LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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.

Convergents are given by the standard recurrences and tail formula

Statement

Let $[a_{0}; a_{1}, \dots, a_{n}]$ be a finite regular continued fraction, and let $p_{k}, q_{k}$ be its convergent numerators and denominators as in Convergents of a regular continued fraction. Then $[a_{0}; a_{1}, \dots, a_{n}] = \frac{p_{n}}{q_{n}} .$

More generally, for every real $t > 0$ one has $[a_{0}; a_{1}, \dots, a_{n}, t] = \frac{t p_{n} + p_{n - 1}}{t q_{n} + q_{n - 1}},$ where $[a_{0}; a_{1}, \dots, a_{n}, t]$ means the finite continued fraction obtained by appending the last tail $t$ .

Facts & Assumptions

Given: A finite regular continued fraction $[a_{0}; a_{1}, \dots, a_{n}]$ , the convergent recurrences for $p_{k}, q_{k}$ , and a real parameter $t > 0$ .

[F1]

A finite regular continued fraction is evaluated recursively by $[a_{n}] = a_{n}$ and $[a_{0}; a_{1}, \dots, a_{n}] = a_{0} + 1 / [a_{1}; \dots, a_{n}]$ , while the convergents satisfy $p_{- 2} = 0$ , $p_{- 1} = 1$ , $q_{- 2} = 1$ , $q_{- 1} = 0$ , and $p_{k} = a_{k} p_{k - 1} + p_{k - 2}$ , $q_{k} = a_{k} q_{k - 1} + q_{k - 2}$ for $k \geq 0$ . (Finite and infinite regular continued fractions, Convergents of a regular continued fraction).

[F2]

If a subset of $N$ contains $0$ and is closed under successor, then it is all of $N$ (The principle of mathematical induction).

Proof

technique · direct

1.1

For $n = 0$ one has. [given, F1, base, algebra] $[a_{0}, t] = a_{0} + \frac{1}{t} = \frac{t a_{0} + 1}{t} = \frac{t p_{0} + p_{- 1}}{t q_{0} + q_{- 1}},$ because $p_{0} = a_{0}$ , $q_{0} = 1$ , $p_{- 1} = 1$ , and $q_{- 1} = 0$ by [F1].

givenF1basealgebra

2.1

Assume the tail formula holds for a fixed length $n$ . [step 1.1, F1, induction, algebra] Put $u : = a_{n + 1} + 1 / t > 0$ . Then $[a_{0}; a_{1}, \dots, a_{n + 1}, t] = [a_{0}; a_{1}, \dots, a_{n}, u] = \frac{u p_{n} + p_{n - 1}}{u q_{n} + q_{n - 1}}$ by the induction hypothesis, and multiplying numerator and denominator by $t$ gives $\frac{(a_{n + 1} t + 1) p_{n} + t p_{n - 1}}{(a_{n + 1} t + 1) q_{n} + t q_{n - 1}} = \frac{t p_{n + 1} + p_{n}}{t q_{n + 1} + q_{n}}$ by the recurrences of [F1].

step 1.1F1inductionalgebra

3.1

Steps 1.1 and 2.1 show, by induction on the length, that. [F2, step 1.1, step 2.1, discharge-induction] $[a_{0}; a_{1}, \dots, a_{n}, t] = \frac{t p_{n} + p_{n - 1}}{t q_{n} + q_{n - 1}}$ for every $n \geq 0$ and every $t > 0$ .

F2step 1.1step 2.1discharge-induction

4.1

Setting $t = a_{n + 1}$ in step 3.1 yields. [step 3.1, F1, algebra] $[a_{0}; a_{1}, \dots, a_{n + 1}] = \frac{a_{n + 1} p_{n} + p_{n - 1}}{a_{n + 1} q_{n} + q_{n - 1}} = \frac{p_{n + 1}}{q_{n + 1}},$ and renaming the index proves the finite-convergent formula.

step 3.1F1algebra∎

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Sources

Peter Hackman, Elementary Number Theory (standard reference, not scraped)
William Stein, Elementary Number Theory: Primes, Congruences, and Secrets (standard reference, not scraped)