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 $\mu(n+1,\alpha)<\mu(n,\alpha)$?
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 $\alpha\in(0,1)$.
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 $\mu(n+1,\alpha)>\mu(n,\alpha)$.
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,
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 $\alpha \in (0,1)$.
