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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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7  
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 $\arccos(1/\sqrt5)$. –  Martin Sleziak Nov 7 '11 at 14:17
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This was also discussed in XKCD forum a little while ago. –  Jyrki Lahtonen Nov 7 '11 at 14:27
    
Wow! Such knowledgeable, clear answers, all while I was in a meeting! :-) Thanks to all!! –  Joseph O'Rourke Nov 7 '11 at 15:06
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BTW: this is the reason why this tiling en.wikipedia.org/wiki/Pinwheel_tiling rotates but never ends in the same orientation ;) –  N. S. Nov 7 '11 at 15:17
    
This seems relevant. –  J. M. Dec 18 '11 at 10:53

2 Answers 2

up vote 14 down vote accepted

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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13  
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. –  Thomas Andrews Nov 7 '11 at 14:59
    
@ThomasAndrews Very nice proof... –  N. S. Nov 7 '11 at 15:06

$\arctan(x)$ 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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Very clean argument, Thomas, with a broad conclusion: the arctan of any rational is not a rational multiple of $\pi$. Thanks! –  Joseph O'Rourke Nov 7 '11 at 15:16
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Other than $1$, $-1$, and $0$. –  Thomas Andrews Nov 7 '11 at 15:35

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