Rational numbers are in 1-1 correspondence with natural numbers. For example, let's consider enumeration mentioned in wikipedia: https://en.wikipedia.org/wiki/Rational_number#Properties (https://en.wikipedia.org/wiki/File:Diagonal_argument.svg)
The above mentioned enumeration makes a sequence: $q_1, q_2, ...$. In other words, $q$ is a function form $\mathbb{N}$ to $\mathbb{Q}$.
Function $q$ is not monotone (i.e. - neither descending nor ascending): it is not necessarily true that for every $i$ $q(i) < q(i+1)$.
At the same time, there exist unique total order on all rational numbers. Because of that I wonder, is there (or why isn't there) a permutation $p: \mathbb{N} \xrightarrow{} \mathbb{N}$ such that $q\circ p$ becomes monotone: $$q_{p_1} < q_{p_2} < ... $$ If such permutation does not exist - why?
Please note, I understand the controversy that if such order would existed, there would have been infinite number of other rationals in between of two subsequent rationals.
Edit: Thanks for the reference to a similar question, I just don't think explanation there is evident enough. It is even more puzzling, because the number of permutation has a next level of cardinality (continuum) and yet - in such a rich set there is no single permutation that will bring order. Or maybe there is?
Edit 2: The initial question was mainly about existence of a permutation. Whether such permutation is incomprehensible to write down or does not exist at all. If it does not exist - why and what can be concluded from that.
The rationals are dense, which means that for any rationals $p,q$, there is a rational $r$ with $p < r < q$. So, for any enumeration of rationals $q_i$, there is an index $j$ with $q_1 < q_j < q_2$, and thus the enumeration cannot be strictly increasing.
Alternatively, the result also follows from the fact that $\mathbb Q$ has no least element, so any enumeration must include an element smaller than the first one.