📂Abstract Algebra

Constructible Numbers

Definition

A number that can be obtained by a finite number of operations including addition, subtraction, multiplication, division, and taking square roots, starting with $1$ is said to be Constructible.

Explanation

The concept of constructibility was originally discussed in the context of ancient Greek geometric arguments, but modern algebra renders the process of drawing circles with compasses and lines with rulers essentially unnecessary. Let’s examine how these operations replace construction.

Addition and Subtraction

Addition and subtraction are achieved by drawing a circle with a radius equal to the number to be added or subtracted from the end of a line segment.

Multiplication and Division

Multiplication and division are achieved using parallel lines and similarity of triangles.

Square Roots

Square roots are obtained using the similarity of right-angled triangles.

Rational Numbers

By adding $1$ a finite number of times, we obtain $\mathbb{N}$, and then by multiplying we obtain $0$, and by subtracting $1$ a finite number of times from $0$, we get $\mathbb{Z}$. Dividing integers yields $\mathbb{Q}$. Therefore, the set of constructible numbers is at least larger than the field of rational numbers, and by allowing roots, we can obtain a slightly larger field. Consequently, all rational numbers are constructible.

Algebraic Numbers and Transcendental Numbers

Irrational numbers can also be constructible. For example, the irrational number $\sqrt{ 1+ \sqrt{3} }$ is constructible because it can be obtained by taking the square root of $3$, adding $1$, and then taking the square root again. However, transcendental numbers like $\pi$ are not constructible.

From the definition, it is easy to infer that constructible numbers are algebraic numbers. For example, $\sqrt{2}$ is obtained by taking the square root of $2$ and is also an algebraic number as a root of $(x^{2} - 2 ) \in \mathbb{Q} [ x ]$. Conversely, a number like $a = \sqrt{ 1+ \sqrt{3} }$ $$\begin{align*} & a^2 = 1 + \sqrt{3} \\ &=a^2 - 1 = \sqrt{3} \\ =& \left( a^2 - 1 \right)^2 = 3 \\ =& a^4 - 2 a^2 - 2 = 0 \end{align*}$$ is also an algebraic number as a root of $\left( a^4 - 2 a^2 - 2 \right) \in \mathbb{Q} [ x ]$.

Theorem

• [1]: Constructible numbers are algebraic numbers.
• [2]: If $\gamma \not\in \mathbb{Q}$ is constructible, then for $i=2, \cdots , n$ $$\left[ \mathbb{Q} \left( a_{1} , \cdots , a_{i-1} , a_{i} \right) : \mathbb{Q} \left( a_{1} , \cdots , a_{i-1} \right) \right] = 2 \\ \mathbb{Q} ( \gamma) = \mathbb{Q} \left( a_{1} , \cdots , a_{n} \right)$$ there exists a finite sequence $\left\{ a_{i} \right\}_{i=1}^{n}$ such that for some $r \in \mathbb{N}$ $$\left[ \mathbb{Q} \left( \gamma \right) : \mathbb{Q} \right] = 2^{r}$$

Proof

[1]

This is self-evident from the definition of constructibility.

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[2]

$$\left[ \mathbb{Q} \left( a_{1} , \cdots , a_{i-1} , a_{i} \right) : \mathbb{Q} \left( a_{1} , \cdots , a_{i-1} \right) \right] = 2$$ means that for some $q \in \mathbb{Q} \left( a_{1} , \cdots , a_{i-1} \right)$, $a_{i} = \sqrt{q}$. Adding such a $a_{i}$ implies that it is possible to add and subtract $a_{i}$ to and from some element $q_{1} a_{1} + \cdots + q_{i-1} a_{i-1}$ existing in the original $\mathbb{Q} \left( a_{1} , \cdots , a_{i-1} \right)$ and multiply by rational numbers, hence $\mathbb{Q} \left( a_{1} , \cdots , a_{i} \right)$ becomes the set of numbers obtained by a finite number of operations and taking square roots on constructible numbers.

That $\gamma$ is constructible means such an operational process exists, hence a finite set $n \in \mathbb{N}$ satisfying $\mathbb{Q} ( \gamma) = \mathbb{Q} \left( a_{1} , \cdots , a_{n} \right)$ also exists.

Properties of Finite Extension Fields: Let $E$ be a finite extension field of $F$, and $K$ be a finite extension field of $E$.

• [2]: $$[E : F] = 1 \iff E = F$$
• [3]: $$[K : F] = [K : E ] [E : F]$$

Then, from the properties of finite extension fields, $$\begin{align*} 2^n =& \left[ \mathbb{Q} \left( a_{1} , \cdots , a_{n} \right) : \mathbb{Q} \right] \\ =& \left[ \mathbb{Q} \left( a_{1} , \cdots , a_{n} \right) : \mathbb{Q} ( \gamma ) \right] \left[ \mathbb{Q} ( \gamma ) : \mathbb{Q} \right] \end{align*}$$ Meanwhile, since $\mathbb{Q} ( \gamma) = \mathbb{Q} \left( a_{1} , \cdots , a_{n} \right)$, $$2^n = \left[ \mathbb{Q} ( \gamma ) : \mathbb{Q} \right]$$

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