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Consider a $2 \times 1$ rectangle split by a diagonal. Then the two angles at a corner are ArcTan(2) and ArcTan(1/2), which are about $63.4^\circ$ and $26.6^\circ$. Of course the sum of these angles is $90^\circ = \pi/2$.

I would like to know if these angles are rational multiples of $\pi$. It doesn't appear that they are, e.g., $(\tan^{-1} 2 )/\pi$ is computed as

0.35241638234956672582459892377525947404886547611308210540007768713728\ 85232139736632682857010522101960

to 100 decimal places by Mathematica. But is there a theorem that could be applied here to prove that these angles are irrational multiples of $\pi$? Thanks for ideas and/or pointers!

(This question arose thinking about Dehn invariants.)

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    $\begingroup$ Aigner, Ziegler: Proofs from the BOOK, 4th Ed., p.40: For every odd integer $n\ge 3$ the number $\frac1{\pi}\arccos\left(\frac1{\sqrt{n}}\right)$ is irrational. If I am not mistaken, your number is just $\arccos(1/\sqrt5)$. $\endgroup$ Nov 7, 2011 at 14:17
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    $\begingroup$ This was also discussed in XKCD forum a little while ago. $\endgroup$ Nov 7, 2011 at 14:27
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    $\begingroup$ BTW: this is the reason why this tiling en.wikipedia.org/wiki/Pinwheel_tiling rotates but never ends in the same orientation ;) $\endgroup$
    – N. S.
    Nov 7, 2011 at 15:17
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    $\begingroup$ This seems relevant. $\endgroup$ Dec 18, 2011 at 10:53
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    $\begingroup$ In this paper: lacim.uqam.ca/~plouffe/articles/plouffe.pdf, it is shown that $\frac1{\pi} \arctan x$ is transcendental if $x$ is rational and $x\neq 0, \pm 1$. $\endgroup$ May 25, 2017 at 20:55

3 Answers 3

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Lemma: If $x$ is a rational multiple of $\pi$ then $2 \cos(x)$ is an algebraic integer.

Proof

$$\cos(n+1)x+ \cos(n-1)x= 2\cos(nx)\cos(x) \,.$$

Thus

$$2\cos(n+1)x+ 2\cos(n-1)x= 2\cos(nx)2\cos(x) \,.$$

It follows from here that $2 \cos(nx)= P_n (2\cos(x))$, where $P_n$ is a monic polynomial of degree $n$ with integer coefficients.

Actually $P_{n+1}=XP_n-P_{n-1}$ with $P_1(x)=X$ and $P_0(x)=1$.

Then, if $x$ is a rational multiple of $\pi$ we have $nx =2k \pi$ for some $n$ and thus, $P_n(2 \cos(x))=1$.


Now, coming back to the problem. If $\tan(x)=2$ then $\cos(x) =\frac{1}{\sqrt{5}}$. Suppose now by contradiction that $x$ is a rational multiple of $\pi$. Then $2\cos(x) =\frac{2}{\sqrt{5}}$ is an algebraic integer, and so is its square $\frac{4}{5}$. But this number is algebraic integer and rational, thus integer, contradiction....

P.S. If $\tan(x)$ is rational, and $x$ is a rational multiple of $\pi$, it follows exactly the same way that $\cos^2(x)$ is rational, thus $4 \cos^2(x)$ is algebraic integer and rational. This shows that $2 \cos(x) \in \{ 0, \pm 1, \pm 2 \}$.....

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    $\begingroup$ Alternative proof of your lemma. Being a rational multiple of $\pi$ means that $(\cos x + i\sin x)^n=1$ for some positive integer $n$. So $\cos x + i\sin x$ is an algebraic integer, and so is $\cos x - i\sin x$. So the sum, $2\cos x$ is an algebraic integer. $\endgroup$ Nov 7, 2011 at 14:59
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    $\begingroup$ @ThomasAndrews Very nice proof... $\endgroup$
    – N. S.
    Nov 7, 2011 at 15:06
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$\arctan(x)$ is a rational multiple of $\pi$ if and only if the complex number $1+xi$ has the property that $(1+xi)^n$ is a real number for some positive integer $n$. It is fairly easy to show this isn't possible if $x$ is an integer with $|x|>1$.

This result essentially falls out of the fact that $\mathbb Z[i]$ is a UFD, and the fact that the only specific primes in $\mathbb Z[i]$ are divisors of their conjugates.

You can actually generalize this for all rationals, $|x|\neq 1$, by noting that $(q+pi)^n$ cannot be real for any $n$ if $(q,p)=1$ and $|qp|> 1$. So $\arctan(\frac{p}q)$ cannot be a rational multiple of $\pi$.

