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Why is any number to the zeroeth power equal to 1? I would think it would be equal to zero, since nothing multiplied by nothing is, well, I would think 0. But it is 1?

Examples: $(-5)^0 = 1$; $0^0 = 1$; $5^0 = 1$;

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    $\begingroup$ So that it satisfies the rule $n^an^b=n^{a+b}$ $\endgroup$ – Gregory Grant Apr 29 '15 at 14:59
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    $\begingroup$ Careful, $0^0\neq 1$. $\endgroup$ – TravisJ Apr 29 '15 at 15:04
  • $\begingroup$ @TravisJ any reference on this? It think one would wish the function $x\mapsto x^\alpha$ to be continuous for any $\alpha\in \mathbb{R}$. $\endgroup$ – Thibaut Dumont Apr 29 '15 at 15:14
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    $\begingroup$ Duplicate question? See math.stackexchange.com/q/135/215011 for discussion of $0^0$ $\endgroup$ – grand_chat Apr 29 '15 at 15:20
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    $\begingroup$ @Evorlor WolframAlpha doesn't define $0^0$. It depends on who you're talking to. Some like to define $0^0≝1$ just because it makes some theorem statements easier to express. E.g., binomial theorem: $(0+a)^2=0^0a^2+2\cdot 0^1\cdot a^1+0^2a^2=0^0 a^2$. Others refuse to give a value to it at all because $f(x,y)=x^y$ is discontinuous at $(0,0)$ ($x=0,y\to 0\,\Rightarrow\, f(x,y)\to 0$ and $x\to 0,y=0\,\Rightarrow\, f(x,y)\to 1$), although we do have $\lim_{x\to 0} x^x=1$, which is another argument some people use for $0^0≝1$. $\endgroup$ – user26486 Apr 29 '15 at 15:49
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Expanding on what has already been said, I want to emphasize that $a^0=1$ is a convention. There is no computation to be made, because there is no obvious meaning to "multiply with itself zero times".

The reason why the convention is reasonable is what has been mentioned by Thorben's answer and Gregory's comment: the relation $a^{m+n}=a^ma^n$ is so nice that it makes sense to extend to the rest of the integers. Once you have $a^0=1$, you also get $a^{-n}a^n=a^{n-n}=1$, so $a^{-n}=1/a^n$.

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    $\begingroup$ It's not a convention. Rather, it is a consequence of the universal property of the adjunction of a neutral element to a monoid. $\endgroup$ – Bill Dubuque Apr 29 '15 at 15:25
  • $\begingroup$ Of course it is a convention. Algebraic structures don't know about notation. $\endgroup$ – Martin Argerami Apr 29 '15 at 15:37
  • $\begingroup$ But notation is meaningless without interpretation, and here the syntax is determined by the semantics. To dismiss it as a convention misses the algebraic essence of the matter. $\endgroup$ – Bill Dubuque Apr 29 '15 at 16:10
  • $\begingroup$ I don't "dismiss it as a convention": I think that it is a great convention, and I agree (and I imply it in my answer) that the reason for the convention is algebraic. $\endgroup$ – Martin Argerami Apr 29 '15 at 17:31
  • $\begingroup$ Do you think that $\,0 + n = n\,$ is a "convention" too? $\endgroup$ – Bill Dubuque Apr 29 '15 at 17:42
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Since $n^k=n^{k+0}=n^k \cdot n^0$. This short computation suggests that $n^0$ should be $1$.

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The way I think about this (which may or may not be write) is that exponentiation (by integers) is viewed as

$$ a^{n}=1\cdot a\cdot a\cdot ... \cdot a$$ where there are $n$ copies of $a$. If $n=0$, then there is only the $1$.

You could also view $a^{n}$ as

$$a^{n} = \prod_{i=1}^{n}a.$$ Then, just note that the empty product is $1$... unlike the empty sum which is $0$. Again, this only works when $n$ is an integer.

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    $\begingroup$ Note that this is "mental heuristic" not mathematical proof. $\endgroup$ – TravisJ Apr 29 '15 at 15:00
  • $\begingroup$ You better add that line in the post. :) $\endgroup$ – Sufyan Naeem Apr 29 '15 at 16:50
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$$\frac{n^x}{n^x}=1, n \neq 0$$ By laws of indices,

\begin{align*}n^{x-x}&=1\\ \implies n^0&=1\end{align*}

So that's how we prove that $n^0=1$

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