I'm constructing a proof by induction using p-adic numbers, but I've never used them before. If $h(x)=\frac{\lvert x\rvert_2^{-1}-x}{4}$
then I want to evaluate: $$\lvert h(h(x))\rvert_2^{-1}$$
Here's my attempt:
If $g(x)=\lvert x\rvert_2^{-1}-x$
i.e. the inverse of the 2-adic metric of $x$, minus $x$.
So for $x=28$, $g(x)=4-28=-24$
Then am I right in thinking that provided $x\notin2^{\mathbb{N}}$ the following is true:
$\lvert g(g(x))\rvert_2=\lvert g(x)\rvert_2$
since it appears to follow trivially by the Euclidean algorithm.
So then...
I want to evaluate if instead $h(x)=\frac{\lvert x\rvert_2^{-1}-x}{4}$
It would seem to me that $\frac{\lvert x\rvert_2^{-1}}{4}=\lvert\frac{x}{4}\rvert_2^{-1}$
So provided $x,h(x)\notin 2^\mathbb{N}$ and $\frac{\lvert x\rvert}{x}=\frac{\lvert f(x)\rvert}{f(x)}$
am I correct in thinking $\lvert h(h(x))\rvert_2^{-1}=\frac{\lvert h(x)\rvert_2^{-1}}{4}$ ?
and $0$ otherwise?
This is a strange problem, and I’m having trouble grasping what’s going on, but I’m pretty sure you won’t be able to use (standard) induction to verify any formula. My personal inclination is for you not to use the standard $2$-adic absolute value in the description of $h$, but rather the $2$-adic order (valuation) function, for which $v_2(2^m\frac rs)$ is $m$, if $r$ and $s$ are both odd. It measures the divisibility of a rational or $2$-adic number by $2$.
Using this function, your $h$ is defined as $$ h(x)=\frac{2^{v(x)}-x}4\,. $$ I would start by showing that $h(2x)=2h(x)$, and then test the behavior of $h\circ h$ for various numbers. For instance, $h(3)=-\frac12$, and $h(-\frac12)=\frac12h(-1)=\frac12$, while $h(17)=-4$ and $h(-4)=2$. I don’t believe that these numbers verify your conjecture.