Gauss and Regular Polygons: Euclidean Constructions Primer

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Introduction

What do we exactly mean by the term "Euclidean Constructions"? Informally the term refers to the geometrical constructions done using the ruler (also called straightedge) and compass. Such constructions are studied as part of high-school (7th to 10th grade) mathematics curriculum and I hope most readers are familiar with the construction of bisection of line segment, bisection of an angle and construction of equilateral triangles.

To formalize the term, we shall allow only the following steps in any Euclidean construction:
  1. Given two points, the ruler can be used to join them to make a line segment. Note that the ruler cannot be used to draw a segment of pre-determined length as we have to treat the ruler as unmarked.
  2. Given two line segments, the ruler can be used to extend them (if needed) to find their point of intersection (if it exists).
  3. Given a point P and a line segment AB, the compass can be used to draw a circle (or an arc) with center P and radius equal to the length of line segment AB. Note that the compass can not be used to draw circles of pre-determined radius.
  4. Given a circle and a line segment, the ruler can be used to extend the line segment (if needed) to find its points of intersection with the circle if any.

Fig 1. Geometrical Constructions with Euclidean Tools

Algebraic Meaning of Constructibility

We now switch to the language of co-ordinate geometry (analytic geometry). Two points O and A are arbitrarily chosen on the plane. O is referred as origin and the distance OA is chosen as the unit of length. The segment OA is extended on both sides to form the X-axis. A line perpendicular to X-axis and passing through origin is drawn and is called Y-axis. The positive half of the X-axis is the one in direction from O to A and other half is negative. Similarly if turn anticlockwise from the positive half of X-axis we reach the positive half of the Y-axis (the other half being negative).

The co-ordinate plane is now set up with both the axes and the origin O and a unit length OA. Now to every point in the plane corresponds a pair of real numbers $ x$ and $ y$ and the point is denoted by $ (x, y)$ where $ x$ (called abcissa or x-coordinate of the point) is the signed distance (positive if it lies towards positive half of X-axis, negative otherwise) of the point from Y-axis and $ y$ (called ordinate or y-coordinate of the point) is the signed distance (positive if it lies towards the positive half of Y-axis, negative otherwise) of the point from X-axis.

Coordinate Plane
Fig 2. Coordinate Plane

Having established a coordinate plane we can now figure out the implications of the Euclidean constructions described above in the language of algebra. First of all we note that starting with unit length OA we can construct a line segment AB whose length is a rational number (Integers are easy, just add many segments equal to OA one after another. For rationals just note that a line segment can be divided into a number of equal parts by drawing some parallel lines and using Thales theorem). Therefore it is possible to locate a point $ P(x, y)$, whose coordinates are rational numbers, by using Euclidean tools only.

As we can see any new point in the plane is introduced in one of the following ways:
  1. Intersection of two lines: Assuming that both lines are obtained as extensions of line segments whose end points have rational coordinates, it follows the equations of the both lines are of the form $ ax + by + c = 0$ where $ a, b, c$ are rational numbers. Consequently the coordinates of the point of intersection of these lines are also rational.
  2. Intersection of a line and a circle: We can assume that the center of circle is having rational coordinates and the radius $ r$ as the length of segment $ AB$ where $ A$ and $ B$ have rational coordinates. In that case $ r^{2}$ is rational and so the equation of the cirle $ {(x - p)}^{2} + {(y - q)}^{2} = r^{2}$ has rational coefficients. Similarly the equation of the given line $ ax + by + c = 0$ has rational coefficients. Therefore the solutions of these equations will ultimately be either rational or be roots of quadratic equations with rational coefficients. Hence the coordinates of the points of intersection will either be rational or be roots of quadratic equations with rational coefficients.
  3. Intersection of two circles: Here the equations of both the circles will be having rational coefficients and therefore as in the previous case the coordinates of the points of intersection will be either rational or be roots of quadratic equations with rational coefficients.Using points so obtained one can calculate new lengths (which could be used as radius of a circle in further constructions). Since the distance formula $ d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$ also involves the extraction of a square root, it is clear that if one starts with points having rational coordinates then after one step of Euclidean construction either one gets points with rational coordinates or the coordinates of the points obtained are roots of quadratic equations with rational coefficients. Same holds true of the distances between any pair of old and new points so constructed.
If we apply one more step of Euclidean construction then we have the following possibilities for the coordinates of new points and distance between a pair of points so obtained:
  1. Either these are rational.
  2. OR, these are roots of quadratic equations with rational coefficients.
  3. OR, these are roots of quadratic equations whose coefficients themselves are roots of quadratic equations with rational coefficients.
Generalizing the above reasoning for $ n$ steps of Euclidean constructions we can say the following about the coordinates of points so obtained and the distances between them:
  1. Either these are rational.
  2. OR these are roots of quadratic equations of the following types:
\begin{align} a_{1}x^{2} + b_{1}x + c_{1} &= 0\tag{1}\\ a_{2}x^{2} + b_{2}x + c_{2} &= 0\tag{2}\\ a_{3}x^{2} + b_{3}x + c_{3} &= 0\tag{3}\\ \cdots &= \cdots \notag\\ a_{n}x^{2} + b_{n}x + c_{n} &= 0\tag{n} \end{align} where the coefficients of equations of type $ (1)$ are rational and for $ i > 1$ the coefficients of equations of type $ (i)$ are the roots of an equation of type $ (i - 1)$.

