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

TravisJ
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Evorlor
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    So that it satisfies the rule $n^an^b=n^{a+b}$ – Gregory Grant Apr 29 '15 at 14:59
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    Careful, $0^0\neq 1$. – TravisJ Apr 29 '15 at 15:04
  • @TravisJ any reference on this? It think one would wish the function $x\mapsto x^\alpha$ to be continuous for any $\alpha\in \mathbb{R}$. – Thibaut Dumont Apr 29 '15 at 15:14
  • @TravisJ When I enter 0^0 into google, it outputs 1. Know where the miscommunication between me and google calculator is? – Evorlor Apr 29 '15 at 15:14
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    Duplicate question? See http://math.stackexchange.com/q/135/215011 for discussion of $0^0$ – grand_chat Apr 29 '15 at 15:20
  • I think it may be because such a definition brings much more convenience and compatibility in other places than if we defined $n^0=0$. – Vim Apr 29 '15 at 15:22
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    @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 $0^0≝1$. – user26486 Apr 29 '15 at 15:49

4 Answers4

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Since $n^k=n^{k+0}=n^k \cdot n^0$. This short computation suggests that $n^0$ should be $1$.

ThorbenK
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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 $a^{-n}=1/a^n$.

Martin Argerami
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    It's not a convention. Rather, it is a consequence of the universal property of the adjunction of a neutral element to a monoid. – Bill Dubuque Apr 29 '15 at 15:25
  • Of course it is a convention. Algebraic structures don't know about notation. – Martin Argerami Apr 29 '15 at 15:37
  • 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. – Bill Dubuque Apr 29 '15 at 16:10
  • 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. – Martin Argerami Apr 29 '15 at 17:31
  • Do you think that $,0 + n = n,$ is a "convention" too? – Bill Dubuque Apr 29 '15 at 17:42
  • It depends on how you define the natural numbers and their addition. But usually yes, you define $0$ as the number that satisfies $0+n=n$ (if you are adjoining $0$ to the monoid), or you define addition by zero by $0+n=n$, if you are defining addition recursively from the Peano axioms: in both cases it is a convention, inspired by the notion of "adding nothing". – Martin Argerami Apr 29 '15 at 18:09
  • To call it a convention misses the algebraic essence of the matter, viz. that it is a special case of the adjunction of a neutral element to a semigroup to obtain a monoid. – Bill Dubuque Apr 29 '15 at 18:13
  • That is one point of view, but not the only one. You can start with ZFC and use the axioms to construct the naturals as the minimal set satisfying the axiom of infinity; in that case $0$ is defined as the empty set, and it is an original member of $\omega$. – Martin Argerami Apr 29 '15 at 18:18
  • I did not know math had "points of view". I always thought it was black and white. Oh my, I may be in over my head with this question. – Evorlor Apr 30 '15 at 02:02
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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 $n^0=1$

pjs36
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Caratheodory_Enthusiast
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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.

TravisJ
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