ATRACTOR

Since \(\cos(\theta)=-\cos(\pi-\theta)\), to complete the previous argument we only have to show that:

If there are rational numbers \(p\) and \(q\) such that \(\theta= \frac{p\,\pi}{q}\), \(\cos(q\,\theta)\) is \(\pm 1\). But what is the relation between \(\cos(q\,\theta)\) and \(\cos(\theta)\)? We will first look at this question.

Given an angle \(x\) and a natural number \(n\), we have that \[ 2\,\cos\,(n+1)\,x =(2\,\cos\,n\,x)\,(2\,\cos\,x) - 2\,\cos\,(n-1)\,x\;\;\;(1)\] from the formulas of the cosine of the addition and of the difference of two angles \[\begin{eqnarray*} \cos\,(z+w) &=& \cos\,z \,\cos w - sin \,z \,sin \, w \\ \cos\,(z-w) &=& \cos\,z \,\cos w + sin \,z \,sin \, w \end{eqnarray*}\] which, when added, show that \[\cos\,(z+w) = 2\,\cos\,z \,\cos w - \cos\,(z-w).\] Now we only have to make \(z=n\,x\), \(w=x\), to obtain \[\cos\,(n+1)\,x =2\,\cos\,n\,x\,\cos\,x - \cos\,(n-1)\,x,\] and then to multiply the equality by \(2\).

From this formula, we can easily deduce another property of \(2\,\cos\,(n+1)\,x\): it can be written as a finite linear combination, with integer coefficients, of powers of \(2\,\cos\,x\). That is, there are integers \(c_{n-1}, \,\cdots,\, c_1,\, c_0\) such that \[ 2\,\cos\,n\,x = (2\,\cos\,x)^n + c_{n-1}\,(2\,\cos\,x)^{n-1} + \cdots + c_1 \,(2\,\cos\,x) + c_0.\;\;\;(2) \] We will prove this equality by induction on \(n \in \mathbb{N}\).

For \(n=1\), the identity is trivial: \(2\,\cos\,x = 2\,\cos\,x\).

For \(n=2\), the identity follows directly from \(\cos\,2\,x=2\,\cos^2\,x - 1\), which can be rewritten as \(2\,\cos\,2\,x = (2\,\cos\,x)^2 -2\).

Suppose now that the equality \((2)\) is true for \(n\) and \(n-1\), where \(n\) is a natural number bigger than 2. That is, we suppose that, \[\begin{eqnarray*} 2\,\cos\,n\,x &=& (2\,\cos\,x)^{n} + a_{n-1}\,(2\,\cos\,x)^{n-1} + \cdots + a_1 \,(2\,\cos\,x) + a_0 \\ 2\,\cos\,(n-1)\,x &=& (2\,\cos\,x)^{n-1} + b_{n-2}\,(2\,\cos\,x)^{n-2} + \cdots + b_1 \,(2\,\cos\,x) + b_0 \end{eqnarray*}\] where the \(a_i's\) and \(b_j's\) are integer numbers. We now need to prove the equality for \(2\,\cos\,(n+1)\,x\). For that, we rewrite the equality \((1)\) in the following way: \[\begin{eqnarray*} 2\,\cos\,(n+1)\,x &=& (2\,\cos\,n\,x)\,(2\,\cos\,x) - 2\,\cos\,(n-1)\,x = \\ &=& 2\,\cos\,x\,[(2\,\cos\,x)^{n} + a_{n-1}\,(2\,\cos\,x)^{n-1} + \cdots + a_0] -\\ &-& [(2\,\cos\,x)^{n-1} + b_{n-2}\,(2\,\cos\,x)^{n-2} + \cdots + b_1 \,(2\,\cos\,x) + b_0] \\ &=& (2\,\cos\,x)^{n+1} + a_{n-1}\,(2\,\cos\,x)^{n} + (a_{n-2}-1)\,(2\,\cos\,x)^{n-1} + \\ &+& (a_{n-3}-b_{n-2})\,(2\,\cos\,x)^{n-2} + \cdots + (a_0-b_1)(2\,\cos\,x) - b_0. \end{eqnarray*}\] Since all the coefficients \(c_k\) of the latter are integers, by the Induction Principle the property \((2)\) holds for \(x\) and for every \(n\). The proof of the equality \((2)\) is now complete.

We can now finish the proof of \(\mathcal{P}_2\).

If \(\theta\) is an angle whose amplitude is a rational multiple of \(\pi\), that is \(\theta= \frac{p\,\pi}{q}\) for some pair of natural numbers \(p,\) \(q\), then \(\cos(2q\,\theta)= 1\). Replacing this value in \((2)\), for \(n=2q\) and \(x=\theta\), we have that \[2\,\cos\,2q\,\theta = (2\,\cos\,\theta)^{2q} + c_{2q-1}\,(2\,\cos\,\theta)^{2q-1} + \cdots + c_1 \,(2\,\cos\,\theta) + c_0\] or equivalently,\[(2\,\cos\,\theta)^{2q} + c_{2q-1}\,(2\,\cos\,\theta)^{2q-1} + \cdots + c_1 \,(2\,\cos\,\theta) + c_0 - 2=0.\] This shows that \(2\,\cos\,\theta\), which is a rational number by hypothesis, is a zero of a polynomial function with integer coefficients, degree \(2q\) and for which the coefficient of the term of bigger degree is \(1\), \[t \in \mathbb{R} \,\, \mapsto \,\, t^{2q} + c_{2q-1}\,t^{2q-1} + \cdots + c_1 \,t + c_0 - 2.\] If \(\frac{r}{s}\) is an irreducible fraction (\(r \in \mathbb{Z}\), \(s \in \mathbb{N}\) and the greatest common divisor of \(r\) and \(s\) is \(1\)) and \(\frac{r}{s}\) is a zero of this polynomial, then \(\frac{r}{s}\) must be an integer, since from the equality \[\left(\frac{r}{s}\right)^{2q} + c_{2q-1}\,\left(\frac{r}{s}\right)^{2q-1} + \cdots + c_1 \,\left(\frac{r}{s}\right) + c_0 - 2=0\] we conclude that \[r^{2q} + c_{2q-1}\,r^{2q-1}\,s + \cdots + c_1 \,r\,s^{2q-1} + (c_0 - 2)s^{2q} = 0\] and, by the fact that \(c_{2q-1}, \,c_{2q-2},\,\cdots,\,c_{1},\,c_0-2\) are integers, we also conclude that the natural \(s\) must divide \(r^{2q}\). Knowing that \(\frac{r}{s}\) is an irreducible fraction, we have \(s=1\). That is, \(2\,\cos\,\theta\) is an integer.

The number \(2\,\cos\,\theta\) is an integer, and, since \(|\cos\,\theta|\leq 1\), one has that \(-2\leq 2\,\cos\,\theta \leq 2\); furthermore, since \(\theta \in\,\,]0,\pi/2[\), \(0 < \cos\,\theta <1\), and then \(2\,\cos\,\theta \in \,\,]0, 2[\); but in this interval there is only one integer, \(1\). Finally, \(2\,\cos\,\theta=1\) and thus \(\theta=\pi/3\).