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Hometheorem on constructible numbers
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theorem on constructible numbers
Theorem 1.
Let $\mathbb{F}$ be the field of constructible numbers and $\alpha\in\mathbb{F}$. Then there exists a nonnegative integer $k$ such that $[\mathbb{Q}(\alpha)\!:\!\mathbb{Q}]=2^{k}$.
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 $a,b\in S$;

$\alpha=ab$ 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 $0\leq\theta<2\pi$.
Lemma 1.
Let $S$ be a subset of $\mathbb{C}$ that contains a nonzero complex number and $\alpha\in\mathbb{C}$. Then $\alpha$ is constructible from $S$ if and only if there exists a finite sequence $\alpha_{1},\dots,\alpha_{n}\in\mathbb{C}$ such that $\alpha_{1}$ is immediately constructible from $S$, $\alpha_{2}$ is immediately constructible from $S\cup\{\alpha_{1}\}$, $\dots$ , and $\alpha$ is immediately constructible from $S\cup\{\alpha_{1},\dots,\alpha_{n}\}$.
Lemma 2.
Let $F$ be a subfield of $\mathbb{C}$ and $\alpha\in\mathbb{C}$. If $\alpha$ is immediately constructible from $F$, then either $[F(\alpha)\!:\!F]=1$ or $[F(\alpha)\!:\!F]=2$.
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 $[\mathbb{Q}(\alpha)\!:\!\mathbb{Q}]=2^{k}$. ∎
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