Prove that the following set is dense

It is hard for me to show that the set $\{\sqrt{m}-\sqrt{n}; m,n\in \Bbb N\}$ is dense in $\Bbb R$. Please help me.

Solution 1:

Let $A=\{\sqrt{m}-\sqrt{n}; m,n\in \Bbb N\}$. We need to show that for every open interval $(a,b)$ we have $(a,b)\cap A \neq \emptyset$. To prove it we'll use two facts:

1. For any $\epsilon \ge 0$ we can find $x=\sqrt{m}-\sqrt{n}$ for some $n,m\in\mathbb{N}$ such that $0< x\le \epsilon$. To see it just consider $$\sqrt{m+1}-\sqrt{m}=\frac{1}{\sqrt{m+1}+\sqrt{m}}$$

2. If $x\in A$ and $k\in\mathbb{Z}$, then $kx\in A$.

Then consider open interval $(a,b)\subset\mathbb{R}$, take $\epsilon=\frac{b-a}{2}$ and by (1) there is $x$ such that $0<x\le\epsilon<b-a$. Now, as $x$ is smaller than length of $(a,b)$, we can find $k\in \mathbb{Z}$ such that $kx\in (a,b)$, and also $kx\in A$ by (2), which shows that $(a,b)\cap A$ is non-empty, and therefore $A$ is dense in $\mathbb{R}$.

Solution 2:

Hint: $\sqrt{k^2n+k^2}-\sqrt{k^2n}=\frac{k}{\sqrt{n}+\sqrt{n+1}}=\frac{k}{2\sqrt{n}+\sqrt{n+1}-\sqrt{n}}=\frac{k}{2\sqrt{n}+\frac{1}{\sqrt{n}+\sqrt{n+1}}}$

$$\sqrt{k^2n+k^2}-\sqrt{k^2n}=\frac{k}{2\sqrt{n}+\frac{1}{\sqrt{n}+\sqrt{n+1}}}$$

To show that the set is dense in $\mathbb{R}$, it suffices to estimate all rationals. To estimate the rational $\frac{p}{q}$:

1. Get a large natural number $m$. Now we consider the rational number $\frac{2pm}{2qm}$
2. Set $k=2pm$ and $n=(qm)^2$. It follows that $2\sqrt{n}=2qm$. Since $m$ is large, therefore $n$ would be large. Thus, the fraction $\frac{1}{\sqrt{n}+\sqrt{n+1}}$ that appears in the denominator would be negligible.