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✓ 14 results · all verified · 2 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 12 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Regular Continued Fractions and Diophantine Approximation

1 · Prerequisites

2 · Summary

Regular continued fractions turn Euclidean division into an approximation machine. This page fixes the low-anchor route from the design block: it proves the integer-part step directly from Archimedeanness and well-ordering, derives convergence from completeness rather than from later sequence pages, and then uses the determinant identity and complete-quotient formula to reach best approximation and Legendre's criterion.

The last third of the page turns to quadratic irrationals. It isolates the two load-bearing ideas separately: an eventually periodic tail satisfies a quadratic equation, while the continued-fraction algorithm on a quadratic irrational runs inside a finite integral state space. Lagrange's theorem is then exactly the equivalence between those two descriptions.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-27Open item page →

Finite and infinite regular continued fractions

Definition

A finite regular continued fraction is an expression $[a_{0}; a_{1}, \dots, a_{n}]$ with $a_{0} \in Z$ and $a_{i} \in Z_{> 0}$ for $i \geq 1$ . Its value is defined recursively in $Q$ (The rationals as equivalence classes of pairs of integers, Arithmetic on the rationals) by $[a_{n}] : = a_{n}, [a_{0}; a_{1}, \dots, a_{n}] : = a_{0} + \frac{1}{[a_{1}; \dots, a_{n}]} (n \geq 1) .$

This recursion is well defined. Indeed, starting from the last digit and working backwards, every tail $[a_{i}; \dots, a_{n}]$ with $i \geq 1$ is a positive rational: the last tail is $a_{n} > 0$ , and $a_{i} + 1 / t > 0$ whenever $a_{i} > 0$ and $t > 0$ . In particular every denominator occurring in the recursion is nonzero.

An infinite regular continued fraction is a digit sequence $(a_{n})_{n \geq 0}$ with $a_{0} \in Z$ and $a_{n} \in Z_{> 0}$ for $n \geq 1$ , written $[a_{0}; a_{1}, a_{2}, \dots] .$ Its value is not assumed by the notation; existence and uniqueness of the value are proved in Every infinite regular continued fraction converges to a unique real number.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-27Open item page →

Convergents of a regular continued fraction

Definition

Let $a_{0}, a_{1}, \dots$ be the digit sequence of a regular continued fraction (Finite and infinite regular continued fractions), terminating at $a_{N}$ in the finite case. Define integers $p_{n}, q_{n}$ by $p_{- 2} = 0, p_{- 1} = 1, q_{- 2} = 1, q_{- 1} = 0,$ and, for every index $n \geq 0$ for which the digit $a_{n}$ exists, $p_{n} = a_{n} p_{n - 1} + p_{n - 2}, q_{n} = a_{n} q_{n - 1} + q_{n - 2} .$

The denominators are positive at every digit index: $q_{0} = 1$ , and if the digit $a_{1}$ exists then $q_{1} = a_{1} > 0$ ; thereafter $q_{n} = a_{n} q_{n - 1} + q_{n - 2} > 0$ by induction because $a_{n} > 0$ and the preceding denominators are nonnegative. Thus the quotient below is defined in $Q$ .

The rational number $p_{n} / q_{n}$ is the $n$ -th convergent. The initial labels $- 2$ and $- 1$ belong only to this recurrence convention; they do not extend the digit sequence itself to negative indices.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27Open item page →

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

1.1givenF1basealgebra

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].

2.1step 1.1F1inductionalgebra

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].

3.1F2step 1.1step 2.1discharge-induction

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 $t > 0$ .

4.1step 3.1F1algebra∎

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.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27Open item page →

Determinant identity for consecutive convergents

Statement

Let $p_{n} / q_{n}$ be the convergents of a regular continued fraction. Then for every $n \geq 0$ , $p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1} .$ Consequently, for every $n \geq 1$ , $\frac{p_{n}}{q_{n}} - \frac{p_{n - 1}}{q_{n - 1}} = \frac{(- 1)^{n - 1}}{q_{n} q_{n - 1}} .$

Facts & Assumptions

Given: A regular continued fraction with convergent sequences $p_{n}, q_{n}$ .

[F1]

The convergents satisfy $p_{- 2} = 0$ , $p_{- 1} = 1$ , $q_{- 2} = 1$ , $q_{- 1} = 0$ , and $p_{n} = a_{n} p_{n - 1} + p_{n - 2}$ , $q_{n} = a_{n} q_{n - 1} + q_{n - 2}$ for $n \geq 0$ . (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

1.1givenF1basealgebra

At $n = 0$ one has. [given, F1, base, algebra] $p_{0} q_{- 1} - p_{- 1} q_{0} = a_{0} \cdot 0 - 1 \cdot 1 = - 1 = (- 1)^{- 1} .$

1.2F1inductionalgebra

If $D_{n} : = p_{n} q_{n - 1} - p_{n - 1} q_{n}$ , then the recurrences of [F1] give. [F1, induction, algebra] $D_{n + 1} = (a_{n + 1} p_{n} + p_{n - 1}) q_{n} - p_{n} (a_{n + 1} q_{n} + q_{n - 1}) = - D_{n} .$ So the sign flips at each successor step.

