Limit point of set $\{\sqrt{m}-\sqrt{n}:m,n\in \mathbb N\} $

Question

How can I calculate the limit points of set $\{\sqrt{m}-\sqrt{n}\mid m,n\in \mathbb N\} $?

Answer 1

Intuition says that every real number is a limit point. So given a real number $a$, we want to show that there are integers $m$ and $n$ such that $\sqrt{m}-\sqrt{n}$ is close to $a$. Without loss of generality we may assume that $a\ge 0$. Given $\epsilon \gt 0$, we want to produce $m$ and $n$ such $|(\sqrt{m}-\sqrt{n})-a|\lt \epsilon$.

One idea is to note that $\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}$. So there is an integer $d=d(\epsilon)$ such that $0\lt \sqrt{d+1}-\sqrt{d} \lt \epsilon$.

Now consider the numbers $k(\sqrt{d+1}-\sqrt{d})=\sqrt{k^2d+k^2}-\sqrt{k^2d}$, as $k$ ranges over the positive integers. For every $a\ge 0$, there is a positive integer $k$ such that $k(\sqrt{d+1}-\sqrt{d})$ is at distance less than $\epsilon$ from $a$.

Answer 2

Let $a$ be a real number. To show that $a$ is a limit point, it's enough to show that for any $N>0$, some $\sqrt{m} - \sqrt{n}$ is within $1\over N$ units of $a$.

• Choose $m=M^2$ large enough so that the consecutive differences in the sequence $M=\sqrt{m},\sqrt{m+1},\sqrt{m+2},\dots,\sqrt{m+2M+1}=M+1$ are all less than $1\over N$. This will be the case if the derivative of $\sqrt{x}$ at $x=m$ is less than ${1\over N}$, or when $M>{\lceil{N\over2}\rceil}$.
• Let $n=(M-\lfloor{a}\rfloor)^2$. Then $\sqrt{m} - \sqrt{n}=\lfloor{a}\rfloor$ and $\sqrt{m+2M+1} - \sqrt{n}=\lfloor{a}\rfloor+1$
• Then $\sqrt{m+i} - \sqrt{n}$ must be within $1\over N$ of $a$ for some $i$.