Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 v0

Statement

A parallelogram has Jordan content det[v w], equal to base times height when v0.

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 v0, this value is v2d(w,Rv) in the convention of Base and perpendicular height for a chosen side of a plane figure.

Facts & Assumptions

Given: A base point pR2 and spanning vectors v,wR2.

[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 v0, v2d(w,Rv)=det[v w] (v2d(w,Rv)=det[v w] for v0 in R2).

[L3]

Proof

technique · direct
1.1

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.

L1L3
2.1

When v0, substitute [L2] into the determinant formula of step 1.1 to obtain content equal to base length times perpendicular height.

step 1.1L2

Depends on

Used by

Dependency tree · two levels

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Sources