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

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

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

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    @cheepychappy: I assume that $k$ is meant to refer to natural numbers only.2012-10-05

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Let $k=\lceil 2^n\sqrt 2\rceil$. Then $2^n\sqrt 2. In fact we have either $2^n\sqrt 2 or $2^n\sqrt 2+\frac12, depending on the ($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 holds. Together with $k\sqrt 2 -1<\lfloor k\sqrt 2\rfloor we find $ 2^n\cdot 2-1 <\lfloor k\sqrt 2\rfloor <2^n\cdot 2+\frac{\sqrt 2}2,$ hence $\lfloor k\sqrt 2\rfloor=2^{n+1}$.

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    Thanks for the answer, but I see a lot of unexplained things, which I don't seem to be able to comprehend :( . (n+1)st binary digit is the one counted from the left end, right? And, how does this **bit** influence the range of k like you posted? Infinitely many k? But k is a natural number, right? How would k take infinitely many values in between 2 real numbers differing by 1/2?2012-10-05