Welcome back to Numberholic! I’m your host, Sejari, and it’s always a pleasure to have you here. Today, we’re diving into the fifteenth episode of Proven Yet Hard to Believe.
This is the second construction problem that has been proven impossible. The problem involves trisecting an arbitrarily given angle, which likely emerged during attempts to construct a regular nonagon.
A regular hexagon can be easily constructed from an equilateral triangle since it is possible to bisect an arbitrary angle. But what about a regular nonagon? To construct it, we need to trisect a 60-degree angle, which is one of the interior angles of an equilateral triangle.
Various attempts were made in ancient Greece. One such attempt was proposed by the mathematical genius of the ancient Greek world, Archimedes. His method involved marking the angle to be trisected and finding a line such that the radius and the segment outside the semicircle are equal. However, this method feels somewhat unsatisfactory.
Another tool, shaped like a tomahawk, appeared in a European book in 1835, though its exact inventor is unknown. This tool consists of a semicircle, its radius, and a perpendicular line. By simply fitting it onto the angle we wish to trisect, it seems to accomplish the task. However, even this tomahawk would not have satisfied Greek mathematicians.
In this video, we will explore why a regular nonagon cannot be constructed. We will also discuss the mathematical genius Carl Friedrich Gauss, who, with his cold, logical mind, discovered the constructibility of the regular heptadecagon (17-sided polygon).
Constructing a regular n-gon ultimately involves dividing the central angle of a circle into n equal parts, meaning we must construct an angle of 360/n degrees. Viewed from an algebraic perspective—specifically, from the perspective of equations—this fascinatingly translates to finding the roots of x^n – 1 = 0. This is due to De Moivre’s theorem, though we will explore it through examples rather than directly proving it.
Before proceeding, we need to introduce the concept of the complex plane. This plane represents the complex number a + bi as the point (a, b).
Let’s solve x^3 – 1 = 0.
Plotting these roots on the complex plane, we see that they form a perfect equilateral triangle.
Now, solving x^4 – 1 = 0.
This forms a perfect square.
Fascinating, isn’t it?
Now, let’s combine this with a concept from the previous video: “A constructible number must be the root of a polynomial whose degree is a power of 2.”
Let’s consider the pentagon.
A quartic polynomial appears. This quartic equation can be solved in two steps:
1. Substituting an appropriate variable T to simplify the equation.
2. Rearranging into quadratic equations and solving.
Since all resulting equations are quadratic, a regular pentagon is constructible.
What about a nonagon?
The first factor, x^3 – 1, relates to constructing an equilateral triangle, while the second factor corresponds to trisecting 60 degrees for the nonagon. However, this cannot be factored into a product of quadratics. Therefore, the regular nonagon is not constructible.
From this, we can infer the following: for example, a regular 60-gon is constructible because a regular triangle and a regular pentagon are constructible, making the regular 15-gon constructible. Following this, we can construct a regular 30-gon and then a 60-gon.
For a prime number n, what must hold true for a regular n-gon to be constructible?
Since x^n – 1 can be factored like this, this polynomial must represent constructible numbers. This means that n – 1 must be a power of 2. The number 17 satisfies this condition.
Now we move on to Gauss’s work. On March 30, 1796, at the age of 19, Gauss discovered the following:
First, factor the polynomial x^17 – 1.
This results in a 16th-degree equation.
Can this be solved in four steps using quadratic equations?
Let’s define one root as A and set x_1 ,x_2.
The hardest part is dividing the terms into two suitable groups. However, for Gauss, who learned to calculate before he could speak, this wasn’t a difficult task. From this, we obtain:
x_1 + x_2 = -1
x_1x_2 = -4,
allowing us to form the quadratic equation.
We now solve for x_1 and x_2, then further divide each x_1 and x_2.
From this, we derive another quadratic equation:
t^2 – x_1t – 1 = 0.
Similarly, with y_3 + y_4 = x_2 and y_3y_4 = -1, leading to another quadratic equation:
t^2 – x_2t – 1 = 0.
Now, onto the third step.
The resulting quadratic equation is:
t^2 – y_1t + y_3 = 0.
One of its roots is z_1. If you have followed the calculation process correctly, z_1 should be like this.
Since A^16 = 1/A, we reach the final quadratic equation,
A^2 – z_1A + 1 = 0.
Thus, the 16th-degree equation was solved using four quadratic equations, proving that a regular 17-gon is constructible. This was Gauss’s brilliant achievement.
That’s all for today’s episode of Proven Yet Hard to Believe! Until next time, keep exploring, keep questioning, and stay Numberholic!