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)?
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.