Triangle Inequality Proof Review for $d(x, y) = |x^2 - y^2|$ I am trying to show that the function $d: \mathbb{N} \times \mathbb{N} \to \mathbb{R}$ given by $d(x, y) = |x^2 - y^2|$ is a metric on the natural numbers $\mathbb{N}$.
For the triangle inequality, the provided proof is as follows:
$d(x, z) = |x^2 - z^2|$ (By the definition of $d$.)
$= |x^2 - y^2 + y^2 - z^2|$ (Adding and subtracting $y^2$.)*
$\le |x^2 - y^2| + |y^2 - z^2|$ (Triangle inequality for $\mathbb{R}$.)
$= d(x, y) + d(y, z)$
What I'm concerned about is the following line of the proof:
$= |x^2 - y^2 + y^2 - z^2|$
This line seems superfluous to me. It really does not seem to be justifying anything, nor does it seem to be adding value in any way to the proof. So why can't we just leave this part out of the proof and go straight to
$\le |x^2 - y^2| + |y^2 - z^2|$ (Triangle inequality for $\mathbb{R}$.)
?
I would greatly appreciate it if people could please take the time to clarify this.
 A: The triangle inequality for $\mathbb{R}$ only tells us that $|x-z|<|x|+|z|$. It doesn't say anything about introducing a third point $y$. This is, in a very technical sense, an application of the triangle inequality, with the missing step being the adding and subtracting of $y$. So the triangle inequality tells us that $|x-y+y-z|<|x-y| + |y-z|$ but it doesn't tell us that $|x-z|=|x-y+y-z|$.
A: That step is actually essential to make the proof a "proof": maybe looking at the process from another angle will help. You know that the triangle inequality holds:
$$|A + B| \leq |A| + |B| \qquad \forall A,B \in \mathbb N $$
and you know that $d(x,z) = |x^2 - z^2|$. Now pick a $y \in \mathbb N$ and let $a, b \in \mathbb N$ such that $a^2 := x^2 - y^2$ and $b^2 = y^2 - z^2$. Then $x^2 - z^2 = a^2 + b^2$, so that
$$d(x,z) = |a^2 + b^2| $$
and you may apply the triangle inequality with $A = a^2$ and $B = b^2$ to obtain
$$d(x,z) \leq |a^2| + |b^2|. $$
Substituting for the values of $a^2$ and $b^2 $ you get 
$$d(x,z) \leq |x^2 - y^2| + |y^2 - z^2| = d(x,y) + d(y,z). $$
