Alphabeta Math
DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)
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.

Finite sums and finite products, by recursion

Definition

Throughout this page R is the complete ordered field (Complete ordered field (least-upper-bound property)), in particular an ordered field (Ordered field) and a field (Field), and N is the set of natural numbers (The natural numbers N (von Neumann)) with successor σ(n)=n+1 (Addition of natural numbers).

Let a:N→R be a sequence of reals, written ak for a(k). Finite sums and finite products of a are defined by recursion on the upper index, which is legitimate because of the recursion theorem (The recursion theorem). That theorem produces a function of one variable, so the running index has to be carried along inside the value: applying it to the set A=N×R, the starting element (0,0) and the function f(n,s)=(σ(n), s+an) gives a unique g:N→N×R with

g(0)=(0,0),g(σ(n))=f(g(n))(n∈N).

Write g(n)=(π1(g(n)), Σn) for its two coordinates.

The first coordinate is the index itself, and that is a small induction, not an observation (The principle of mathematical induction). Indeed π1(g(0))=0; and if π1(g(n))=n, then g(σ(n))=f(π1(g(n)),Σn)=(σ(π1(g(n))), Σn+aπ1(g(n)))=(σ(n), Σn+an), so π1(g(σ(n)))=σ(n). By induction π1(g(n))=n for every n∈N. Only now may the second coordinate of the two displayed clauses be read off, and doing so gives

Σ0=0,Σσ(n)=Σn+an.

Σ is moreover the unique function N→R with those two properties: if Σ′ also has them then n↦(n,Σn′) satisfies the two clauses defining g, hence equals g by the uniqueness clause of The recursion theorem, so Σ′=Σ.

We write ∑k<nak:=Σn. The same construction with starting element (0,1) and f(n,p)=(σ(n), p⋅an), with the same induction on the first coordinate and the same uniqueness argument, gives the unique Π:N→R with

Π0=1,Πσ(n)=Πn⋅an,

and we write ∏k<nak:=Πn.

Notation. For m,n∈N we abbreviate

∑k=0nak:=∑k<n+1ak,∏k=0nak:=∏k<n+1ak,

and, for a general lower index m with m≤n+1, writing d=n+1−m for the number of terms,

∑k=mnak:=∑j<dam+j,∏k=mnak:=∏j<dam+j.

When m=n+1 we have d=0 and the sum is empty, with value 0, while the empty product has value 1. In the same spirit ∑k=0−1ak is notation for the empty sum Σ0=0 and ∏k=0−1ak for the empty product Π0=1; the index −1 never occurs as an element of N and is only a way of writing "no terms".

Only finitely many values of a enter ∑k<nak, so the notation ∑k<nak and ∏k<nak is also used for a list a0,…,an−1 of reals given without reference to any extension of the list to all of N: extend the list by ak=0 (respectively ak=1) for k≥n and apply the definition above.

Remarks

  • Why recursion and not "a0+a1+⋯+an−1". The dots are not a definition: they presuppose that the displayed pattern determines a value for every n, which is exactly what the recursion theorem (The recursion theorem) supplies, and its uniqueness clause is what makes ∑k<nak a single well-determined real rather than a family of choices. Associativity and commutativity of addition are not used in the definition; they are used in the laws proved from it (Laws of finite sums and finite products).
  • Naturals and rationals inside R (a convention used on the whole page). A natural number n and a rational number r are not literally elements of R: they enter R through the canonical embedding ι:Q→R, which is an injective, order-preserving field homomorphism (The unique embedding of ℚ into an ordered field), restricting on positive naturals to n↦n⋅1R=1R+⋯+1R (Canonical naturals are positive and strictly increasing). Following ordinary practice, and only where no confusion is possible, we write n for ι(n) and r for ι(r); so, for instance, 1n∑k<nak means ι(n)−1⋅∑k<nak, which makes sense because ι(n)>0 for n≥1. Exponents are the one place where the identification is deliberately NOT made: in an and ar the exponent stays a natural, an integer or a rational (Integer powers am, Rational powers ar of a positive base), never a real.
  • The two indexings are related by ∑k=0nak=∑k<n+1ak, so a statement proved for one is available for the other. Sums over k<n are the primitive form here because Σ0, the empty sum, is then the base case of every induction, and no index outside N is ever needed.

Depends on

Used by

…and 217 more results.

Dependency tree · two levels

23 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