2.1F2step 1.1step 1.2discharge-induction

Steps 1.1 and 1.2 imply by induction that. [F2, step 1.1, step 1.2, discharge-induction] $D_{n} = (- 1)^{n - 1}$ for every $n \geq 0$ .

3.1step 2.1algebra∎

For $n \geq 1$ . [step 2.1, algebra] $\frac{p_{n}}{q_{n}} - \frac{p_{n - 1}}{q_{n - 1}} = \frac{p_{n} q_{n - 1} - p_{n - 1} q_{n}}{q_{n} q_{n - 1}} = \frac{(- 1)^{n - 1}}{q_{n} q_{n - 1}}$ by step 2.1.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27Open item page →

Convergents are reduced fractions

Statement

Every convergent $p_{n} / q_{n}$ of a regular continued fraction is in lowest terms. Moreover, for each $n \geq 0$ the two vectors $(q_{n}, p_{n}), (q_{n + 1}, p_{n + 1})$ form a $Z$ -basis of $Z^{2}$ .

Facts & Assumptions

Given: A regular continued fraction and its convergents $p_{n} / q_{n}$ .

[F1]

Consecutive convergents satisfy $p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1}$ for $n \geq 0$ . (Determinant identity for consecutive convergents).

[F2]

If integers $a, b$ are not both $0$ , then $\gcd (a, b)$ is an integer linear combination of $a$ and $b$ (Bézout's identity: for integers $a, b$ not both zero, $\gcd (a, b)$ is the least positive element of ${a x + b y : x, y \in Z}$ ; in particular $a x + b y = \gcd (a, b)$ has an integer solution).

Proof

technique · direct

1.1F1F2given

Let $d$ be a common divisor of $p_{n}$ and $q_{n}$ . [F1, F2, given] Then $d$ divides every integer linear combination of $p_{n}$ and $q_{n}$ , in particular $p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1}$ by [F1]. Hence $d$ divides $1$ , so $d = 1$ and $p_{n} / q_{n}$ is reduced.

1.2F1algebra

The determinant of the matrix with columns $(q_{n}, p_{n})$ and $(q_{n + 1}, p_{n + 1})$ is. [F1, algebra] $q_{n} p_{n + 1} - p_{n} q_{n + 1} = (- 1)^{n}$ by [F1]. Therefore $(\begin{array}{c} u \\ v \end{array}) = (- 1)^{n} (u p_{n + 1} - v q_{n + 1}) (\begin{array}{c} q_{n} \\ p_{n} \end{array}) + (- 1)^{n} (v q_{n} - u p_{n}) (\begin{array}{c} q_{n + 1} \\ p_{n + 1} \end{array})$ for every $(u, v) \in Z^{2}$ , so the two columns span $Z^{2}$ over $Z$ .

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 are exactly the two assertions.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-27Open item page →

Complete quotients in the continued-fraction algorithm

Definition

Work in the complete ordered field of real numbers (Complete ordered field (least-upper-bound property)). For a real number $α$ , set $α_{0} : = α .$ Given $α_{n}$ , the Archimedean property (Every complete ordered field is Archimedean) and the well-ordering principle (The well-ordering principle) produce a unique integer $a_{n}$ with $a_{n} \leq α_{n} < a_{n} + 1.$

For completeness, here is that construction. Choose positive integers $r, s$ with $- r < α_{n} < s$ by Archimedeanness, and let $T : = {k \in N : α_{n} < - r + k} .$ The set $T$ is nonempty because $r + s \in T$ , so it has a least element $k_{0}$ . Since $- r < α_{n}$ , one has $k_{0} > 0$ ; write $k_{0} = j + 1$ . Minimality gives $- r + j \leq α_{n} < - r + j + 1$ , so $a_{n} : = - r + j$ works. If two integers satisfied the displayed inequalities, discreteness of the integer order would put one at least $1$ above the other and contradict the upper inequality, proving uniqueness. If $α_{n} \neq a_{n}$ , define the next complete quotient $α_{n + 1} : = \frac{1}{α_{n} - a_{n}} .$

Thus the continued-fraction algorithm associates to $α$ its integer parts $a_{n}$ and its successive complete quotients $α_{n}$ . Whenever $α_{n + 1}$ is defined, one has $0 < α_{n} - a_{n} < 1$ and therefore $α_{n + 1} > 1$ , so every later digit $a_{n + 1}, a_{n + 2}, \dots$ is positive.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27Open item page →

Complete-quotient tail formula

Statement

Let $α$ be a real number, let $a_{n}$ and $α_{n}$ be its continued-fraction digits and complete quotients, and let $p_{n} / q_{n}$ be the convergents attached to $a_{0}, a_{1}, \dots$ . Whenever $α_{n + 1}$ is defined, $α = \frac{α_{n + 1} p_{n} + p_{n - 1}}{α_{n + 1} q_{n} + q_{n - 1}} .$

Facts & Assumptions

Given: A real number $α$ , its complete quotients $α_{n}$ , and its continued-fraction digits $a_{n}$ .

