In the book of Real algebraic geometry by Bochnak-Coste-Roy, in page $7$ (and yes, I have stuck on the first page of the book :) ), it is given that
There is exactly one ordering $\mathbb{R} (X)$ such that $X$ is positive and smaller than any positive real number.If $$P(X) = a_n X^n + a_{n-1} X^{n-1} + ... + a_k X^k \quad with \quad a_k \not = 0,$$ then $P(X) > 0 $ for this ordering iff $a_k > 0$. [...]. Note that with this ordering the field $\mathbb{R} (X)$ is not arhimedean.It contains infinitely small elements (i.e positive and smaller than 1/n, for every $n \in \mathbb{N}, n \not = 0$)such as X, and also infinitely large elements (i.e bigger than n, for every $n \in \mathbb{N}$), such as $1/X$
My question is since the ordering is done by the coefficient of the least non-zero degree term, in the "infinitely small" case;
Let $p(x) = x$, and $g(x) = 1/n$, so we want to show that for any $a\in \mathbb{N},$ $a * p(x) < g(x) \Rightarrow a*p(x) - g(x) < 0$
So, since $$h(x) = a*x - 1/n < 0$$(i.e since -1/n < 0), we have p(x) is infinitely small.
However, in this case any positive constant function would work instead of 1/n, so is there any reason for the choice of 1/n in this case?
A side question: Amost all the sources that I have looked orders the set of rational functions with the leading coefficient of the rational function, whereas in this book author chooses the order the set with the coefficient of the least nonzero element, so is there any particular apparent reason why did author has chosen this particular choice ?