Lagmental Vicfred

Continued-Fraction Convergents Beat One over q Squared by Vicfred

Every irrational has infinitely many convergents whose error is smaller than the inverse square of the denominator. I will separate the object being defined from the consequence being claimed.

Statement

A continued fraction \([a_0;a_1,a_2,\ldots]\) produces convergents \(p_n/q_n\) with exceptional rational approximation. Quadratic irrationals are exactly the eventually periodic cases.

$$ \alpha\in\mathbf R\setminus\mathbf Q $$

The formulas should not be merged too early. The datum \(\mathsf D\), the conclusion \(\mathsf C\), and the bridge \(\Longrightarrow\) have three different logical jobs.

$$ \left|\alpha-\frac{p_n}{q_n}\right|<\frac1{q_n^2} $$

Worked algebra

The middle display is intentionally dense: it is where signs, bounds, multiplicities, or normalising factors are most likely to be lost.

$$ \left|\alpha-\frac{p_n}{q_n}\right|=\frac1{q_n(q_{n+1}+q_n\theta_{n+1})}\qquad(0<\theta_{n+1}<1) $$

Conceptual compression

The abstraction earns its keep by explaining why the same computation reappears. The notation compresses repeated reasoning without erasing the hypothesis that licenses it.

$$ \begin{aligned} \mathsf{D}\;&:\quad \alpha\in\mathbf R\setminus\mathbf Q,\\[5pt] \mathsf{C}\;&:\quad \left|\alpha-\frac{p_n}{q_n}\right|<\frac1{q_n^2}. \end{aligned} $$

Caveat

Good approximation does not mean arbitrary denominator. The convergents are special because their determinants alternate between plus and minus one.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \left|\alpha-\frac{p_n}{q_n}\right|<\frac1{q_n^2} \end{gathered}} $$

A symbolic summary is trustworthy only because the example and limitation remain visible beside it. The box compresses the conclusion without hiding its origin.

This article was posted on Mon 31 May 2021. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.