Mathematical-Reasoning · Unit 24 · Video 3 · Interactive Practice

Swapping the Pairs: The Inverse of a Function as a Set

IKey Formulas

Statement	Meaning
(1) for every $a \in A$ there is $b \in B$ with $(a, b) \in f$	Every element of $A$ is an input
(2) if $(a, b) \in f$ and $(a, c) \in f$ , then $b = c$	No input has two different outputs; with (1), a subset $f \subseteq A \times B$ satisfying both is a function from $A$ to $B$
$f^{-1} = \{ (b, a) \in B \times A \mid f(a) = b \}$	The inverse of $f : A \to B$ : every pair of $f$ with its coordinates swapped; a subset of $B \times A$ , not automatically a function
$y = \ln x \iff x = e^y$ $\ (x > 0)$	$(e, 1) \in \ln$ and $(1, e) \in \exp$ : the logarithm's pairs swapped are the exponential's pairs

Key Insight: To decide whether $f^{-1}$ is a function from $B$ to $A$ , apply the same two conditions with $B$ as the first set: every element of $B$ must be a first coordinate of $f^{-1}$ exactly once.

IITesting the Swapped Pairs

An arrow landing twice on an output, or never, becomes a failed condition for $f^{-1}$ .

Function

IIISwapping Coordinates Reflects the Graph

The pair $(x, \ln x)$ of the logarithm and the swapped pair $(\ln x, x)$ of the exponential are mirror images across $y = x$ .

x e

IVQuiz Questions

Problem 1 · Writing the Inverse

Given: $A = \{1, 2, 3\}$ , $B = \{p, q\}$ and $f = \{ (1, p), (2, q), (3, p) \}$ . Which set is $f^{-1}$ ?

✅ Correct! Every pair of

f

is swapped, including both pairs ending in

p

❌ Nothing was swapped.

f^{-1}

is a subset of

B \times A

, so its first coordinates are

p

and

q

❌ A pair is missing. The inverse contains the swap of every pair of

f

, even when that makes it fail to be a function.

❌ Not quite. Swap each pair of

f

exactly:

(3, p)

becomes

(p, 3)

Show solution

By definition $f^{-1} = \{ (b, a) \in B \times A \mid (a, b) \in f \}$ . Swapping each pair: $(1, p) \to (p, 1)$ , $(2, q) \to (q, 2)$ , $(3, p) \to (p, 3)$ . So

f^{-1} = \{ (p, 1), (q, 2), (p, 3) \}.

Problem 2 · Which Condition Fails?

Given: the same $f$ and $f^{-1} = \{ (p, 1), (q, 2), (p, 3) \}$ , tested as a possible function from $B = \{p, q\}$ to $A$ . Which conditions does $f^{-1}$ fail?

✅ Correct! Both

p

and

q

are first coordinates, so (1) holds; but

(p, 1)

and

(p, 3)

break (2).

❌ Check condition (1) again. The first set is

B = \{p, q\}

, and both

p

and

q

appear as first coordinates.

❌ Not quite. Look at the two pairs that start with

p

Show solution

Condition (1): $p$ appears in $(p, 1)$ and $q$ in $(q, 2)$ , so every element of $B$ is a first coordinate. Condition (2): $(p, 1) \in f^{-1}$ and $(p, 3) \in f^{-1}$ but $1 \ne 3$ , so it fails. In arrow terms, $f$ sends two inputs to $p$ (it is not an injection) but reaches every element of $B$ (it is a surjection).

Problem 3 · Finding the Witness

Given: $h : \{a, b, c, d\} \to \{p, q, r\}$ with $h(a) = r$ , $h(b) = p$ , $h(c) = r$ , $h(d) = q$ . Which two pairs of $h^{-1}$ show that $h^{-1}$ is not a function from $\{p, q, r\}$ to $\{a, b, c, d\}$ ?

✅ Correct! The input

r

h^{-1}

has two different outputs,

a

and

c

, so condition (2) fails.

❌ These are pairs of $h$ . They are allowed in

h

(outputs may repeat); swap them to see the problem in

h^{-1}

❌ Different first coordinates. Two pairs only break condition (2) when they share a first coordinate.

❌ Not quite. Look for two pairs of

h^{-1}

with the same first coordinate.

Show solution

h^{-1} = \{ (r, a), (p, b), (r, c), (q, d) \}.

Condition (1) holds: $p$ , $q$ , $r$ are all first coordinates. Condition (2) fails: $(r, a), (r, c) \in h^{-1}$ with $a \ne c$ . This comes from $h(a) = h(c) = r$ : $h$ is a surjection but not an injection.

Problem 4 · A Point on the Inverse's Graph

Given: $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = x^3 + 1$ , so $(2, 9) \in f$ . Which ordered pair must belong to $f^{-1}$ ?

✅ Correct!

(a, b) \in f

exactly when

(b, a) \in f^{-1}

, and

f(2) = 9

❌ $f^{-1}$ is not a reciprocal. The exponent

-1

names the inverse, which swaps coordinates; it does not mean

\tfrac{1}{f}

❌ That reflects through the origin. Swapping coordinates reflects across the line

y = x

; also

f(-2) = -7

, not

-9

Show solution

$f(2) = 2^3 + 1 = 9$ , so $(2, 9) \in f$ , and by the definition of the inverse, $(9, 2) \in f^{-1}$ . (Here $f$ is a bijection, so $f^{-1}$ is a function and $f^{-1}(9) = 2$ ; the swap would hold even if it were not.)

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