theorem on constructible numbers

Before proving this theorem, some preliminaries must be addressed.

First of all, within this entry, the following nonconventional definition will be used:

Let $S$ be a subset of $\mathbb{C}$ that contains a nonzero complex number and $\alpha\in\mathbb{C}$ . Then $\alpha$ is immediately constructible from $S$ if any of the following hold:

• $\alpha=a+b$ for some $a,b\in S$ ;
• $\alpha=a-b$ for some $a,b\in S$ ;
• $\alpha=ab$ for some $a,b\in S$ ;
• $\alpha=a/b$ for some $a,b\in S$ with $b\neq 0$ ;
• $\alpha=\sqrt{|z|}e^{\frac{i\theta}{2}}$ for some $z\in S$ with $z \neq 0$ and $\theta=\operatorname{arg}(z)$ with $0\le\theta <2\pi$ .

The following lemmas are clear from this definition:

Now to prove the theorem.

Proof. By the first lemma, there exists a finite sequence $\alpha_1,\dots ,\alpha_n\in\mathbb{C}$ such that $\alpha_1$ is immediately constructible from $\mathbb{Q}$ , $\alpha_2$ is immediately constructible from $\mathbb{Q} \cup \{\alpha_1\}$ , $\dots$ , and $\alpha$ is immediately constructible from $\mathbb{Q} \cup \{\alpha_1,\dots ,\alpha_n\}$ . Thus, $\alpha_2$ is immediately constructible from $\mathbb{Q}(\alpha_1)$ , $\dots$ , and $\alpha$ is immediately constructible from $\mathbb{Q}(\alpha_1,\dots ,\alpha_n)$ . By the second lemma, $[\mathbb{Q}(\alpha_1)\!:\!\mathbb{Q}]$ is equal to either $1$ or $2$ , $[\mathbb{Q}(\alpha_1,\alpha_2)\!:\!\mathbb{Q}(\alpha_1)]$ is equal to either $1$ or $2$ , $\dots$ , and $[\mathbb{Q}(\alpha_1, \dots ,\alpha_n,\alpha)\!:\!\mathbb{Q}(\alpha_1, \dots ,\alpha_n)]$ is equal to either $1$ or $2$ . Therefore, there exists a nonnegative integer $m$ such that $[\mathbb{Q}(\alpha_1, \dots ,\alpha_n,\alpha)\!:\!\mathbb{Q}]=2^m$ . Since $\mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\alpha_1, \dots ,\alpha_n,\alpha)$ , it follows that there exists a nonnegative integer $k$ such that $[\mathbb{Q}(\alpha)\!:\!\mathbb{Q}]=2^k$ .