I am trying to prove the following:
Let $n \in \mathbb{N}$. Then among any $n$ real numbers, there exists a pair of them, say $r_1, r_2 \in \mathbb{R}$, such that $||r_1 - r_2| - k| \le \frac{1}{n}$ for some $k \in \mathbb{Z}$.
(In other words, there's a pair that's within $\frac{1}{n}$ of an integer).
If we just have any $2$ real numbers, it's clear to me that their difference must either be an integer, or some decimal that clearly lies within $\frac{1}{2}$ of an integer (gotten by either rounding up or down). Trouble is, I'm having trouble seeing the reason why this extends to longer sets of reals, much less actually proving it formally... Nonetheless, here's my attempt.
If we consider $3$ real numbers, say $\{r_1, r_2, r_2\}$, then we know that each of the 3 combinations of pairs of reals have a difference that's within $\frac{1}{2}$ of a real number. By considering only the decimal part of these differences and imagine a number line going from $0$ to $1$, I can see that we can spread these "decimals" by at most $\frac{1}{3}$ from each other, and that trying to make the distance between any two decimals bigger shrinks the difference between another. My best attempt at a proof would be to use the exact same argument as above, just with $k$ instead of $3$.
Is there a less hand-wavy way of proving this, or is my approach more or less okay?
Let the numbers be $x_1, \ldots, x_n$.
Since only the differences mod $1$ are important, wlog we can suppose $x_1 = 0$ and $x_2, \ldots, x_n \in [0,1]$. Sort them in increasing order. If no pair are within distance $1/n$, we'd have $x_2 > 1/n$, $x_3 > x_2 + 1/n > 2/n$, ..., $x_n > (n-1)/n$. But then $|x_n - x_1| - 1 < 1/n$.