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Let $0<r<1$ a real number which is not a fraction of the form $p/2^n$ for any integers $p,n$. Now, for every integer $n\ge 1$ we can find the closest fraction of the form $p/2^n$ to $r$, which will be on the left or on the right of $r$. For example, for $r=2/3$, those fractions will be: $1/2$, $3/4$, $5/8$, $11/16$ etc.

We will now look at the sequence $a_n=(3/2)^n\pmod 1$, which itself contains fractions of the form $p/2^n$ from the interval $(0,1)$, and could, for any given $r$, potentially "hit" one of those "closest" fractions to $r$.

Now, my question is this: can it be proven that, for every $r$, $0<r<1$, not of the form $p/2^n$, the sequence $a_n$ above contains at least one of those "closest" fractions to $r$? It seems intuitive to me that this should be true.

Stinking Bishop
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    Do you mandate $x$ to be an integer, or at least a fraction? (Obviously, if not, this is easy: take e.g. $x=\log_{3/2}(7/4)$, for example and $(3/2)^x\equiv 3/4\pmod 1$ etc.) – Stinking Bishop Sep 08 '21 at 20:56
  • x must be a positive integer. – user965964 Sep 08 '21 at 21:15
  • Please make the body of your MSE questions self-contained: don't rely on the title for important information. In this case, you should copy the title in as the first sentence in the body. – Rob Arthan Sep 08 '21 at 21:23
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    @user965964: What do you mean by *"a fraction immediately next to a real number"*? And what do you mean by *"the real number generated by $(3/2) ^n \pmod 1$ when $n \in \mathbb N$"*? – Alex M. Sep 08 '21 at 21:24
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 08 '21 at 21:25
  • ... further to @AlexM's comment, the role of $r$ in the statement of your question is also unclear. – Rob Arthan Sep 08 '21 at 21:26
  • @user965964 I have substantially edited your question to make it clearer, please check if I have done it right. In addition, it won't harm if you could add any attempts you've had at solving it, or any clues about what motivated you to ask it in the first place. I've also voted to reopen it, it seems interesting to me. Good luck! – Stinking Bishop Sep 09 '21 at 06:42

1 Answers1

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The fraction $a_1=3/2\pmod 1 = 1/2$ is the "closest" to all the numbers in $(1/4, 3/4)$. The fraction $a_2=9/4\pmod 1 = 1/4$ is the "closest" to all numbers in $(1/8, 3/8)$ etc. Generally, the fraction $a_n$ will be the closest to all the numbers in $(a_n-1/2^{n+1}, a_n+1/2^{n+1})$. So the numbers $r$ for which one of $a_n$'s is the closest are precisely those contained in the union of those intervals $(a_n-1/2^{n+1}, a_n+1/2^{n+1})$.

Now, the length of the interval $(a_n-1/2^{n+1}, a_n+1/2^{n+1})$ is $1/2^n$ so the total size (measure) of the union cannot exceed the sum of all of the lengths, which is $1/2+1/4+1/8+\ldots=1$. However, obviously, if any two of those intervals have a "nontrivial" intersection (however small, of size $\alpha>0$) the total measure of the union won't exceed $1-\alpha<1$.

Finally, it is enough to notice that the first two intervals $(1/4, 3/4)$ and $(1/8,3/8)$ already have a nontrivial intersection. Thus, the total measure of the union is strictly smaller than $1$, and so there must be a number $r$ which does not belong to this union. This $r$ will consequently never have any of the $a_n$'s as the "closest" to it.

In short, the statement you are trying to prove is not true.

Stinking Bishop
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