Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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A parallelogram has Jordan content ∣det⁡[v w]∣, equal to base times height when v≠0

Statement

A parallelogram has Jordan content ∣det⁡[v w]∣, equal to base times height when v≠0.

More precisely, P(p;v,w) (Parallelograms and triangles in R2) is Jordan measurable and

cont⁡(P(p;v,w))=∣det⁡[v w]∣,

including the singular case. If v≠0, this value is ∥v∥2d(w,Rv) in the convention of Base and perpendicular height for a chosen side of a plane figure.

Facts & Assumptions

Given: A base point p∈R2 and spanning vectors v,w∈R2.

[L1]

For column vectors v1,…,vn, the closed parallelepiped is Jordan measurable with content equal to the absolute determinant; this includes the singular case, when the content is zero (The Jordan content of the parallelepiped spanned by the columns of a square real matrix is the absolute value of its determinant).

[L2]

For v≠0, ∥v∥2d(w,Rv)=∣det⁡[v w]∣ (∥v∥2 d(w,Rv)=∣det⁡[v w]∣ for v≠0 in R2).

[L3]

Proof

technique · direct
1.1L1L3

Specialize [L1] to n=2 and use [L3] to translate the origin-based parallelepiped by p; it gives Jordan measurability and cont⁡(P(p;v,w))=∣det⁡[v w]∣, including dependent or zero spanning vectors.

2.1step 1.1L2∎

When v≠0, substitute [L2] into the determinant formula of step 1.1 to obtain content equal to base length times perpendicular height.

Depends on

Used by

Dependency tree · two levels

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Sources