Mathematical-Reasoning · Unit 22 · Video 5 · Interactive Practice

One-to-One and Onto for Functions on ℝ × ℝ

IKey Formulas

Formula	Name	When it applies
$(p, q) = (u, v) \iff p = u \text{ and } q = v$	Equality of ordered pairs	Any ordered pairs; one equation between pairs is a system of two equations
$f(a, b) = f(c, d) \Rightarrow (a, b) = (c, d)$ for all $(a, b), (c, d) \in A \times B$	$f : A \times B \to C$ is an injection	Inputs are ordered pairs; the codomain $C$ may be a set of numbers or of pairs
for every $y \in C$ there is $(a, b) \in A \times B$ with $f(a, b) = y$	$f : A \times B \to C$ is a surjection	Construct $(a, b)$ by working backward, then verify $f(a, b) = y$ forward
$f(a, b) = (r, s) \iff (a, b) = \left(\dfrac{r + s}{3}, \dfrac{r - 2s}{3}\right)$	Preimage under $f(a, b) = (2a + b, a - b)$	$f : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ , any $(r, s) \in \mathbb{R} \times \mathbb{R}$

Key Insight: a target pair gives two equations in two unknowns, while a target number gives only one equation, which a whole line of pairs can satisfy.

IIOne Target Pair, One Preimage

For $f(a, b) = (2a + b, a - b)$ , each coordinate equation is a line in the domain plane, and the preimage of $(r, s)$ is where they cross.

Domain and codomain

domain · inputs (a, b)

codomain · targets (r, s)

💡 Challenge: in the $\mathbb{Z} \times \mathbb{Z}$ setting, find the condition on $r + s$ that decides whether the target $(r, s)$ has a preimage.

IIIOne Target Number, a Line of Preimages

For $g : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ , $g(x, y) = 2x + y$ , every pair on the line $2x + y = z$ is sent to the same number $z$ .

IVQuiz Questions

Problem 1 · Classify a Map on Integer Pairs

Given: $h : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , $h(m, n) = 3m - 5n$ . Which statement is true?

✅ Correct!

h(2k, k) = k

reaches every integer, while

h(5, 3) = 0 = h(0, 0)

shows two different pairs sharing an output.

❌ Not quite. Onto is right, but one equation

3m - 5n = t

in two unknowns has many integer solutions:

h(5, 3) = 15 - 15 = 0 = h(0, 0)

❌ Not quite. Both halves fail:

h(5, 3) = h(0, 0)

, and every integer

k

is an output, since

h(2k, k) = 6k - 5k = k

❌ Not quite. No integer is missed:

h(2k, k) = 6k - 5k = k

for every

k \in \mathbb{Z}

Show solution

Not one-to-one. $(5, 3) \ne (0, 0)$ , but $h(5, 3) = 3 \cdot 5 - 5 \cdot 3 = 0$ and $h(0, 0) = 0$ .

Onto. Let $k \in \mathbb{Z}$ . Work backward from $3m - 5n = k$ : since $3 \cdot 2 - 5 \cdot 1 = 1$ , try $(m, n) = (2k, k)$ , which is in $\mathbb{Z} \times \mathbb{Z}$ because $2k$ and $k$ are integers. Verify forward:

h(2k, k) = 3(2k) - 5k = 6k - 5k = k.

So every integer is an output: $h$ is a surjection that is not an injection.

Problem 2 · A Valid Disproof of Onto

Given: $k : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , $k(m, n) = 4m + 6n$ . Which argument correctly proves that $k$ is not a surjection?

✅ Correct! A single unreached integer disproves onto, and the evenness of every output shows

1

is unreached for all pairs at once.

❌ Not quite. The conclusion happens to be true, but checking only

m = 0

says nothing about the other pairs; a disproof of onto must rule out every

(m, n) \in \mathbb{Z} \times \mathbb{Z}

❌ Not quite. Two pairs with the same output show

k

is not one-to-one; that is a different property from onto.

❌ Not quite. The domain contains negative integers:

k(1, -1) = 4 - 6 = -2

, so

-2

is an output.

Show solution

To prove $k$ is not a surjection, exhibit one $t \in \mathbb{Z}$ such that $k(m, n) \ne t$ for every $(m, n) \in \mathbb{Z} \times \mathbb{Z}$ .

For any integers $m$ and $n$ , $k(m, n) = 4m + 6n = 2(2m + 3n)$ , and $2m + 3n$ is an integer, so every output is even. Since $1$ is odd, $k(m, n) \ne 1$ for all $(m, n)$ , and $k$ is not onto.

