FsReveal

Basic Arithmetic

First of all we need to understand that the set of points we can construct form an algebraic structure.
We think of the plane as the complex numbers and I claim that if we have constructed any two points, then we can add and multiply them as complex numbers. (To be more precise it is a subfield)
To show this, it is enough to prove it for real numbers, i.e. points on the $x$ -axis. This will make the pictures with the constructions less 'crowded' with shapes. (For more details on why this works click down.)

For arithmetic to even make sense we need to fix a frame of reference. We have seen from the construction of the square that we can draw the coordinate axis. See the second helper slide for a more details.

Some Quick Notes on Complex Numbers I

Real numbers (denoted $\mathbb{R}$ ) are numbers as most people would understand them in everyday life. They can be positive or negative, whole numbers like $1,2,-7$ , fractions like $\frac{1}{2}$ or more complicated things like $\sqrt{2}$ or $\pi$ .

A complex number $z$ consists of two parts, a real and imaginary part. You can think of the real part as the $x$ -direction and of the imaginary part as the $y$ -direction. That is why we can identify points on a plane with complex numbers.

Some Quick Notes on Complex Numbers II

Because we can easily 'rotate' every number by 90°, we can construct the complex number $a + bi$ from the two real numbers $a$ and $b$ with ruler and compass. And therefore it is enough to only consider basic arithmetic on the real number line.

On the picture below we see how we can transform $a$ to $a\cdot i$ or vice versa.

This construction can also be seen on the next slide.

Addition

Let's say, from the two points we started with, we have constructed two additional arbitrary real numbers $a$ and $b$ .

We can add them in the following way:

First rotate $b$ up to the $y$ -axis by drawing the circle.
Draw a perpendicular line to the $y$ -axis through $bi$ .
Draw a perpendicular line to the $x$ -axis through $a$ .
Where the two lines meet is the point $a+bi$ .
Drawing the circle around $a$ through $a+bi$ gives us $a+b$ as one of the intersections with the $x$ -axis.

Multiplication

This time we start with $b$ already on the $y$ -axis. To construct the product of two arbitary real numbers $a$ and $b$ we:

Draw the line between the points $(0,1)$ (or $i$ in the complex numbers) and $(a,0)$ .
We draw the parallel line through $(0,b)$ (or $bi$ in the complex numbers).
The point where it crosses the $x$ -axis is the point $(a\cdot b, 0)$ .

Try to prove that this is the case. A solution is on the helper slides.

Proof that the Constructed Point is actually the Product

The line between $(0,1)$ and $(a,0)$ satisfies the equation $y=-\frac{1}{a}x + 1$ . To see this noe that we are going $a$ units to the left and one unit down. That is where the $-\frac{1}{a}$ comes from. The $+1$ is because it intersects the $y$ -axis in $1$ .

Now the parallel line has the same slope so its equation is $y=-\frac1a x + b$ . Solving this for zero gives $\begin{align*} -\frac{1}{a} x + b &\overset{!}{=} 0\\ \frac{1}{a} x &= b\\ x &= ab \end{align*}$

Inverses

With everything we've done so far, you might want to try to figure out how to construct $-a$ and $\frac{1}{a}$ from $a$ .

Here are two hints:

$-a$ can be defined as the number such that $0$ is the midpoint between it and $a$ .
$\frac{1}{a}$ is the number that gives $1$ when multiplied by $a$ . So on the previous slide try to go backwards by assuming $ab=1$ and then finding $b$ .

Solutions are on the helper slides

Multiplicative Inverse

If $a=1$ we are done. Assume $a\neq 1$ and of course $a\neq 0$ .

Draw the line between $a$ and $i$ as in the construction for multiplication.
Construct the parallel line that goes through $1$ on the $x$ -axis.
This line will intersect the $y$ -axis in $\frac{1}{a}i$ . By rotating by 90° clockwise we get the desired point.

To prove that this works consider:
The first line satisfies $y=-\frac{1}{a}x + 1$ . The new one has the same slope but an unkown intersection with the $y$ -axis, let's call it $b$ . So the equation is $y = -\frac{1}{a}x + b$ . We know that at $x=1$ the $y$ -value is $0$ , so $-\frac{1}{a}\cdot 1 + b = 0$ . The result follows.

Attacking the Pentagon

Remember: Multiplying two complex numbers means adding their angles and multiplying their lengths. But $\zeta$ has length $1$ since it is on the unit circle, so we just add the angles.

Arithmetically that means the verteces of the pentagon we try to construct are $1,\zeta,\zeta^2,\zeta^3,\zeta^4$ .

Why does Multiplication Correspond to Adding Angles?

