Rational Numbers are Well-Orderable/Proof 2

Proof
The rational numbers are arranged thus:


 * $\dfrac 0 1, \dfrac 1 1, \dfrac {-1} 1, \dfrac 1 2, \dfrac {-1} 2, \dfrac 2 1, \dfrac {-2} 1, \dfrac 1 3, \dfrac 2 3, \dfrac {-1} 3, \dfrac {-2} 3, \dfrac 3 1, \dfrac 3 2, \dfrac {-3} 1, \dfrac {-3} 2, \dfrac 1 4, \dfrac 3 4, \dfrac {-1} 4, \dfrac {-3} 4, \dfrac 4 1, \dfrac 4 3, \dfrac {-4} 1, \dfrac {-4} 3 \ldots$

It is clear that every rational number will appear somewhere in this list.

Thus it is possible to set up a bijection $f: \Q \to \N$ between each rational number and its position in the list, which is an element of $\N$.

We set up an ordering $\preccurlyeq$ defined as:
 * $\forall x, y \in \Q: x \preccurlyeq y \iff \map f x \le \map f y$

where $\le$ is the usual ordering on $\N$.

It remains to demonstrate that $\preccurlyeq$ is a well-ordering.