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CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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∥⋅∥1 on R2 violates the parallelogram law, so no symmetric bilinear form induces it

Statement refuted

Refuted claim: every norm on R2 arises from a symmetric bilinear form, that is, for every norm N there is a function B:R2×R2→R that is symmetric and additive and homogeneous in each argument, with N(x)=B(x,x) for every x.

The witness is N:=∥⋅∥1 on R2 (The p-norms ∥x∥p for rational p≥1, and ∥x∥∞), and the obstruction is the parallelogram law, which every such N satisfies (Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation clause 3 is the instance for the Euclidean form, and the general computation is two lines of bilinearity, done below) and which ∥⋅∥1 fails at x=e0, y=e1.

What is and is not claimed. What is refuted is the displayed claim, whose hypothesis is a symmetric bilinear form on R2 written out in full. The general converse — that a norm satisfying the parallelogram law is induced by an inner product, the Jordan-von Neumann theorem — is not proved here and is not used here; nor is any abstract theory of inner product spaces, which belongs to a page of this library earlier in the plan order that is not yet built (Conventions of this page, the standing n≥1 hypothesis, and what is taken up elsewhere in the reading order).

Facts & Assumptions

[A1]

The refuted claim at N=∥⋅∥1: there is a symmetric B:R2×R2→R, additive and homogeneous in each argument, with ∥x∥1=B(x,x) for every x∈R2.

[L2]

Absolute values: ∣1∣=∣−1∣=1 and ∣0∣=0 (Absolute value in an ordered field, Basic properties of the absolute value).

[L3]

Square roots: c is the unique nonnegative s with s2=c, so (B(x,x))2=B(x,x) whenever B(x,x)≥0 (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Integer powers am).

[L4]

Canonical naturals are strictly increasing and positive and carry sums to sums and products to products, so ι(2)2=ι(4), ι(4)+ι(4)=ι(8), 2⋅1+2⋅1=ι(4) and ι(8)≠ι(4) (The canonical natural ι(n)=n⋅1F of a field, Canonical naturals are positive and strictly increasing).

Counterexample

technique · direct
1.1

Assume [A1] and write Q(x):=B(x,x), so ∥x∥12=Q(x) for every x.

A1L3
1.2

By symmetry and additivity and homogeneity in each argument, Q(x+y)=B(x+y,x+y)=Q(x)+2B(x,y)+Q(y) and Q(x−y)=Q(x)−2B(x,y)+Q(y), hence Q(x+y)+Q(x−y)=2Q(x)+2Q(y) for all x,y.

A1
1.3

Computing: ∥(1,1)∥1=∣1∣+∣1∣=ι(2) and ∥(1,−1)∥1=∣1∣+∣−1∣=ι(2), while ∥e0∥1=∥e1∥1=1.

L1L2
2.1

Instantiate step 1.2 at x=e0, y=e1: the left side is ∥(1,1)∥12+∥(1,−1)∥12 and the right side is 2∥e0∥12+2∥e1∥12.

step 1.1step 1.2
3.1

So the left side of step 2.1 is ι(2)2+ι(2)2=ι(4)+ι(4)=ι(8) and the right side is 2⋅1+2⋅1=ι(4), giving ι(8)=ι(4), which contradicts the strict increase of ι.

step 2.1step 1.3L4
4.1

Hence [A1] is false: no symmetric bilinear form on R2 induces ∥⋅∥1, and in particular ∥⋅∥1≠∥⋅∥2.

step 1.1step 3.1A1
5.1

The parallelogram law does hold for ∥⋅∥2, which is induced by the Euclidean inner product, so the failure above is a genuine separation between the two norms and not a defect of the computation.

L5∎

Remarks

  • Equivalence of norms says nothing about inner products. By For n≥1 all norms on Rn are equivalent the norms ∥⋅∥1 and ∥⋅∥2 on R2 are equivalent: they have the same open sets, the same convergent sequences and the same Cauchy sequences. What the computation above shows is that they are nevertheless different norms, and that one of them cannot be written as B(⋅,⋅) for any symmetric bilinear B. Equivalence is a metric statement; the parallelogram law is not.

  • Only one instance of the law is needed. The claim is refuted by a single pair (e0,e1), and the arithmetic is ι(8)≠ι(4). No general theory is required, which is exactly why this item can be stated on a page that has no abstract inner products.

  • The converse direction is a different theorem. That a norm satisfying the parallelogram law is induced by an inner product is the Jordan-von Neumann theorem, proved by polarisation; Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation clause 4 contains the polarisation identity for the Euclidean form, but the general theorem needs the abstract theory and is not asserted anywhere in this library.

Depends on

Used by

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Sources