LemmaStatement: Literature-sourcedProof: AI-adapted

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.

An additive $f : R \to R$ satisfies $f (0) = 0$ , $f (- x) = - f (x)$ and $f (q x) = q f (x)$ for every rational $q$ and every real $x$ ; in particular $f (q) = q f (1)$ at every rational $q$

Statement

Let $f : R \to R$ be additive (Cauchy's functional equation $f (x + y) = f (x) + f (y)$ , and the additive functions $R \to R$ ), and identify $N \subseteq Z \subseteq Q \subseteq R$ along the canonical embeddings (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals), writing each natural directly for its canonical value in $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field). Then, for every real $x$ :

$f (0) = 0$ ;
$f (- x) = - f (x)$ ;
$f (n x) = n f (x)$ for every $n \in N$ ;
$f (m x) = m f (x)$ for every integer $m$ ;
$f (q x) = q f (x)$ for every rational $q$ .

In particular, taking $x = 1$ in claim 5, $f (q) = q f (1)$ at every rational $q$ : an additive function is determined on $Q$ by its value at $1$ .

Facts & Assumptions

Given: An additive $f : R \to R$ , so $f (x + y) = f (x) + f (y)$ for all reals $x, y$ .

[A1]

$f (x + y) = f (x) + f (y)$ for all reals $x, y$ (Cauchy's functional equation $f (x + y) = f (x) + f (y)$ , and the additive functions $R \to R$ ).

[L1]

Induction on $N$ (The principle of mathematical induction).

[L2]

The canonical inclusion sends zero to the real zero and a successor to the preceding value plus $1$ ; the real value of $n$ agrees with the additive multiple $n \cdot 1_{R}$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, In a field, the additive multiple $n \cdot 1_{F}$ is the canonical natural $ι (n)$ : the additive power of the group-power definition and the canonical natural are the same function, both being the unique one given by the recursion $ι (0) = 0_{F}$ , $ι (σ (n)) = ι (n) + 1_{F}$ , Canonical naturals are positive and strictly increasing).

[L3]

The embeddings $N \subseteq Z \subseteq Q \subseteq R$ preserve arithmetic; the image of $N$ in $Z$ is exactly the nonnegative integers, and positive canonical naturals are nonzero (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals, Canonical naturals are positive and strictly increasing).

[L5]

Every rational is represented by a pair $(a, b)$ of integers with $b \neq 0$ , where $(a, b) \sim (c, d)$ means $a d = c b$ ; rational multiplication and inversion are the pair operations (The rationals as equivalence classes of pairs of integers, Arithmetic on the rationals).

[L6]

The integers form a totally ordered commutative ring (The integers form a totally ordered ring).

[L4]

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

Proof

technique · induction

1.1A1L4

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

1.2ih

Claim 3, inductive hypothesis: suppose $f (n x) = n f (x)$ for a given $n \in N$ and every real $x$ .

2.1step 1.1A1L4

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

2.2step 1.1L2base

Claim 3, base case $n = 0$ : $f (0 x) = f (0) = 0 = 0 f (x)$ .

2.3step 1.2A1L2L4

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

3.1step 2.2step 2.3L1

Claim 3 holds for every $n \in N$ and every real $x$ , by induction on $n$ from steps 2.2 and 2.3.

4.1step 2.1step 3.1L3L6

Claim 4: by the total order and addition compatibility in [L6], either $m \geq 0$ or $- m \geq 0$ ; by [L3] the nonnegative one is the image of some natural $n$ . Thus $m = n$ or $m = - n$ . In the first case claim 3 applies directly. In the second, $f (m x) = f (- (n x)) = - f (n x) = - n f (x) = m f (x)$ .

5.1step 4.1L3L4L5L6algebra

Claim 5: by [L5], write $q = [(a, b)]$ with $a, b \in Z$ and $b \neq 0$ . By [L6], either $b > 0$ or $- b > 0$ ; in the second case replace $(a, b)$ by $(- a, - b)$ , which represents the same rational because $a (- b) = (- a) b$ . The resulting positive denominator is the image of a natural $n \geq 1$ by [L3]; call the resulting numerator $m$ . The pair formulas in [L5] give $[(m, n)] = [(m, 1)] [(1, n)] = j (m) j (n)^{- 1}$ , where $j$ is the integer embedding, so after the field embedding into $R$ this is $q = m / n$ . By [L3], $n \neq 0$ . Applying claim 4 with the integer $n$ to the real $q x$ gives $n f (q x) = f (n q x) = f (m x) = m f (x)$ , and division by $n$ gives $f (q x) = (m / n) f (x) = q f (x)$ .

6.1step 1.1step 2.1step 3.1step 4.1step 5.1discharge-induction∎

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

Remarks

What this does not say. Claim 5 is $Q$ -homogeneity, not $R$ -homogeneity: nothing here gives $f (λ x) = λ f (x)$ for irrational $λ$ , and that is exactly the gap that FALSE: every additive $f : R \to R$ is of the form $x \mapsto c x$ for a single real $c$ shows cannot be closed without a regularity hypothesis.
The induction is on $N$ and everything else is algebra. Only claim 3 needs induction; claims 4 and 5 are obtained from it by the two field operations, and claims 1 and 2 are two substitutions into the equation. The base case is $n = 0$ , where the real value of $n$ is $0$ and the identity reads $f (0) = 0$ ; it is a genuine case and not a convention, since $N$ contains $0$ .
This is the whole of the algebraic theory. Every regularity theorem about Cauchy's equation (Six regularity conditions each force an additive $f : R \to R$ to be $x \mapsto f (1) x$ : 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 $R^{2}$ ) works by combining claim 5 with density of $Q$ in $R$ : the value of $f$ is pinned on a dense set, and a regularity hypothesis is what forbids the values off that set from being arbitrary.

Depends on

Used by

Dependency tree · two levels

50 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Cauchy's functional equation (Wikipedia) (standard reference, not scraped)

An additive f:R→Rf:R→R satisfies f(0)=0f(0)=0, f(−x)=−f(x)f(−x)=−f(x) and f(qx)=q f(x)f(qx)=qf(x) for every rational qq and every real xx; in particular f(q)=q f(1)f(q)=qf(1) at every rational qq