Fuller proof:

If $q+pi=z\in \mathbb Z[i]$, and $z^n$ is real, with $(p,q)=1$, then if $z=u\pi_1^{\alpha_1} ... \pi_n^{\alpha_n}$ is the Gaussian integer prime factorization of $z$ (with $u$ some unit,) $z^n = u^n \pi_1^{n\alpha_1}...\pi_n^{n\alpha_n}$. But if a Gaussian prime $\pi_i$ is a factor of a rational integer, $z^n$, then the complement, $\bar{\pi}_i$ must also be a factor of $z^n$, and hence must be a factor of $z$.

But if $\pi_i$ and $\bar{\pi}_i$ are relatively prime, that means $\pi_i\bar{\pi}_i=N(\pi_i)$ must divide $z$, which means that $N(\pi_i)$ must divide $p$ and $q$, so $p$ and $q$ would not be relatively prime.

So the only primes which can divide $q+pi$ can be the primes which are multiples of their complements. But the only such primes are the rational integers $\equiv 3\pmod 4$, and $\pm1\pm i$. The rational integers are not allowed, since, again, that would mean that $(p,q)\neq 1$, so the only prime factors of $z$ can be $1+i$ (or its unit multiples.) Since $(1+i)^2 = 2i$, $z$ can have at most one factor of $1+i$, so that means, finally, that $z\in\{\pm 1 \pm i, \pm 1, \pm i\}$.

But then $|pq|=0$ or $|pq|=1$.

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    $\begingroup$ Very clean argument, Thomas, with a broad conclusion: the arctan of any rational is not a rational multiple of $\pi$. Thanks! $\endgroup$ Nov 7, 2011 at 15:16
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    $\begingroup$ Other than $1$, $-1$, and $0$. $\endgroup$ Nov 7, 2011 at 15:35
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There's actually an incredible way of proving it that is quite elementary.

First, we note that $\arctan(2)$ can be represented in the complex number $1 + 2i$ as $\sqrt{5} e^{i \arctan(2)}$

The question of proving whether $\arctan(2)$ is a rational multiple of $\pi$ is now framed as checking if the complex number $(1 + 2i)^n$ is ever real for all $n \in \mathbb{N}$. If it is real, then the argument of $\sqrt{5} e^{i \cdot n *\arctan(2)}$ must be a multiple of pi. In essence, if $(1 + 2i)^n$ is ever real, then $\arctan(2) \cdot n = m \cdot \pi \implies \arctan(2) = \frac{m}{n} \pi$.

Why was it useful to phrase the problem in this manner? Let's try and define a recurrence relation for the terms of $(1 + 2i)^n$. Suppose $a_1 = 1$ and $a_2 = 2$, then define $a_n$ to be the real part of $(1 + 2i)^n$ and $b_n$ to be the imaginary part.

$a_{n+1} + ib_{n+1} = (1+2i) \cdot (1 + 2i)^n = (1+2i) \cdot (a_n + ib_n) \\ a_{n+1} = a_n - 2b_n \\ b_{n+1} = 2a_n + b_n$

Okay, I don't like our recursion having two variables in it. Let's try and mess with it a bit.

$a_n = a_{n-1} - 2b_{n-1} \\ b_n = 2a_{n-1} + b_{n-1}$

Substituting the first equation into the $b_{n+1}$ equation, we get $b_{n+1} = 2a_{n-1} - 4b_{n-1} + b_n$

Now we want to get rid of $a_{n-1}$, so let's substitute the second equation after solving for $a_{n-1}$ $b_{n+1} = 2b_n - 5b_{n-1}$

Nice, we got it all in one variable. It may look tempting to solve the recurrence relation, but we can do better. Let's take both sides mod 5.

$b_{n+1} \equiv 2b_n \pmod 5$

Why did we do this? Well, we know the first term, $b_1$, is 2. Therefore, every term after that will just be a power of 2 (mod 5).

$b_n \equiv 2^n \pmod 5$

If there exists some term $b_k = 0$, then it will be 0 (mod 5), because 5 divides 0. (If $b_k = 0$, then that means there exists a natural number $k$ such that $(1+2i)^k$ is real)

But powers of 2 can never be congruent to 0 (mod 5). Therefore, $(1+2i)^n$ is never real and thus $\arctan(2)$ is not a rational multiple of $\pi$.

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  • $\begingroup$ This is also the exact method used in this Michael Penn video: youtube.com/… $\endgroup$ May 1 at 16:07

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