Let x be an irrational number. Show that for some positive integer j not exceeding the positive integer n, the absolute value of the difference between j x and the nearest integer to j x is less than 1/n.
I've referred to its solution but not able to understand. Can someone please explain how can I solve this question ? Thanks in advance.
Let $a_j=jx-\lfloor jx\rfloor\in[0,1)$ for $j\in\mathbb{Z}^+,j\le n$. As $x\notin\mathbb{Q}$, $a_j\notin\{0,\frac{1}{n},\frac{2}{n},\dots,\frac{n-1}{n},1\}$.
Let $I_k=(\frac{k-1}{n},\frac{k}{n})$ for $k=1,2,\dots,n$. By the Pigeonhole principle, at least one of the following two statements is true.
(1) For each $k\in\{1,2,\dots,n\}$, there exist an $i\in\{1,2,\dots,n\}$ such that $a_i\in I_k$.
(2) There exist a $k\in\{1,2,\dots,n\}$ such that $a_i,a_j\in I_k$ for some $i,j\in\{1,2,\dots,n\}$ with $i<j$.
If (1) is true, let $a_l\in I_1$. Then the difference between $lx$ and the nearest integer to $lx$ is $a_l<\frac{1}{n}$.
If (2) is true, then $a_{j-i}\in I_1$ and by the above argument, the difference between $(j-i)x$ and the nearest integer to $(j-i)x$ is less than $\frac{1}{n}$.
Edited:
if $n=1$, then no positive number $j$ meets $j<n$. But that the difference between $x$ and the nearest integer to $x$ is less than $1$ is trivial when $x\in\mathbb{Q}$.