$\mathbb{R}(x)$ as non o-minimal structure

Question

Let us consider the field of real rational functions in one variable $\mathbb{R}(x)$ as an ordered field with $ x > r$ for all $r \in \mathbb{R}$. have to show that $\mathbb{R}(x)$ is not o-minimal. I imagine that it is meant that the language of the structure is $ \lbrace 0, 1, \cdot, +,-, < \rbrace $, but the text of the exercise is not explicit on this.

Answer 1

Consider the formula $\exists y\, y^2 = z$. Does this formula define a finite union of points and intervals in $\mathbb{R}(x)$? (Note that every positive real number is a square, but $x$ is not.)

More generally, no ordered field in which a positive element fails to be a square can be o-minimal. It's a theorem that every o-minimal field is real closed. Real closed ordered fields can be characterized by the properties that every positive element is a square and every odd-degree polynomial has a root.