[F1]

The complete quotients satisfy $α_{n} = a_{n} + 1 / α_{n + 1}$ whenever $α_{n + 1}$ is defined. (Complete quotients in the continued-fraction algorithm).

[F2]

For every $t > 0$ , $[a_{0}; a_{1}, \dots, a_{n}, t] = \frac{t p_{n} + p_{n - 1}}{t q_{n} + q_{n - 1}} .$ (Convergents are given by the standard recurrences and tail formula).

Proof

technique · direct

1.1givenF1algebra

Repeatedly substituting the identities of [F1] yields. [given, F1, algebra] $α = [a_{0}; a_{1}, \dots, a_{n}, α_{n + 1}]$ whenever $α_{n + 1}$ is defined.

2.1step 1.1F1F2∎

Since every complete quotient after the first is $> 1$ , in particular $α_{n + 1} > 0$ . [step 1.1, F1, F2] So step 1.1 and [F2] give $α = \frac{α_{n + 1} p_{n} + p_{n - 1}}{α_{n + 1} q_{n} + q_{n - 1}} .$

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27Open item page →

Every infinite regular continued fraction converges to a unique real number

Statement

Let $[a_{0}; a_{1}, a_{2}, \dots]$ be an infinite regular continued fraction, and let $c_{n} : = p_{n} / q_{n}$ be its convergents. Then:

$c_{0} < c_{2} < c_{4} < \dots$ and $c_{1} > c_{3} > c_{5} > \dots$ ;
every even convergent is below every odd convergent; and
there is a unique real number $x$ such that both subsequences $(c_{2 m})_{m \geq 0}$ and $(c_{2 m + 1})_{m \geq 0}$ converge to $x$ .

This real number is the value of the infinite regular continued fraction.

Facts & Assumptions

Given: An infinite regular continued fraction and its convergents $c_{n} = p_{n} / q_{n}$ .

[F1]

Consecutive convergents satisfy $c_{n} - c_{n - 1} = \frac{(- 1)^{n - 1}}{q_{n} q_{n - 1}}$ for $n \geq 1$ (Determinant identity for consecutive convergents).

[F2]

Every complete ordered field is Archimedean (Every complete ordered field is Archimedean).

[F3]

The constructed real field has the least-upper-bound property (The Cauchy-sequence reals have the least-upper-bound property, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1givenF1inductionalgebra

Since every partial quotient after the first is at least $1$ . [given, F1, induction, algebra] The denominators satisfy $q_{n + 1} = a_{n + 1} q_{n} + q_{n - 1} \geq q_{n} + q_{n - 1} > q_{n} > 0$ for $n \geq 1$ , with $q_{0} = 1$ and $q_{1} = a_{1} \geq 1$ . Thus $(q_{n})$ is nondecreasing from $q_{0}$ onward and strictly increasing from $q_{1}$ onward, and induction on $n$ gives $q_{n} \geq n$ for $n \geq 1$ .

2.1F1step 1.1algebra

By [F1], the signs of $c_{n} - c_{n - 1}$ alternate, and by step 1.1 their absolute values strictly decrease. [F1, step 1.1, algebra] Hence $c_{2 m} < c_{2 m + 1}, c_{2 m + 2} = c_{2 m + 1} - (c_{2 m + 1} - c_{2 m + 2}) > c_{2 m},$ and similarly $c_{2 m + 3} = c_{2 m + 2} - (c_{2 m + 2} - c_{2 m + 3}) < c_{2 m + 1} .$ So the even convergents increase, the odd convergents decrease, and every even convergent is below every odd convergent.

2.2F2step 1.1algebra

Put $d_{n} : = 1 / (q_{n} q_{n + 1})$ . [F2, step 1.1, algebra] Step 1.1 gives $d_{n} \leq 1 / (n (n + 1))$ for $n \geq 1$ , and the Archimedean property [F2] therefore implies $d_{n} \to 0$ . For $ε > 0$ choose $N \geq 1$ with $1 / N < ε$ , then for $n \geq N$ , $0 < d_{n} \leq \frac{1}{n (n + 1)} \leq \frac{1}{n} \leq \frac{1}{N} < ε .$

3.1step 2.1given

Let $E : = {c_{2 m} : m \geq 0}$ . [step 2.1, given] By step 2.1 the set $E$ is nonempty and bounded above by $c_{1}$ , so [F3] gives a real number $x : = \sup E$ .