Complex numbers can be written, as $r(\cos(\alpha) + i\sin(\alpha))$ for an angle $\alpha$ and a length $r$ . We assume $r=1$ . Now if you have two numbers with two angles, $\alpha$ and $\beta$ , you get

$(\cos(\alpha) + i\sin(\alpha))\cdot(\cos(\beta) + i\sin(\beta)) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta) + i(\cos(\alpha)\sin(\beta) + \sin(\alpha) \cos(\beta))$

Now you can check Wikipedia to convince yourself that

$\begin{align*} \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta) &= \cos(\alpha + \beta) \text{ and }\\ \cos(\alpha)\sin(\beta) + \sin(\alpha) \cos(\beta) &= \sin(\alpha + \beta) \end{align*}$

Therefore multiplying the complex numbers on the unit circle with angles $\alpha$ and $\beta$ gives the number with angle $\alpha + \beta$

Studying $X^5 - 1 = 0$

The equation $X^5 - 1 = 0$ has one easy solution, namely $1$ . Another one is $\zeta$ , that's how we stumbled upon this equation in the first place.

In fact the other solutions are the powers of $\zeta$ , since $(\zeta^k)^5 = \zeta^{5k} = (\zeta^5)^k = 1^k = 1$ . So constructing a pentagon is the same as constructing the solutions of this equation.

But we are starting with $1$ already given, so in a sense it is already constructed. Can we find a equation that encodes only the $4$ remaining points? In fact with Polynomial Division we can. The result is $X^4+X^3+X^2+X+1$ .

Polynomial Division

For a more thorough treatment see Wikipedia. I will only give a short version here.

One can show that if $P(X)$ and $Q(X)$ are polynomials, then $P(X)=Q(X)\cdot A(X) + R(X)$ for some polynomials $A(X)$ and $R(X)$ , where $R(X)$ has lower degree than $Q$ . If $Q(X) = (X-x_0)$ for a root of $P$ , then $P(x_0)=0$ and $Q(x_0)=0$ . It follows that $R(x_0)=0$ . But since $Q(X)$ has degree one, and the degree of $R(X)$ is lower it must be a constant. So putting that all together it is $R(X) = 0$ a constant polynomial.

That is how I knew to divide $X^5-1$ by $(X-1)$ .

The Big Picture Idea

We got from a geometric picture to an algebraic equation. Now we will play around with the equations, and that will help us find which circles and lines we need to draw. The way to translate algebra into geometry will be through the basic circle and line equations. In general they are $\begin{align*} (X-x_0)^2 &+ (Y - y_0)^2 = r^2\\ aX &+ bY = c \end{align*}$ for a circle or a line, respectively.

So we will aim to find equations that have these forms, and then we know what circle or line they correspond to.

A geometric intermediate step

Here I chose to construct the real part. But the same methods could be used to go for the imaginary part.

The real part of $\zeta$ is $\frac{\zeta + \overline{\zeta}}{2} = \frac{\zeta + \zeta^4}{2}$ . However the $2$ in the denominator will be a bit annoying. Therefore I will construct $\zeta+\zeta^4$ , and we know from the beginning that we can divide by $2$ .

Since we know what the algebraic definition of $\zeta$ is we can use that to find it for $\zeta + \zeta^4$

Finding a Circle for $\zeta + \zeta^4$

We need to find a circle that intersects the $x$ -axis in $\zeta + \zeta^4$ . The equation of the $x$ -axis is simply $Y=0$ . The general equation for a circle is $(X-x_0)^2 + (Y - y_0)^2 = r^2$ . So the intersection is $X^2 \color{red}{-2x_0}X + \color{blue}{x_0^2 + y_0^2} = \color{blue}{r^2}.$ We want to transform this into $X^2 + \color{red}{1}X + \color{blue}{(- 1)} = 0$ , so $\begin{align*} -2x_0 &= 1 \left(\Rightarrow x_0 = -\frac12 \right)\\ x_0^2 + y_0^2 -r^2 &= -1 \left(\Rightarrow y_0^2 - r^2 = -\frac54 \right) \end{align*}$

One possible solution is $y_0 = 1$ and $r=3/2$ . The helper slide has some details.

How to $y_0$ and $r$ .

First of all, these aren't the only solutions. But we want rational solutions, since we know from the arithmetic part before, that we can construct all rational numbers.

Next we need $r>0$ since otherwise it wouldn't be a geometric circle.

As far as actual algebra goes, I would just suggest multiplying with $4$ , noticing that $4=2^2$ and then get $(2y_0)^2 - (2r)^2 = -5$ So we need two square numbers that differ by $5$ , and $(4,9)$ can be easily found by hand, and then transformed to the solution on the slide above.

Constructing a Pentagon from First Principles