Show $\{e^{2\pi ik\sqrt{2}}:k\in\mathbb{Z}\}$ is dense in the unit circle.
This fact was mentioned to me in the comment on another post here. I've tried but I cannot prove it. I was hoping to work out for myself whether this is the unique power of $1$ having this property... but I fell before even the first hurdle.
I began by taking roots of unity and raising to the power of $\frac{1}{\sqrt{2}}$; the results are invariably on the unit circle as would be expected. But I rather thought given statement implied every root of unit raised to the power of $\frac{1}{\sqrt{2}}$ would equal $1$; which I did not find. Clearly I am missing something (or many things).
First note that
the mapping $\euler^{\ie (.)}: \mathbb{R} \to \mathbb{C}$ is uniformly continuous $$|\euler^{\ie t} - \euler^{\ie s}| \leq \sqrt{2} |t-s|.$$
For every irrational number $\alpha$, for every $\epsilon>0$, and for every $C>0$ there exists a fraction $\frac{n}{m}$ such that $0<|\alpha - \frac{n}{m}|< \epsilon$, $m > C$ and $\gcd(n,m) = 1$.
For every $a,b\in \mathbb{Z}$ and $\epsilon>0$ there exists a $C>0$ such that for every $d \in \mathbb{Z}$, $d\geq C$ there exists a $c \in \mathbb{Z}$ such that $$\left|\frac{a}{b}-\frac{c}{d}\right| < \epsilon$$
$\newcommand{\z}{\zeta}$Now let $\z \in \mathbb{T}$ then there exists a $t \in \mathbb{R}$ such that $\z = \euler^{\ie t} = \euler^{2\ie\pi \frac{t}{2\pi}}$ and fix $\epsilon>0$. Furthermore there is a fraction $\frac{a}{b} \in \mathbb{Q}$ such that $\left|\frac{a}{b} - \frac{t}{2\pi}\right|< \epsilon$. Let $C>0$ be such that is satisfies point three of the stated facts for $a,b$. Since $\sqrt{2}$ is irrational we use fact two and obtain a fraction $\frac{n}{m}$ such that $$m>C,\quad \gcd(m,n)=1 \quad\text{and}\quad 0<\left|\sqrt{2} -\frac{n}{m}\right|< \epsilon$$ Since $\gcd(m,n)=1$ there are by the euclidean algorithm $r_1,r_2 \in \mathbb{Z}$ such that $$r_1 n + r_2 m = 1$$ From fact point three there is a $c$ such that $\left|\frac{a}{b} - \frac{c}{m}\right|< \epsilon$.
Choose $k=cr_1$ then $$\euler^{2\ie\pi k \frac{n}{m}} = \euler^{2\ie\pi \frac{cr_1 n}{m}} = \euler^{2\ie\pi \frac{c-cr_2 m}{m}} = \euler^{2\ie\pi \frac{c}{m}} \underbrace{\euler^{2\ie\pi \frac{cr_2 m}{m}}}_{=1} = \euler^{2\ie\pi \frac{c}{m}} $$ From fact point one and the triangular inequality we obtain $$\left|\euler^{2\ie\pi k \sqrt{2}}-\z\right| \leq \left|\euler^{2\ie\pi k \sqrt{2}}-\euler^{2\ie\pi k \frac{n}{m}}\right| + \left|\euler^{2\ie\pi k \frac{n}{m}}-\euler^{2\ie\pi \frac{a}{b}}\right| + \left|\euler^{2\ie\pi \frac{a}{b}}-\z\right| \leq 3\sqrt{2} \epsilon$$ which proves the claim.