Prove that $\operatorname{Spec}\sqrt2$ contains infinitely many powers of $2$.

Question

$\newcommand{\spec}{\operatorname{Spec}}\spec\sqrt2=\{\lfloor k\sqrt2\rfloor: k \ge 0\}$.

I have no idea of how I can prove the statement in the question.

Prove that $\spec\sqrt2$ contains infinitely many powers of $2$.

Answer 1

Let $k=\lceil 2^n\sqrt 2\rceil$. Then $2^n\sqrt 2<k<2^n\sqrt 2+1$. In fact we have either $2^n\sqrt 2<k<2^n\sqrt 2+\frac12$ or $2^n\sqrt 2+\frac12<k<2^n\sqrt 2+1$, depending on the ($n+1)$st binary digit of $\sqrt 2$ (which becomes the first digit of $2^n\sqrt 2$). Since $\sqrt 2$ is irrational, there are infinitely many $n$ (and correspondingly infinitely many $k$) such that $$2^n\sqrt 2<k<2^n\sqrt 2+\frac12$$ holds. Together with $k\sqrt 2 -1<\lfloor k\sqrt 2\rfloor <k\sqrt 2$ we find $$ 2^n\cdot 2-1 <\lfloor k\sqrt 2\rfloor <2^n\cdot 2+\frac{\sqrt 2}2,$$ hence $\lfloor k\sqrt 2\rfloor=2^{n+1}$.