Mathematical-Reasoning · Unit 21 · Video 4 · Interactive Practice

Functions of Two Variables: Preimages in ℤ × ℤ

IKey Formulas

Formula	Name	What it says
$f : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , $f(m, n)$ for $f\big((m, n)\big)$	Function of two variables	One input, the ordered pair $(m, n)$ ; one output in $\mathbb{Z}$
$(a, b) = (c, d) \iff a = c \text{ and } b = d$	Equal ordered pairs	Order matters: $(2, 3) \ne (3, 2)$
$\{(m, n) \in \mathbb{Z} \times \mathbb{Z} \mid f(m, n) = y\}$	Preimages of $y$	All solutions of one equation in two unknowns; a subset of the domain
$\{(m, n) \in \mathbb{Z} \times \mathbb{Z} \mid 2m + n = 0\} = \{(m, n) \in \mathbb{Z} \times \mathbb{Z} \mid n = -2m\}$	Two descriptions, one set	Equivalent conditions pick out the same pairs

Key Insight: Each pair $(m, n)$ has exactly one output, but nothing stops many pairs from sharing an output. The set of preimages of $y$ can be infinite, finite or empty, and its elements are always ordered pairs.

IIOne Pair In, One Integer Out

Every pair $(m, n)$ has exactly one output; several pairs may share it, and swapping the coordinates can change it.

Function m 3 n −2

IIIHunting for Preimages

The preimages of $t$ are the solutions of one equation in two unknowns: infinitely many, finitely many, or none.

Function Target t 5

💡 Challenge: for $k(m, n) = |m| + |n|$ , find the target $t$ whose set of preimages has exactly $8$ elements, and list them.

IVQuiz Questions

Problem 1 · Evaluate at an Ordered Pair

Given: $p : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , $p(m, n) = 3m - 2n$ . Find $p(-2, 5)$ and $p(5, -2)$ .

What is p(−2, 5)?

What is p(5, −2)?

✅ Correct!

p(-2, 5) = -16

and

p(5, -2) = 19

: the pairs

(-2, 5)

and

(5, -2)

are different inputs, and here they have different outputs.

❌ Check the sign. The second term is

-2n = -2 \cdot 5 = -10

, so

p(-2, 5) = -6 - 10

, not

-6 + 10

❌ Order matters.

19

p(5, -2)

. In

p(-2, 5)

the first coordinate is

m = -2

and the second is

n = 5

❌ Not quite.

3(-2) + 5 = -1

uses

+n

instead of

-2n

. The rule subtracts twice the second coordinate.

❌ Check the sign. With

n = -2

, the term

-2n

-2 \cdot (-2) = +4

, so

p(5, -2) = 15 + 4

❌ Order matters.

-16

p(-2, 5)

. Swapping the coordinates gives a different input, and here a different output.

❌ Not quite.

15 + (-2) = 13

uses

+n

instead of

-2n

; the term is

-2 \cdot (-2) = 4

Show solution

Substitute the first coordinate for $m$ and the second for $n$ :

p(-2, 5) = 3(-2) - 2(5) = -6 - 10 = -16

p(5, -2) = 3(5) - 2(-2) = 15 + 4 = 19

Since $(-2, 5) \ne (5, -2)$ , these are two different elements of the domain $\mathbb{Z} \times \mathbb{Z}$ , and nothing forces their outputs to agree.

Problem 2 · Preimages in Set-Builder Form

Given: $q : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , $q(m, n) = m - n^2$ . Which set is the set of all preimages of $3$ ?

✅ Correct!

m - n^2 = 3 \iff m = n^2 + 3

, and every integer

n

gives exactly one preimage

(n^2 + 3, n)

❌ Check the sign. Adding

n^2

to both sides of

m - n^2 = 3

gives

m = n^2 + 3

. This set contains

(-3, 0)

, but

q(-3, 0) = -3

❌ The coordinates have swapped roles. Here the square is on

n

, not

m

. This set contains

(2, 1)

, but

q(2, 1) = 2 - 1 = 1

❌ Not all of them. Each listed pair is a preimage of

3

, but so is

(12, 3)

, and in fact

(n^2 + 3, n)

for every integer

n

: the set is infinite.

Show solution

The preimages of $3$ are the pairs $(m, n) \in \mathbb{Z} \times \mathbb{Z}$ with $q(m, n) = 3$ , that is, the integer solutions of $m - n^2 = 3$ . Solving for $m$ :

m - n^2 = 3 \iff m = n^2 + 3.

For each integer $n$ , $n^2 + 3$ is an integer, so each choice of $n$ gives exactly one preimage, and there are infinitely many:

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

Check one element: $(7, -2)$ gives $7 - (-2)^2 = 7 - 4 = 3$ . The set is a subset of the domain $\mathbb{Z} \times \mathbb{Z}$ .

