Whether a function$d(m,n)=\left\vert\frac{1}{m}-\frac{1}{n}\right\vert$ metrics I saw in a magazine the following example"
Whether a function $d(m,n)=\left\vert\frac{1}{m}-\frac{1}{n}\right\vert,$ where $m,n\in\mathbb{N}$ metrics.
I know that map $d:XxX\rightarrow\mathbb{R}$ that has property:
$M1)$ $d(x,y)=0\Leftrightarrow x=y$   
$M2)$ $d(x,y)=d(y,x)$
$M3)$ $d(x,y)\leq d(x,z)+d(z,y),$ where $x,y,z\in X$ called a metric.
I've found:
$M1)$ $$d(m,n)=\left\vert\frac{1}{m}-\frac{1}{n}\right\vert=0\Leftrightarrow \frac{1}{m}-\frac{1}{n}=0\Leftrightarrow\frac{1}{m}=\frac{1}{n}\Leftrightarrow m=n,$$ using $\vert a\vert=0\Leftrightarrow a=0, and \frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc$
$M2)$ $$d(m,n)=\left\vert\frac{1}{m}-\frac{1}{n}\right\vert \cdot 1=\left\vert\frac{1}{m}-\frac{1}{n}\right\vert \cdot \vert -1 \vert=\left\vert(-1)\left(\frac{1}{m}-\frac{1}{n}\right)\right\vert=\left\vert\frac{1}{n}-\frac{1}{m}\right\vert=d(n,m)$$ 
I know not to try property $M3)$. Can someone please help me to prove $M3)$ and if I'm wrong during the confirmation plase tell me. Thanks for your help and your attention.

Of course this is a metric, although I don't like proving it directly. The abundance of fractions only serves to cloud what is really happening.
So, here is the hand-waving approach that I prefer. Suppose $f: A \to B$ is an injective map, and $d_B$ is a metric on $B$. Define a function $d_A: A \times A \to \mathbb{R}$ by
$$
d_A(x, y) = d_B(f(x), f(y)).
$$
Exercise: prove that $d_A$ is a metric on $A$. This way, with all the concrete specifics out of the way, it becomes very easy.
Now, that $d(m, n) = \left|\frac{1}{m} - \frac{1}{n}\right|$ is a metric follows from this exercise if we set $A = \mathbb{N}$, $B = \mathbb{R}$, $f(n) = \frac{1}{n}$ and $d_B$ the standand metric on $\mathbb{R}$: $d_B(x, y) = |x-y|$.
