Using distance what would be the simplest way to prove that the
abs(-a) = abs(a) for all a
I'm probably going to get into trouble saying this (because my comment is over-generalized) but here goes:
The way I introduce the absolute value function is to define it as the distance from the point x = a to the origin. Note that when you get to the modulus of a complex number this same exact definition will hold. (This idea is my own, but is spurred on by the fact that the absolute value and the complex modulus operators are given using the same symbol. At least in the texts I've seen. So presumably someone else out there might have had similar ideas.)
-Dan
In my Real Multivariable Class we define $\displaystyle | \bold{x} | = \sqrt{x_1^2+...+x_2^2}$. For any $\displaystyle \bold{x}\in \mathbb{R}^n$. Hence $\displaystyle \bold{x-y}=\bold{y-x}$ is trivial. And so $\displaystyle |a|=|-a|$ because this norm on 1 dimensional space is $\displaystyle \sqrt{x_1^2}=|x_1|$. Maybe this is what you are saying (because the complex plane is just $\displaystyle C=\mathbb{R}^2$).