Non-Constructible Lengths

[Doubling a Cube] [Trisecting an Angle] [Constructing a Regular Heptagon] [Squaring a Circle] [Result]

From the previous section we know that there are certain constructions that cannot be created using just a compass and a straight-edge. The first of which is doubling a cube.

Doubling a Cube

Lets start with a cube whose edge is of unit length 1. The cube's volume would be the cubic unit. We are trying to find the edge x of a cube that is twice this volume that we just found to be 1 cubic unit. So the volume of the new cube a^3 would be 2. So given a real number a such that a^3 = 2, its minimal polynomial over the rationals is p(x) = x^3 - 2. 3 is not a power of 2, so a is not constuctible.

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Trisecting a 60 degree Angle

Let's start with a 60 degree angle. If we could trisect this angle then we would be able to construct a right triangle with hypotenuse length 1 where one leg a = cos 20°. Our goal is to show that a is not constructible. By using the triple angle formula (cos3ß = 4cos^3(ß) - 3cosß), cos(3·20°) = cos60° = 1/2. Let a = cosß, then 4a^3 - 3a = 1/2, 8a^3 - 6a = 1, 8a^3 - 6a - 1 = 0. If b = 2a then b^3 - 3a - 1 = 0 and if a is constructible so is c. The minimal polynomial where p(b) = 0 is p(x) = x^3 - 3x - 1. So the degree of the minimal polynomial is 3 but 3 is not a power of 2, so b is not constructible. Therefore a is not constructible. If a is not constructible then the angle measuring 20° is not constructible and the angle measuring 60° cannot be trisected using a compass and a straight-edge.

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Constructing a Regular Heptagon Inscribed in a Unit Circle

Let's assume that we can construct a regular heptagon inscribed in a unit circle. In order to prove otherwise we will try to find the length of a side.The vertices of the heptagon are found by the roots of the equation z^7 - 1 = 0 where z = x + yi. One root of the equation is 1, 1^7-1 = 0. So (z^7 - 1)/(z-1) = z^6 + z^5 + z^4 + z^3 + z^2 + z + 1. Now divide by z^3 and get, z^3 + z^2 + z + 1 + 1/z + 1/z^2 + 1/z^3, (z + 1/z)^3 + (z + 1/z)^2 + (z + 1/z) + 1. Let x = (z + 1/z), then we get x^3 + x^2 + x + 1 to be the minimal polynomial of degree 3. But 3 is not a power of 2 so the vertices cannot be constructed on the unit circle, so the sides of the heptagon can't be created using a straight-edge and thus an inscribed regular heptagon cannot be constructed.

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Squaring a Circle

The area of a given circle is (pi)r^2. Let the radius of a given circle be 1. So the area of this cirlce will be pi. Can we construct a square whose area is pi? In order to construct a square with area pi, its sides must be length sqrt(pi). Therefore we must construct sqrt(pi). However the sqrt(pi) is not an algebraic number so are unable to construct length sqrt(pi). If we cannot construct the length of the side of the square with a compass and a straight-edge then we will be unable to construct the square.

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Non-constructible Regular Polygons

Result: Now that we know there are certain constructions aren't possible, we must take a closer look at how these constructions or lack there of affect our ability to construct regular polygons.

As seen above there are certain real numbers that we are unable to construct using a compass and a straight-edge. The same result holds true when we go to construct regular n-gons where n is a nonnegative integer.

Carl Friedrich Gauss discovered that "A regular n-gon is constructible iff the odd prime factors of n are distinct Fermat primes"(wikipedia).

This means that a regular n-gon is constructible if n = (2^m)·p where m is a nonegative integer and p is a unique Fermat primes. A Fermat prime is not just any prime number, it is of the form p = 2^(2^n) + 1 where n is a nonnegative integer.

Our result in the inability to construct a heptagon inscribed in a circle illustrates the result of this theorem.

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