PROOF 124

Prove.

The discriminant and the number of x-intercepts

On the real numbers ℝ.

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

The graph G of f(x) = ax² + bx + c (a ≠ 0) is the set of all ordered pairs (x, f(x)).

Proof.

(1)A graph is not a picture but a set of ordered pairs. a ≠ 0 is assumed in every block below.convention

(2)D means b² − 4ac.convention

∎

Definition D2

The x-axis is the set of all points whose y-coordinate is 0. That is, (x, y) ∈ x-axis ⟺ y = 0

Proof.

The x-axis is also treated not as a picture but as a set of points.convention

∎

Definition D3

A point belonging to both G and the x-axis is called an intersection point of the graph and the x-axis.

Proof.

The number of intercepts means the number of all such points.convention

∎

Theorem T1

x₀ is a root of f(x) = 0 if and only if (x₀, 0) is an intersection point. Moreover, x₀ ↦ (x₀, 0) is a one-to-one correspondence between all roots and all intersection points.

Proof.

(1)Suppose x₀ is a root. Then f(x₀) = 0.assumption, A1

(2)(x₀, f(x₀)) is an element of G, and f(x₀) = 0, so (x₀, f(x₀)) = (x₀, 0).(1), D1, A2

(3)The y-coordinate of (x₀, 0) is 0, so that point lies on the x-axis.D2

(4)Therefore (x₀, 0) is an intersection point.(2), (3), D3

(5)Conversely, suppose P is an intersection point and write P = (x₀, y₀).assumption, D3

(6)P is an element of G, so y₀ = f(x₀).(5), D1, A2

(7)P lies on the x-axis, so y₀ = 0.(5), D2

(8)f(x₀) = y₀ and y₀ = 0, so f(x₀) = 0, and x₀ is a root.(6), (7), A3, A1

(9)If x₀ ≠ x₁, then (x₀, 0) ≠ (x₁, 0), so distinct roots give distinct intersection points.A2

(10)Every intersection point has its x-coordinate as a root, so it is obtained by the correspondence of (4).(5), (6), (7), (8)

(11)Therefore all roots and all intersection points correspond one-to-one, and their numbers are equal.(4), (9), (10)

∎

Theorem T2

If D > 0, there are exactly two intersection points; if D = 0, exactly one.

Proof.

(1)The number of intersection points equals the number of roots.T1

(2)If D > 0, there are exactly two roots, so exactly two intersection points.(1), A4

(3)If D = 0, there is exactly one root, so exactly one intersection point.(1), A5

∎

Theorem T3

If D < 0, the graph and the x-axis have no intersection point.

Proof.

(1)The number of intersection points equals the number of roots.T1

(2)If D < 0, there are no real roots at all.A6

(3)Therefore there are no intersection points at all.(1), (2)

∎

Corollary C

The graph meets the x-axis if and only if D ≥ 0.

Proof.

(1)D ≥ 0 means D > 0 or D = 0, and in both cases there is an intersection point.convention, T2

(2)If D < 0, there is no intersection point.T3

(3)By trichotomy, D falls into exactly one of the three cases.A7

(4)Therefore having an intersection point is equivalent to D ≥ 0.(1), (2), (3)

∎

Remark

Can A6 be removed from this proof?

Separate the lines linking roots to intersection points from the lines counting them. Even among the counting lines, only one uses this axiom.