# Complete metric space on unique metric

Consider the metric $$d(m,n) = \frac{|m-n|}{mn}$$

Is this metric on the natural numbers $$(1,2,\ldots)$$ complete?

I'm struggling but heres an idea I have from reading other similar questions.

The sequence given by $$n^2 = \{1,4,9,16,\ldots\}$$ is Cauchy in this metric space but the only way it converges is if there exists a natural number $$m$$ such that $$d(n^2,m)=0$$. But this implies $$n=m=1$$. This is a contradiction, right? So it is not a complete metric space?

It is true that $$(\mathbb N,d)$$ is not a complete metric space. Consider the sequence $$1,2,3,4,\ldots$$, which is a Cauchy sequence. However, it doesn't converge. Fix $$N\in\mathbb N$$. Then the distance from $$N$$ to any other $$M\in\mathbb N$$ (distinct from $$N$$) is at least $$\frac1{N(N+1)}$$. So, no injective sequence can possible converge to $$N$$.
No, not quite. It's perfectly possible for convergent (and hence Cauchy) sequences to never attain their limit. The fact that $$d(m, n^2) > 0$$ is not sufficient to show that $$n^2$$ fails to converge.
Here's another way to look at it. Note that the metric can be expressed like so: $$d(m, n) = \left|\frac{1}{n} - \frac{1}{m}\right|.$$ If we let $$f : \Bbb{N} \to [0, 1] : m \mapsto \frac{1}{m},$$ where the domain has the metric $$d$$ and the set of reciprocals has the usual absolute value metric, then $$f$$ is an isometry, that is, $$|f(m) - f(n)| = d(m, n)$$.
Note that the sequence $$n^2$$ maps to $$\frac{1}{n^2}$$, which converges in $$[0, 1]$$ to $$0$$. This makes $$n^2$$ Cauchy, since $$f$$ is an isometry. But $$0$$ is not in the range of $$f$$; if there were some $$L \in \Bbb{N}$$ such that $$n^2 \to L$$, then $$f(n^2) \to f(L) \neq 0$$. This would contradict the uniqueness of limits. Thus $$n^2$$ fails to converge.