Talk:Niven's theorem

This article states that Niven's Theorem can be extended to tangents with the result that there is no rational tan(X°) for rational X other than for X=0° and X=90°. May I inform you that tan(45°) is also rational, since it is equal to 1? This is the result of both sin(45°) and cos(45°) being the same irrational number. As far as I am aware, although I'm not certain, this is the only other rational value of tan(X°) for a rational X besides 0° and 90° for 0°<=X<=90° (apologies for not using the proper less-than-or-equal-to signs, I'm not sure how to get them on my phone) CherubAgent1440 (talk) 11:59, 11 October 2024 (UTC)

Proof sketch

I don't have access to Galois theory texts, so this would be from memory.

Suppose $0\leq r\leq {\tfrac {1}{4}}$ is a rational number, and $\cos(2\pi r)$ is rational. Then $z=e^{2\pi ir}$ is of index at most two over $\mathbb {Q}$, satisfying the equation $z^{2}-2\cos(2\pi r)z+1=0.$ However, if $r={\tfrac {p}{q}}$ in lowest terms, then the index of $e^{2\pi ir}$ over $\mathbb {Q}$ is $\varphi (q)$, where $\varphi$ represents Euler's totient function. Hence, if r is a candidate, then $\varphi (q)\leq 2$, and q is 1, 2, 3, 4, or 6. The only possible values for r in the target range are ${\tfrac {0}{1}}=0$, ${\tfrac {1}{4}}$, and ${\tfrac {1}{6}}$, which correspond to 0, 90°, and 60°, corresponding to 90°, 0°, and 30° in the original problem. — Arthur Rubin (talk) 01:03, 4 December 2011 (UTC)

If $0\leq r<{\tfrac {1}{2}}$ is a rational number, and $s=\tan(\pi r)$ is rational, then $e^{\pi ir}={\frac {1+is}{\sqrt {1+s^{2}}}}$, and $e^{2\pi ir}={\frac {\left({1+is}\right)^{2}}{1+s^{2}}}.$ Hence, if $r={\tfrac {p}{q}}$ is a candidate, then $\varphi (q)\leq 2$, and q is 1, 2, 3, 4, or 6. The only possible values of r in the target range are ${\tfrac {0}{1}}=0$, ${\tfrac {1}{6}}$, ${\tfrac {1}{4}}$, and ${\tfrac {1}{3}}$.

But,

$\tan 0\pi =0$

$\tan {\tfrac {1}{6}}\pi ={\frac {1}{\sqrt {3}}}$

$\tan {\tfrac {1}{4}}\pi =1$

$\tan {\tfrac {1}{3}}\pi ={\sqrt {3}}$

So, the only solutions are ${\tfrac {0}{1}}=0$ and ${\tfrac {1}{4}}$, which correspond to 0 and 45°.

But, I'm sure there's a better proof. — Arthur Rubin (talk) 08:31, 13 December 2011 (UTC)

Another proof:

It is easy to see by induction that the polynomials $2T_{n}\left({\frac {x}{2}}\right)-1$ (when $T_{n}$) is the chebyshev polynomial of the first kind) are monic polynomials with integer coefficent for $n\geq 1$, so any rational root of one of those polynomials must be integer. Suppose $\cos \left({\frac {2\pi k}{n}}\right)$ is rational, so $2\cos \left({\frac {2\pi k}{n}}\right)$ is a rational root of $2T_{n}\left({\frac {x}{2}}\right)-1$, so it must be an integer between -2 and 2 inclusive (because cosinus is between -1 and 1 inclusive) and you can get all of the possibilities. --87.68.255.144 (talk) 17:19, 31 December 2011 (UTC)

A beautiful visual proof is shown in Mathologer: What does this prove? Some of the most gorgeous visual "shrink" proofs ever invented, referencing articles in Elemente der Mathematik (1946): Elemente der Mathematik (1946), p. 97: W. Scherrer: Die Einlagerung eines regulären Vielecks in ein Gitter Elemente der Mathematik (1946), p. 98: Hugo Hadwinger: Über die rationalen Hauptwinkel der Goniometrie — Preceding unsigned comment added by 2001:A61:3A00:3001:B85D:4814:9B6C:FB3F (talk) 21:15, 26 July 2020 (UTC)

history (from Wikipedia talk:WikiProject Mathematics)

Niven's theorem states that the only rational multiples of π whose sine is rational are the ones you learned in childhood: sin 0°, ±sin 30°, and ±sin 90°. The article is new. Work on it if you are so moved. (Currently it does not include a proof and there are just two references. Also, there may be more articles that ought to link to it than currently do.) Michael Hardy (talk) 18:48, 3 December 2011 (UTC)

Are you sure the theorem is due to Niven? Hermite proved that exp(r) is irrational when r is rational; I was sure that irrationality of sin(r) was proved around the same time. Sasha (talk) 21:34, 3 December 2011 (UTC) Quotation from Niven (which is consistent with my claim):

The central result, Theorem 3.9, was proved by D. H. Lehmer,

Amer. Math. Monthly, 40 (1933), 165-166. The extension to the tangent function in Theorem 3.11 has not been given elsewhere, so far as we know. A proof of Corollary 3.12 independent of Theorems 3.9 and 3.11 was given by J. M. H. Olmsted, Amer. Math. Monthly, 52 (1945), 507-508. The topic is a recurring one in the popular literature: as examples we cite B. H. Arnold and Howard Eves, Amer. Math. Monthly, 56 (1949), 20-21; R. W. Hamming, Amer. Math. Monthly, 52 (1945), 336-337; E. Swift, Amer. Math. Monthly, 29 (1922), 404-405; R. S. Underwood, Amer. Math. Monthly, 28

(1921), 374-376.

Here "Cor. 3.12" is what Michael called Niven's theorem, it follows from Thm. 3.9. Here is a link to Olmsted for your convenience. Sasha (talk) 22:38, 3 December 2011 (UTC)

PS The only place I have found where the result is called Niven's thm. is mathworld, which is not necessarily reliable (see discussion above).

I agree about Mathworld, especially regarding neologisms, but something like it (but with tan instead of sin) is called Niven's thm here. —David Eppstein (talk) 01:31, 4 December 2011 (UTC)

google does not show me this page in your ref, what is the statement there? In any case, my main argument is that Niven himself cites an earlier reference for this result. Sasha (talk) 05:32, 4 December 2011 (UTC)

The relevant part of that page says "...we see that $\tan(\pi /t)$ is rational; it follows by Niven's theorem that $t=4$." —David Eppstein (talk) 06:06, 4 December 2011 (UTC)

I think we should add a ref a) to Lehmer (whose theorem JSTOR 2301023 implies the result immediately) and to Olmsted. Both preceded Niven. Sasha (talk) 16:28, 4 December 2011 (UTC)