Mathematical-Reasoning · Unit 22 · Video 3 · Interactive Practice

Surjections and Bijections: Is 3m + 2 Onto?

IKey Formulas

Statement	Meaning
$f: A \to B$ is a surjection provided $\text{range}(f) = \text{codom}(f)$ , that is, $\forall y \in B, \exists x \in A, f(x) = y$	Onto: every element of $B$ receives at least one arrow; the input $x$ may depend on the target $y$
$f: A \to B$ is not a surjection provided $\exists y \in B, \forall x \in A, f(x) \ne y$	One named element of the codomain that no input reaches is a complete disproof
$f: A \to B$ is a bijection provided $f$ is an injection and a surjection	One-to-one and onto: every element of $B$ receives exactly one arrow, so $f$ pairs off the elements of $A$ with the elements of $B$
To prove $f: A \to B$ onto: let $y \in B$ ; solve $f(x) = y$ for $x$ (scratch work); check $x \in A$ ; check $f(x) = y$	For $f(x) = 3x + 2$ the scratch input is $x = \frac{y - 2}{3}$ , a real number for every real $y$ but an integer only for some integers $y$

Key Insight: The rule $3x + 2$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$ , but from $\mathbb{Z}$ to $\mathbb{Z}$ it is one-to-one and not onto, because $0$ is never an output. Whether a function is a surjection depends on its domain and codomain, not on its formula alone.

IISolve Backward, Check Forward

The scratch input $\frac{y - 2}{3}$ always exists, but it proves $y$ is reached only when it lies in the domain.

Sets target y 0

IIICounting Arrows into the Codomain

Onto means every element of $B$ receives at least one arrow; a bijection means every element receives exactly one.

Function

IVQuiz Questions

Problem 1 · Construct the Input

Given: $g: \mathbb{Q} \to \mathbb{Q}$ defined by $g(x) = 4x - 7$ . A proof that $g$ is a surjection begins "Let $y \in \mathbb{Q}$ ." Which input $x$ should the proof produce?

✅ Correct!

\frac{y + 7}{4}

is rational whenever

y

is, and

g\left(\frac{y + 7}{4}\right) = (y + 7) - 7 = y

❌ Not quite. That applies the rule to

y

instead of undoing it:

g(4y - 7) = 16y - 35

, which is not

y

(for

y = 0

it is

-35

❌ Check the sign. Solving

4x - 7 = y

adds

7

to both sides. Here

g\left(\frac{y - 7}{4}\right) = y - 14

❌ Not quite. The

7

must be divided by

4

too:

g\left(\frac{y}{4} + 7\right) = y + 28 - 7 = y + 21

Show solution

Scratch work. We need $4x - 7 = y$ , so $4x = y + 7$ and $x = \frac{y + 7}{4}$ .

Check 1: $x$ is in the domain. $y + 7$ is rational, and a rational number divided by the nonzero integer $4$ is rational, so $x \in \mathbb{Q}$ .

Check 2: $x$ maps to $y$ .

g\left(\frac{y + 7}{4}\right) = 4 \cdot \frac{y + 7}{4} - 7 = (y + 7) - 7 = y.

So for every $y \in \mathbb{Q}$ there is an $x \in \mathbb{Q}$ with $g(x) = y$ , and $g$ is a surjection. Since $4x_1 - 7 = 4x_2 - 7$ forces $x_1 = x_2$ , $g$ is also an injection, so $g$ is a bijection.

Problem 2 · Same Scratch Work, Smaller Domain

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

✅ Correct! One integer target with no integer input is a complete disproof.

❌ Not quite. That is only the scratch work. The proof must also check that

m

is in the domain

\mathbb{Z}

, and for

y = 0

it gives

m = \frac{3}{5}

, which is not an integer.

❌ Not quite.

2

is reached:

h(1) = 5 - 3 = 2

. Test the target with the scratch input:

\frac{2 + 3}{5} = 1 \in \mathbb{Z}

❌ Not quite.

h

is an injection:

5m_1 - 3 = 5m_2 - 3

gives

m_1 = m_2

. And the two properties are independent, so failing one would not settle the other.

Show solution

Scratch work. $5m - 3 = y$ gives $m = \frac{y + 3}{5}$ , an integer only when $y + 3$ is a multiple of $5$ . The target $y = 0$ fails that test.

Proof that $0 \notin \text{range}(h)$ . Suppose some $m \in \mathbb{Z}$ has $h(m) = 0$ . Then $5m - 3 = 0$ , so $5m = 3$ and $m = \frac{3}{5}$ . But $\frac{3}{5}$ is not an integer, which contradicts $m \in \mathbb{Z}$ . So for every $m \in \mathbb{Z}$ , $h(m) \ne 0$ .