3.2step 2.1step 2.2F1

If $n = 2 m + 1$ is odd, then $c_{n - 1} \in E$ and step 2.1 gives. [step 2.1, step 2.2, F1] $c_{n - 1} \leq x \leq c_{n}$ , so $0 \leq c_{n} - x \leq c_{n} - c_{n - 1} = d_{n - 1} .$ If $n = 2 m$ is even, then $c_{n} \in E$ and step 2.1 gives $c_{n} \leq x \leq c_{n + 1}$ , so $0 \leq x - c_{n} \leq c_{n + 1} - c_{n} = d_{n} .$ Since the right-hand sides tend to $0$ by step 2.2, both subsequences converge to $x$ .

4.1step 3.2algebra∎

If $y$ were another real with both subsequences converging to $y$ , then for every $m$ . [step 3.2, algebra] $∣ x - y ∣ \leq ∣ x - c_{2 m} ∣ + ∣ c_{2 m} - y ∣,$ and the right-hand side tends to $0$ as $m \to \infty$ by step 3.2. Hence $x = y$ , so the value is unique.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27Open item page →

The continued-fraction algorithm for real numbers

Statement

Let $α$ be a real number. The complete-quotient algorithm of Complete quotients in the continued-fraction algorithm either terminates with a finite regular continued fraction equal to $α$ , or produces an infinite regular continued fraction whose convergents converge to $α$ .

Facts & Assumptions

Given: A real number $α$ , its complete quotients $α_{n}$ , its digits $a_{n}$ , and its convergents $p_{n} / q_{n}$ .

[F1]

Whenever $α_{n + 1}$ is defined, $α = \frac{α_{n + 1} p_{n} + p_{n - 1}}{α_{n + 1} q_{n} + q_{n - 1}} .$ (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]

Every infinite regular continued fraction has a unique value, namely the common limit of its even and odd convergent subsequences. (Every infinite regular continued fraction converges to a unique real number).

Proof

technique · direct

1.1givenalgebra

If $α_{N} = a_{N}$ for some $N$ , then repeated substitution of the identities $α_{n} = a_{n} + 1 / α_{n + 1}$ for $n < N$ yields. [given, algebra] $α = [a_{0}; a_{1}, \dots, a_{N}] .$ For $N \geq 1$ the last digit $a_{N} = α_{N}$ is positive because every complete quotient after the first is greater than $1$ , so the output is a finite regular continued fraction.

1.2F1F2algebra

Suppose the algorithm never terminates. Then every $α_{n + 1}$ is defined and satisfies $α_{n + 1} > a_{n + 1} \geq 1$ , so [F1] and [F2] give. [F1, F2, algebra] $α - \frac{p_{n}}{q_{n}} = \frac{p_{n - 1} q_{n} - p_{n} q_{n - 1}}{q_{n} (α_{n + 1} q_{n} + q_{n - 1})} = \frac{(- 1)^{n}}{q_{n} (α_{n + 1} q_{n} + q_{n - 1})} .$ Hence $∣ α - \frac{p_{n}}{q_{n}} ∣ < \frac{1}{q_{n} (a_{n + 1} q_{n} + q_{n - 1})} = \frac{1}{q_{n} q_{n + 1}} .$

2.1F3step 1.2algebra

The infinite digit sequence $a_{0}, a_{1}, \dots$ is therefore a regular continued fraction, so by [F3] its convergents $p_{n} / q_{n}$ converge to some real number $x$ . [F3, step 1.2, algebra] Step 1.2 shows $α - p_{n} / q_{n} \to 0$ , and therefore $∣ α - x ∣ \leq ∣ α - \frac{p_{n}}{q_{n}} ∣ + ∣ \frac{p_{n}}{q_{n}} - x ∣ \to 0.$ Thus $α = x$ .

3.1step 1.1step 2.1∎

Step 1.1 handles the terminating case and step 2.1 the nonterminating case, so the algorithm always reconstructs the original real number.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27Open item page →

The continued-fraction algorithm terminates exactly on rational numbers

Statement

For a real number $α$ , the continued-fraction algorithm terminates if and only if $α$ is rational. When $α$ is rational, the continued-fraction digits are exactly the successive quotient digits of the Euclidean algorithm.

Facts & Assumptions

Given: A real number $α$ and its continued-fraction algorithm.

[F1]

If $u, v \in Z$ with $v > 0$ , then there are unique integers $a, r$ with $u = a v + r, 0 \leq r < v .$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[F2]

A terminating continued-fraction expansion is a finite regular continued fraction, and every finite regular continued fraction equals its last convergent $p_{n} / q_{n} \in Q$ (The continued-fraction algorithm for real numbers, Convergents are given by the standard recurrences and tail formula).

[F3]

Every complete quotient after the first is greater than $1$ (Complete quotients in the continued-fraction algorithm).

Proof

technique · direct

1.1givenF1algebra

If $α$ is an integer, the algorithm stops immediately. Otherwise write $α = u_{0} / v_{0}$ with integers $v_{0} > 0$ and $\gcd (u_{0}, v_{0}) = 1$ . Applying [F1] to. [given, F1, algebra] $u_{0}$ and $v_{0}$ gives $u_{0} = a_{0} v_{0} + r_{0}, 0 < r_{0} < v_{0},$ so $α = a_{0} + \frac{r_{0}}{v_{0}}, α_{1} = \frac{1}{α - a_{0}} = \frac{v_{0}}{r_{0}} .$ Thus the next numerator is the previous denominator.