We say that a real number is constructible if one can construct a line segment of that length starting from a line segment of unit length using Euclidean constructions only. It is now clear that all constructible numbers must be roots of quadratic equations of the above types. Since solution of a quadratic equation involves the extraction of a square root, it is clear that constructible numbers can be expressed as a finite combination of rationals with the following operations : addition, subtraction, multiplication, division and square roots. Moreover, given two line segments of lengths $ a$ and $ b$ we can always get line segments of lengths $ a \pm b$, $ ab, a / b$ and $ \sqrt{a}$ using Euclidean constructions only. Thus any number which can be expressed finitely using rationals and operations of "+, -, *, /" and square roots is constructible.

So we have the following result:
A real number $ \alpha$ is constructible if and only if can be expressed finitely as a combination of rationals with the following operations: addition, subtraction, multiplication, division and square root extraction.

It is also clear from the above discussion that finding the expression for a constructible number will involve solving a series of quadratic equations where solutions of one equation will be used as coefficients in the next equation. Thus if we have only two equations to solve and we wish to remove the irrationalities involved in the coefficients of 2nd equation then we need to eliminate them from both the equations. This will always result in either a quadratic equation or a quartic equation (degree 4) with rational coefficients and we shall never have 3rd degree equation involved. (Readers are requested to check this using some real equations and elimination). The logic behind this conclusion looks obvious after we check a few examples, however a proper proof would require us to develop field theory and lot of "Abstract Algebra" stuff.

An informal reasoning would suffice here. Roughly equations of type $ (1)$ will lead to numbers having only one square root in their expression and when such numbers are added to the set of rationals (in the sense we can combine them with rationals using usual operations of +, -, *, /) then it will form a field say $F_{1}$. If an equation of type $ (2)$ splits as a factor of two linear polynomials having coefficients from the field $F_{1}$, then the roots of this equation will also be in $F_{1}$, and in this case the elimination of coefficients from equations of type $ (1)$ and type $ (2)$ will result in a quadratic equation with rational coefficients. If this is not the case (i.e. an equation of type $ (2)$ is irreducible over the field $F_{1}$) then the elimination of irrational coefficients will lead to a quartic equation with rational coefficients.

Proceeding in this fashion one can conclude the following:
A constructible number $ \alpha$ is algebraic (i.e. it is the root of a polynomial equation with rational coefficients) and moreover, the minimal polynomial (the polynomial of least degree with rational coefficients satisfied by the number in question) for $ \alpha$ will have a degree which is power of 2.

This is a very powerful theorem and its exact proof is somewhat cumbersome and full of discussions of extensions fields and irreducibility and other stuff which we don't need here. Needless to say that the formal proof does not contain any new idea apart from those used in our informal reasoning and therefore we will be content with the informal version only.

The power of this theorem is exhibited in a negative fashion i.e. it can be used to prove that certain numbers are not constructible. For example consider $ \alpha = \sqrt[3]{2}$. Clearly we have $ {\alpha}^{3} - 2 = 0$ and therefore $ \alpha$ is a root of the polynomial equation $ x^{3} - 2 = 0$. It is easy to establish that $ x^{3} - 2$ is irreducible over the rationals and therefore it is the minimal polynomial for $ \alpha = \sqrt[3]{2}$. Since the degree of this polynomial is 3 which is not a power of 2, therefore $ \alpha = \sqrt[3]{2}$ is not constructible.

Construction of angles

We next focus on the problem of construction of angles. Suppose we wish to construct an angle $ \theta$ using Euclidean tools. We know this is possible if $ \theta = \pi / 2, \pi / 4, \pi / 3, \pi / 6$. To address the problem for general $ \theta$, let's assume that angle $ \theta$ can be constructed in this fashion. We can then have a line segment AB of unit length and we can construct a ray AX such that $ \angle BAX = \theta$. From $ B$ a perpendicular is drawn to ray $ AX$ which intersects it at point $ C$. Then it clear that $$ AC = \cos\theta, \,\, BC = \sin\theta$$ so that $ \sin\theta$ and $ \cos\theta$ are constructible numbers.
Construction of Angles 
Fig 3. Construction of Angles

Conversely if we assume that $ \sin\theta$ is constructible, then $ \cos\theta = \sqrt{1 - \sin^{2}\theta}$ is also constructible and consequently the point $ P(\cos\theta, \sin\theta)$ can be located on the coordinate plane using Euclidean tools and thus we can construct $ \angle POX = \theta$ ($ X$ being any point on X-axis). We thus have the following result:
An angle $ \theta$ is constructible using Euclidean tools if and only if $ \sin\theta$ (or $ \cos\theta$) is a constructible number.

