# Show $\sqrt[3]{2}\notin F$

Suppose that $F$ is the infinite extension of $\mathbb{Q}$ obtained by adjoining the square root of every integer (positive or negative). I'm trying to show that $\sqrt[3]{2}\notin F$. I have no idea how to prove it. Anyone can help me?

• I guess you have to say that beside $F$ being an infinite extension of $\mathbb{Q}$, if $a \in F$ then there is some $n$ such that $a \in \mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{3}, \ldots,\sqrt{n})$ Apr 8 '16 at 10:13

$F$ is the union of adjoining of finitely many quadratic extension which have degree over $\mathbb{Q}$ a power of 2. Then your element must be in one of those finite adjoining, but it has degree 3 over $\mathbb{Q}$ and that's impossible