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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?

Ronald Becerra
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    it is the opposite : $0^0$ is well-defined only as a limit. for example $\lim_{\epsilon\to 0^+} 0^\epsilon = 0$ and $\lim_{\epsilon\to 0^+} \epsilon^0 = 1$ – reuns Sep 16 '16 at 23:24
  • one way to define exponentiation of natural numbers is saying that a^0=1. And one way to define powers is saying that $0^n = 0$ for any $n$. You can't take $0$ as an absolute (whatever that means) in both base and exponent. In other words, you need to use limits. – dxiv Sep 16 '16 at 23:24
  • I used "absolute" to specify that the values are not numbers tending to zero, but zero properly. – Ronald Becerra Sep 16 '16 at 23:27
  • If the values are taken as zero properly then you end up with a contradiction, as pointed in both prior comments. – dxiv Sep 16 '16 at 23:29
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    Possible duplicate of: Is there any good reason not to define 0^0 =1, such as contradictions in algebra or arithmetic?. P.S. Previous link was the top hit for $0^0$ on my first try of the 3rd party search engine announced here. – dxiv Sep 16 '16 at 23:37
  • @user1952009 Well defined by whom? What does mean "well defined" as opposed to just "defined"? And why that who cannot well define 0^0=1? – Anixx Sep 16 '16 at 23:57
  • @Anixx hint : $f(x,y) = x^y$ isn't continuous at $(0,0)$, but it is continuous in $x$ when $y = 0$, and in $y$ when $x =0 $ (see my comment) – reuns Sep 17 '16 at 00:01
  • @user1952009 so what? – Anixx Sep 17 '16 at 00:01
  • @Anixx you don't understand why $f(0,0)$ isn't well-defined when $f(x,y)$ is continuous everywhere except at $(0,0)$ ? well-defined means "no need for a special/careful definition, since it is obvious from the context" – reuns Sep 17 '16 at 00:04
  • @user1952009 your definition of "well defined" contradicts what Wikipedia says: "A function is well-defined if it gives the same result when the representation of the input is changed without changing the value of the input." https://en.wikipedia.org/wiki/Well-defined A function can be specially defined at a point and still be well defined. – Anixx Sep 17 '16 at 00:08
  • @Anixx useless discussion – reuns Sep 17 '16 at 00:10
  • Related: https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 – Henry Jan 08 '24 at 07:04

2 Answers2

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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.

Michael Hardy
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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.

Oscar Lanzi
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