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We know that we can't define division by zero "in any mathematical system that obeys the axioms of a field", because it would be inconsistent with such axioms.

(1) Why can we define $a^0$ ($a\neq 0$) to be $1$? Is it possible to prove that such definition is consistent with any rule of arithmetic? How to conclude that to define $a^0$ ($a\neq 0$) we don't need abolish any other basic rule of arithmetic?

(2) More generally, how to know if a definition is consistent with a given mathematical theory?

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There is no general algorithm for determining when a theory is consistent. That is a huge topic which includes Godel's incompleteness theorems. But your specific question is easier.

In Peano arithmetic (with axioms stated using $+,\times$) an exponential function $x^y$ can be defined by recursion $x^0=1$ and $x^{s(y)}=x\times x^{y}$. The axioms prove that functions can be defined recursion. So if you believe (as nearly everyone does) that Peano arithmetic (with axioms stated using $+,\times$) is consistent, then you must believe the extension with that exponential function is consistent.

Since your question mentions basic rules of arithmetic I answered in terms of Peano Arithmetic. If you merely want consistency with the field axioms the question is simpler yet: The field of integers modulo 2 proves consistency of those axioms plus $x^1=x$ and $x^0=1$, by giving a finite model. But this includes very little of arithmetic and notably does not include $x^{(y+z)}=x^y\times x^z$. See "finite field" on Wikipedia.

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  • $\begingroup$ If there is no obvious problem with a definition (like there is with $0/0=1$), we can't say anything about its consistency with respect a given set of axioms? There is no mathematical reason (rather than belief) to argue that the definition $x^0=1$ is consistent with field axioms? $\endgroup$ – Pedro Aug 17 '14 at 5:04
  • $\begingroup$ @Pedro Of course there is a great deal to say about consistency of theories that have no obvious problems; and logic journals annually publish thousands of pages of research on such questions. I mentioned that is a huge topic. And I gave you a mathematical proof that if Peano Arithmetic with $+,\times$ is consistent then so is it with exponentiation including $x^0=1$. What more do you want to know? Sadly, I must say that if you do not believe PA is consistent -- then you should not believe PA with exponentiation is either. $\endgroup$ – Colin McLarty Aug 17 '14 at 6:16
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We don't actually define $1^0$ to be 1. That's its value. Likewise $0^0=1$ is a derived value, the supposed indefinite values all rely on the same $0/0$ proof that lets $0=1$. If you take the limit of $x^{ax}$ as x-> 0, the limit is 1 for all a.

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    $\begingroup$ Wait--what? You can't consistently define $0^0$. The problem comes when you consider $\lim_{x\rightarrow 0+}x^0$ versus $\lim_{x\rightarrow 0+}0^x$. The first limit is $1$ but the second limit is $0$. $\endgroup$ – MPW Aug 17 '14 at 1:26
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    $\begingroup$ Well, if you require that an exponential function $y^x$ for real numbers extend the usual notion of natural number powers $x^2,x^3,\dots,$ and satisfy the multiplication rule $y^{(x+z)}=y^x\times y^z$, then you must say $1^0=1$. And if you require it to be continuous for real arguments $x,y$ then you cannot define it for $x=y=0$. $\endgroup$ – Colin McLarty Aug 17 '14 at 1:38
  • $\begingroup$ @MPW demonstrates the effect of 0/0 in his answer. By taking the limit along the line y=0 for y^x, we see in effect 0/0, which you can prove anything really. But there are lots of other lines that pass through x=0, y=0, and following any of these universally point to 0^0=1. For more, you can show that a.0^0 = a, because there are zero numbers equal to zero on the LHS, and thus 0^0 is the identity element of multilication. Since there can only be one of these, 0^0=1. $\endgroup$ – wendy.krieger Aug 17 '14 at 2:31
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    $\begingroup$ @wendy.krieger if the argument is that $x^y \to 1$ along any "reasonable" path such that $x \to 0$ and $y \to 0$, I'd argue $x(t) = e^{-1/t}$ and $y(t) = t$ (shamelessly stolen from wikipedia) contradicts this. I don't think arguing for $0^0 = 1$ strictly from limiting behavior is really very appealing; combinatoric arguments seem more convincing but many would argue it is better to just give $0^0$ a context-dependent definition and in most contexts $0^0 = 1$ is appropriate. $\endgroup$ – guy Aug 17 '14 at 3:00
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    $\begingroup$ @wendy.krieger you are begging the question when you state that you can multiply north sides by $b^0$ for all $b$ and retain equality; this assumes that $b^0$ is a well defined number for all $b$, which is precisely the matter under dispute. If you are arguing that any reasonable definition of $x^y$ where $x$ and $y$ are reals, together with the axioms of the real field, logically imply $0^0 = 1$ then this simply isn't true. For example, the definition $x^y = \sup_{y > q \in \mathbb Q_+} x^q$ clearly doesn't apply, nor does $x^y = \exp(y \log x)$. $\endgroup$ – guy Aug 17 '14 at 15:15

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