I was asked to prove that on the unit circle, the rational points are dense. My idea is to first show that the point (3/5,4/5) corresponds to an irrational angle (more precisely, its ratio with respect to $\pi$), then by rotation, its image will be dense on the circle. This is Kronecker's theorem?
So I have to show that the angle is irrational.
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$\begingroup$You can prove the result you actually want without any extra tools like Kronecker's theorem.
Given a point $(x_0,y_0)$ on the unit circle, it will lie on the line $y = mx - 1$, where $m = \frac{y_0+1}{x_0}$. If you replace $m$ by a rational approximation to $m$, you get a line that passes through $(0,-1)$ and a rational point on the unit circle close to $(x_0,y_0)$. The rational point can be made arbitrarily close to $(x_0,y_0)$ by making the rational approximation to $m$ sufficiently good.
$\endgroup$ $\begingroup$By Niven's theorem the only rational values taken by $\cos$ or $\sin$ on rational multiples of $\pi$ are $\pm 1$, $\pm 1/2$ and $0$. In particular, $3/5$ and $4/5$ are not the sine or cosine of a rational multiple of $\pi$.
$\endgroup$ $\begingroup$Hint: $\alpha/\pi$ is rational iff $e^{i \alpha}$ is an $n$th root of unity for some $n \in \Bbb{Z}$. Maybe you can show that, if $$z = \frac{3 + 4i}{5},$$ then $z^n \neq 1$ for any $n \geq 2$.
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