# 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 $a,b\in S$;

• $\alpha=a-b$ 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 $0\leq\theta<2\pi$.

The following lemmas are clear from this definition:

###### 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 $[\mathbb{Q}(\alpha)\!:\!\mathbb{Q}]=2^{k}$. ∎

Title theorem on constructible numbers TheoremOnConstructibleNumbers 2013-03-22 17:16:28 2013-03-22 17:16:28 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Theorem msc 12D15 ConstructibleNumbers ClassicalProblemsOfConstructibility immediately constructible from