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Let $\mathbb{N}$ denote the set of positive integers. For $\alpha\in \; ]0,1[\;$, let $$\mu(n,\alpha) = \min\big\{|\alpha-\frac{b}{n}|: b\in\mathbb{N}\cup\{0\}\big\}.$$ (Note that we could have written $\inf\{\ldots\}$ instead of $\min\{\ldots\}$, but it is easy to see that the infimum is always a minimum.)

Is there an $\alpha\in \; ]0,1[$ such that for all $n\in\mathbb{N}$ we have $\mu(n+1,\alpha)<\mu(n,\alpha)$?

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3 Answers 3

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There is no such $\alpha$.

If $\alpha\in\mathbb Q$, there is $n$ such that $\mu(n,\alpha)=0$, thus $\mu(n+1,\alpha)<\mu(n,\alpha)$ is impossible.

If $\alpha\notin\mathbb Q$, a classical result in Diophantine approximation says that there are infinitely many $n$ such that $$\mu(n,\alpha)<\frac1{\sqrt5n^2}.$$ If then $$\mu(n+1,\alpha)<\mu(n,\alpha),$$ let $a/n$ and $b/(n+1)$ be the respective closest approximations of $\alpha$. We have $$\left|\frac an-\frac b{n+1}\right|<\frac2{\sqrt5n^2}<\frac1{n(n+1)},$$ while $$\left|\frac an-\frac b{n+1}\right|=\frac{|a(n+1)-bn|}{n(n+1)}\ge\frac1{n(n+1)}$$ unless $a(n+1)=bn$, i.e., the approximating fractions are $0$ or $1$. Since this happens for infinitely many $n$, this is impossible for $\alpha\in(0,1)$.

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No. If $\alpha$ is rational, set $n$ to the denominator of $\alpha$. Otherwise set $n$ to the denominator of the third convergent to $\alpha$. In both cases, we get $\mu(n+1,\alpha)>\mu(n,\alpha)$.

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It's easy to proof that $\alpha$ shouldn't be a rational number.

Now, let $\frac{1}{n-1}>\alpha>\frac{1}{n}, n>1$ and $\alpha-\frac{1}{n} < \frac{1}{n-1}-\alpha$.

Then, $\mu(\alpha, k+1)< \mu(\alpha ,k)$ for all $k=1,2..., n-1$.

If $\mu(\alpha, n+1)<\mu(\alpha, n)$,

then either,

  1. $\frac{1}{n-1}>\alpha >\frac{b}{n+1} >\frac{1}{n}$ for some $b \in \mathbb N, b>1$. or,
  2. $\frac{1}{n-1}>\frac{b}{n+1} >\alpha >\frac{1}{n} $ for some $b \in \mathbb N, b>1$ (With $\frac{b}{n+1}+\frac{1}{n}>2\alpha$).

Both of these implies

$1+\frac{2}{n-1}>b>1+\frac{1}{n} \Rightarrow n=2$

To satisfy $\mu(\alpha, 4)<\mu(\alpha, 3)<\mu(\alpha, 2)<\mu(\alpha, 1)$ we need, $\frac{3}{4}>\alpha>\frac{17}{24}$.

But, $\frac{4}{5}>\frac{3}{4}$ and $\frac{3}{4}-\frac{17}{24}<\frac{17}{24}-\frac{3}{5}$, hence, $\mu(\alpha, 5)>\mu(\alpha, 4)$.

So, there can't be any such $\alpha \in (0,1)$.

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    $\begingroup$ Perhaps some kind of an argument like this can be made to work, but as far as I can see, it is not correct as written: if $17/24<\alpha<3/4$, then $n=2$, but $\mu(1,\alpha)>\mu(2,\alpha)>\mu(3,\alpha)>\mu(4,\alpha)$. $\endgroup$ Aug 27, 2020 at 16:13
  • $\begingroup$ Thank you for correcting me. I missed that. $\endgroup$
    – Alapan Das
    Aug 27, 2020 at 16:27

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