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a long time ago, when watching a video about continued fractions, I saw something interesting, all continued fractions in that video (all that were non-transcendental) had a rational-looking fraction. In other words, when writing a continued fraction in terms of $a$, then made a decimal from all $a$ terms, then the number would be rational. $$\text{Fraction: }a_1+\cfrac1{a_2+\cfrac1{a_3+\cfrac1{a_4+\cfrac1{a_5+\cfrac1{a_6+\cfrac1{\ddots}}}}}}$$ $$\text{decimal: }\sum_{k=1}^\infty \frac{a_k}{10^k}$$ but, all transcendental numbers shown in the video gave (I suspect) irrational decimals.

So, can you prove that an irrational, non-transcendental number, will always give a rational "continued fraction decimal"? or that a transcendental number sometimes has an algebraic "continued fraction decimal"?

is there a proof already?


PS: $a_k \in \Bbb Z$

and, if iterating the process (take the decimal you got and use it as your new number). could you create a tier list,(how many iterations does it take to get a transcendental number to become non-transcendental)?

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    $\begingroup$ It's quite easy to show this, actually. If a transcendental has a rational decimal expansion, that means that the expansion repeats at some point. That means that we can write the continued fraction as $X = a_1+\frac1{a_2+\frac1{...+\frac1X}}$, and then once you un-continued fraction this expression, you get an algebraic equation, of which $X$, the actual value of the continued fraction, is one of the roots. As such, $X$ is algebraic, a contradiction. $\endgroup$ – Don Thousand Oct 13 '19 at 22:26
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    $\begingroup$ A number has a periodic expansion as an infinite simple continued fraction if and only if it's a quadratic irrational. So the simple continued fraction expansion of a transcendental number like $\pi$ or a higher degree algebraic irrational like $2^{1/3}$ will be aperiodic. $\endgroup$ – bof Oct 13 '19 at 22:54
  • $\begingroup$ I mean, what is your "decimal" if, say, $a_k=k$ for all $k$? (The continued fraction is then equal to $I_0(2)/I_1(2)$, the transcendence of which is not as well known as of $e$; the series you give is $\sum_{k=1}^{\infty}k/10^k=10/81$.) Or don't you allow such continued fractions? $\endgroup$ – metamorphy Oct 13 '19 at 23:03
  • $\begingroup$ @metamorphy $a_k \in \Bbb Z$. and my decimal is stated above as an equation involving all $a$ terms. though I guess I could write it as a summation, as you showed. $\endgroup$ – spydragon Oct 13 '19 at 23:09
  • $\begingroup$ @bof I just clarified my question, sorry if it confused you $\endgroup$ – spydragon Oct 13 '19 at 23:39
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yes,
if you are given an irrational non-transcendental number $X$, it's continued fraction must simplify at some point so that you can turn it into an algebraic equation, and because it can simplify, that means that at some point the terms start to repeat. in other words: $a_1=a_n,\space a_2=a_{n+1},\space\cdots\space a_k=a_{n+(k-1)}$.

as for transcendental numbers, they cannot be simplified into an algebraic equation which means, the terms will never repeat in a cyclic pattern.

in summery:
transcendental --> irrational (unknown if transcendental)
and
irrational non-transcendental --> rational repeating --> rational terminating

as for the tier list of transcendental numbers, I still don't know.

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