Niven’s theorem proof.

I couldn’t find any proof on the internet. I would appreciate an elementary proof , but any help would be appreciated! Thank you.

Solution 1:

A non-elementary proof goes like this. Suppose $r$ is rational. Then $\exp(\pm i\pi r)$ are roots of unity, therefore algebraic integers, and since the algebraic integers form a ring, $2\sin(\pi r) = i \exp(-i\pi r) - i \exp(i\pi r)$ is an algebraic integer. But the only algebraic integers that are rational are ordinary integers. Thus the only cases where $r$ and $\sin(\pi r)$ are both rational is where $2 \sin(\pi r)$ is an integer (and that integer can only be $0$, $\pm 1$ or $\pm 2$).

EDIT: Unpacked a little, the proof goes like this. It's slightly more convenient to use $\cos$ instead of $\sin$, but $\cos(\theta) = \sin(\theta + \pi/2)$, so this is equivalent. Suppose $r= m/n$ (in lowest terms) is rational, and let $w = \exp(i \pi r) = \cos(\pi r) + i \sin(\pi r)$. By de Moivre's theorem we have $w^{2n} = (\cos(\pi r) + i \sin(\pi r))^{2n} = \exp(2 i \pi m)= 1$, i.e. $w$ is a root of the polynomial $X^n - 1$, and similarly $\overline{w} = \exp(-i\pi r)$ is also a root of $X^n - 1$. Thus $w$ and $\overline{w}$ are eigenvalues of the $n \times n$ matrix $$ M = \pmatrix{0 & \ldots & 0 & 1\cr 1 & \ldots & 0 & 0\cr \ldots & \ldots & \ldots & \ldots\cr 0 & \ldots & 1 & 0\cr} $$ (i.e. the matrix with $1$ in the top right corner and just below the main diagonal, and $0$ everywhere else). Let $u$ and $v$ be eigenvectors of this matrix for $w$ and $\overline{w}$ respectively. Consider the $n^2 \times n^2$ matrix $$ A = (M \otimes I) + (I \otimes M)$$ where $\otimes$ is the Kronecker product. Then $u \otimes v$ is an eigenvector of $A$ for eigenvalue $w + \overline{w} = 2 \cos(\pi r)$, i.e. $$A (u \otimes v) = (M u \otimes v) + (u \otimes Mv) = w (u \otimes v) + \overline{w} (u \otimes v) = (w + \overline{w})(u \otimes v) $$ So $2 \cos(\pi r)$ is a root of the characteristic polynomial of $A$, which is a monic polynomial $P(X) = X^{n^2} + \sum_{j=0}^{n^2-1} c_j X^j$ with integer coefficients. But if $x$ is a rational number with denominator $d > 1$ (in lowest terms), $P(x)$ has denominator $d^{n^2}$ and is not an integer. Therefore the only cases where $r$ and $\cos(\pi r)$ are both rational are when $2 \cos(\pi r)$ is an integer.