Is $0 ^ 0 = 1$ a valid result?

Question

Most people say $0 ^ 0$ is indeterminate, but that's in the context of limits. I mean $0 ^ 0$ when the zeros are absolute.

I've seen that one way to define exponentiation of natural numbers is saying that $a ^ 0 = 1$ for any natural $a$ (including zero), and $a^{1+n} = a * a ^ n$

Under this definition there is no problem, but my concern arises because in the construction of natural numbers, as in Von Neumann's using sets, natural are not the same non-negative integers. Integers are constructed as pairs of natural numbers.

So what would be the value of this expression $0 ^ 0 + (-4)$? You can not just replace $0 ^ 0$ by $1$ using definition above and say the result is $-3$, because you can not operate a natural with an integer. They are different structures. For this expression been properly defined it is needed to be a definition of exponentiation in integers where $0 ^ 0 = 1$

Or perhaps it is not correct to define exponentiation in integers, since it would not be closed. But it would be closed in the complex numbers. What is the definition of exponentiation in the complex, using set theory?

Is it possible that $0 ^ 0$ is equal to $1$ only with natural zeros but not with the zeros of other numerical sets?

Answer 1

It depends on the definition of exponentiation one uses. There are different definitions of exponentiation based on multiplication, set theory, logarithms, even solution of algebraic equations (e.g. defining $a^{1/2}$ as the nonnegative solution to $x^2=a$). Some definitions extend naturally to give $0^0=1$, others do not.

Surely thus matter has been discussed on MSE before, so I recommend a search.

Answer 2

It is indeterminate in the sense that if $f(x)\to 0$ and $g(x)\to 0$ as $x\to\text{something}$ then the limit of $f(x)^{g(x)}$ could be any positive number or $0$ or $+\infty$ depending on which functions $f$ and $g$ are.

However, if $(f(x),g(x))$ approaches $(0,0)$ from within any sector bounded by lines $y=ax$ and $y=bx$ with $0<a<b<\infty$, then the limit is $1$.

But it is equal to $1$ because it's an empty product. That is important when one recalls that $$ e^z = \sum_{n=0}^\infty \frac{z^n}{n!} $$ and that $e^0=1$. The first term in the series when $z=0$ is $\dfrac{0^0}{0!}$ and all the other terms are $0$. So $0^0/0!=1$.

Or one can recall that the Poisson distribution with expected value $0$ is equal to $0$ with probability $1$.

That $0^0=1$ can also matter in combinatorics.