Why is $x^{-1} = \frac{1}{x}$?

So there is a subtle question here, We can verify that $x^n * x^m = x^{n+m}$ when $n,m$ are natural numbers and $x$ is a natural number. Now the "natural generalization" here is to just decide that that formula applies for all numbers. From there we can compute it for negative values. Of course there are other, much less practical versions of the exponent, one could make up where this doesn't hold, and the addition rule holds only when n,m are natural numbers.

What that effectively amounts to is finding functions

$$f(x) | f(n+m) = f(n)f(m)$$ When $n,m$ are natural numbers. Exponentiation, in the traditional sense is one of the smoother solutions to this, but I'm sure you can invent some freakish counterexample which does very very unpredictable things on inputs that aren't natural numbers.

For example consider the functional equation:

$$ f(n+m) = (f(n) + frac(n)+ 0.5 \delta(x) )(f(m) + frac(m) + 0.5\delta(x))$$

On positive integers this reduces to

$$ f(n+m) = f(n)f(m)$$

But elsewhere it does something WILDLY different. So this function if looked at from just the natural numbers would like identical to our exponentiation, but outside of there would be very very strange looking, in principle $f(-1)$ would not be anything like $\frac{1}{x}$

But, the guiding principle here then isn't "WHAT IS MATHEMATICALLY CORRECT", it is, what makes most sense to use. And the most natural tool then is the original exponential the way we are used to.

Given $a,b,c$ from any set $Y$ with a multiplication defined and $1$ is the corresponding multiplication identity. By definition,

• $\frac{a}{b}$ is the unique $y \in Y$ (if exists) such that $yb = a = by$.
• $c^{-1}$ is the unique $y \in Y$ (if exists) such that $yc = 1 = cy$.

Substitute $a$ by $1$ and $b, c$ by $x$, we find both definition of $\frac{1}{x}$ and $x^{-1}$ reduce to the unique $y$ in $Y$ (if exists) such that $yx = 1 = xy$. As a result, $\frac{1}{x} = x^{-1}$ whenever they make sense.