Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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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.

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An additive f:RRf : \mathbb{R} \to \mathbb{R} satisfies f(0)=0f(0) = 0, f(x)=f(x)f(-x) = -f(x) and f(qx)=qf(x)f(qx) = q\,f(x) for every rational qq and every real xx; in particular f(q)=qf(1)f(q) = q\,f(1) at every rational qq

Statement

Let f:RRf : \mathbb{R} \to \mathbb{R} be additive (Cauchy's functional equation f(x+y)=f(x)+f(y)f(x+y) = f(x) + f(y), and the additive functions RR\mathbb{R} \to \mathbb{R}), and identify NZQR\mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R} along the canonical embeddings (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals), writing ι(n)\iota(n) for the canonical natural of nn in R\mathbb{R} (The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field). Then, for every real xx:

  1. f(0)=0f(0) = 0;
  2. f(x)=f(x)f(-x) = -f(x);
  3. f(ι(n)x)=ι(n)f(x)f(\iota(n)\,x) = \iota(n)\,f(x) for every nNn \in \mathbb{N};
  4. f(mx)=mf(x)f(m x) = m\,f(x) for every integer mm;
  5. f(qx)=qf(x)f(q x) = q\,f(x) for every rational qq.

In particular, taking x=1x = 1 in claim 5, f(q)=qf(1)f(q) = q\,f(1) at every rational qq: an additive function is determined on Q\mathbb{Q} by its value at 11.

What this does not say. Claim 5 is Q\mathbb{Q}-homogeneity, not R\mathbb{R}-homogeneity: nothing here gives f(λx)=λf(x)f(\lambda x) = \lambda f(x) for irrational λ\lambda, and that is exactly the gap that FALSE: every additive f:RRf : \mathbb{R} \to \mathbb{R} is of the form xcxx \mapsto cx for a single real cc shows cannot be closed without a regularity hypothesis.

Facts & Assumptions

Given: An additive f:RRf : \mathbb{R} \to \mathbb{R}, so f(x+y)=f(x)+f(y)f(x+y) = f(x) + f(y) for all reals x,yx, y.

[L1]

Induction on N\mathbb{N} (The principle of mathematical induction).

[L3]

Every integer is ι(n)\iota(n) or ι(n)-\iota(n) for a natural nn, and every rational is m/ι(n)m/\iota(n) with mm an integer and nn a natural 1\ge 1; the embeddings preserve sums and products, and ι(n)0\iota(n) \ne 0 for n1n \ge 1 (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals, The integers as equivalence classes of pairs of naturals, Canonical naturals are positive and strictly increasing).

[L4]

R\mathbb{R} is a field, so cancellation, distributivity and inverses of nonzero elements are available (Complete ordered field (least-upper-bound property)).

Proof

technique · induction
1.1

Claim 1: taking x=y=0x = y = 0 in the functional equation gives f(0)=f(0)+f(0)f(0) = f(0) + f(0), and adding f(0)-f(0) to both sides gives f(0)=0f(0) = 0.

A1L4
1.2

Claim 3, inductive hypothesis: suppose f(ι(n)x)=ι(n)f(x)f(\iota(n)x) = \iota(n)f(x) for a given nNn \in \mathbb{N} and every real xx.

ih
2.1

Claim 2: taking y=xy = -x gives 0=f(0)=f(x)+f(x)0 = f(0) = f(x) + f(-x), so f(x)=f(x)f(-x) = -f(x).

step 1.1A1L4
2.2

Claim 3, base case n=0n = 0: ι(0)=0\iota(0) = 0, so f(ι(0)x)=f(0)=0=ι(0)f(x)f(\iota(0)x) = f(0) = 0 = \iota(0)f(x).

step 1.1L2base
2.3

Claim 3, inductive step: ι(n+1)x=ι(n)x+x\iota(n+1)x = \iota(n)x + x, so f(ι(n+1)x)=f(ι(n)x)+f(x)=ι(n)f(x)+f(x)=(ι(n)+1)f(x)=ι(n+1)f(x)f(\iota(n+1)x) = f(\iota(n)x) + f(x) = \iota(n)f(x) + f(x) = (\iota(n)+1)f(x) = \iota(n+1)f(x).

step 1.2A1L2L4
3.1

Claim 3 holds for every nNn \in \mathbb{N} and every real xx, by induction on nn from steps 2.2 and 2.3.

step 2.2step 2.3L1
4.1

Claim 4: an integer mm is ι(n)\iota(n) or ι(n)-\iota(n) for some natural nn. In the first case claim 3 applies directly. In the second, f(mx)=f((ι(n)x))=f(ι(n)x)=ι(n)f(x)=mf(x)f(mx) = f(-(\iota(n)x)) = -f(\iota(n)x) = -\iota(n)f(x) = m f(x).

step 2.1step 3.1L3
5.1

Claim 5: let qq be rational and write q=m/ι(n)q = m/\iota(n) with mm an integer and nn a natural 1\ge 1, so ι(n)0\iota(n) \ne 0. Applying claim 4 with the integer ι(n)\iota(n) to the real qxqx gives ι(n)f(qx)=f(ι(n)qx)=f(mx)=mf(x)\iota(n) f(qx) = f(\iota(n) q x) = f(mx) = m f(x), and dividing by ι(n)\iota(n) gives f(qx)=(m/ι(n))f(x)=qf(x)f(qx) = (m/\iota(n)) f(x) = q f(x).

step 4.1L3L4
6.1

Taking x=1x = 1 in claim 5 gives f(q)=qf(1)f(q) = q f(1) for every rational qq, and all five claims are proved.

step 1.1step 2.1step 3.1step 4.1step 5.1discharge-induction

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