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As we were taught in grade school, that $i = \sqrt{-1}$. This opened up math so that we could start solving real or complex functions for solutions.

As such, I was wondering if there exists a similar concept for extending the absolute value to allow for an inverse-like function. In otherwords, let $ι = |-1|^{-1}$ where $|x|^{-1}$ is the inverse of the absolute value function. Then $|ι| = -1$ and if we take the identity $|ab|=|a||b|$ then we also get $|±ι|=|-1||ι| = -1$ and with $|±1| = 1$ we have our definitions of the base units. We can expand it further to where given a positive real number $a$ we can then say $|±aι| = -a$.

But where I struggle to come up with a consistent definition for this idea is when I try to expand it to add real numbers.

How would I come up with a complete definition for $|a±bι| =?$ and $|a±bι|^{-1} = ?$

Edit: The reason I need this is that I am trying to solve ray-intersect complicated absolute value function by plugging in the ray equation $P=O+tD$ and solving for $t$. Which will let me come up systematically with equation to evaluate in a program.

Here is the equation:

let $a_x,a_y,b_x,b_y,d,x,y \in \mathbb{R}$

$a=\left(a_x,a_y\right)$

$b=\left(b_x,b_y\right)$

$C=\frac{\left(a+b\right)}{2}$

$q=b-a$

$M=\sqrt{q.x\cdot q.x+q.y\cdot q.y}$

$R=\frac{M}{2}\ -\ d$

$Q=\frac{q}{M}$

$K=\left(-Q.y,Q.x\right)$

$P=\left(x,y\right)$

$v\ =\ P-C$

$S\ =v.x\cdot Q.x+v.y\cdot Q.y$

$B\ =v.x\cdot K.x+v.y\cdot K.y$

$\frac{1}{2}\left|S^{2}-d^{2}\ \right|-d\left|S-d\ \right|-d\left|S+d\ \right|\ +\frac{1}{2}S^{2}+\frac{3}{2}d^{2}+\ B^{2}\ =\ R^{2}$

So then I would plug in $P=O+tD$ into $P=\left(x,y\right)$ and solve for $t$.

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    $\begingroup$ $\iota=|-1|^{-1}=1^{-1}=1$. How do you deduce $|\iota|=-1$? $\endgroup$ Commented May 28 at 6:27
  • $\begingroup$ @ChristopheBoilley it is by definition, just like how we define $i = \sqrt{-1}$. I don't see how $|−1|^{−1}=1^{−1}$. Since $|x|^{-1}$ isn't the absolute value applied to $-1$ it is the inverse of the absolute value. Thats what I meant by my definition above. $\endgroup$ Commented May 28 at 7:14
  • $\begingroup$ Related: math.stackexchange.com/questions/1345191/…, math.stackexchange.com/questions/259584/… $\endgroup$ Commented May 28 at 7:53
  • $\begingroup$ OK, so you want to extend the absolute value on $\mathbb R$ to a multiplicative function over a polynomial ring $\mathbb R[\iota]$ such that $|\iota|=-1$. You need to define your function on all irreducible polynomial. Some continuity condition imply that $|\iota+x|\sim|x|$ when $x\to\pm\infty$. $\endgroup$ Commented May 28 at 8:30
  • $\begingroup$ @ChristopheBoilley exactly. This extension should be complete in the sense that I can plug in any value in the form $a±bι$ and be able to evaluate it in both the absolute value and the inverse of the absolute value $\endgroup$ Commented May 29 at 4:19

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