Since $0$ is in the codomain and not in the range, $h$ is not a surjection. The range consists of the integers that are $3$ less than a multiple of $5$ : $\ldots$ , $-8$ , $-3$ , $2$ , $7$ , $12$ , $\ldots$ The function $h$ is one-to-one but not onto, so it is not a bijection.

Problem 3 · A Piecewise Rule on the Integers

Given: $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(n) = \begin{cases} \dfrac{n}{2}, & \text{if } n \text{ is even;} \\[4pt] \dfrac{n - 1}{2}, & \text{if } n \text{ is odd.} \end{cases}$ For example, $f(4) = 2$ and $f(5) = 2$ .

Is f a surjection?

Is f a bijection?

✅ Correct! Every integer

y

is reached (by

2y

, and also by

2y + 1

), so

f

is onto; two inputs per output make it not one-to-one.

❌ Not quite. Odd outputs occur:

f(2) = 1

and

f(-2) = -1

❌ Not quite.

\frac{y}{2}

is not an integer when

y

is odd, and even when it is, it does not map to

y

: for

y = 4

f(2) = 1

❌ Not quite.

2y

is even for every integer

y

, so the even case of the rule applies and

f(2y) = y

for odd

y

too, e.g.

f(6) = 3

❌ Not quite. A bijection must be one-to-one as well as onto. Compare

f(0)

and

f(1)

❌ Not quite. Exactly one output per input is what makes

f

a function. A bijection needs every element of the codomain to receive exactly one arrow.

❌ Not quite.

f

is a surjection: every

y \in \mathbb{Z}

equals

f(2y)

. The failure is elsewhere.

Show solution

A few outputs: $f(-2) = -1$ , $f(-1) = -1$ , $f(0) = 0$ , $f(1) = 0$ , $f(2) = 1$ , $f(3) = 1$ . Each integer seems to be hit twice.

Surjection. Let $y \in \mathbb{Z}$ and set $n = 2y$ . Then $n \in \mathbb{Z}$ and $n$ is even, so the first case applies:

f(2y) = \frac{2y}{2} = y.

So every integer is in the range, and $f$ is a surjection.

Not an injection. $0$ is even and $1$ is odd, so $f(0) = \frac{0}{2} = 0$ and $f(1) = \frac{1 - 1}{2} = 0$ . Two different inputs share an output, so $f$ is not an injection.

Onto but not one-to-one, $f$ is not a bijection.

Problem 4 · The Floor Function

Given: for a real number $x$ , $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ ; for example $\lfloor 2.7 \rfloor = 2$ and $\lfloor -1.5 \rfloor = -2$ . Define $F: \mathbb{R} \to \mathbb{Z}$ by $F(x) = \lfloor x \rfloor$ .

Which input proves that a target y in ℤ is in range(F)?

Is F a bijection?

✅ Correct! Every integer

y

is its own floor, so

F

is onto; but the whole interval from

y

up to (not including)

y + 1

maps to

y

, so

F

is not one-to-one.

❌ Not quite.

y + 1

is an integer, so

\lfloor y + 1 \rfloor = y + 1 \ne y

❌ Not quite. The greatest integer below

y - \frac{1}{2}

y - 1

, so

F\left(y - \frac{1}{2}\right) = y - 1

❌ Not quite.

\lfloor -y \rfloor = -y

, which equals

y

only when

y = 0

❌ Not quite. Exactly one floor per real number is what makes

F

a function. A bijection must also be one-to-one, and many inputs share each output.

❌ Not quite. The codomain is

\mathbb{Z}

, so

\frac{1}{2}

is not a target at all. (For the same rule with codomain

\mathbb{R}

, this argument would be right.)

❌ Not quite. The definition asks only for one-to-one and onto;

k: \{a, b\} \to \{1, 2\}

with

k(a) = 1

k(b) = 2

is a bijection between different sets.

Show solution

Surjection. Let $y \in \mathbb{Z}$ and take $x = y$ . Since $\mathbb{Z} \subseteq \mathbb{R}$ , $x$ is in the domain. The greatest integer less than or equal to the integer $y$ is $y$ itself, so $F(y) = \lfloor y \rfloor = y$ . Hence $F$ is a surjection. (The input is not unique: $F\left(y + \frac{1}{2}\right) = y$ as well.)

Not an injection. $F(0) = \lfloor 0 \rfloor = 0$ and $F\left(\frac{1}{2}\right) = \left\lfloor \frac{1}{2} \right\rfloor = 0$ , with $0 \ne \frac{1}{2}$ . So $F$ is not an injection, and therefore not a bijection.

The codomain matters. The same rule viewed as a function $\mathbb{R} \to \mathbb{R}$ is not a surjection: its outputs are all integers, so the real number $\frac{1}{2}$ is in the codomain but not in the range.

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