1.2F2

Conversely, if the algorithm terminates, then by [F2] the original number is a finite regular continued fraction and hence rational.

1.3F1F3inductionalgebra

Let $n \geq 1$ , and suppose. [F1, F3, induction, algebra] $α_{n} = \frac{u_{n}}{v_{n}} (v_{n} > 0, \gcd (u_{n}, v_{n}) = 1)$ is not an integer. Since $n \geq 1$ , fact [F3] gives $α_{n} > 1$ , so $u_{n} > v_{n} > 0$ . Applying [F1] gives $u_{n} = a_{n} v_{n} + r_{n}, 0 < r_{n} < v_{n}, α_{n + 1} = \frac{v_{n}}{r_{n}} .$ Any common divisor of $v_{n}$ and $r_{n} = u_{n} - a_{n} v_{n}$ also divides $u_{n}$ , so $\gcd (v_{n}, r_{n}) = 1$ . Hence this is already the reduced form $α_{n + 1} = \frac{u_{n + 1}}{v_{n + 1}} = \frac{v_{n}}{r_{n}},$ and therefore $u_{n + 1} = v_{n} < u_{n}$ .

2.1step 1.1step 1.3induction

When $α$ is a nonintegral rational, step 1.1 gives $α_{1} = \frac{v_{0}}{r_{0}} > 1.$ So step 1.3 applies successively to $α_{1}, α_{2}, \dots$ as long as they remain nonintegral. Their positive integer numerators $u_{1}, u_{2}, u_{3}, \dots$ then form a strictly decreasing sequence, which cannot continue forever. Therefore some complete quotient is an integer, and the algorithm terminates.

3.1step 1.1step 1.3F1∎

The equations in steps 1.1 and 1.3 are exactly the Euclidean divisions of the successive numerator-denominator pairs, so the continued-fraction digits are. [step 1.1, step 1.3, F1] the Euclidean quotient digits.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27Open item page →

Normalized finite regular continued fractions are unique

Statement

Every rational number has a unique normalized finite regular continued fraction: either the expansion has length $0$ , or its last digit is at least $2$ . If the rational is not an integer, then it has exactly one other finite regular continued-fraction expansion, obtained by replacing the last digit $a_{n} \geq 2$ by the pair $a_{n} - 1, 1$ . An integer $m$ has exactly the two expansions $[m]$ and $[m - 1; 1]$ .

Facts & Assumptions

Given: A rational number and its finite regular continued-fraction expansions.

[F1]

The continued-fraction algorithm terminates exactly on rational numbers, and its digit list is the Euclidean-algorithm digit list. (The continued-fraction algorithm terminates exactly on rational numbers).

[F2]

For every finite regular continued fraction, $[a_{0}; a_{1}, \dots, a_{n}, t] = \frac{t p_{n} + p_{n - 1}}{t q_{n} + q_{n - 1}},$ so in particular $[a_{0}; a_{1}, \dots, a_{n}, 1] = [a_{0}; a_{1}, \dots, a_{n} + 1] .$ (Convergents are given by the standard recurrences and tail formula).

Proof

technique · direct

1.1F1given

By [F1], every rational number has a terminating continued-fraction expansion. If the rational is not an integer, the last complete quotient is a positive integer greater than $1$ . [F1, given] Because every complete quotient after the first is greater than $1$ , the algorithm already produces a normalized finite expansion.

1.2F2algebra

If $a_{n} \geq 2$ , then [F2] gives. [F2, algebra] $[a_{0}; a_{1}, \dots, a_{n}] = [a_{0}; a_{1}, \dots, a_{n} - 1, 1] .$ For an integer $m$ , the same identity reads $[m] = [m - 1, 1] = [m - 1; 1] .$ So every normalized finite expansion produces a second finite expansion.

1.3F2induction

Conversely, any finite expansion with last digit $1$ and length at least $1$ can be shortened by the identity. [F2, induction] $[a_{0}; a_{1}, \dots, a_{n - 1}, 1] = [a_{0}; a_{1}, \dots, a_{n - 1} + 1] .$ Repeating this collapse removes every terminal $1$ and ends at a normalized expansion. Thus every finite expansion is obtained from a normalized one by at most one final split of the last digit.

2.1F1step 1.2step 1.3∎

The normalized expansion is unique because the continued-fraction algorithm on a rational has unique digits at each step. [F1, step 1.2, step 1.3] Each digit is the unique integer part of the current complete quotient, and [F1] says the process terminates. Therefore nonintegers have exactly two finite expansions, while integers have exactly the two listed in step 1.2.

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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

1.1F1F2F3F4algebra

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 $(- 1)^{n}$ .

