PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Pythagoras, the parallelogram identity, and the real and complex polarisation identities

Statement

For vectors $u, v$ in an inner product space:

if $⟨ u, v ⟩ = 0$ , then $∥ u + v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2}$ ;
$∥ u + v ∥^{2} + ∥ u - v ∥^{2} = 2 ∥ u ∥^{2} + 2 ∥ v ∥^{2}$ ;
over $R$ , $⟨ u, v ⟩ = \frac{1}{4} (∥ u + v ∥^{2} - ∥ u - v ∥^{2})$ ;
over $C$ with the linear-first convention, $⟨ u, v ⟩ = \frac{1}{4} (∥ u + v ∥^{2} - ∥ u - v ∥^{2} + i ∥ u + i v ∥^{2} - i ∥ u - i v ∥^{2}) .$

Facts & Assumptions

Given: Vectors $u, v$ in a real or complex inner product space.

[L1]

The inner product is linear first, conjugate-linear second, and conjugate symmetric (Real and complex inner product spaces, with the inner product linear in the first argument).

[L2]

Squared norm is $∥ w ∥^{2} = ⟨ w, w ⟩$ (The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product).

Proof

technique · direct

1.1L1L2algebra

Expanding by [L1] and [L2] gives $∥ u + v ∥^{2} = ∥ u ∥^{2} + ⟨ u, v ⟩ + \overline{⟨ u, v ⟩} + ∥ v ∥^{2}$ . Orthogonality removes the middle terms and proves Pythagoras.

2.1step 1.1L1L2algebra

Expanding $∥ u - v ∥^{2}$ changes the signs of both middle terms. Adding this expansion to step 1.1 proves the parallelogram identity; subtracting gives $4 Re ⟨ u, v ⟩$ .

3.1step 2.1L1L2algebra∎

Over $R$ , the real part is the scalar itself, giving claim 3. Over $C$ , the same expansion with $i v$ gives $∥ u + i v ∥^{2} - ∥ u - i v ∥^{2} = 4 Im ⟨ u, v ⟩$ under the linear-first convention. Combining real and imaginary parts gives claim 4.

Pythagoras, the parallelogram identity, and the real and complex polarisation identities

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources