# Zero to the zero power - is $0^0=1$?

Could someone provide me with good explanation of why $0^0 = 1$?

My train of thought:

$x > 0$

$0^x = 0^{x-0} = 0^x/0^0$, so

$0^0 = 0^x/0^x = ?$

1. $0^0 \cdot 0^x = 1 \cdot 0^x$, so $0^0 = 1$
2. $0^0 = 0^x/0^x = 0/0 = \text{undefined}$

PS. I've read the explanation on mathforum.org, but it isn't clear to me.

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@Nuno: I agree in general, and in this case. I've edited the tags; Stas said in his comments on my answer that he meant the question in "terms of discrete mathematics", so the discrete-mathematics tag is appropriate, and (powers), (zeros), and (number-theory) are all inappropriate. – Arturo Magidin Nov 21 '10 at 3:15
Can you pls link to the explanation that you read on mathforum.org? – Lazer Nov 21 '10 at 6:48
+1 for good elementary question. – tia Nov 21 '10 at 8:41
BTW, this is comprehensively covered on Wikipedia (or see today's version), along with pointers to history and treatment in many systems. – ShreevatsaR Nov 22 '10 at 12:34
@Stas : actually, $0^0$ is generally considered "undefined". But when people use power series, they routinely treat $0^0$ as $1$ without a second thought. If a function $f$ is defined by a power series as $f(x) = \sum_{n=0}^\infty a_n (x-b)^n$, then everyone agrees that $f(b) = a_0$, even though plugging in $x = b$ into the series involves $0^0$. I hope this adds something to the dozens of other comments and previous answers. – Stefan Smith Apr 15 '13 at 16:26

In general, there is no good answer as to what $0^0$ "should" be, so it is usually left undefined.

Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$?

Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus and algebra, we just don't want to give it any value, we just declare it undefined.

However, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote "size" (cardinalities), then the "$A^B$" is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there is one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we must define $0^0$ as $1$. So if we are talking about cardinal exponentiation, then the only possible definition is $0^0=1$, and we define it that way, period.

Added 2: the same holds in Discrete Mathematics, when we are mostly interested in "counting" things. In Discrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is really the same thing as "maps from $\{1,2,\ldots,m\}$ to $\\{1,2,\ldots,n\\}$" when interpreted appropriately, so it is again the same thing as in set theory).

So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly one way of doing that: just sit and do nothing! So we make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you cannot do it, so the answer is that $0^1=0$).

Your "train of thoughts" don't really work: If $x\neq 0$, then $0^x$ means "the number of ways to make $x$ choices from $0$ possibilities". This number is $0$. So for any number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ "should" be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ is the number of ways of making no selections from no choices.

Coda. In the end, it is a matter of definition and utility. In Calculus and algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem or via power series, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions.

We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that "useful and natural" way should be, so we define it that way.

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@Sivam: yes, in the sense that there is exactly one function from any infinite set to the one-element set. This does not say anything about the status of 1^{infty} as an indeterminate form. The two expressions mean different things in different contexts; that they are designated using the same notation is a convenience, not a necessity. – Qiaochu Yuan Nov 20 '10 at 23:27
@Sivam: exactly as Qiaochu says. Note that I said that $0^0=1$ in cardinal exponentiation is the only sensible answer, but "cardinal exponentiation" is not the same as real number exponentiation; when doing real number exponentiation, $0^0$ is most properly undefined/indeterminate. – Arturo Magidin Nov 20 '10 at 23:30
Just a small note: the answer depends on whether you think of exponentiation as a discrete operation (as in set theory, algebra, combinatorics, number theory) or as a continuous operation over spaces like real/complex numbers (as in analysis). – Kaveh Nov 21 '10 at 5:47
"In Calculus and algebra, there is no reasonable definition", should you really include algebra there? Aren't polynomials considered part of algebra, and don't people say that evaluating the polynomial $\sum_{k=0}^n a_k x^k$ at $0$ gives $a_0$? – Omar Antolín-Camarena Jun 8 '13 at 20:28
I wish I could downvote this answer several times. Saying that in algebra there's no reasonable definition is absurd. – egreg Nov 19 '14 at 11:49

This is merely a definition, and can't be proved via standard algebra. However, two examples of places where it is convenient to assume this:

1) The binomial formula: $(x+y)^n=\sum_{k=0}^n {n\choose k}x^ky^{n-k}$. When you set $y=0$ (or $x=0$) you'll get a term of $0^0$ in the sum, which should be equal to 1 for the formula to work.

2) If $A,B$ are finite sets, then the set of all functions from $B$ to $A$, denoted $A^B$, is of cardinality $|A|^{|B|}$. When both $A$ and $B$ are the empty sets, there is still one function from $B$ to $A$, namely the empty function (a function is a collection of pairs satisfying some conditions; an empty collection is a legal function if the domain $B$ is empty).

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You don't need to appeal to the binomial formula. Anytime you write a polynomial as f(x) = sum a_i x^i you need x^0 = 1 to keep your notation consistent, so you need 0^0 = 1 so that f(0) = a_0. – Qiaochu Yuan Nov 21 '10 at 1:12
Yes. I think it is reasonable to define $0^0=1$ (because that seems to be the most useful definition) with the caveat that the function $x^y$ on $\mathbb{R}^{+}\!\!\times\mathbb{R}$ is not continuous at $(0,0)$. – robjohn Nov 13 '13 at 11:28

$0^{0}$ is just one instance of an empty product, which means it is the multiplicative identity 1.

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$0^0$ is undefined. It is an Indeterminate form.

