Mathematical-Reasoning · Unit 22 · Video 1 · Interactive Practice

Shared and Missed Outputs: Why 5x + 3 Has Neither

IKey Formulas

Formula	Name	Meaning
$\forall x, y \in A,\; x \neq y \Rightarrow f(x) \neq f(y)$	Different inputs, different outputs	For $f\colon A \to B$ : no two different inputs share an output (a function with this property is an injection)
$\forall x, y \in A,\; f(x) = f(y) \Rightarrow x = y$	Contrapositive form	Logically equivalent to the row above; a proof of it starts from the equation $f(x) = f(y)$
$\forall y \in B,\; \exists x \in A \text{ with } f(x) = y$	Every element of the codomain is an output	For $f\colon A \to B$ : the same as $\operatorname{range}(f) = B$ (a function with this property is a surjection)
$\exists x, y \in A,\; x \neq y \wedge f(x) = f(y)$ $\exists y \in B,\; \forall x \in A,\; f(x) \neq y$	The two negations	For nonempty $A$ and $B$ : two different inputs share an output; some element of $B$ is never an output

Key Insight: The definition of a function only demands that each input have exactly one output. It allows two inputs to share an output and allows elements of the codomain to be missed; the two properties above are extra conditions a function may or may not meet.

IIVisualization 1 — Shared Outputs and Missed Outputs

Each input has exactly one arrow, yet two arrows may land together and some elements of the codomain may receive none.

Codomain

💡 Challenge: make both lines under the diagram hold at once. Only one of the three codomains allows it.

IIIVisualization 2 — Negating the Two Properties

A negation turns $\forall$ into $\exists$ , $\exists$ into $\forall$ , and $P \Rightarrow Q$ into $P \wedge \neg Q$ .

Statement Step 1

Statement

\forall x, y \in A,\; x \neq y \Rightarrow f(x) \neq f(y)

Negate the whole statement

\neg\big(\forall x, y \in A,\; x \neq y \Rightarrow f(x) \neq f(y)\big)

Rule $\neg\forall \equiv \exists\neg$

\exists x, y \in A,\; \neg\big(x \neq y \Rightarrow f(x) \neq f(y)\big)

Rule $\neg(P \Rightarrow Q) \equiv P \wedge \neg Q$

\exists x, y \in A,\; x \neq y \wedge \neg\big(f(x) \neq f(y)\big)

Simplify, then find a witness

\exists x, y \in A,\; x \neq y \wedge f(x) = f(y)

For $g$ (with $g(1) = a$ , $g(2) = b$ , $g(3) = a$ ): $x = 1$ and $y = 3$ give $1 \neq 3$ and $g(1) = g(3) = a$ , so the negation is true for $g$ .

Statement

\forall y \in B,\; \exists x \in A \text{ with } f(x) = y

Negate the whole statement

\neg\big(\forall y \in B,\; \exists x \in A \text{ with } f(x) = y\big)

Rule $\neg\forall \equiv \exists\neg$

\exists y \in B,\; \neg\big(\exists x \in A \text{ with } f(x) = y\big)

Rule $\neg\exists \equiv \forall\neg$

\exists y \in B,\; \forall x \in A,\; \neg\big(f(x) = y\big)

Simplify, then find a witness

\exists y \in B,\; \forall x \in A,\; f(x) \neq y

For $f$ (with $f(1) = a$ , $f(2) = b$ , $f(3) = c$ ): $y = d$ works, since $f(1)$ , $f(2)$ and $f(3)$ all lie in $\{a, b, c\}$ , so the negation is true for $f$ .

💡 Negating only the conclusion of the first statement gives $\forall x, y \in A,\; x \neq y \Rightarrow f(x) = f(y)$ , which says every two different inputs share an output: that forces $f$ to be constant, and it is not the negation.

IVVisualization 3 — Why $5x + 3$ Has Neither a Shared nor a Missed Output

For $g(x) = 5x + 3$ from $\mathbb{R}$ to $\mathbb{R}$ , no two inputs share an output and no real number is missed.

