5
$\begingroup$

I've been reading Wikipedia's article on continued fractions.

A few examples are given for the continued-fraction representation of irrational numbers:

  • $\sqrt{19}=[4;2,1,3,1,2,8,2,1,3,1,2,8,\dots]$ (OEIS).

    The pattern repeats indefinitely with a period of $6$.

  • $\phi=[1;1,1,1,1,1,1,1,1,1,1,1,\dots]$ (OEIS).

    The pattern repeats indefinitely.

  • $e=[2;1,2,1,1,4,1,1,6,1,1,8,\dots]$ (OEIS).

    The pattern repeats indefinitely with a period of $3$, with $2$ added to the 3rd term in each cycle.

  • $\pi=[3;7,15,1,292,1,1,1,2,1,3,1,\dots]$ (OEIS).

    The terms in this representation are apparently random.


The first two are algebraic irrational, and the last two are transcendental.

We differentiate between these two types of irrational numbers by definition:

A number is algebraic if and only if it is a root of some non-zero rational-coefficient polynomial.

This gives me the impression that transcendental numbers are "more irrational" than algebraic irrational numbers, so to speak (please excuse me for the non-mathematical notation here).

The continued-fraction examples above give me an additional impression that some transcendental numbers are yet even "more irrational" than others (i.e., $\pi$ is "more irrational" than $e$ and $\phi$).

Has any work been made towards differentiating between those transcendental numbers which can be represented with a "patterned" continued-fraction, and those which cannot?


Moreover, as with algebraic numbers which consist only a "small" countable part of all real numbers (where transcendental numbers consist of the "much larger" uncountable part), is it possible that the "patterned" transcendental numbers are countable and the "random" transcendental numbers are uncountable?

This question can be reduced to a distinction between "patterned" sequences of natural numbers and "random" sequences of natural numbers. I think that the former set is countable and the latter set is uncountable, but I am not sure how it can be proved (mostly because I am not sure how to mathematically differentiate between "patterned" and "random").

$\endgroup$
  • $\begingroup$ must the continued fraction be unique? what are the constraints put on the numerators and denominators? do all the numerators have to be 1, denominators positive? can the signs alternate? aren't there different kinds of continued fraction forms? $\endgroup$ – abel Dec 26 '14 at 14:15
  • $\begingroup$ @abel: According to Wikipedia (see enclosed link at the top of my question), all numerators have to be $1$, the denominator are integers, and the signs are always positive (I guess one can equally say that the denominators are positive integers and the signs are both positive and negative). $\endgroup$ – barak manos Dec 26 '14 at 14:17
  • 1
    $\begingroup$ The periodic (simple) continued fractions are quadratic irrationals. While some sufficient criteria have been given for a number to be transcendental based on faster than expected convergence of their simple continued fraction expansion, this doesn't seem to match with your "patterned" vs. "random" distinction. $\endgroup$ – hardmath Dec 26 '14 at 14:19
  • 1
    $\begingroup$ I remember a theorem that I studied when graduating. Sadly I don't remember the name (not sure, but I think that the theorem has a name of some mathematician). It says that the sequence of the terms of the continued fraction is periodic if and only if the number is irrational and it's the solution of some quadratic equation with integer coefficients. $\endgroup$ – ajotatxe Dec 26 '14 at 14:20
  • 2
    $\begingroup$ This is sometimes called Euler's theorem or the Euler-Lagrange theorem which states that $x$ has an eventually periodic continued fraction representation if and only if it is the root of a quadratic polynomial over the integers (Euler proved the right-to-left implication, Lagrange the other). With regard to the question, we really know very little about continued fractions beyond this classification of quadratic irrationals and rationals. It is a long standing open question even of whether the cubic irrationals have 'bounded partial quotients' $\endgroup$ – Dan Rust Dec 26 '14 at 15:02

Your Answer

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

Browse other questions tagged or ask your own question.