Discuss the density of $\left\{\sqrt{m}-\sqrt{n} \mid (m,n)\in \mathbb{N}^2 \right\}$ It is clear that this set is dense in $\mathbb{R}$. To prove it, we have to find that for every $\epsilon >0, x\in\mathbb{R}$, there exist a pair $(n_x,m_x)$, such that $\left|\sqrt m_x-\sqrt n_x-x\right|\le \epsilon$.
 A: We start with the rather trivial observation that, for any real $x,k$ with $k \geq \max(-x,0)$, we have
$$\sqrt{(x+k)^2} - \sqrt{k^2} = |x+k| - |k| = (x+k) - k = x$$
We may now assume $k$ is an integer to make $n_x = k^2$ for some integer $k \geq \max(-x,0)$.

Using the above as a substitution for both $x$ and $n_x$ in your inequality, we find we are looking for natural numbers $k\geq -x$ and $m_x$ such that $$|\sqrt{m_x}-\sqrt{n_x}-x| \leq \epsilon \\ \left|\sqrt{m_x}-\sqrt{k^2}-\sqrt{(x+k)^2}+\sqrt{k^2}\right| \leq \epsilon \\ \left|\sqrt{m_x} - \sqrt{(x+k)^2}\right|\leq \epsilon$$
We note $m_x \approx (x+k)^2$, so we set $m_x = \lceil(x+k)^2\rceil$ to write
$$\left|\sqrt{m_x} - \sqrt{(x+k)^2}\right| = \frac{m_x-(x+k)^2}{\sqrt{m_x}+\sqrt{(x+k)^2}} \leq \frac{1}{2\sqrt{(x+k)^2}} = \frac{1}{2(x+k)}$$
Setting $\dfrac{1}{2(x+k)} \leq \epsilon$ yields a sufficient condition on $k$ given by $k \geq \dfrac{1}{2\epsilon}-x$.

We conclude that, for any real $x$ and any $\epsilon > 0$, we may set $$k = \max\left(\left\lceil\frac{1}{2\epsilon}-x\right\rceil,0\right), \\ n_x = k^2 \\ m_x = \left\lceil(x+k)^2\right\rceil$$ to find $(m_x,n_x) \in \mathbb{N}^2$ satisfying $|\sqrt{m_x}-\sqrt{n_x}-x| \leq \epsilon.$
