Constructible number

Revision as of 22:26, 20 August 2009 by Jam (talk | contribs) (Statement of theorem)

We say that a real number $x$ is constructible if a segment of length $|x|$ can be constructed with a straight edge and compass starting with a segment of length $1$.

We say that a complex number $z = x+yi$ is constructible if $x$ and $y$ are both constructible (we also say that the point $(x,y)$ is constructible). It is easy to show that $x+yi$ is constructible iff the point $(x,y)$ can be constructed with a straight edge and compass in the cartesian plane starting with the points $(0,0)$ and $(1,0)$. (Notice that our two definitions coincide when $z$ is a real number.)

Characterization Theorem

It is possible to completely characterize the set of all constructible numbers:

A complex number $\alpha$ is constructible iff it can be formed from the rational numbers in a finite number of steps using only the operations addition, subtraction, multiplication, division, and taking square roots.

For instance, this means one can construct segments of length: $2,\frac{5}{2},\sqrt{6}$ and $\frac{2\sqrt{7}+\sqrt{\frac{4}{3}+\sqrt{87}}}{9+\sqrt{3}+\sqrt{\sqrt{5}+2\sqrt{19}}}$, but one cannot construct a segment of length $\sqrt[3]{3}$.

This condition can be rephrased in terms of field theory as follows:

A complex number $\alpha$ is constructible iff there is a chain of field extensions $\mathbb Q = K_0\subseteq K_1\subseteq \cdots\subseteq K_n = \mathbb Q(\alpha)$ such that each extension $K_i\subseteq K_{i+1}$ is quadratic (i.e. $[K_{i+1}:K_i]=2$).

This is equivalent because the field extension $F\subseteq K$ is quadratic iff $K = F(\sqrt{a})$ for some $a\in F$ with $\sqrt{a}\not\in F$, so taking a square root in the above construction is equivalent to taking at most a quadratic extension of a field, while adding, subtracting, multiplying or dividing does not add anything to the field. (Does someone else want to phrase that better?)

Using this second characterization (and the tower law) we get the necessary (but not sufficient) condition that $[\mathbb Q(\alpha):\mathbb Q] = 2^n$ for some nonnegative integer $n$, or equivalently that $\alpha$ is algebraic and it's minimal polynomial has degree $2^n$.

Using this theorem one can easily answer many classical construction problems, such as the three Greek problems of antiquity and the question of which regular polygons are constructible.

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