I would be grateful if anyone can tell if there are any methods to classify real numbers using continued fraction. eg: Suppose $[a_0;a_1,a_2,\ldots,a_n]$ is the representation of some real number $\alpha$ then

1) if $n$ is finite then it represents a rational number.

2) if $n$ is infinite then it is an irrational number.

3) if $\alpha$ is a quadratic irrational then the pattern repeats indefinitely with a period of some number $t$. What we can say about the converse?

4) for 2) case how we can distinguish between algebraic and transcendental numbers?

Thanks for your time.

  • $\begingroup$ This is all standard stuff. You're on to something. Have a look around for the continued fraction algorithm; in particular, look for a proof that it works. That'll answer most (if not all) of your questions. $\endgroup$ – Shaun Jun 18 '14 at 12:40

If $n$ is finite:


This is because a finite continued fraction is the result of a finite amount of additions and divisions of integers.

If $n$ is infinite and periodic

$n$ is an irrational root of $ax^2+bx+c=0$ where $a,b,c\in\mathbb{Q}$

If $n$ is infinite and non-periodic

$n$ is not expressible as a root of a second degree or lower polynomial.

Algebraic vs transcendental do not show any differences as far as I figured out.

Transcendental numbers can show patterns, for example:


| cite | improve this answer | |
  • $\begingroup$ So you are saying that converse of 3) holds? $\endgroup$ – schzan Jun 19 '14 at 14:12
  • $\begingroup$ @shzan It does. $\endgroup$ – Alice Ryhl Jun 20 '14 at 6:04

I believe there is no presently known "reasonably explicit" description (i.e. an $n$-term recursion formulation for some positive integer $n)$ of continued fraction expansions that can be used to characterize even just the cubic irrationalities, although a recent paper by Nadir Murru---On the Hermite problem for cubic irrationalities---appears to give such a description for cubic irrationalities when multidimensional continued fraction expansions are used.

I've known about this for maybe a year, but I have not yet looked into what this all means, so hopefully an expert can provide more details and correct any inaccuracies I may have made.

| cite | improve this answer | |
  • $\begingroup$ Thanks for mentioning the paper. $\endgroup$ – schzan Jun 19 '14 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.