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



Approximating by incrementing positive integers
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 $\alpha\in(0,1)$.


