# Contruction of irrational angles with ruler and compass

I have the following question on constructible numbers.

I want to know which angles are constructible using only ruler and compass. I will write my angles always as a multiple of $2\pi$. I already know that the angle $2\pi r$ is constructible as an angle if the complex number $e^{2\pi ir}$ is constructible as a complex number.

I know that if $r\in\mathbb Q$, then $e^{2\pi i r}$ is constructible if and only if the regular $m$-gon is constructible, where $m$ is the denominator of $r$ after cancelling out common factors. And then you only have to check if $m$ is a product of a power of two and some pairwise distinct Fermat primes.

But what if $r$ is irrational? Is it true that $e^{2\pi i r}$ is automatically non-constructible? Is it even true that $e^{2\pi i r}$ is transcendent?

The only theorem in that direction which is known to me, is that $e^z$ is transcendental if $z\neq0$ is algebraic. This would answer my question in the special case of the irrational number $r=\frac1 \pi$ (and all cases where $r$ is an algebraic mulitple of $\frac1\pi$ like $r=\frac{\sqrt2}{\pi}$ or something like that.

What about other irrational numbers like $r=\sqrt2$ ?

Is the number $e^{2\pi i r}=e^{2\pi i \sqrt2}$ algebraic? Is it constructible?

My conjecture would be $e^{2\pi i r}$ is transcendent for all $r\in\mathbb R \setminus\mathbb Q$. Is that correct?

Thanks for your help!

• Possible duplicate of Constructible angles – lulu Mar 3 '17 at 13:02
• Note: that prior question only concerns angles of finite order, so it's not a duplicate. If you make a Pythagorean right triangle, such as $\{\frac 35,\frac 45,1\}$ then the angles involved are clearly constructible (as making the triangle achieves that) and the "$r$" in your equation is not algebraic. – lulu Mar 3 '17 at 13:12
• Thanks for the idea with the Pythagorean triangle. Is there an easy way yo see that r=ArcCos[3/5]/(2pi) is not algebraic? – Neslihan Mar 3 '17 at 13:42
• Easy? No. In fact, the quick proof I had in mind is not correct. I had thought that Lindemann-Weierstrass immediately implied this, but at the moment I am not convinced this is true. I expect my claim is true, however...let me think to see if I can actually prove it. – lulu Mar 3 '17 at 14:00
• One certainly expects that with the obvious exceptions, any angle $\alpha$ with rational cosine would have $\alpha/\pi$ transcendental, but I would have no idea of how to show such a proposition. – Lubin Mar 3 '17 at 14:22

## 1 Answer

We will find a transcendental $r$ such that $e^{2\pi i r}$ is a constructible.

Specifically: consider the Pythagorean right triangle $\{\frac 35,\frac 45, 1\}$ and let $r$ be associated with one of the angles therein. Such an angle is manifestly constructible, we must show that the $r$ we find is transcendental.

Note first that $r$ is not rational. Indeed, the only rational multiples of $\pi$ with rational sine have $\sin(x)\in \{\pm 1,\pm \frac 12,0\}$. See, e.g., this question

To do so we invoke the Gelfond-Schnieder Theorem. This tells us that, if $a$ is algebraic (and $a\neq 0,1$) and if $b$ is algebraic and irrational then $a^b$ is transcendental. Here we take $a=e^{\pi i}=-1$ and $b=2r$, where $r$ is as in our example. As $r$ is not rational, neither is $2r$. Thus if $r$ were algebraic the theorem would tell us that $e^{2\pi i r}$ was transcendental, which it is not.

• Good, thanks very much for this. – Lubin Mar 3 '17 at 16:35
• Thanks very much for this nice answer! So my conjecture is false and I guess there is no nice characterisation for which irrational r the angle e^(2pi i r) is constructible... – Neslihan Mar 6 '17 at 11:07