2.1step 1.1F3F4algebra

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 $∣ α - \frac{p_{n}}{q_{n}} ∣ < \frac{1}{q_{n} q_{n + 1}} .$

3.1F3F4step 2.1algebra∎

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

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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.

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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.

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Quadratic irrationals

Definition

A real number $α$ (The real numbers) is a quadratic irrational if it is irrational and satisfies a quadratic equation $A α^{2} + B α + C = 0$ with $A, B, C \in Q$ (The rationals as equivalence classes of pairs of integers) and $A \neq 0$ .

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Eventually periodic regular continued fractions

Definition

An infinite regular continued fraction $[a_{0}; a_{1}, a_{2}, \dots]$ (Finite and infinite regular continued fractions) is eventually periodic if there are integers $N \geq 0$ and $h \geq 1$ such that $a_{n + h} = a_{n} for every n \geq N .$ The smallest such $h$ is the period of the eventual tail. It exists because the admissible positive integers $h$ form a nonempty subset of $N$ , to which The well-ordering principle applies.

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Eventually periodic regular continued fractions are quadratic irrationals

Statement

The value of every eventually periodic regular continued fraction is a quadratic irrational.

Facts & Assumptions

Given: An eventually periodic regular continued fraction $α = [a_{0}; a_{1}, a_{2}, \dots] .$

[F1]

There are integers $N \geq 0$ and $h \geq 1$ such that $a_{n + h} = a_{n}$ for every $n \geq N$ (Eventually periodic regular continued fractions).

[F2]

If $p_{n} / q_{n}$ are the convergents of a finite prefix, then for every $t > 0$ , $[a_{0}; a_{1}, \dots, a_{n}, t] = \frac{t p_{n} + p_{n - 1}}{t q_{n} + q_{n - 1}} .$ (Convergents are given by the standard recurrences and tail formula).

[F3]

For a finite regular continued fraction whose digits are all positive, the convergents satisfy $q_{0} = 1$ , $p_{0} = a_{0} > 0$ , and, when $n \geq 1$ , $p_{n}, q_{n} > 0$ (Convergents of a regular continued fraction).

[F4]

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

[F5]

The finite convergents of an infinite regular continued fraction converge to its value (Every infinite regular continued fraction converges to a unique real number).

Proof

technique · direct

1.1F1F2F3F5algebra

By increasing $N$ if necessary, we may still assume. [F1, F2, F3, F5, algebra] $a_{n + h} = a_{n} (n \geq N)$ with $N \geq 1$ . Let $x = [a_{N}; a_{N + 1}, \dots, a_{N + h - 1}, a_{N}, a_{N + 1}, \dots]$ be the purely periodic tail. For $r \geq 1$ , let $x_{r}$ be the finite continued fraction formed from $r$ copies of this period. By [F5], $x_{r} \to x$ , and [F2] gives $x_{r + 1} = \frac{x_{r} p_{h - 1} + p_{h - 2}}{x_{r} q_{h - 1} + q_{h - 2}},$ where the convergents are taken for the digit block $a_{N}, \dots, a_{N + h - 1}$ . The denominators are positive by [F3], so a direct difference calculation shows that the displayed fractional-linear expression preserves the limit $x_{r} \to x$ . Hence $x = \frac{x p_{h - 1} + p_{h - 2}}{x q_{h - 1} + q_{h - 2}} .$ Clearing denominators yields the quadratic equation $q_{h - 1} x^{2} + (q_{h - 2} - p_{h - 1}) x - p_{h - 2} = 0.$

2.1step 1.1F3F4algebra

The discriminant of the quadratic in step 1.1 is. [step 1.1, F3, F4, algebra] $Δ = (q_{h - 2} - p_{h - 1})^{2} + 4 q_{h - 1} p_{h - 2} = (p_{h - 1} + q_{h - 2})^{2} - 4 (p_{h - 1} q_{h - 2} - p_{h - 2} q_{h - 1}) .$ Putting $m : = p_{h - 1} + q_{h - 2} > 0$ , fact [F4] gives $Δ = m^{2} - 4 (- 1)^{h - 2} .$ If $h$ is odd, then $Δ = m^{2} + 4$ , which cannot be a square because $(n - m) (n + m) = 4$ has no solution in integers with $m > 0$ . If $h$ is even, then $h \geq 2$ , fact [F3] gives $p_{h - 1} \geq p_{1} \geq 2$ and $q_{h - 2} \geq q_{0} = 1$ , so $m \geq 3$ ; then $Δ = m^{2} - 4$ , which cannot be a square because $(m - n) (m + n) = 4$ has no solution with $m \geq 3$ . Therefore $Δ$ is not a square, so the root $x$ from step 1.1 is irrational.