Question

VQuiz Questions

Problem 1 · Read the Table

Given: $f\colon \{1, 2, 3, 4\} \to \{p, q, r, s, t\}$ with $f(1) = q$ , $f(2) = s$ , $f(3) = p$ , $f(4) = s$ — find a pair of different inputs that share an output, and the range of $f$ .

Which pair of different inputs shares an output?

What is $\operatorname{range}(f)$ ?

✅ Correct!

2 \neq 4

but

f(2) = f(4) = s

, and the outputs actually produced are

p

q

and

s

. So

f

fails both properties: two inputs share an output, and

r

and

t

are missed.

❌ Not quite. Look at the outputs again:

s

appears twice in the table.

❌ Not quite.

f(1) = q

and

f(3) = p

are different outputs, so this pair does not share one.

❌ Not quite. The two inputs must be different.

f(4) = f(4)

holds for every function and says nothing.

❌ Not quite. That is the domain. The range is a subset of the codomain: the outputs

f(x)

actually produced.

❌ Not quite. That is the codomain. The range contains only the letters that some input is sent to, and nothing is sent to

r

t

❌ Not quite. Those are the letters that are missed. The range is the set of letters that are reached.

Show solution

Shared output. The outputs in order are $q, s, p, s$ . The letter $s$ appears for the inputs $2$ and $4$ , so $x = 2$ , $y = 4$ satisfy $x \neq y$ and $f(x) = f(y)$ . No other pair shares an output.

Range. The outputs produced are $q$ , $s$ , $p$ , so $\operatorname{range}(f) = \{p, q, s\}$ . It is a proper subset of the codomain: $r$ and $t$ are never outputs.

Answers: (b) and (d).

Problem 2 · Negate the Contrapositive Form

Given: nonempty sets $A$ , $B$ and $f\colon A \to B$ , and the statement $\forall x, y \in A,\; f(x) = f(y) \Rightarrow x = y$ — which is its negation?

✅ Correct!

\neg\forall

becomes

\exists

, and

\neg(P \Rightarrow Q)

becomes

P \wedge \neg Q

with

P

f(x) = f(y)

and

Q

x = y

. It is the same condition as the negation of the first form: two different inputs share an output.

❌ Not quite. This negates only the conclusion and keeps

\forall

. It can never be true: taking

y = x

gives

f(x) = f(x)

but not

x \neq x

❌ Not quite. The negation of

P \Rightarrow Q

P \wedge \neg Q

: the hypothesis

f(x) = f(y)

stays as it is. Here it was negated too.

❌ Not quite. The quantifier is right, but the negation of a conditional is not a conditional:

\neg(P \Rightarrow Q) \equiv P \wedge \neg Q

Show solution

Write the statement as $\forall x, y \in A,\; P(x, y) \Rightarrow Q(x, y)$ with $P$ : $f(x) = f(y)$ and $Q$ : $x = y$ .

\neg\big(\forall x, y \in A,\; P \Rightarrow Q\big) \equiv \exists x, y \in A,\; \neg(P \Rightarrow Q) \equiv \exists x, y \in A,\; P \wedge \neg Q

Substituting: $\exists x, y \in A$ such that $f(x) = f(y)$ and $x \neq y$ . This is the same condition as the negation of $\forall x, y \in A,\; x \neq y \Rightarrow f(x) \neq f(y)$ , as it must be, since the two statements are contrapositives of each other.

Answer: (c).

Problem 3 · A Missed Output, Written and Found

Given: nonempty sets $A$ , $B$ and $f\colon A \to B$ , and the function $k\colon \{1, 2, 3, 4, 5\} \to \{p, q, r, s\}$ with $k(1) = s$ , $k(2) = q$ , $k(3) = s$ , $k(4) = p$ , $k(5) = q$ — negate the statement $\forall y \in B,\; \exists x \in A \text{ with } f(x) = y$ , then find a witness for $k$ .

Which is the negation?

Which $y$ makes the negation true for $k$ ?

✅ Correct! Both quantifiers flip and the equation is negated. For

k

the outputs are

s, q, s, p, q

, so

k(x) \neq r

for every

x

r

is never an output.

