Proposition 4

— Let $K$ be an ordered field and let $f \in K[X]$. There exists an interval in $K$, in the complement of which $f$ takes the same sign as its highest degree term.

We can immediately reduce ourselves to the case of a monic polynomial; then one can write $f(x) = x^n(1 + a_1 x^{-1} + \ldots + a_n x^{-n})$ for $x \neq 0$. Let

For $|x| > M$ we have $1 + a_1 x^{-1} + \ldots + a_n x^{-n} > 0$, which completes the proof of the proposition.