2.2F1F2F5step 1.1algebra

Append the finite tails $x_{r}$ of step 1.1 after the prefix. [F1, F2, F5, step 1.1, algebra] $a_{0}, \dots, a_{N - 1}$ . These are a subsequence of the convergents of $α$ , so [F5] makes their values tend to $α$ . By [F2] their values are $\frac{x_{r} p_{N - 1} + p_{N - 2}}{x_{r} q_{N - 1} + q_{N - 2}} .$ The same positive-denominator difference calculation used in step 1.1 lets $r \to \infty$ and gives $α = \frac{x p_{N - 1} + p_{N - 2}}{x q_{N - 1} + q_{N - 2}} .$ Clearing denominators and substituting the quadratic equation from step 1.1 shows that $α$ also satisfies a quadratic equation over $Q$ .

3.1step 2.1step 2.2F4algebra

Suppose that $α$ were rational. Step 2.2 gives. [step 2.1, step 2.2, F4, algebra] $x (α q_{N - 1} - p_{N - 1}) = p_{N - 2} - α q_{N - 2} .$ If $α q_{N - 1} - p_{N - 1} = 0$ , then also $p_{N - 2} - α q_{N - 2} = 0$ , and eliminating $α$ yields $p_{N - 1} q_{N - 2} - p_{N - 2} q_{N - 1} = 0,$ contrary to [F4]. Therefore $x = \frac{p_{N - 2} - α q_{N - 2}}{α q_{N - 1} - p_{N - 1}}$ is rational, contradicting step 2.1. Hence $α$ is irrational.

4.1step 2.2step 3.1∎

Steps 2.2 and 3.1 show that $α$ is an irrational real root of a quadratic equation over $Q$ , which is exactly the definition of a quadratic irrational (Quadratic irrationals).

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Complete quotients of a quadratic irrational lie in a finite state space

Statement

Let $α$ be a quadratic irrational, and let $α_{n}$ be the complete quotients produced by the continued-fraction algorithm. Then only finitely many distinct complete quotients occur.

More precisely, if $a α^{2} + b α + c = 0$ with $a, b, c \in Z$ and discriminant $D : = b^{2} - 4 a c > 0$ , then for every $n \geq 1$ the complete quotient $α_{n}$ is a root of an integer quadratic equation $A_{n} x^{2} + B_{n} x + C_{n} = 0$ whose discriminant is also $D$ .

Facts & Assumptions

Given: A quadratic irrational $α$ and its complete quotients $α_{n}$ .

[F1]

A quadratic irrational is an irrational real root of some quadratic equation $A α^{2} + B α + C = 0$ with $A, B, C \in Q$ and $A \neq 0$ . (Quadratic irrationals).

[F2]

An irrational real number never terminates under the continued-fraction algorithm, so every complete quotient $α_{n}$ is defined (The continued-fraction algorithm terminates exactly on rational numbers).

[F3]

For every $n \geq 1$ one has $α = \frac{α_{n} p_{n - 1} + p_{n - 2}}{α_{n} q_{n - 1} + q_{n - 2}} .$ (Complete-quotient tail formula).

[F4]

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

[F5]

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

[F6]

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 $a_{n + 1} \geq 1$ , so $q_{n} \leq q_{n + 1}$ for every $n \geq 0$ (Convergents of a regular continued fraction).

Proof

technique · direct

1.1F1givenalgebra

By [F1], after multiplying by a common denominator we may choose integers. [F1, given, algebra] $a, b, c$ with $a \neq 0$ and $a α^{2} + b α + c = 0.$ Its discriminant $D : = b^{2} - 4 a c$ is positive because $α$ is real and nonsquare because $α$ is irrational.

2.1F2F3algebra

Because $α$ is irrational, [F2] says the continued-fraction algorithm. [F2, F3, algebra] never terminates, so every $α_{n}$ is defined. For each $n \geq 1$ , substituting the expression from [F3] into the quadratic equation from step 1.1 and clearing denominators yields $A_{n} α_{n}^{2} + B_{n} α_{n} + C_{n} = 0,$ where $A_{n} : = a p_{n - 1}^{2} + b p_{n - 1} q_{n - 1} + c q_{n - 1}^{2},$ $B_{n} : = 2 a p_{n - 1} p_{n - 2} + b (p_{n - 1} q_{n - 2} + p_{n - 2} q_{n - 1}) + 2 c q_{n - 1} q_{n - 2},$ $C_{n} : = a p_{n - 2}^{2} + b p_{n - 2} q_{n - 2} + c q_{n - 2}^{2} .$ These coefficients are integers.

