Let $n$ be a positive integer, and $r:=\frac{p}{q}<1$ where $\mathrm{gcd}(p,q)=1$.


I am interested in the product $n\cdot r$


Whenever $n$ is a multiple of $q$, a property of rational numbers is that
$$
\{n\cdot r\}=0
$$
where $\{.\}$ represents the fractional part of a number.


I am interested in studying the local and global minima of the fractional part of $n\cdot r$ as $n$ increases. If we were to plot $\{n\cdot r\}$ vs $n$, we would observe a periodic pattern.


My question is as follow: For what values of $r$ or what is the condition on $r$ such that all the local minima of $\{n\cdot r\}$ are global? Has this been studied before?


For example $r=0.75$ and $n$ is an increasing positive integer. We notice that all the local minima in the fractional part are zero, where as for $r=0.55$, we see that there are local minima that are not global.



On the existence of locals that are global in $np\;\mathrm{mod}\;q$
Local minima which are not global minima will exist for all rational numbers $r=\frac{p}{q}$ as long as $p\neq 1,q-1$. Indeed, by [Bézout's identity](https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity), there is some $n$ such that $\{nr\}=\frac{1}{q}$, and except in those two cases above, both $(n-1)r$ and $(n+1)r$ will have greater fractional part, as it will be of the form $\frac{k}{q}$ for $k$ which cannot be $0$ or $1$.


Though this is beyond the scope of your question, let me mention for completeness that such non-global local minima will also occur whenever $r$ is irrational. Indeed, for irrational $r$ we have that $\{nr\}$ is dense in $(0,1)$, so in particular it has no global minima, and it will take value between $0$ and $\min(r,1-r)$, which you can check will necessarily be a local minimum.


