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How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$?

Should I use some geometrical approach or apagoge?

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    $\begingroup$ Niven establishes this. See this as well. $\endgroup$ – J. M. is a poor mathematician Oct 11 '11 at 12:33
  • $\begingroup$ You may be interested to see this: $\text{Recurrent proofs of the irrationality of certain trigonometric values}$ here at this link: arxiv.org/abs/0911.1933 $\endgroup$ – user9413 Oct 11 '11 at 14:36
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If $\cos \frac{2 \pi}{7}$ is rational, then $i\sin \frac{2 \pi}{7}=\sqrt{\cos^2 \frac{2 \pi}{7}-1}$ is a quadratic irrational, and hence so is $\cos \frac{2 \pi}{7}+i\sin \frac{2 \pi}{7}$. But $\cos \frac{2 \pi}{7}+i\sin \frac{2 \pi}{7}$ is a primitive $7$th root of $1$, and so has minimal polynomial $x^6+x^5+x^4+x^3+x^2+x+1$, which is not a quadratic.

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    $\begingroup$ There is a ${}^{2}$ missing under the $\sqrt{}$ ... $\endgroup$ – martini Oct 11 '11 at 12:48
  • $\begingroup$ @martini: yes, thanks $\endgroup$ – Chris Eagle Oct 11 '11 at 12:51
  • $\begingroup$ Does anyone else see the polynomial $x^6+x^5+\ldots+1$ at the top of the post instead of where it should be? This only seems to happen when I'm using Chrome, not in Opera or Firefox... $\endgroup$ – user1729 Oct 11 '11 at 14:48
  • $\begingroup$ @Swlabr: Take a screenshot and please make a meta post. $\endgroup$ – J. M. is a poor mathematician Oct 11 '11 at 15:03
  • $\begingroup$ @J. M.: Done... $\endgroup$ – user1729 Oct 11 '11 at 15:21
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Since Chebyshev polynomials have the property $T_n\left( \cos \theta \right) = \cos(n \theta)$, we can use this property with $n=7$ and $\theta = \frac{2 \pi}{7}$.

Let $z = \cos\left( \frac{2\pi}{7} \right)$. Then $T_7(z) = 1$

Let $p(x) = T_7(x)-1 = 64 x^7 - 112 x^5 +56 x^3 - 7x - 1 = (x-1)( 8 x^3 + 4 x^2 - 4 x -1 )^2.$

So $z$ must be a root of $q(x) = (2 x)^3 + (2 x)^2 - 2 (2 x) - 1$. Observe that $2z$ is a root of a monic polynomial, and thus is an algebraic integer. If $z$ were rational, then $2z$ must be an integer. Since $-2 \le 2 \cos \theta \le 2$, there are only 5 possibilities to consider, which are ruled out because $q(1) = 7$, $q(-1) = q(0) = q(1/2) = -1$, $q(-1/2) = 1$.

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  • $\begingroup$ It looks quite like the argument here. Nice! $\endgroup$ – J. M. is a poor mathematician Oct 11 '11 at 13:19
  • $\begingroup$ $T_7(z)=1$, not $-1$. $\endgroup$ – Chris Eagle Oct 11 '11 at 13:23
  • $\begingroup$ Following Chris, the polynomial considered should then be $T_7(x)-1=(x-1)(8x^3+4x^2-4x-1)^2$... $\endgroup$ – J. M. is a poor mathematician Oct 11 '11 at 13:33
  • $\begingroup$ @Chris and J.M. Thanks for catching the typo. I have edited the post to correct it. $\endgroup$ – Sasha Oct 11 '11 at 14:54
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Here is a slightly different way of phrasing Chris Eagle's argument:

If $\cos(2\pi/7)$ were rational, $X^2-2\cos(2\pi/7)\,X+1$ would be a factor of $$X^6+X^5+X^4+X^3+X^2+X+1$$ in $\mathbb Q[X]$, contradiction.

EDIT. Let $n$ be an integer $\ge3$. Then the distinct conjugates of $\cos(2\pi/n)$ are the $\cos(2k\pi/n)$, where $k$ runs over the integers prime to $n$ lying between $1$ and $n/2$.

In particular the degree of $\cos(2\pi/n)$ over $\mathbb Q$ is $\varphi(n)/2$, where $\varphi$ is Euler's totient function.

This results immediately from the following well known facts.

If $\zeta$ is a primitive $n$-th root of $1$ in $\mathbb C$, then $\mathbb Q(\zeta)/\mathbb Q$ is Galois of degree $\varphi(n)$.

Let $G$ be the Galois group and $k$ an integer prime to $n$. Then there is a unique $\sigma_k$ in $G$ satisfying $\sigma_k(\zeta)=\zeta^k$.

In particular $k\mapsto\sigma_k$ induces an isomorphism from the multiplicative group of $\mathbb Z/n\mathbb Z$ onto $G$.

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  • $\begingroup$ This is a very nice way of putting it. $\endgroup$ – Eric Naslund Oct 11 '11 at 15:03
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Here is another answer which uses some machinery:

Given $n$, consider $\zeta_n =e^{2\pi i/n}$. Then $$\frac{1}{2}\left(\zeta_n +\zeta_n^{n-1}\right)=\cos \left(\frac{2\pi}{n}\right).$$

The rational numbers are those elements of $\mathbb{Q}[\zeta_n]$ which are fixed by the action of the Galois group. The Galois Group of $\mathbb{Q}[\zeta_n]$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^\times$ and is given by $\zeta_n\rightarrow \zeta_n^a$ for integers $a$ satisfying $(a,n)=1$. From this, we see that $\frac{1}{2}\left(\zeta_n +\zeta_n^{n-1}\right)$ is fixed by the Galois group only when all of the elements either fix $\zeta_n$ and $\zeta_n^{n-1}$ or swap their places. But if there exists $a\neq -1,1$ and $(a,n)=1$ then $\zeta_n\rightarrow \zeta_n^a$ won't fix this element, so that it must be irrational. This only occurs if $|(\mathbb{Z}/n\mathbb{Z})^\times|>2$, or in other words if $n\neq 3,4,6$. In particular we see that for $n=7$ it is irrational.

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As a mirror of Sasha's answer, consider the Chebyshev polynomial of the second kind, $U_6(x)=(8x^3+4x^2-4x-1)(8x^3-4x^2-4x+1)$. That $\cos\dfrac{2\pi}{7}$ is a root is established from expressing $U_n(x)$ as $\dfrac{\sin((n+1)\arccos\,x)}{\sqrt{1-x^2}}$. We can then test for rational roots in the two cubic factors of $U_6(x)$, and find that they have no rational roots at all...

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