# Dense set of numbers

Show that the set $$\{\sqrt{n}-[ \sqrt{n}]; n\in \Bbb N\}$$ is dense in $$\Bbb [0,1]$$.

I showed that for every real number $$a$$ such that $$\ 0 there is at least one number $$x$$ from the set such that $$\ a< x$$.

And for every real number $$b$$ such that $$\ 0 there is at least one number $$y$$ from the set such that $$\ b>y$$

But I don't know how to prove that: for every real number $$a$$ such and every real number $$b$$ such that $$\ 0 there is at least one number $$z$$ from the set such that $$\ a

• Welcome to MSE. This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – user296602 Oct 20 '18 at 21:16

If suffices to show that for every $$\alpha \in \mathbb{Q} \cap [0,1]$$ there exists a sequence $$n_k$$ such that $$\sqrt{n_k} - \lfloor \sqrt{n_k} \rfloor$$ converges to $$\alpha$$. Let $$q_k, p_k$$ such that: (a) $$q_k \nearrow \infty$$, (b) $$\alpha = \frac{p_k}{2q_k}$$. Note that $$p_k \leq 2q_k$$. Define $$n_k = q_k^2 + p_k$$. It is easy to verify that $$q_k^2 \leq n_k < (q_k + 1)^2$$, so $$\lfloor \sqrt{n_k} \rfloor = q_k$$. Moreover, $$\sqrt{n_k} - q_k = \frac{(\sqrt{n_k} - q_k)(\sqrt{n_k} + q_k)}{\sqrt{n_k} + q_k} = \frac{p_k}{\sqrt{n_k} + q_k},$$ and $$\frac{p_k}{2q_k + 1} \leq \frac{p_k}{\sqrt{n_k} + q_k} \leq \frac{p_k}{2q_k} = \alpha.$$
Finally, $$\alpha - \left( \sqrt{n_k} - q_k \right) = \alpha - \frac{p_k}{\sqrt{n_k} + q_k} \leq \alpha - \frac{p_k}{2q_k + 1} = \frac{p_k}{2q_k(2q_k+1)} = \frac{\alpha}{2q_k+1} \leq \frac{1}{2q_k+1} \to 0.$$