Proving a set X is dense in [0,1] equivalence relation [duplicate]

Question

Let the relation in $\mathbb{R}: x \equiv y \ \mbox{mod} \ \mathbb{Z}$, when $x-y \in \mathbb{Z}$. For each $n \in \mathbb{N}$, let $x_n \in [0,1)$ such that $x_n \equiv \sqrt{n} \ \mbox{mod} \ \mathbb{Z}$. Prove $X = \{x_n : n \in \mathbb{N} \}$ is dense in $[0,1]$. I don't know where to start, but I have doubts about using the greatest integer.

Answer 1

Hint: For any $k \in \Bbb N$: if $k^2 \leq n < (k+1)^2$, then we find that $$ \sqrt{(k+1)^2} - \sqrt{(k+1)^2 - 1} \leq \sqrt{n+1} - \sqrt{n} \leq \sqrt{k^2 + 1} - \sqrt{k}. $$ Note that the upper bound $$ \sqrt{k^2 + 1} - \sqrt{k} = \frac{1}{\sqrt{k^2 + 1} + \sqrt{k}} $$ decreases towards $0$ as $k \to \infty$.

Answer 2

Suppose $I$ is an interval with $I\subseteq (0,1),$ and with length $l(I)=(\sup I)-(\inf I)>0.$ Suppose $S$ is a finite subset of $[0,1]$ with $0\in S$ and $1\in S,$ such that if $s,s'$ are two $consecutive$ members of $S$ then $|s-s'|<l(I).$ Then we have $I\cap S\ne\emptyset.$

So take $k\in \Bbb N$ with $1/2k<l(I).$ For integer $j$ with $0\le j\le 2k+1,$ let $\sqrt {k^2+j}=k+D(j).$ Let $S=\{D(j):0\le j\le 2k+1\}.$