input

-1

has no output.

❌ Close, but… Condition (1) fails at negative inputs, and condition (2) fails at

x = 3

(3, 3) \in H

and

(3, -3) \in H

❌ Not quite. That holds for

y = |x|

, with the roles of the coordinates the other way round. Here

x

is the absolute value:

(3, 3)

and

(3, -3)

are both in

H

❌ Not quite. Test a negative input and a positive one.

Show solution

Condition (1) fails. For $x = -1$ , a pair $(-1, y) \in H$ would need $|y| = -1$ , which is impossible because $|y| \ge 0$ . So $-1$ is not a first coordinate of any pair in $H$ .

Condition (2) fails. $|3| = 3$ and $|-3| = 3$ , so $(3, 3) \in H$ and $(3, -3) \in H$ , with $3 \neq -3$ .

So $H$ fails both conditions; it is a relation on $\mathbb{R}$ but not a function from $\mathbb{R}$ to $\mathbb{R}$ .

Problem 4 · A Relation on the Integers

Given: $R = \{ (x, y) \in \mathbb{Z} \times \mathbb{Z} \mid x = 2y \}$ . Is $R$ a function from $\mathbb{Z}$ to $\mathbb{Z}$ ?

✅ Correct! An odd input such as

1

has no partner, since

1 = 2y

has no integer solution; and

2y = 2z

forces

y = z

, so no input has two outputs.

❌ Not quite. For

x = 1

, the output

\tfrac{1}{2}

is not an integer, so no pair

(1, y)

lies in

R

❌ Not quite. If

(x, y) \in R

and

(x, z) \in R

, then

2y = x = 2z

, so

y = z

: condition (2) holds. It is condition (1) that fails, at every odd

x

❌ Close, but… Condition (1) fails at odd inputs, but condition (2) holds:

2y = 2z

gives

y = z

❌ Not quite. Check whether the input

1

has an integer partner, and whether one input can have two.

Show solution

Condition (1) fails. For $x = 1$ , a pair $(1, y) \in R$ would need $1 = 2y$ , so $y = \tfrac{1}{2} \notin \mathbb{Z}$ . So $1 \notin \operatorname{dom}(R)$ ; in fact $\operatorname{dom}(R)$ is the set of even integers.

Condition (2) holds. Let $x, y, z \in \mathbb{Z}$ with $(x, y) \in R$ and $(x, z) \in R$ . Then $x = 2y$ and $x = 2z$ , so $2y = 2z$ and $y = z$ .

So $R$ is not a function from $\mathbb{Z}$ to $\mathbb{Z}$ , because the inputs that are odd have no output.

Solved: 0 / 4