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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.

Depends on

Used by

Lagrange's theorem for regular continued fractions Theorem

Dependency tree · two levels

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

Sources

Peter Hackman, Elementary Number Theory (standard reference, not scraped)
Bruce Ikenaga, Periodic Continued Fractions (standard reference, not scraped)