Vrftjkry

Starting with an transcendental number like $e$, it is known that we can write it as a sum of infinitely many rational numbers of the form :

$e= sum limits _{n=0}^{infty }{dfrac {1}{n!}}={frac {1}{1}}+{frac {1}{1}}+{frac {1}{1cdot 2}}+{frac {1}{1cdot 2cdot 3}}+cdots $

we can see in this form each element here is defined and is related to other elements by a known order.

a question comes to mind here about existance of a sequence of positive rational numbers ${omega _n}subsetmathbb{Q}^+$ satisfying conditions below:

1. $ e= sum limits _{n=1}^{infty }omega_n$


2. each element $omega_i$ is not predefined (means it is unknown to us, we just know it is positive rational number).


3. there be no known order or trend (except the well ordering assortment) between elements of the sequence ${omega _n}$ (means there is no defined relation like $sim$ between elements of any subset of ${omega _n}$).


4. 
   

   there be no defined algorithm by which ${omega _n}$ be constructed from ${dfrac {1}{n!}}$ or any other known sequence with sum of $e$.

   
   
   

   note: here conditions 3 and 4 may correlate eachother but I mentioned both for more accurate explanation.

   
   

It seems that we should investigate first whether this question is undecidable or not under the ZFC axiomatic system but I don't know how?

the question can be extended and ask about the cardinality of such those sequences as subsets of $mathbb{Q}$ if there exist some of it. also what can we say about other irrational numbers except e?

Edit: The question has been edited since I posted this answer to remove any reference to cardinality. Hopefully this answer is still of some use and/or interest.

What it looks like you're asking for is simply the cardinality of the set of sequences ${ omega_n }_{n in mathbb{N}} subseteq mathbb{Q}^+$ such that $sum_{n in mathbb{N}} omega_n = e$. Your conditions (2), (3) and (4) are wrapped up in this since we haven't specified any additional information about the sequences ${ omega_n }_{n in mathbb{N}}$ other than that they sum to $e$.

This set has cardinality $2^{aleph_0}$. To see this, note that there are $2^{aleph_0}$-many real numbers $a$ such that $0<a<e$. The numbers $a$ and $e-a$ both have sequences of positive rationals whose sum converges to them—say ${alpha_n}_{n in mathbb{N}} subseteq mathbb{Q}^+$ and ${beta_n}_{n in mathbb{N}} subseteq mathbb{Q}^+$ satisfy $sum_n alpha_n = a$ and $sum_n beta_n = e-a$. (For example, you could take $alpha_n$ to be the truncation of the decimal expansion of $a$ after its $n^{text{th}}$ digit, and likewise for $beta_n$.)

The sequence ${omega_n}_{n in mathbb{N}}$ defined by $omega_{2n}=alpha_n$ and $omega_{2n+1} = beta_n$ then satisfies $sum_n omega_n = e$, and the function $a mapsto {omega_n}_{n in mathbb{N}}$ is injective, so that your set has cardinality $ge 2^{aleph_0}$. It also has cardinality $le 2^{aleph_0}$ since it is a set of subsets of $mathbb{Q}$, which has cardinality $aleph_0$.

This argument applies just as well with $e$ replaced by any positive real number, transcendental or otherwise.

Just an idea. For $n = 0, 1, 2, ldots$ define $e_n = 1/0! + 1/1! + cdots + 1/n!$. Define $s_0 = 0$. For $n = 1, 2, ldots$ choose randomly a rational $s_n$ such that $e_{n-1} < s_n < e_n$, and then define
$omega_n = s_n - s_{n - 1}$. Then $omega_1 + omega_2 + cdots = e$. We could get a computer to choose the $s_n$ without telling us what they are. Not sure if this is the sort of thing you had in mind. Of course, making infinitely many random choices is a bit problematic.