TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27

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Legendre's criterion for convergents

Statement

Let $α$ be an irrational real number. If $r, s \in Z$ satisfy $s > 0$ , $\gcd (r, s) = 1$ , and $∣ α - \frac{r}{s} ∣ < \frac{1}{2 s^{2}},$ then $r / s$ is a convergent of $α$ .

Facts & Assumptions

Given: An irrational real number $α$ with convergents $p_{n} / q_{n}$ , and a reduced rational number $r / s$ with $s > 0$ .

[F1]

The convergent denominators satisfy $q_{0} = 1$ , $q_{1} = a_{1} \geq 1$ , and $q_{n + 1} = a_{n + 1} q_{n} + q_{n - 1}$ with $a_{n + 1} \geq 1$ . Hence they are strictly increasing from $q_{1}$ onward. Moreover $q_{n + 2} \geq q_{n + 1} + q_{n} \geq q_{n} + 1$ , so they are unbounded (Convergents of a regular continued fraction).

[F2]

For $n \geq 1$ , the contrapositive of the best-approximation theorem says that if $s < q_{n + 1}$ , then $∣ s α - r ∣ \geq ∣ q_{n} α - p_{n} ∣ .$ (Convergents are best rational approximations of the first kind).

[F3]

For an irrational $α$ , the first complete quotient is defined and $α = a_{0} + \frac{1}{α_{1}}, α_{1} > a_{1} = q_{1} .$ (Complete quotients in the continued-fraction algorithm).

Proof

technique · direct

1.1F3givenalgebra

First suppose $s < q_{1}$ . By [F3]. [F3, given, algebra] $0 < α - a_{0} = \frac{1}{α_{1}} < \frac{1}{q_{1}} \leq \frac{1}{2} .$ If $r / s \neq a_{0}$ , then $∣ r - s a_{0} ∣ \geq 1$ . Since the integers satisfy $q_{1} \geq s + 1$ , the reverse triangle inequality gives $∣ α - \frac{r}{s} ∣ \geq ∣ \frac{r}{s} - a_{0} ∣ - ∣ α - a_{0} ∣ > \frac{1}{s} - \frac{1}{q_{1}} \geq \frac{1}{s} - \frac{1}{s + 1} = \frac{1}{s (s + 1)} \geq \frac{1}{2 s^{2}},$ contrary to the hypothesis. Hence $r / s = a_{0} = p_{0} / q_{0}$ is the zeroth convergent.

1.2F1given

It remains to suppose $q_{1} \leq s$ . Since the denominators are unbounded. [F1, given] and strictly increase from $q_{1}$ onward, [F1] gives an index $n \geq 1$ with $q_{n} \leq s < q_{n + 1} .$

2.1step 1.2F2givenalgebra

Assume $r / s \neq p_{n} / q_{n}$ . Since $s < q_{n + 1}$ , [F2] and the hypothesis give. [step 1.2, F2, given, algebra] $∣ q_{n} α - p_{n} ∣ \leq ∣ s α - r ∣ < \frac{1}{2 s} .$

3.1step 2.1givenalgebra∎

Since $r q_{n} - s p_{n}$ is a nonzero integer when $r / s \neq p_{n} / q_{n}$ , one has. [step 2.1, given, algebra] $1 \leq ∣ r q_{n} - s p_{n} ∣ = s q_{n} ∣ \frac{r}{s} - \frac{p_{n}}{q_{n}} ∣ .$ Using the triangle inequality and step 2.1, $∣ r q_{n} - s p_{n} ∣ \leq q_{n} ∣ r - s α ∣ + s ∣ q_{n} α - p_{n} ∣ \leq (q_{n} + s) ∣ s α - r ∣ < \frac{q_{n} + s}{2 s} \leq 1,$ a contradiction. Therefore $r / s = p_{n} / q_{n}$ , so $r / s$ is a convergent.

Depends on

Used by

A generalized Pell solution need not be a convergent Counterexample
The constant 1/2 in Legendre's criterion cannot be replaced by 3/4 Counterexample
The fractions 22/7, 333/106, and 355/113 as approximations to pi Example
Negative Pell is soluble exactly for odd period length Theorem

Dependency tree · two levels

17 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

Peter Hackman, Elementary Number Theory (standard reference, not scraped)
Bruce Ikenaga, Approximation by Rational Numbers (standard reference, not scraped)