Why is the metric on $\mathbb{N}$ defined as the following? This is from Muscat's Functional Analysis:http://staff.um.edu.mt/jmus1/metrics.pdf
Show that $d(m,n) = |\dfrac{1}{m} - \dfrac{1}{n}|$, $m,n \in \mathbb{N}$
So the first two properties of the metric are trivial. The last one is also trivial:
$d(m,r) = |\dfrac{1}{m} - \dfrac{1}{r}| =  |\dfrac{1}{m} - \dfrac{1}{n} + \dfrac{1}{n} - \dfrac{1}{r}| \leq d(m,n) + d(n,r)$
Then I realized, this is the exactly the proof to show that $(\mathbb{R}, |x-y|)$ is a metric space
Does anyone have any special insight as to why this particular metric is defined for the natural numbers? 
 A: This is a good observation. It can be explained by noticing the metric space $(\mathbb N,d)$ given to you is basically the same as the metric space $(S,d')$ where $S$ is the defined as
$$S=\left\{\frac{1}n:n\in\mathbb N\right\}$$
and $d'(x,y)=|x-y|$ is the usual metric on $\mathbb R$.
To notice this, just see that the map $f(n)=\frac{1}n$ is a bijection between $\mathbb N$ and $S$ and satisfies that
$$d(n,m)=d'(f(n),f(m))$$
which basically means that we're just identifying $\mathbb N$ with a subset $S$ of $\mathbb R$ - and since we already knew that the latter was a metric space, so must be the former. More generally, if $f$ is an injective function from a set $A$ to a set $B$ and $(B,d')$ is a metric space, then $d(x,y)=d'(f(x),f(y))$ is a metric on $A$.
A: I think that the thing you're seeing is that the map 
$$
h : \mathbb N \to \mathbb R : k \mapsto \frac{1}{k} 
$$
is an isometry onto its image, where the target space $\mathbb R$ is given the usual metric, and the domain is given your unusual metric. 
A: According to your question: "Does anyone have any special insight as to why this particular metric is defined for the natural numbers? " one answer is exactly as in the answer of Milo Brandt. Another reason is the following: metrics always induce topology and the most natural topology on $\mathbb{N}$ is the discrete topology. Why it is natural? It is so because the induced topology from $\mathbb{R}$ is discrete. And it is not hard to see that your metric also induces discrete topology (since if $n_k \to n$ in your metric means that this sequence is constant from some moment). But please note that in your metric $\mathbb{N}$ is not complete(for example the sequence $(n)_n$ is Cauchy but not convergent)-opposite to the standard discrete metric defined by $d(n,m)=1$ for $n \neq m$ and $d(n,n)=0$. 
