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

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Convergents are best rational approximations of the first kind

Statement

Let $α$ be an irrational real number with convergents $p_{n} / q_{n}$ , and let $n \geq 1$ . If $r, s \in Z$ with $s > 0$ and $∣ s α - r ∣ < ∣ q_{n} α - p_{n} ∣,$ then $s \geq q_{n + 1} .$ Consequently, no rational number with denominator at most $q_{n}$ approximates $α$ more closely than $p_{n} / q_{n}$ .

Facts & Assumptions

Given: An irrational real number $α$ , an index $n \geq 1$ , its convergents $p_{n} / q_{n}$ , and integers $r, s$ with $s > 0$ .

[F1]

Since $n \geq 1$ , the corollary Convergents are reduced fractions applied at the index $n - 1$ shows that the vectors $(q_{n}, p_{n}), (q_{n - 1}, p_{n - 1})$ form a $Z$ -basis of $Z^{2}$ .

[F2]

The complete-quotient algorithm produces a unique integer $a_{n + 1}$ with $a_{n + 1} \leq α_{n + 1} < a_{n + 1} + 1,$ and every later complete quotient satisfies $α_{n + 1} > 1$ (Complete quotients in the continued-fraction algorithm).

[F3]

If $α_{n + 1}$ is the next complete quotient, then $α = \frac{α_{n + 1} p_{n} + p_{n - 1}}{α_{n + 1} q_{n} + q_{n - 1}} .$ (Complete-quotient tail formula).

[F4]

The convergent denominators satisfy $q_{- 1} = 0, q_{0} = 1, q_{n + 1} = a_{n + 1} q_{n} + q_{n - 1} .$ (Convergents of a regular continued fraction)

[F5]

The convergent errors satisfy $α - \frac{p_{n}}{q_{n}} = \frac{(- 1)^{n}}{q_{n} (α_{n + 1} q_{n} + q_{n - 1})}, ∣ α - \frac{p_{n}}{q_{n}} ∣ < \frac{1}{q_{n} q_{n + 1}} .$ (Convergent error bound).

Proof

technique · direct

1.1F3F5algebra

From [F3] and [F5] one obtains. [F3, F5, algebra] $q_{n} α - p_{n} = \frac{(- 1)^{n}}{α_{n + 1} q_{n} + q_{n - 1}}, q_{n - 1} α - p_{n - 1} = - α_{n + 1} (q_{n} α - p_{n}) .$ So the consecutive errors have opposite signs and satisfy $∣ q_{n - 1} α - p_{n - 1} ∣ = α_{n + 1} ∣ q_{n} α - p_{n} ∣ .$

2.1F1step 1.1algebra

By the basis statement in [F1], there are unique integers $u, v$ with. [F1, step 1.1, algebra] $(\begin{array}{c} s \\ r \end{array}) = u (\begin{array}{c} q_{n} \\ p_{n} \end{array}) + v (\begin{array}{c} q_{n - 1} \\ p_{n - 1} \end{array}) .$ Subtracting $r$ from $s α$ gives $s α - r = u (q_{n} α - p_{n}) + v (q_{n - 1} α - p_{n - 1}) = (u - v α_{n + 1}) (q_{n} α - p_{n})$ by step 1.1.

3.1step 2.1F2algebra

Assume $∣ s α - r ∣ < ∣ q_{n} α - p_{n} ∣$ . Step 2.1 gives. [step 2.1, F2, algebra] $∣ u - v α_{n + 1} ∣ < 1.$ If $v < 0$ , then $u < 0$ as well, because otherwise $u - v α_{n + 1} \geq 1 + α_{n + 1} > 1$ ; but then $s = u q_{n} + v q_{n - 1} < 0$ , impossible. If $v = 0$ , then $∣ u ∣ < 1$ , so $u = 0$ and again $s = 0$ , impossible. Therefore $v > 0$ . Since $u$ is an integer and [F2] gives $α_{n + 1} > a_{n + 1}$ , the inequality above implies $u > v α_{n + 1} - 1 > v a_{n + 1} - 1,$ hence $u \geq v a_{n + 1}$ .

4.1step 3.1F2F4algebra∎

Now. [step 3.1, F4, algebra] $s = u q_{n} + v q_{n - 1} \geq v (a_{n + 1} q_{n} + q_{n - 1}) = v q_{n + 1} \geq q_{n + 1},$ which is the first claim. Because $n \geq 1$ , fact [F2] gives $a_{n + 1} \geq 1$ , and [F4] then gives $q_{n + 1} = a_{n + 1} q_{n} + q_{n - 1} > q_{n} .$ For the consequence, suppose $s \leq q_{n}$ and $∣ α - \frac{r}{s} ∣ < ∣ α - \frac{p_{n}}{q_{n}} ∣ .$ Then $∣ s α - r ∣ = s ∣ α - \frac{r}{s} ∣ < s ∣ α - \frac{p_{n}}{q_{n}} ∣ \leq q_{n} ∣ α - \frac{p_{n}}{q_{n}} ∣ = ∣ q_{n} α - p_{n} ∣,$ contradicting the first claim because $s < q_{n + 1}$ . Thus no denominator at most $q_{n}$ gives a closer rational approximation.

Depends on

Used by

Dependency tree · two levels

18 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)