3.1step 2.1F4algebra

The discriminant of that quadratic is. [step 2.1, F4, algebra] $B_{n}^{2} - 4 A_{n} C_{n} = (b^{2} - 4 a c) (p_{n - 1} q_{n - 2} - p_{n - 2} q_{n - 1})^{2} = D,$ because [F4] gives $(p_{n - 1} q_{n - 2} - p_{n - 2} q_{n - 1})^{2} = 1.$

3.2F5F6step 1.1algebra

Put. [F5, F6, step 1.1, algebra] $δ_{n} : = α q_{n - 1} - p_{n - 1} = q_{n - 1} (α - \frac{p_{n - 1}}{q_{n - 1}}) .$ Facts [F5] and [F6] give $∣ δ_{n} ∣ < \frac{1}{q_{n}} \leq 1, ∣ q_{n - 1} δ_{n} ∣ < \frac{q_{n - 1}}{q_{n}} \leq 1.$ Since $p_{n - 1} = α q_{n - 1} - δ_{n}$ , the coefficient $A_{n}$ from step 2.1 becomes $A_{n} = a (α q_{n - 1} - δ_{n})^{2} + b (α q_{n - 1} - δ_{n}) q_{n - 1} + c q_{n - 1}^{2} = - (2 a α + b) q_{n - 1} δ_{n} + a δ_{n}^{2},$ because $a α^{2} + b α + c = 0$ . Therefore $∣ A_{n} ∣ \leq ∣ 2 a α + b ∣ + ∣ a ∣,$ so only finitely many integers can occur as $A_{n}$ .

4.1step 2.1step 3.2step 3.1algebra∎

For $n \geq 2$ , the formulas in step 2.1 give $C_{n} = A_{n - 1}$ , so the integers. [step 2.1, step 3.2, step 3.1, algebra] $C_{n}$ also range over a finite set. Then step 3.1 yields $B_{n}^{2} = D + 4 A_{n} C_{n},$ so only finitely many integers $B_{n}$ can occur as well. Thus only finitely many triples $(A_{n}, B_{n}, C_{n})$ arise, and each $α_{n}$ with $n \geq 1$ is one of the at most two roots of one of those finitely many quadratics. Together with the single value $α_{0} = α$ , this proves that only finitely many complete quotients occur.

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Lagrange's theorem for regular continued fractions

Statement

A real number has an eventually periodic regular continued fraction if and only if it is a quadratic irrational.

Facts & Assumptions

Given: A real number $α$ and its continued-fraction algorithm.

[F1]

Every eventually periodic regular continued fraction has quadratic- irrational value (Eventually periodic regular continued fractions are quadratic irrationals).

[F2]

If $α$ is quadratic irrational, then only finitely many complete quotients occur in its continued-fraction algorithm (Complete quotients of a quadratic irrational lie in a finite state space).

[F3]

The continued-fraction algorithm is deterministic: each digit $a_{n}$ is the unique integer with $a_{n} \leq α_{n} < a_{n} + 1$ , and whenever $α_{n} \neq a_{n}$ the next complete quotient is $α_{n + 1} = 1 / (α_{n} - a_{n})$ (Complete quotients in the continued-fraction algorithm).

Proof

technique · direct

1.1F1given

If the continued fraction of $α$ is eventually periodic, then [F1] shows that $α$ is a quadratic irrational.

1.2F2given

Suppose now that $α$ is a quadratic irrational. By [F2], only finitely. [F2, given] many complete quotients $α_{n}$ occur, so there exist indices $m < n$ with $α_{m} = α_{n} .$

2.1step 1.2F3induction

From $α_{m} = α_{n}$ and the determinism in [F3], the next digits agree: $a_{m} = a_{n}$ , and then the next complete quotients agree: $α_{m + 1} = α_{n + 1}$ . Repeating this argument inductively gives $a_{m + j} = a_{n + j} for every j \geq 0,$ so the continued-fraction digits repeat with period $n - m$ from the index $m$ onward. Thus the continued fraction is eventually periodic.

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove both directions of the equivalence.

5 · Examples, counterexamples and false statements

None yet.

Regular Continued Fractions and Diophantine Approximation

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Finite and infinite regular continued fractions

Definition

Convergents of a regular continued fraction

Definition

Convergents are given by the standard recurrences and tail formula

Statement

Facts & Assumptions

Proof

Determinant identity for consecutive convergents

Statement

Facts & Assumptions

Proof

Convergents are reduced fractions

Statement

Facts & Assumptions

Proof

Complete quotients in the continued-fraction algorithm

Definition

Complete-quotient tail formula

Statement

Facts & Assumptions

Proof

Every infinite regular continued fraction converges to a unique real number

Statement

Facts & Assumptions

Proof

The continued-fraction algorithm for real numbers

Statement

Facts & Assumptions

Proof

The continued-fraction algorithm terminates exactly on rational numbers

Statement

Facts & Assumptions

Proof

Normalized finite regular continued fractions are unique

Statement

Facts & Assumptions

Proof

Convergent error bound

Statement

Facts & Assumptions

Proof

Convergents are best rational approximations of the first kind

Statement

Facts & Assumptions

Proof

Legendre's criterion for convergents

Statement

Facts & Assumptions

Proof

Quadratic irrationals

Definition

Eventually periodic regular continued fractions

Definition

Eventually periodic regular continued fractions are quadratic irrationals

Statement

Facts & Assumptions

Proof

Complete quotients of a quadratic irrational lie in a finite state space

Statement

Facts & Assumptions

Proof

Lagrange's theorem for regular continued fractions

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources