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

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.
