Let us suppose that the real numbers are structured as follows:$$\mathbb{Q} \subset \mathcal{S} \subset \mathcal{A} \subset \mathbb{R}$$ We have seen that $\mathbb{Q}$ and the constructible numbers, or the surds, $\mathcal{S}$, are subfields of $\mathbb{R}$. The set $\mathcal{A}$ of Algebraic real numbers is also a sub-field of $\mathbb{R}$. We want to do a partial proof of this by doing the following:
Given that $s \in \mathcal{A}$, start with a polynomial $f(x)$, and for simplicity assume $f(x) = a_0 + a_1x + a_2x^2 +a_3x^3$, with $a_i ∈ \mathbb{Q}$, that accepts $s$ as a root, construct another polynomial $g(x)$ that can accept $−s$ as a root.