PROOF 29

Prove.

The graph of the inverse function is symmetric about y = x

On real functions.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

What you may use

A1sets with the same elements are equal
Equality of sets
A2∀b, the a with f(a)=b is unique
f is a bijection (hypothesis)
A3midpoint = ((a+c)/2, (b+d)/2)
Midpoint of a segment
A4slope = (y₂−y₁)/(x₂−x₁)
Slope (x₁ ≠ x₂)
A5perpendicular ⟺ m₁m₂ = −1
Perpendicularity condition (Proof 22)
A6a + b = b + a
Commutativity of addition
A7t ≠ 0 ⟹ (−t)/t = −1
Sign and division
A8u − v = −(v − u)
Sign rule
A9u − v = 0 ⟺ u = v
Difference and equality
A10a × 1 = a
Multiplicative identity
A11ab = ba
Commutativity of multiplication

D1(a,b)∈G(f) ⟺ b=f(a)
Definition of the graph
D2f⁻¹(b)=a ⟺ f(a)=b
Definition of the inverse function
D3σ(x,y) = (y,x)
The coordinate-swapping map
L1σ is reflection in y=x
Lemma 1
T1(a,b)∈G ⟺ (b,a)∈G⁻¹
Theorem 1 (G⁻¹ = G(f⁻¹))

Nothing outside this list may be cited.

Definition D1

The graph G(f) of a function f is the set of all ordered pairs (a, f(a)). That is, (a,b) ∈ G(f) ⟺ b = f(a)

Proof.

A graph is not a picture but a set of ordered pairs. This definition cannot be applied to a correspondence that is not a function.convention

∎

Definition D2

When f is a bijection, f⁻¹(b) = a is defined by f(a) = b.

Proof.

(1)For each b, the a with f(a) = b is unique, so this correspondence is a function.A2

(2)Therefore f⁻¹ also has a graph G(f⁻¹).(1), D1

∎

Definition D3

Let σ be the map sending the point (x, y) to (y, x). The line y = x is the set of all points whose two coordinates are equal.

Proof.

σ is not yet called a reflection. Whether it is one is verified in L1.convention

∎

Lemma L1

For P = (a, b), σ(P) is the mirror image of P in y = x. That is, the midpoint of segment Pσ(P) lies on y = x, and the segment is perpendicular to y = x.

Proof.

(1)σ(P) = (b, a)D3

(2)If a = b, then σ(P) = P, and since the two coordinates are equal, P lies on y = x.(1), D3

(3)Now suppose a ≠ b.assumption

(4)The midpoint of segment Pσ(P) is ((a+b)/2, (b+a)/2).(1), A3

(5)a + b = b + a, so the midpoint's two coordinates are equal and the midpoint lies on y = x.(4), A6, D3

(6)b − a ≠ 0, so the slope of Pσ(P) is (a−b)/(b−a).(3), A9, A4

(7)a − b = −(b − a), so that slope is −1.(6), A8, A7

(8)(0,0) and (1,1) are points on y = x, and the slope of y = x is (1−0)/(1−0) = 1.D3, A4

(9)1 × (−1) = (−1) × 1 = −1, so the two lines are perpendicular.(7), (8), A11, A10, A5

∎

Theorem T1

(a, b) ∈ G(f) ⟺ (b, a) ∈ G(f⁻¹)

Proof.

(1)(a,b) ∈ G(f) ⟺ b = f(a)D1

(2)b = f(a) ⟺ f⁻¹(b) = aD2

(3)f⁻¹(b) = a ⟺ (b,a) ∈ G(f⁻¹)D2, D1

(4)Chaining the three equivalences gives (a,b) ∈ G(f) ⟺ (b,a) ∈ G(f⁻¹)(1), (2), (3)

∎

Corollary C

The image of G(f) under σ is G(f⁻¹), and the two graphs are symmetric about y = x.

Proof.

(1)If (a,b) ∈ G(f), then σ(a,b) = (b,a) ∈ G(f⁻¹)D3, T1

(2)Conversely, if (b,a) ∈ G(f⁻¹), then (a,b) ∈ G(f) and σ(a,b) = (b,a).T1, D3

(3)The two sets have the same elements, so σ(G(f)) = G(f⁻¹)(1), (2), A1

(4)σ is the reflection in y = x.L1

(5)Therefore the two graphs are symmetric about y = x.(3), (4)

∎

Remark

Can A2 be removed from this proof?

Folding the plane and calling the folded result a function rest on different axioms. Check which line rests on which.