PROOF 30

Prove.

The graph of a linear function is a line

On the coordinate plane.

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.

Definition D1

For distinct points P(x₁, y₁), Q(x₂, y₂) with x₁ ≠ x₂, the slope is m(P, Q) = (y₂ − y₁)/(x₂ − x₁).

Proof.

(1)If x₁ = x₂, the denominator is 0 and the slope is undefined.convention

(2)u − v means u + (−v).convention

∎

Definition D2

Three distinct points P, Q, R are said to lie on one line when m(P, Q) = m(Q, R).

Proof.

In the coordinate plane, “on one line” is determined by this equation, not by a picture.convention

∎

Definition D3

For a point A and a number m, ℓ(A, m) is the set consisting of A itself and all points whose slope with A is m.

Proof.

P ∈ ℓ(A, m) ⟺ P = A or m(A, P) = mconvention, D1

∎

Lemma L1

Let G be the graph of y = ax + b. Then the slope of any two distinct points of G is always a.

Proof.

(1)Let the two points be P₁(x₁, ax₁ + b), P₂(x₂, ax₂ + b) with P₁ ≠ P₂.assumption

(2)If x₁ = x₂, then ax₁ + b = ax₂ + b and the two points coincide. Therefore x₁ ≠ x₂.(1), E

(3)y₂ − y₁ = (ax₂ + b) + (−(ax₁ + b))D1

(4)−(ax₁ + b) = (−(ax₁)) + (−b)A7

(5)y₂ − y₁ = (ax₂ + (−(ax₁))) + (b + (−b)) = ax₂ + (−(ax₁))(3), (4), A2, A3, A5, A4

(6)a(x₂ − x₁) = a(x₂ + (−x₁)) = ax₂ + a(−x₁)D1, A1

(7)a(−x₁) = −(ax₁), so a(x₂ − x₁) = ax₂ + (−(ax₁))(6), A8

(8)y₂ − y₁ = a(x₂ − x₁)(5), (7)

(9)x₂ − x₁ ≠ 0(2), A9

(10)m(P₁, P₂) = (y₂ − y₁)/(x₂ − x₁) = a(8), (9), D1, A6

∎

Theorem T1

Any three distinct points of G lie on one line.

Proof.

(1)Let the three points be P, Q, R.assumption

(2)m(P, Q) = aL1

(3)m(Q, R) = aL1

(4)m(P, Q) = m(Q, R), so the three points lie on one line.(2), (3), D2

∎

Theorem T2

Let A = (0, b). Then G = ℓ(A, a). That is, the graph is the line through A with slope a, and every point of that line lies on the graph.

Proof.

(1)a × 0 + b = 0 + b = b, so A = (0, b) lies on G.A10, A3, A4

(2)Let P(p, ap + b) ∈ G with P ≠ A.assumption

(3)m(A, P) = a(1), (2), L1

(4)Therefore G ⊆ ℓ(A, a).(1), (3), D3

(5)Conversely, let Q(p, q) ∈ ℓ(A, a) with Q ≠ A.assumption

(6)m(A, Q) = (q − b)/(p − 0) = (q − b)/p = a and p ≠ 0.(5), D3, D1, A4

(7)q − b = ap(6), A6

(8)Adding b to both sides gives q = ap + b, so Q ∈ G.(7), A2, A3, A5, A4

(9)Therefore ℓ(A, a) ⊆ G.(1), (5), (8)

(10)Both inclusions hold, so G = ℓ(A, a)(4), (9)

∎

Remark

Can A1 be removed from this proof?

That a point of the line satisfies the equation and that the points satisfying the equation form one line are not the same thing. Check which direction rests on what.