Construction of Regular Polygons

Construction of a regular polygon is ultimately linked to the construction of its central angle (angle subtended by a side at the center of its circumcircle). If a regular polygon is constructible using Euclidean tools then we can find its center by drawing bisectors of two adjacent sides and finding their point of intersection. Joining this center with two adjacent vertices we can construct the central angle. For a regular polygon of $ n$ sides the measure of the central angle is $ \theta = 2\pi / n$.

Again on the other hand if we assume that an angle of measure $ \theta = 2\pi / n$ is constructible by Euclidean tools then we can draw a circle of unit radius and draw two rays from its center such that angle between them is $ \theta = 2\pi / n$. These rays will then intersect the circle in two points, say $ A$ and $ B$. $ A$ and $ B$ will then act as adjacent vertices of the regular polygon. Using $ AB$ as the radius and center $ B$ we can draw an arc to cut the original circle at $ C$ which is the 3rd vertex of the polygon. Continuing in this fashion all the vertices of the regular polygon can be obtained and the vertices can be joined to form the desired regular polygon.

We can now conclude (using the result of the previous section):
A regular polygon of $ n$ sides can be constructed by using Euclidean tools if and only if $ \cos(2\pi / n)$ (or equivalently $ \sin(2\pi / n)$) is a constructible number.

We shall investigate the constructibility of $ \cos(2\pi / n)$ along the lines of Gauss in the forthcoming posts.

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5 comments :: Gauss and Regular Polygons: Euclidean Constructions Primer

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  1. Could you please explain what you mean by:

    “Thus we if have only two equations to solve and we wish to remove the irrationalities involved in the coefficients of 2nd equation then we need to eliminate them from both the equations”

    and

    “If an equation of type splits as a factor of two linear polynomials having coefficients from the field F1, then the roots of this equation will also be in F1, and in this the elimination of coefficients from equations of type and type will result in a quadratic equation with rational coefficients.”

    The key word in question for me is “eliminated.” Is this simply a mathematical operation, or does one redo the construction so that the irrationality disappears, or is it some other process? Could you please show an example of what you are referring to?

    Thanks!

  2. @rasraster

    Let me explain by using a simple example. If $x = \sqrt{2} - 1$ then $x^{2} + 2x - 1 = 0$.

    Let us then assume that $x_{1}, x_{2}$ are roots of $x^{2} + 2x - 1 = 0$ and $x_{1} > x_{2}$. Next suppose we need to solve the following equation:

    $y^{2} + x_{1}y - 1 = 0$

    Then the approach is to replace $x_{1}$ by $x$ in the above equation and then we have the following system of equations:

    $x^{2} + 2x - 1 = 0\,\,\,\,\cdots (1)$
    $y^{2} + xy - 1 = 0\,\,\,\,\cdots (2)$

    From the second equation above we can write $x = (1 - y^{2})/y$ and putting this value of $x$ in $(1)$ we get
    $\displaystyle \left(\frac{1 - y^{2}}{y}\right)^{2} + 2\cdot\frac{1 - y^{2}}{y} - 1 = 0$

    This is what I have termed as “elimination” in my post and the objective here is a find an equation for $y$ with rational coefficients only.

    After simplification we get an equation of degree 4 and it may happen that it gets factorized as a production of two quadratic expressions over rationals (this is the case I am referring to as “the elimination of coefficients from equations of type and type will result in a quadratic equation with rational coefficients”) or it may remain irreducible. However it will never happen that it gets split as a product of a linear factor and a cubic factor over rationals. The reason is simple that the linear factor will make $y$ rational and hence $x = (1 - y^{2})/y$ will also turn out to be rational which is not the case.

  3. @paramanand

    Thanks for the response. I follow it until the last sentence. Having a factor that is linear in y and a rational constant means that there is at least one rational root, but I don’t see how it implies that all values of y must be rational. Could you please explain? Thanks.

  4. Hello rasraster,
    Since there will be at least one value of $y$ which is rational, it will lead to at least one value of $x = (1 - y^{2})/y$ as rational. But from the equation $x^{2} + 2x - 1 = 0$ we see that both the values of $x$ are irrational. That’s why we can’t have even a single value of $y$ as rational.

  5. Thanks, paramanand! When you switched from $x_{1}$ to $x$ my thought process switched from specific to general, but really the discussion stems from a specific example so I should have realized that you are exposing a contradiction in the case at hand.