❌ Not quite. Only the equation was negated; neither quantifier was flipped. Negating "for all

y

" gives "there exists

y

", and the inner

\exists x

must then become

\forall x

❌ Not quite. The

\exists x

must flip to

\forall x

. "Some input misses

y

" is true for every

f

as soon as

B

has two elements, even when every element of

B

is reached; the negation needs every input to miss

y

❌ Not quite. The inner

\exists x

correctly became

\forall x

, but the leading

\forall y

must also flip to

\exists y

; here it stayed

\forall

. This version says no element of

B

is an output at all, which is false whenever

A

is nonempty.

❌ Not quite.

s

is an output twice (

k(1) = k(3) = s

). That is a shared output, which is the other property; a witness here must be an element no input reaches.

❌ Not quite.

5

is an input, not an element of the codomain

\{p, q, r, s\}

. The witness

y

must come from the codomain.

❌ Not quite. List the outputs:

s, q, s, p, q

. The letter

r

never appears.

Show solution

Negation. $\neg\forall y$ becomes $\exists y$ , then $\neg\exists x$ becomes $\forall x$ , and $\neg(f(x) = y)$ becomes $f(x) \neq y$ :

\exists y \in B,\; \forall x \in A,\; f(x) \neq y.

Witness for $k$ . The outputs are $k(1) = s$ , $k(2) = q$ , $k(3) = s$ , $k(4) = p$ , $k(5) = q$ , so $\operatorname{range}(k) = \{p, q, s\}$ . For $y = r$ , $k(x) \neq r$ for all five inputs, so the negation is true for $k$ : not every element of the codomain is an output.

Answers: (d) and (a).

Problem 4 · Changing the Codomain

Given: the rule $F(1) = p$ , $F(2) = q$ , $F(3) = r$ , $F(4) = p$ on the domain $\{1, 2, 3, 4\}$ — choose a codomain $C$ so that $F\colon \{1, 2, 3, 4\} \to C$ is a function with every element of $C$ an output, then decide whether different inputs give different outputs.

Which codomain $C$ works?

With that codomain, do different inputs give different outputs?

✅ Correct! Shrinking the codomain to the range

\{p, q, r\}

removes the missed outputs, but it cannot separate

1

and

4

: they share the output

p

for every choice of codomain.

❌ Not quite. No input is sent to

s

, so

s

is a missed output.

❌ Not quite.

F(3) = r

is not in

\{p, q\}

, so

F

is not even a function from

\{1, 2, 3, 4\}

to this set: the codomain must contain every output.

❌ Not quite. No input is sent to

s

t

, so both are missed outputs.

❌ Not quite. The two properties are independent. Every element of

C

being an output says nothing about whether two inputs share one.

❌ Not quite. The condition is about all pairs of different inputs, including the pair

1

and

4

❌ Not quite. The conclusion is right but the reason is not: an output reached once causes no failure. The failure is the pair

1 \neq 4

with

F(1) = F(4)

Show solution

Codomain. The outputs are $p, q, r, p$ , so the range is $\{p, q, r\}$ . A codomain must contain every output, which rules out $\{p, q\}$ . Every element of $C$ is an output exactly when $C = \operatorname{range}(F) = \{p, q, r\}$ ; the larger sets contain the missed letters $s$ (and $t$ ).

Different inputs. The negation $\exists x, y,\; x \neq y \wedge F(x) = F(y)$ is witnessed by $x = 1$ , $y = 4$ : $1 \neq 4$ and $F(1) = F(4) = p$ . This pair does not depend on the codomain, so no choice of $C$ gives $F$ the first property.

Answers: (a) and (c).

Solved: 0 / 4

Shared and Missed Outputs: Why 5x + 3 Has Neither

IKey Formulas

IIVisualization 1 — Shared Outputs and Missed Outputs

IIIVisualization 2 — Negating the Two Properties

IVVisualization 3 — Why 5x+35x + 35x+3 Has Neither a Shared nor a Missed Output

VQuiz Questions

IVVisualization 3 — Why $5x + 3$ Has Neither a Shared nor a Missed Output