Problem 3 · Finite and Infinite Preimage Sets

Given: $h : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , $h(m, n) = mn$ . Find how many preimages $6$ has, and the set of preimages of $0$ .

How many preimages does 6 have?

Which set is the set of preimages of 0?

✅ Correct!

6

has exactly

8

preimages, while the preimages of

0

fill both coordinate axes of the grid: infinitely many.

❌ Missing the negatives.

(1, 6)

(2, 3)

(3, 2)

(6, 1)

are preimages, but so is

(-2, -3)

, since

(-2)(-3) = 6

❌ Not here. One equation in two unknowns can have finitely many integer solutions. If

mn = 6

, then

m

divides

6

, so

m

has only

8

possible values, each with one

n

❌ Order matters, and so do signs.

(2, 3)

and

(3, 2)

are different ordered pairs, and negative factors also multiply to

6

❌ Too few.

(0, 0)

is a preimage, but so is

(0, 7)

0 \cdot 7 = 0

. A product is

0

whenever either factor is

0

❌ That is the condition for a sum.

m = -n

means

m + n = 0

. This set contains

(1, -1)

, but

h(1, -1) = -1

❌ Only half. These pairs are preimages, but so are the pairs

(0, n)

, such as

(0, 4)

, which are not in this set.

Show solution

Preimages of 6. We need integers with $mn = 6$ . Then $m$ divides $6$ , so $m \in \{-6, -3, -2, -1, 1, 2, 3, 6\}$ , and each such $m$ determines $n = 6/m$ :

$(1, 6)$ , $(2, 3)$ , $(3, 2)$ , $(6, 1)$ , $(-1, -6)$ , $(-2, -3)$ , $(-3, -2)$ , $(-6, -1)$ .

That is exactly $8$ preimages: a finite set.

Preimages of 0. For integers, $mn = 0$ exactly when $m = 0$ or $n = 0$ . So

\{(m, n) \in \mathbb{Z} \times \mathbb{Z} \mid mn = 0\} = \{(m, n) \in \mathbb{Z} \times \mathbb{Z} \mid m = 0 \text{ or } n = 0\},

which contains every pair $(m, 0)$ and every pair $(0, n)$ : infinitely many.

Problem 4 · Counting Lattice Points on a Circle

Given: $s : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , $s(m, n) = m^2 + n^2$ . Find how many preimages $25$ has, and the set of preimages of $3$ .

How many preimages does 25 have?

Which set is the set of preimages of 3?

✅ Correct!

25

has exactly

12

preimages, and

3

has none: its set of preimages is the empty set, a perfectly good subset of

\mathbb{Z} \times \mathbb{Z}

❌ Too few.

(0, \pm 5)

and

(\pm 5, 0)

are preimages, but so is

(3, 4)

9 + 16 = 25

❌ Too few. The

8

pairs

(\pm 3, \pm 4)

(\pm 4, \pm 3)

are preimages, but so is

(5, 0)

25 + 0 = 25

❌ Not here. If

m^2 + n^2 = 25

then

m^2 \le 25

, so

-5 \le m \le 5

; the same holds for

n

, leaving only finitely many pairs to check.

❌ Check the arithmetic.

s(1, 1) = 1 + 1 = 2

, not

3

❌ Not in the domain.

\sqrt{2}

is not an integer, so

(1, \sqrt{2}) \notin \mathbb{Z} \times \mathbb{Z}

; preimages must be elements of the domain.

❌ The coordinates are squared.

s(0, 3) = 0 + 9 = 9

, not

3

Show solution

Preimages of 25. If $m^2 + n^2 = 25$ , then $m^2 \le 25$ and $n^2 \le 25$ . The squares of integers up to $25$ are $0, 1, 4, 9, 16, 25$ , and the ordered ways to write $25$ as a sum of two of them are $0 + 25$ , $25 + 0$ , $9 + 16$ , $16 + 9$ . With all sign choices:

$(0, 5)$ , $(0, -5)$ , $(5, 0)$ , $(-5, 0)$ , $(3, 4)$ , $(3, -4)$ , $(-3, 4)$ , $(-3, -4)$ , $(4, 3)$ , $(4, -3)$ , $(-4, 3)$ , $(-4, -3)$ .

That is $2 + 2 + 4 + 4 = 12$ preimages.

Preimages of 3. If $m^2 + n^2 = 3$ , then $m^2 \le 3$ and $n^2 \le 3$ , so $m^2, n^2 \in \{0, 1\}$ and $m^2 + n^2 \in \{0, 1, 2\}$ . None of these is $3$ , so

\{(m, n) \in \mathbb{Z} \times \mathbb{Z} \mid m^2 + n^2 = 3\} = \emptyset.

In other words, $3$ is an element of the codomain that is not in the range of $s$ .

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