The following question was asked by one of my students.He didn't know the conclusion was correct. I thought for a long time and felt right.Because we seem to be using Kronecker theorem
It follows from Kronecker's density theorem stating that if $\theta$ is an irrational real number, $\alpha$ is a real number, and $\varepsilon>0$ is any positive real number, then there exist integers $h,k$ with $0<k$ such that $$ |k\theta-h-\alpha|<\varepsilon. $$
Problem:
Suppose $c\in (0,1)$ and irrational numbers $\alpha,\beta$ are such that $$(\{n\alpha\}-c)(\{n\beta\}-c)\ge 0,\forall n\in N^{+}.$$ Prove or disprove $$\{\alpha\}=\{\beta\}.$$
We indeed must have $\{\alpha\} = \{\beta\}$. Otherwise, as the comments point out, we may assume $\alpha = \beta+\frac{p}{q}$ for some $p \ge 1,q \ge 2, \gcd(p,q) = 1$.
First assume $c > 1/q$. Take $0 < \epsilon < \min(1-c,\frac{1}{q})$ and $n \equiv p^{-1} \pmod{q}$ with $\{n\beta\} \in (c-\frac{1}{q},c-\frac{1}{q}+\epsilon)$, which we may do since $q\beta$ is irrational. Then $\{n\alpha\} = \{n\beta\}+\frac{1}{q} > c$.
If $c \le \frac{1}{q}$, then $c+\frac{1}{q} \le 1$. Take $0 < \epsilon < c$ and $n \equiv -p^{-1} \pmod{q}$ with $\{n\beta\} \in (c+\frac{1}{q}-\epsilon,c+\frac{1}{q})$. Then $\{n\alpha\} = \{n\beta\}-\frac{1}{q} < c$.