Prove $\alpha \in\mathbb R$ is irrational, when $\cos(\alpha \pi) = \frac{1}{3}$

Question

I am trying to prove:

If $\cos(\pi\alpha) = \frac{1}{3}$ then $\alpha \in \mathbb{R} \setminus \mathbb{Q}$

So far, I've tried making it into an exponential, since exponentials are easier to manipulate (at least for me), when compared to $\cos$ or $\sin$. So:

$$\cos(\pi\alpha)^2 + \sin(\pi\alpha)^2 = 1$$ $$\Big(\frac{1}{3}\Big)^2 + \sin(\pi\alpha)^2 = 1$$ $$\sin(\pi\alpha)^2 = \frac{8}{9}$$ $$\sin(\pi\alpha) = \frac{\sqrt{8}}{3}$$ $$\sin(\pi\alpha) = \frac{2\sqrt{2}}{3}$$

Then we can use Euler's formula:

$$e^{i\pi\alpha} = \cos(\pi\alpha) + i\sin(\pi\alpha) = \frac{1}{3} + i\frac{2\sqrt{2}}{3} = \frac{1+i2\sqrt{2}}{3}$$

Now take the log of both sides:

$$ \ln(e^{i\pi\alpha}) = \ln(\frac{1+i2\sqrt{2}}{3}) = \ln(1+i2\sqrt{2}) - \ln(3) = i\pi\alpha$$

$$\therefore \alpha = \frac{\ln(1+i2\sqrt{2}) - \ln(3)}{i\pi}$$

But, here I get stuck, and don't know how to show it is irrational, any ideas? (Even though it looks pretty darn irrational to me...) Should I be trying something else? I'm so lost...

Answer 1

Just in case it is not assumed that $\alpha\in\mathbb{R}$, let $\alpha=a+bi$ with $a,b\in\mathbb{R}$. Then $$\begin{align} \cos(a\pi+b\pi i)&=\cos(a\pi)\sin(b\pi i)+\sin(a\pi)\cos(b\pi i)\\ \frac13&=i\cos(a\pi)\sinh(b\pi)+\sin(a\pi)\cosh(b\pi)\\ \end{align}$$ Since the left side is real, either $\cos(a\pi)$ or $\sinh(b\pi)$ is $0$. The former implies $\frac13=\pm\cosh(b\pi)$, which is impossible, So $\sinh(b\pi)=0$, which implies $b=0$, and so $\alpha\in\mathbb{R}$.

Now assume that $\alpha$ is rational: $\alpha = \frac{a}{b}$ with $a,b\in \mathbb{Z}$ and consider

$$\left(e^{i\alpha\pi}\right)^b = e^{i\pi a} = (-1)^a$$

Now use that $e^{i\alpha\pi} = \frac{1}{3} \pm i\frac{2\sqrt{2}}{3}$ to derive a contradiction.

Answer 2

Another possible approach. Assume, by contradiction, that $\alpha=\frac{p}{q}\in\mathbb{Q}$. Consider the polynomial: $$ g(x)=T_{2q}(x)-1 $$ where $T_{2q}(x)$ is a Chebyshev polynomial of the first kind. We have that $\frac{1}{3}$ is a rational root of $g(x)$, but this contradicts the rational root theorem, since $g(0)=-2$ and the leading coefficient of $g(x)$ is a power of $2$.

Answer 3

A different approach is to use Dirichlet's well-known discontinuous function

$$ f(x)=\lim_{k\to\infty}\left(\lim_{j\to\infty}(\cos(k!\pi x))^{2j}\right), $$

which is $1$ if $x$ is rational and $0$ if $x$ is irrational.

Since $\cos(\pi\alpha)=\frac{1}{3}$ implies $\alpha=\pm\frac{1}{\pi}\arccos\frac{1}{3}+2n$, with $n$ an arbitrary integer, we get

$$ f(\alpha)=\lim_{k\to\infty}\left(\lim_{j\to\infty}[\cos(\pm k!\arccos(1/3))]^{2j}\right). $$

Next, we use the fact that $\cos(nx)=T_{n}(\cos x)$, where $T_n$ denotes the $n$-th Chebyshev polynomial of the first kind. We see that $$ \cos(\pm k!\arccos(1/3))]=T_{k!}(1/3) $$

Using the recurrence relation $T_{n+1}(1/3)=\frac{2}{3}T_n(1/3)-T_{n-1}(1/3)$ (with $T_0(1/3)=1$ and $T_1(1/3)=1/3$), it is not difficult to prove that $|T_n(1/3)|<1$ for all $n\geq 1$. (I did it by solving the recurrence explicitly.) It is then trivial to show that both the outer and the inner limit equal $0$.