I know that the rational numbers form an ordered field, but I would like to order them differently.
If I use the function $f:Q->Q$, where f(x)=$\ \begin{cases} 0 & x=0 \\ 1/x & x\neq0 \\ \end{cases} $
and then following the naturally induced ordering $x_1<x_2$ iff $f(x_1)<f(x_2)$, will this effectively re-order the rationals?
I have read that ordered fields are ordered "linearly." Intuitively, I think of this as just viewing the elements from least to greatest on the real life, from -$\infty$ to $\infty$.
In the perspective of the new ordering created by the function, will this intuition be translated as , in order of least to greatest, (0,-$\infty$), {0}, ($\infty$,0)?
I apologize for the abuse of notation here.
One of the axioms of an ordered field is:
For any $a,b,c$, if $a\le b$, then $a+c\le b+c$.
However, taking $a=-1, b=-3, c=2$, we get a problem with your proposed ordering. $a\le b$, since $\frac{1}{-1}\le \frac{1}{-3}$. However, $a+c>b+c$, since $\frac{1}{1}> \frac{1}{-1}$.
Hence, you have imposed a different total order, but that order does not respect the field axioms, so it is no longer an ordered field with this new order.