Accumulation points of $\{ \sqrt{n} - \sqrt{m}: m,n \in \mathbb{N} \}$ This is my first post on MSE, so, pardon me if I'm not used to the site's rule yet.
I'm trying to prepare myself for competitions in the future and I'm trying to improve my problem solving skills. I've come accross this problem from the book "Problems in mathematical analysis" by Witkowski and Piotr. It asks us to find the set of accumulation points of $$A=\{ \sqrt{n}-\sqrt{m}: m,n \in \mathbb{N} \}$$
Well, I guess $0$ is one of the accumlation points. Because if we set $n=m+1$ then we'll find a non-constant subsequence that converges $0$. 
I've been trying to find other non-constant sequences that converge to other real numbers, but I haven't succeeded so far. I'm very hesitant to think that $0$ is the only accumulation point of $A$. The main problem is that $n$ and $m$ must be natural numbers, that is very restricting for my intuition.
I tried several subsequences of $\sqrt{n}-\sqrt{m}$ by hand or by using wolfram alpha but either they diverged to infinity or they converged to $0$.
So, please, give me some hints. I'm NOT looking for a full solution YET! Just give me some ideas about how a problem solver might attack this problem. If this problem shows up in a competition, how should I analyze it?
Should I start by trying several subsequences of it to get a picture or I should think more abstractly? If you're giving a hint, please try to explain how that idea has come to you.
Thanks in advance.
 A: I don't know the answer (I have a conjecture, however), but in looking at
$$\sqrt{n} - \sqrt{m} \; = \; \frac{n-m}{\sqrt{n} + \sqrt{m}},$$
I notice that if for each positive integer $m$ we choose a value of $n$ whose distance from $m + \sqrt{m}$ is at most $1\;$ (that is, $\;m + \sqrt{m} - 1 \leq n \leq m + \sqrt{m}+1)\;$ then, using the fact that $f(x) = \frac{x - m}{\sqrt{x} + \sqrt{m}} = \sqrt{x} - \sqrt{m}$ is an increasing function for $x>0,$ we get
$$\frac{\left(m + \sqrt{m} - 1\right) \; - \; m}{\sqrt{m + \sqrt{m} - 1} \; + \; \sqrt{m}} \;\; \leq \;\; \frac{n - m}{\sqrt{n} \; + \; \sqrt{m}}  \;\; \leq \;\; \frac{\left(m + \sqrt{m} + 1\right) \; - \; m}{\sqrt{m + \sqrt{m} + 1} \; + \; \sqrt{m}} $$
$$\frac{\sqrt{m} - 1}{\sqrt{m + \sqrt{m} - 1} \; + \; \sqrt{m}} \;\; \leq \;\; \sqrt{m} - \sqrt{n} \;\; \leq \;\; \frac{\sqrt{m} + 1}{\sqrt{m + \sqrt{m} + 1} \; + \; \sqrt{m}} $$
Now notice that letting $m \rightarrow \infty$ for these choices of $m$ and $n,$ we get $\sqrt{n} - \sqrt{m} \rightarrow \frac{1}{2}$ by the squeeze theorem (also called the sandwich theorem) for limits.
A: A good first approach to any problem is to learn what you can from simple cases. In my experience, only after you've exhausted this avenue should you try other methods. In this particular problem, a natural family of simple cases is to let $n$ and $m$ be functions of some other parameter, and then see what you can find. You've already tried $n=k+1$ and $m=k$ to conclude that zero is an accumulation point. What are other natural functions to try?
