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

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Convergent error bound

Statement

Let $α$ be an irrational real number, and let $p_{n} / q_{n}$ be its continued- fraction convergents. Then, for every $n \geq 0$ , $∣ α - \frac{p_{n}}{q_{n}} ∣ < \frac{1}{q_{n} q_{n + 1}} \leq \frac{1}{q_{n}^{2}} .$ Moreover, $α - \frac{p_{n}}{q_{n}}$ has sign $(- 1)^{n}$ , so the convergents alternate around $α$ .

Facts & Assumptions

Given: An irrational real number $α$ , its continued-fraction digits $a_{n}$ , its complete quotients $α_{n}$ , and its convergents $p_{n} / q_{n}$ .

[F1]

An irrational real does not terminate under the continued-fraction algorithm, so every complete quotient $α_{n + 1}$ is defined and $α = \frac{α_{n + 1} p_{n} + p_{n - 1}}{α_{n + 1} q_{n} + q_{n - 1}} .$ (The continued-fraction algorithm terminates exactly on rational numbers, Complete-quotient tail formula).

[F2]

Consecutive convergents satisfy $p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1}$ . (Determinant identity for consecutive convergents).

[F3]

The convergent denominators satisfy $q_{- 1} = 0$ , are positive for every index $n \geq 0$ , and obey $q_{n + 1} = a_{n + 1} q_{n} + q_{n - 1}$ (Convergents of a regular continued fraction).

[F4]

For irrational $α$ , the algorithm does not terminate, so $α_{n + 1} \neq a_{n + 1}$ ; the defining floor inequality therefore gives $1 \leq a_{n + 1} < α_{n + 1}$ (The continued-fraction algorithm terminates exactly on rational numbers, Complete quotients in the continued-fraction algorithm).

Proof

technique · direct

1.1

By [F1] and [F2]. [F1, F2, F3, F4, algebra] $α - \frac{p_{n}}{q_{n}} = \frac{q_{n} (α_{n + 1} p_{n} + p_{n - 1}) - p_{n} (α_{n + 1} q_{n} + q_{n - 1})}{q_{n} (α_{n + 1} q_{n} + q_{n - 1})} = \frac{(- 1)^{n}}{q_{n} (α_{n + 1} q_{n} + q_{n - 1})} .$ The denominator is positive by [F3] and [F4], so the sign is $(- 1)^{n}$ .

F1F2F3F4algebra

2.1

Fact [F4] gives $α_{n + 1} > a_{n + 1}$ , and [F3] gives. [step 1.1, F3, F4, algebra] $α_{n + 1} q_{n} + q_{n - 1} > a_{n + 1} q_{n} + q_{n - 1} = q_{n + 1} .$ Taking absolute values in step 1.1 yields $∣ α - \frac{p_{n}}{q_{n}} ∣ < \frac{1}{q_{n} q_{n + 1}} .$

step 1.1F3F4algebra

3.1

Facts [F3] and [F4] give $q_{n + 1} \geq q_{n} > 0$ , so the second inequality is immediate. [F3, F4, step 2.1, algebra] $\frac{1}{q_{n} q_{n + 1}} \leq \frac{1}{q_{n}^{2}}$

F3F4step 2.1algebra∎

Depends on

Used by

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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)