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

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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

1.1

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∎

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

1.1

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

givenF1basealgebra

1.2

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.

F1inductionalgebra

2.1

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 $n \geq 0$ .

F2step 1.1step 1.2discharge-induction

3.1

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.

step 2.1algebra∎

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

1.1

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.

F1F2given

1.2

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 $Z$ .

F1algebra

2.1

Steps 1.1 and 1.2 are exactly the two assertions.

step 1.1step 1.2∎

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

1.1

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.

givenF1algebra

2.1

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 $α = \frac{α_{n + 1} p_{n} + p_{n - 1}}{α_{n + 1} q_{n} + q_{n - 1}} .$

step 1.1F1F2∎

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

1.1

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 $n \geq 1$ .

givenF1inductionalgebra

2.1

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.

F1step 1.1algebra

2.2

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 $n \geq N$ , $0 < d_{n} \leq \frac{1}{n (n + 1)} \leq \frac{1}{n} \leq \frac{1}{N} < ε .$

F2step 1.1algebra

3.1

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 $x : = \sup E$ .

step 2.1given

3.2

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 $x$ .

step 2.1step 2.2F1

4.1

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.

step 3.2algebra∎

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

1.1

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.

givenalgebra

1.2

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 $∣ α - \frac{p_{n}}{q_{n}} ∣ < \frac{1}{q_{n} (a_{n + 1} q_{n} + q_{n - 1})} = \frac{1}{q_{n} q_{n + 1}} .$

F1F2algebra

2.1

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 $α = x$ .

F3step 1.2algebra

3.1

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

step 1.1step 2.1∎

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

1.1

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.

givenF1algebra

1.2

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

1.3

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 $u_{n + 1} = v_{n} < u_{n}$ .

F1F3inductionalgebra

2.1

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.

step 1.1step 1.3induction

3.1

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.

step 1.1step 1.3F1∎

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

1.1

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.

F1given

1.2

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.

F2algebra

1.3

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.

F2induction

2.1

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.

F1step 1.2step 1.3∎

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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∎

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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

1.1

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 $∣ q_{n - 1} α - p_{n - 1} ∣ = α_{n + 1} ∣ q_{n} α - p_{n} ∣ .$

F3F5algebra

2.1

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.

F1step 1.1algebra

3.1

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 $u \geq v a_{n + 1}$ .

step 2.1F2algebra

4.1

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.

step 3.1F2F4algebra∎

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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

1.1

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.

F3givenalgebra

1.2

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 $q_{n} \leq s < q_{n + 1} .$

F1given

2.1

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

step 1.2F2givenalgebra

3.1

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.

step 2.1givenalgebra∎

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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

1.1

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 $q_{h - 1} x^{2} + (q_{h - 2} - p_{h - 1}) x - p_{h - 2} = 0.$

F1F2F3F5algebra

2.1

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.

step 1.1F3F4algebra

2.2

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 $Q$ .

F1F2F5step 1.1algebra

3.1

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.

step 2.1step 2.2F4algebra

4.1

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).

step 2.2step 3.1∎

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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

1.1

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.

F1givenalgebra

2.1

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.

F2F3algebra

3.1

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 $(p_{n - 1} q_{n - 2} - p_{n - 2} q_{n - 1})^{2} = 1.$

step 2.1F4algebra

3.2

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 $A_{n}$ .

F5F6step 1.1algebra

4.1

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.

step 2.1step 3.2step 3.1algebra∎

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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

1.1

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

F1given

1.2

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 $α_{m} = α_{n} .$

F2given

2.1

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.

step 1.2F3induction

3.1

Steps 1.1 and 2.1 prove both directions of the equivalence.

step 1.1step 2.1∎

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