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Let $\alpha \in \mathbb{R} $\ $\mathbb{Q}$, then I claim that there exist positive integers $m,k$ such that $2^m > k\alpha > 2^m - 1$.

I tried many elementary approaches but all of them either reformulated the problem or cycled back on itself. This area of maths is uncharted territory for me and so I don't know any advanced approaches I could use, but I am open to anything.

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  • $\begingroup$ Do you want to assume also $\alpha>0$? [guess that's necessary, given $m,k$ positive integers]. $\endgroup$ – coffeemath Feb 26 '17 at 23:20
  • $\begingroup$ Yes, in fact you can assume $\alpha > 1$, since the problem for $0 < \alpha < 1$ is simple. $\endgroup$ – TCiur Feb 26 '17 at 23:34
  • $\begingroup$ @TCiur have you tried Kronecker's approximation theorem? $\endgroup$ – dezdichado Mar 24 '17 at 21:15
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The claim is false; there exist $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ that cannot be approximated in this way.

To see this, rewrite the desired inequality as $$\frac{k}{2^m} < \frac1\alpha < \frac{k}{2^m-1} \tag{1}$$ If the right-hand inequality holds at all, it holds when $k$ is chosen as large as it can be while the left-hand inequality holds: $k = \left\lfloor \frac{2^m}{\alpha} \right\rfloor$. In other words, an equivalent task is to find $m$ such that $$\frac1\alpha < \frac{\lfloor 2^m/\alpha \rfloor}{2^m - 1} \tag{2}$$ We'll assume that $\alpha > 1$, so that $\frac1\alpha \in (0,1)$. Suppose that the binary expansion of $\frac1\alpha$ is $0.b_1b_2b_3b_4\dots$. In that case, $k = \lfloor 2^m/\alpha \rfloor$ is the integer whose binary expansion is $b_1b_2b_3\dots b_m$, and $\frac{\lfloor 2^m/\alpha \rfloor}{2^m - 1}$ is the real number whose binary expansion repeats as follows: $$0.b_1b_2b_3\dots b_mb_1b_2b_3\dots b_mb_1b_2b_3\dots b_m \dots$$ So to rule out the possibility that $(2)$ holds for any $m$, it is sufficient to choose $\frac1\alpha$ to satisfy the following:

  1. $b_1 = b_2 = 0$.
  2. In the binary expansion of $\frac1\alpha$, we never have $b_i = b_{i+1} = 0$ for any $i>1$.

Then for any $m$, $\frac1\alpha$ and $\frac{\lfloor 2^m/\alpha \rfloor}{2^m - 1}$ agree in the first $m$ bits, but the $(m+1)$-th and $(m+2)$-th bits of $\frac{\lfloor 2^m/\alpha \rfloor}{2^m - 1}$ are both $0$ to match $b_1$ and $b_2$, whereas at least one of $b_{m+1}$ and $b_{m+2}$ is $1$. Therefore $\frac{\lfloor 2^m/\alpha \rfloor}{2^m - 1} < \frac1\alpha$ and $(2)$ does not hold.

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