Suppose $0<x<1$. Let $k$ is any integer, and $n$ any positive integer. Fix $k,n$. If $q$ is such integer that $qn\le k<(q+1)n$, then how to show that there exists positive integer $p$ such that $k<px+qn<k+1$?
Context: I am reading solution to Rudin exercise 25, ch 4. There it is mentioned.
First note that we must have $n>0$ because of $q_n \leq k < (q+1)n$.
Case 1: If $qn=k$, we can take $p=1$, because $$k<x+k=x+qn < k+1.$$
Case 2: Now, let $k >qn$, then take $p = \lceil (k - qn)/(x-\delta) \rceil$. Here, we have $$k<\frac{(k-qn)}{x-\delta}x+qn \leq px+qn < \frac{(k-qn)}{x-\delta}x +x+qn < \frac{x}{x-\delta}k +x < k+1,$$ where the last estimate is satisfied if we take $\delta >0$ small enough.