Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)
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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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FALSE: every additive f:R→R is of the form x↦cx for a single real c

Statement

FALSE. Every additive f:R→R (Cauchy's functional equation f(x+y)=f(x)+f(y), and the additive functions R→R) is of the form x↦c x for a single real c.

What is true is the Q-linear part of it, f(qx)=qf(x) for rational q (An additive f:R→R satisfies f(0)=0, f(−x)=−f(x) and f(qx)=q f(x) for every rational q and every real x; in particular f(q)=q f(1) at every rational q), and the conditional statements of Six regularity conditions each force an additive f:R→R to be x↦f(1)x: continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in R2, each of which adds a regularity hypothesis. The claim above asserts the conclusion with no hypothesis at all, and it is false.

The refutation assumes the Axiom of Choice (The Axiom of Choice), which it uses through Assuming the Axiom of Choice, R has a Hamel basis over Q: there is B⊆R such that every real is a finite Q-linear combination of elements of B in exactly one way, and each basis vector carries a well-defined Q-linear coefficient map and hence through Zorn's lemma. The hypothesis is carried explicitly in the Facts below and in every step that needs it. It is an axiom already adopted in this library, so the refutation is a refutation and not a conditional one; what it does not settle is whether a counterexample exists without choice, and nothing here bears on that question.

Facts & Assumptions

Given: The Axiom of Choice, and Q denoting the canonical copy of the rationals inside R (The rationals embed densely in the reals).

[A1]

The Axiom of Choice (The Axiom of Choice, Zorn's lemma).

[L1]

Assume the Axiom of Choice. Then there is B⊆R, a basis of R as a vector space over Q by restriction of scalars, and for each b⋆∈B a map Λb⋆:R→Q with Λb⋆(x+y)=Λb⋆(x)+Λb⋆(y) for all reals x,y, with Λb⋆(b⋆)=1, and with range the whole of Q (Assuming the Axiom of Choice, R has a Hamel basis over Q: there is B⊆R such that every real is a finite Q-linear combination of elements of B in exactly one way, and each basis vector carries a well-defined Q-linear coefficient map, claims 1 and 4, A field is a vector space over itself, and over any subfield K⊆F every F-vector space is a K-vector space by restricting the scalars, Vector space over a field, Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S).

[L2]

A function f:R→R is additive when f(x+y)=f(x)+f(y) for all reals x,y (Cauchy's functional equation f(x+y)=f(x)+f(y), and the additive functions R→R).

[L3]

There exists an irrational real, that is a real not lying in Q: the irrationals are dense in R and in particular nonempty (Both Q and R∖Q are dense in R, and every nonempty open subset of R is uncountable).

[L4]

R is a field, so a nonzero real is invertible (Complete ordered field (least-upper-bound property)).

Refutation

technique · direct
1.1

Assume the Axiom of Choice and fix a Hamel basis B of R over Q together with an element b⋆∈B; such an element exists because B spans R, which is not {0}, so B is nonempty. Put f:=Λb⋆, regarded as a function R→R.

A1L1construct
2.1

f is additive: Λb⋆(x+y)=Λb⋆(x)+Λb⋆(y) for all reals x,y is one of the properties of the coefficient map.

step 1.1L1L2
2.2

Every value of f is rational, and f(b⋆)=1.

step 1.1L1
3.1

Suppose there were a real c with f(x)=c x for every real x. Then c b⋆=f(b⋆)=1, so c≠0 and c is invertible.

step 1.1step 2.2L4
4.1

Take an irrational real θ and put x0:=c−1θ. Then f(x0)=c x0=θ, which is irrational; but every value of f is rational by step 2.2. This is impossible, so no such c exists.

step 2.2step 3.1L3L4
5.1

So f is an additive function R→R that is not of the form x↦c x for any real c, and the claim in the Statement is false.

step 2.1step 4.1discharge-construct∎

Remarks

Depends on

Used by

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Sources