Prove that $A$ is dense in $\bf R$ Let  $A =\{\sqrt{m} - \sqrt{n }:     \text{$n$ and $m$ are positive natural numbers}\}$.
Prove that $A$ is dense in $\bf R$, 
I really can't find two natural numbers between $x$ and $y$.
I tried using $E(x)$ and $E(y)$ but with no good result.
 A: Suppose you have any interval $(a,b)$ and want to show that it contains an element of $A$. Assume without loss of generality (why?) that the interval lies in the positive reals.
Let $\delta=b-a$, and select $n$ large enough that $\sqrt{k+1}-\sqrt{k} < \delta$ for all $k\ge n$. (This is always possible because the difference between successive square roots converges to $0$ -- consider for example the derivative of the square root function and the mean value theorem).
Now, as you increase $m$ from $n$ and upwards in steps of $1$, the value of $\sqrt m-\sqrt n$ increases in increments of less than $\delta$. Therefore, at least one of these differences must fall within our initial interval of length $\delta$.
A: Assume that $x\gt0$ and, for every $n\geqslant0$, consider $$m_x(n)=\lceil(x+\sqrt{n})^2\rceil,$$ then $\sqrt{m_x(n)}-\sqrt{n}\geqslant x$ and $m_x(n)\lt(x+\sqrt{n})^2+1$ hence $$\sqrt{m_x(n)}-\sqrt{n}\leqslant\sqrt{(x+\sqrt{n})^2+1}-\sqrt{n}\leqslant\frac{(x+\sqrt{n})^2+1-n}{\sqrt{(x+\sqrt{n})^2+1}+\sqrt{n}}\leqslant\frac{2x\sqrt{n}+x^2+1}{2\sqrt{n}},$$ that is, $$\sqrt{m_x(n)}-\sqrt{n}\leqslant x+\frac{x^2+1}{2\sqrt{n}}.$$ Thus, for every positive $\varepsilon$, if $n$ is large enough, then $$x\leqslant\sqrt{m_x(n)}-\sqrt{n}\leqslant x+\varepsilon.$$
If $x\lt0$, exchange $m$ and $n$.
