Editing Constructible number (section)

== Impossible constructions ==
{{multiple image|total_width=600
|image1=01-Würfelverdoppelung-Menaichmos-1.svg|caption1=A cube and its double
|image2=01 Dreiteilung-Winkel Einleitungsbild.svg|caption2=An angle and its trisection
|image3=SquareCircle.svg|caption3=Circle and square with equal areas}}
The [[ancient Greece|ancient Greeks]] thought that certain problems of [[straightedge and compass construction]] they could not solve were simply obstinate, not unsolvable.{{sfnp|Stewart|1989|p=51}}  However, the non-constructibility of certain numbers proves that these constructions are logically impossible to perform.{{sfnp|Klein|1897|p=3}} (The problems themselves, however, are solvable using methods that go beyond the constraint of working only with straightedge and compass, and the Greeks knew how to solve them in this way. One such example is [[Archimedes|Archimedes']] [[Neusis construction]] solution of the problem of [[Angle trisection]].)<ref>The description of these alternative solutions makes up much of the content of {{harvp|Knorr|1986}}.</ref>

In particular, the algebraic formulation of constructible numbers leads to a proof of the impossibility of the following construction problems:

;[[Doubling the cube]]
:The problem of doubling the unit square is solved by the construction of another square on the diagonal of the first one, with side length <math>\sqrt2</math> and area <math>2</math>. Analogously, the problem of doubling the cube asks for the construction of the length <math>\sqrt[3]{2}</math> of the side of a cube with volume <math>2</math>. It is not constructible, because the [[minimal polynomial (field theory)|minimal polynomial]] of this length, <math>x^3-2</math>, has degree 3 over <math>\Q</math>.{{sfnmp|Klein|1897|1p=13|Fraleigh|1994|2pp=429–430}} As a cubic polynomial whose only real root is irrational, this polynomial must be irreducible, because if it had a quadratic real root then the [[Conjugate (square roots)|quadratic conjugate]] would provide a second real root.<ref>{{harvp|Courant|Robbins|1996|pp=134–135|loc=Section III.3.1: Doubling the cube}}</ref>
;[[Angle trisection]]
:In this problem, from a given angle <math>\theta</math>, one should construct an angle <math>\theta/3</math>. Algebraically, angles can be represented by their [[trigonometric function]]s, such as their [[sine]]s or [[cosine]]s, which give the Cartesian coordinates of the endpoint of a line segment forming the given angle with the initial segment. Thus, an angle <math>\theta</math> is constructible when <math>x=\cos\theta</math> is a constructible number, and the problem of trisecting the angle can be formulated as one of constructing <math>\cos(\tfrac{1}{3}\arccos x)</math>. For example, the angle <math>\theta=\pi/3=60^\circ</math> of an equilateral triangle can be constructed by compass and straightedge, with <math>x=\cos\theta=\tfrac12</math>. However, its trisection <math>\theta/3=\pi/9=20^\circ</math> cannot be constructed, because <math>\cos\pi/9</math> has minimal polynomial <math>8x^3-6x-1</math> of degree 3 over <math>\Q</math>. Because this specific instance of the trisection problem cannot be solved by compass and straightedge, the general problem also cannot be solved.<ref>{{harvp|Fraleigh|1994|pp=429–430}}; {{harvp|Courant|Robbins|1996|pp=137–138|loc=Section III.3.3: Trisecting the angle}}.</ref>
;[[Squaring the circle]]
:A square with area <math>\pi</math>, the same area as a [[unit circle]], would have side length <math>\sqrt\pi</math>, a [[transcendental number]]. Therefore, this square and its side length are not constructible, because it is not algebraic over <math>\Q</math>.{{sfnp|Fraleigh|1994|pp=429–430}}
;[[Constructible polygon|Regular polygons]]
:If a regular <math>n</math>-gon is constructed with its center at the origin, the angles between the segments from the center to consecutive vertices are <math>2\pi/n</math>. The polygon can be constructed only when the cosine of this angle is a trigonometric number. Thus, for instance, a 15-gon is constructible, but the regular [[heptagon]] is not constructible, because 7 is prime but not a Fermat prime.{{sfnp|Fraleigh|1994|p=504}} For a more direct proof of its non-constructibility, represent the vertices of a regular heptagon as the complex roots of the polynomial <math>x^7-1</math>. Removing the factor <math>x-1</math>, dividing by <math>x^3</math>, and substituting <math>y=x+1/x</math> gives the simpler polynomial <math>y^3+y^2-2y-1</math>, an irreducible cubic with three real roots, each two times the real part of a complex-number vertex. Its roots are not constructible, so the heptagon is also not constructible.<ref>{{harvp|Courant|Robbins|1996|pp=138–139|loc=Section III.3.4: The regular heptagon}}.</ref>
;[[Alhazen's problem]]
:If two points and a circular mirror are given, where on the circle does one of the given points see the reflected image of the other? Geometrically, the lines from each given point to the point of reflection meet the circle at equal angles and in equal-length chords. However, it is impossible to construct a point of reflection using a compass and straightedge. In particular, for a unit circle with the two points <math>(\tfrac16,\tfrac16)</math> and <math>(-\tfrac12,\tfrac12)</math> inside it, the solution has coordinates forming roots of an irreducible degree-four polynomial <math>x^4-2x^3+4x^2+2x-1</math>. Although its degree is a power of two, the [[splitting field]] of this polynomial has degree divisible by three, so it does not come from an iterated quadratic extension and Alhazen's problem has no compass and straightedge solution.<ref>{{harvp|Neumann|1998}}. {{harvp|Elkin|1965}} comes to the same conclusion using different points and a different polynomial.</ref>