F. Ruler and Compass Constructions.

Suppose there is given in the plane a finite number of elementary geometric figures, that is, points, straight lines and circles. We seek to construct others which satisfy certain conditions in terms of the given figures.

Permissible steps in the construction will entail the choice of an arbitrary point interior to a given region, drawing a line through two points and a circle with given center and radius, and finally intersec­ting pairs of lines, or circles, or a line and circle.

Since a straight line, or a line segment, or a circle is determined by two points, we can consider ruler and compass constructions as con­structions of points from given points, subject to certain conditions.

If we are given two points we may join them by a line, erect a perpendicular to this line at, say, one of the points and, taking the dis­tance between the two points to be the unit, we can with the compass lay off any integer n on each of the lines. Moreover, by the usual method, we can draw parallels and can construct m/n. Using the two lines as axes of a cartesian coordinate system, we can with ruler and compass construct all points with rational coordinates.

IfajbjC,... are numbers involved as coordinates of points which determine the figures given, then the sum, product, difference and quotient of any two of these numbers can be constructed. Thus, each

element of the field R( a, b, c, . . .) which they generate out of the rational numbers can be constructed.

It is required that an arbitrary point is any point of a given region. If a construction by ruler and compass is possible, we can always choose our arbitrary points as points having rational coordinates. If we join two points with coefficients in R( a, b, c, . . . ) by a line, its equa­tion will have coefficients in R( a, b, c, . . .) and the intersection of two such lines will be a point with coordinates in R( a, b, c, . . . ). The equa­tion of a circle will have coefficients in the field if the circle passes through three points whose coordinates are in the field or if its center and one point have coordinates in the field. However, the coordinates of the intersection of two such circles, or a straight line and circle, will involve square roots.

It follows that if a point can be constructed with a ruler and com­pass, its coordinates must be obtainable from R( a, b, c, . . . ) by a formula only involving square roots, that is, its coordinates will lie in a field Rs D Ra l D . . . R1 = R(a,b,c, ...) where each field Rj is splitting field over R._j of a quadratic equation x2 - a = 0. It follows (Theorem 6, p. 21) since either Ri = RM or ( R/RM ) = 2, that (Rs/Rj ) is a power of two. If x is the coordinate of a constructed point, then (Rj( x)/R1 ) ■ ( Rs/Rj (x)) = (Ra/Rj ) = 2U SO that R^ x)/R} must also be a power of two.

Conversely, if the coordinates of a point can be obtained from R( a, b, c, . . . ) by a formula involving square roots only, then the point can be constructed by ruler and compass. For, the field operations of addition, subtraction, multiplication and division may be performed by ruler and compass constructions and, also, square roots using 1: r = r : fj to obtain r - y r^ may be performed by means of ruler and compass instructions.

As an illustration of these considerations, let us show that it is impossible to trisect an angle of 604 Suppose we have drawn the unit circle with center at the vertex of the angle, and set up our coordinate system with X-axis as a side of the angle and origin at the vertex.

Trisection of the angle would be equivalent to the construction of the point (cos 20°, sin 20°) on the unit circle. From the equation cos SO - 4 cos3 0 - 3 cos 0, the abscissa would satisfy 4x3 - 3x = 1/2. The reader may readily verify that this equation has no rational roots, and is therefore irreducible in the field of rational numbers. But since we may assume only a straight line and unit length given, and since the 60° angle can be constructed, we may take R( a, b, c,, . . ..) to be the field R of rational numbers. A root a of the irreducible equation 8x3 - 6x - 1 = 0 is such that (R(a)/R) = 3, and not a power of two.