Option (b) tests only the pairs with $m = 0$ ; option (c) disproves one-to-one instead; option (d) is false because $k(1, -1) = -2$ .

Problem 3 · Solve for the Preimage

Given: $F : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ , $F(x, y) = (x + y, x - 2y)$ . Find the preimage of $(5, -1)$ , then the preimage of an arbitrary $(r, s) \in \mathbb{R} \times \mathbb{R}$ .

Which pair satisfies $F(x, y) = (5, -1)$ ?

Which pair satisfies $F(x, y) = (r, s)$ ?

✅ Correct! Every

(r, s)

has exactly the one preimage

\left(\frac{2r + s}{3}, \frac{r - s}{3}\right)

, and

(r, s) = (5, -1)

gives

(3, 2)

❌ Not quite. The coordinates are swapped:

F(2, 3) = (5, 2 - 6) = (5, -4)

❌ Not quite. Check the sign of

y

F(3, -2) = (1, 3 + 4) = (1, 7)

. Subtracting

x - 2y = -1

from

x + y = 5

gives

3y = 6

❌ Not quite. The target is an output, not an input:

F(5, -1) = (4, 7)

❌ Not quite. The

y

-coordinate is right, but then

x = r - y = \frac{3r - (r - s)}{3} = \frac{2r + s}{3}

. Test at

(5, -1)

: your formula gives

(1, 2)

, and

F(1, 2) = (3, -3)

❌ Not quite. Subtracting the second equation from the first gives

3y = r - s

, not

s - r

. At

(5, -1)

your formula gives

(3, -2)

, and

F(3, -2) = (1, 7)

❌ Not quite. Averaging

r

and

s

does not solve the system; at

(5, -1)

your formula gives

(2, 2)

, and

F(2, 2) = (4, -2)

❌ Not quite. Set the coordinates equal:

x + y = 5

and

x - 2y = -1

❌ Not quite. Solve

x + y = r

and

x - 2y = s

for

x

and

y

Show solution

Specific target. $F(x, y) = (5, -1)$ means $x + y = 5$ and $x - 2y = -1$ . Subtracting the second equation from the first gives $3y = 6$ , so $y = 2$ and $x = 5 - 2 = 3$ . Check: $F(3, 2) = (3 + 2, 3 - 4) = (5, -1)$ .

Arbitrary target. Work backward from $x + y = r$ and $x - 2y = s$ . Subtracting gives $3y = r - s$ , so $y = \frac{r - s}{3}$ , and $x = r - y = \frac{2r + s}{3}$ . Both are real numbers, so the pair is in $\mathbb{R} \times \mathbb{R}$ . Verify forward:

x + y = \frac{2r + s + r - s}{3} = \frac{3r}{3} = r, \qquad x - 2y = \frac{2r + s - 2r + 2s}{3} = \frac{3s}{3} = s.

So $F$ is a surjection. The same subtraction shows that $F(a, b) = F(c, d)$ forces $3b = 3d$ and then $a = c$ , so $F$ is also an injection, hence a bijection.

Problem 4 · Sum and Product

Given: $G : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ , $G(x, y) = (x + y, xy)$ . Which statement is true?

✅ Correct! Swapping the coordinates keeps the sum and the product, and the target

(0, 1)

is never reached.

❌ Not quite. Two equations in two unknowns need not have exactly one solution when one of them is not linear:

G(1, 2) = (3, 2) = G(2, 1)

❌ Not quite. Not one-to-one is right, but try the target

(0, 1)

x + y = 0

forces

y = -x

, and then

xy = -x^2 \le 0

, never

1

❌ Not quite. Not onto is right, but

(1, 2) \ne (2, 1)

while

G(1, 2) = (3, 2) = G(2, 1)

Show solution

Not one-to-one. $(1, 2) \ne (2, 1)$ , but $G(1, 2) = (1 + 2, 1 \cdot 2) = (3, 2)$ and $G(2, 1) = (3, 2)$ .

Not onto. Consider the target $(0, 1) \in \mathbb{R} \times \mathbb{R}$ . If $G(x, y) = (0, 1)$ , then $x + y = 0$ and $xy = 1$ . The first equation gives $y = -x$ , so $xy = -x^2$ . Since $x^2 \ge 0$ for every real $x$ , $-x^2 \le 0 < 1$ , a contradiction. So no pair is sent to $(0, 1)$ .

Hence $G$ is neither an injection nor a surjection.

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