Why do we do mathematical induction only for positive whole numbers? After reading a question made here, I wanted to ask "Why do we do mathematical induction only for positive whole numbers?"
I know we usually start our mathematical induction by proving it works for $0,1$ because it is usually easiest, but why do we only prove it works for $k+1$?
Why not prove it works for $k+\frac12$, assuming it works for $k=0$.
Applying some limits into this, why don't we prove that it works for $\lim_{n\to0}k+n$?
I want to do this because I realized that mathematical induction will only prove it works for $x=0,1,2,3,\dots$. assuming we start at $x=0$, meaning it is unknown if it will work for $0<x<1$ for all $x$.
And why not do $k-1$?  This way we can prove it for negative numbers as well, right?
What's so special about our positive whole numbers when doing mathematical induction?
And then this will only work for real numbers because we definitely can't do it for complex numbers, right?
And what about mapping values so that it becomes one of the above?  That is, change it so that we have $x\to\frac x2$?  Then proving for $x+1$ becomes a proof for all $x$ that is a multiple of $\frac12$!?

All of the contexts in which you propose to apply induction are implicitly using the positive integers anyway.
For instance, suppose we have a property $P$, and we prove that $P(x)\implies P(x+\frac 1 2)$. Then if we know $P(0)$, we can prove $P(0.5)$, $P(1)$, $P(1.5)$, etc. But you're implicitly using the natural numbers here. You're proving that if $P(n\frac 1 2)$ holds, then $P((n+1)\frac 1 2)$ holds, and thus concluding that $P(n\frac 1 2)$ holds for all $n$.
Or suppose we prove that $P(x)\implies P(x-1)$. Then from $P(0)$ we could deduce $P(n)$ for any negative $n$. But here you're implicitly proving that if $P(-n)$ holds then $P(-(n+1))$ holds and thus proving that $P(-n)$ holds for all positive $n$.
