🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Since $E$ is a vector space of degree 2 over $\mathbb{Q}$ then it has some basis $(1,k)$. It must be that $k\notin \mathbb{Q}$ because if $k\in \mathbb{Q}$ then $(1,k)$ would no longer be linearly independent. Moreover we can write $k^2\in E$ as a linear combination of the basis vectors so that we have some $c,d\in \mathbb{Q}$ such that $k^2=-c-dk$ which is to say that $k^2+dk+c=0$ which means that $k$ is a solution to the polynomial $x^2+dx+c=0$.

The quadratic formula determines that $k$ is of the form $\frac{-(-d)\pm \sqrt{d^2-4·1·c}}{2}$, now if $d^2-4c$ is a square, then the entire expression yields a rational, but $k$ is not rational, so we know that $d^2-4c$ is not a square. Let $m:=d^2-4c\in \mathbb{Q}$ so that $k=\frac{d}{2}\pm \frac{\sqrt{m}}{2}⟺\sqrt{m}=\pm (2k-d)$. Recall that $k\notin \mathbb{Q}$ so that $2k-d\notin \mathbb{Q}$ so that $\sqrt{m}\notin \mathbb{Q}$

For any number $x\notin \mathbb{Q}$ we know that $\mathbb{Q}\left(x\right)=\mathbb{Q}(ax+b)$ where $a,b\in \mathbb{Q}$ then we know that $\mathbb{Q}\left(k\right)=\mathbb{Q}\left(\sqrt{m}\right)$ so we've proven the statement true.