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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∎

Depends on

Used by

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Hackman, Elementary Number Theory (standard reference, not scraped)
William Stein, Elementary Number Theory: Primes, Congruences, and Secrets (standard reference, not scraped)