# 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...

• Write $\alpha$ as $a+bi$ and use the angle sum formula for $\cos$. That should quickly let you establish that $\alpha\in\mathbb{R}$. Sep 1 '14 at 22:02
• If it was rational, $e^{i\pi a}=\dfrac13+i\dfrac{2\sqrt2}3$ would have to be a root of unity (why?). However, I'm not sure where to go from there. Sep 1 '14 at 22:03
• As for not being in $\mathbb{Q}$, there is this theorem. Sep 1 '14 at 22:05
• @alex.jordan We want to show that it's irrational. We already know it's real. Sep 1 '14 at 22:05
• I think $\mathbb R \setminus \mathbb Q$ is correct for "$\alpha$ is an irrational number", which is after all a much stronger statement than merely "$\alpha$ is not a rational number" (since neither is, say, a brick). Sep 2 '14 at 10:08

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.

• Classical approach (looking what a fraction as solution would imply). Nice.
– mvw
Sep 1 '14 at 22:07
• It's not clear that "the opposite" is that $\alpha$ is rational. OP seems to not be assuming that $\alpha$ is real, and needs that established too (which is easy enough). Sep 1 '14 at 22:08
• @alex.jordan: Honestly, I thought it had to be real...
– Dair
Sep 1 '14 at 22:09
• Hope you don't mind me editing your answer. Sorry if I overstepped---I won't be upset if you roll back. Sep 1 '14 at 22:17
• @alex.jordan Fair enough. I took it for granted that $\alpha$ was real here. Yeah, no problem, saves me the work:) Sep 1 '14 at 22:17

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 $2$.

• Interesting! I have never heard of a Chebyshev polynomial! I will definitely look into this when I get further along in my mathematical career! :D
– Dair
Sep 1 '14 at 23:16
• Heard about it but nice to see the application for this! Feb 7 '16 at 15:18

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 $0$.