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CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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1\lVert\cdot\rVert_1 on R2\mathbb{R}^{2} violates the parallelogram law, so no symmetric bilinear form induces it

Statement refuted

Refuted claim: every norm on R2\mathbb{R}^{2} arises from a symmetric bilinear form, that is, for every norm NN there is a function B:R2×R2RB : \mathbb{R}^{2}\times\mathbb{R}^{2} \to \mathbb{R} that is symmetric and additive and homogeneous in each argument, with N(x)=B(x,x)N(x) = \sqrt{B(x,x)} for every xx.

The witness is N:=1N := \lVert\cdot\rVert_1 on R2\mathbb{R}^{2} (The pp-norms xp\lVert x\rVert_p for rational p1p \ge 1, and x\lVert x\rVert_\infty), and the obstruction is the parallelogram law, which every such NN satisfies (Cauchy-Schwarz x,yx2y2\lvert\langle x,y\rangle\rvert \le \lVert x\rVert_2\lVert y\rVert_2 with its equality case, the triangle inequality for 2\lVert\cdot\rVert_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\lVert\cdot\rVert_1 fails at x=e0x = e_0, y=e1y = e_1.

What is and is not claimed. What is refuted is the displayed claim, whose hypothesis is a symmetric bilinear form on R2\mathbb{R}^{2} 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 n1n \ge 1 hypothesis, and what is taken up elsewhere in the reading order).

Facts & Assumptions

[A1]

The refuted claim at N=1N = \lVert\cdot\rVert_1: there is a symmetric B:R2×R2RB : \mathbb{R}^{2}\times\mathbb{R}^{2}\to\mathbb{R}, additive and homogeneous in each argument, with x1=B(x,x)\lVert x\rVert_1 = \sqrt{B(x,x)} for every xR2x \in \mathbb{R}^{2}.

[L2]

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

[L3]

Square roots: c\sqrt{c} is the unique nonnegative ss with s2=cs^{2} = c, so (B(x,x))2=B(x,x)\bigl(\sqrt{B(x,x)}\bigr)^{2} = B(x,x) whenever B(x,x)0B(x,x) \ge 0 (Square roots exist: a unique a0\sqrt{a} \ge 0 with (a)2=a(\sqrt{a})^2 = a; the positives are {x2:x0}\{x^2 : x \neq 0\}, Integer powers ama^m).

[L4]

Canonical naturals are strictly increasing and positive and carry sums to sums and products to products, so ι(2)2=ι(4)\iota(2)^{2} = \iota(4), ι(4)+ι(4)=ι(8)\iota(4)+\iota(4) = \iota(8), 21+21=ι(4)2\cdot 1 + 2\cdot 1 = \iota(4) and ι(8)ι(4)\iota(8) \ne \iota(4) (The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, Canonical naturals are positive and strictly increasing).

Counterexample

technique · direct
1.1

Assume [A1] and write Q(x):=B(x,x)Q(x) := B(x,x), so x12=Q(x)\lVert x\rVert_1^{2} = Q(x) for every xx.

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)Q(x+y) = B(x+y,x+y) = Q(x) + 2B(x,y) + Q(y) and Q(xy)=Q(x)2B(x,y)+Q(y)Q(x-y) = Q(x) - 2B(x,y) + Q(y), hence Q(x+y)+Q(xy)=2Q(x)+2Q(y)Q(x+y)+Q(x-y) = 2Q(x)+2Q(y) for all x,yx,y.

A1
1.3

Computing: (1,1)1=1+1=ι(2)\lVert (1,1)\rVert_1 = |1|+|1| = \iota(2) and (1,1)1=1+1=ι(2)\lVert (1,-1)\rVert_1 = |1|+|-1| = \iota(2), while e01=e11=1\lVert e_0\rVert_1 = \lVert e_1\rVert_1 = 1.

L1L2
2.1

Instantiate step 1.2 at x=e0x = e_0, y=e1y = e_1: the left side is (1,1)12+(1,1)12\lVert (1,1)\rVert_1^{2} + \lVert (1,-1)\rVert_1^{2} and the right side is 2e012+2e1122\lVert e_0\rVert_1^{2} + 2\lVert e_1\rVert_1^{2}.

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)\iota(2)^{2}+\iota(2)^{2} = \iota(4)+\iota(4) = \iota(8) and the right side is 21+21=ι(4)2\cdot 1 + 2\cdot 1 = \iota(4), giving ι(8)=ι(4)\iota(8) = \iota(4), which contradicts the strict increase of ι\iota.

step 2.1step 1.3L4
4.1

Hence [A1] is false: no symmetric bilinear form on R2\mathbb{R}^{2} induces 1\lVert\cdot\rVert_1, and in particular 12\lVert\cdot\rVert_1 \ne \lVert\cdot\rVert_2.

step 1.1step 3.1A1
5.1

The parallelogram law does hold for 2\lVert\cdot\rVert_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

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