You might want to look at this post.

Why is $1^{\infty}$ considered to be an indeterminate form

As you said, $0^0$ has many possible interpretations and hence it is an indeterminate form.

For instance,

$\displaystyle \lim_{x \rightarrow 0^{+}} x^{x} = 1$.

$\displaystyle \lim_{x \rightarrow 0^{+}} 0^{x} = 0$.

$\displaystyle \lim_{x \rightarrow 0^{-}} 0^{x} =$ not defined.

$\displaystyle \lim_{x \rightarrow 0} x^{0} = 1$.

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Is $\lim_{x\to0}0^x$ really defined? It can only be approached from the positive side. – kennytm Nov 21 '10 at 19:52
@KennyTM: Accepted and edited accordingly. – user17762 Nov 22 '10 at 11:47
0^0 is undefined by whom? I saw somebody defined it, can I now say it is defined? – Anixx Nov 4 '12 at 22:58
As I commented to Gadi A, I think it is quite reasonable to define $0^0=1$, as that seems to be the most useful definition, but note that the function $x^y$ on $\mathbb{R}^{+}\!\!\times\mathbb{R}$ is not continuous at $(0,0)$. – robjohn Nov 13 '13 at 11:32
This answer is wrong. It is indeed an indeterminate form, but it is defined and equal to $1$. See my answer also posted here. – Michael Hardy Mar 5 '14 at 18:06

I'm surprised that no one has mentioned the IEEE standard for $0^0$. Many computer programs will give $0^0=1$ because of this. This isn't a mathematical answer per se, but it's worth pointing out because of the increasingly computational nature of modern mathematics, so that one doesn't run afoul of anything.

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The use of positive integer exponents appears in arithmetic as a shorthand notation for repeated multiplication. The notation is then extended in algebra to the case of zero exponent. The justification for such an extension is algebraic. Furthermore, in abstract algebra, if $G$ is a multiplicative monoid with identity $e$, and $x$ is an element of $G$, then $x^0$ is defined to be $e$. Now, the set of real numbers with multiplication is precisely such a monoid with $e=1$. Therefore, in the most abstract algebraic setting, $0^0=1$.

Continuity of $x^y$ is irrelevant. While there are theorems that state that if $x_n \to x$ and $y_n \to y,$ then $(x_n + y_n) \to x+y$ and $(x_n)(y_n) \to xy$, there is no corresponding theorem that states that $(x_n)^(y_n) \to x^y$. I don't know why people keep beating this straw man to conclude that $0^0$ can't or shouldn't be defined.

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Downvote, because Onez focuses on a very narrow view of the exponentiation operation and its applications, and writes as if that is the only view. Continuity is of rather significant importance in a wide variety of situations, and the requirements of an exponentiation function of continuous arguments are rather different than those limited to integer or rational exponents, and $0^0$ runs into that difference. Another example where the needs differ are $(-1)^{1/3}$ – Hurkyl Oct 26 '11 at 22:06
I hardly consider the whole domain of algebra to be narrow. YMMV. In any case, when extending the domain of functions, one may ask: Is the extension useful? Defining 0^0 to be 1 is useful in Combinatorics, Set Theory, and Algebra. Indeed, it is even useful in calculus when using summation notation for polynomial functions and infinite series. Perhaps you care to list several advantages of leaving 0^0 undefined? Particularly in light of the fact that many definitions require the additional caveat that a,b,x,y etc. not be equal to zero in order for them to be true. – Onez Oct 27 '11 at 1:13
By your argument, we should not define (-2)^0 = 1, but leave it undefined since this extension of exponentiation fails your second case (non-positive base) and your third case (no continuous extension of x^y to all of C). You still haven't supplied any advantages to leaving 0^0 undefined. Economy of notation (the reason exponential notation was developed in the first place) is gained by defining 0^0 = 1. – Onez Oct 27 '11 at 21:11
Since 0 is not in the range of the exponential function, I take it you have issue with 0^y being defined even for y>0. The argument seems to hinge on whether one is to define 0^0=1 and economize several definitions and theorems from algebra, combinatorics, and analysis, at the expense of one caveat for a single function, OR to leave 0^0 undefined, have several caveats so as to preserve the continuity on the domain of definition of a single function, namely x^y. Where is the greatest economization to be had? Who has the narrow view? – Onez Oct 28 '11 at 3:29
You missed the more relevant option -- acknowledge the multiple exponentiation operations that come up in mathematics, rather than conflating them. – Hurkyl Oct 28 '11 at 6:18

Some indeterminates forms $0^{0}, \displaystyle\frac{0}{0}, 1^{\infty}, \infty − \infty, \displaystyle\frac{\infty}{\infty}, 0 × \infty,$ and $\infty^{0}$

Futhermore,

$$\lim_{ x \rightarrow 0+ }x^{0}=1$$ and

$$\lim_{ x \rightarrow 0+ }0^{x}=0$$

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One should note that indeterminate is not the same as undefined. – Hagen von Eitzen Jul 10 '13 at 16:27
@Hagen, what's the difference? – JMCF125 Nov 22 '13 at 23:12
@JMCF125: $\lim\limits_{(x,y)\to(0,0)}x^y$ is indeterminate. This actually frees us to define $0^0$ to be whatever value is most useful. In almost every practical case, that is $0^0=1$. – robjohn Feb 10 '14 at 6:23
@robjohn, ah, I see. Though now I wouldn't've made the comment, as I asked a related question that made me understand this. – JMCF125 Feb 10 '14 at 11:16
@JMCF125: That's why the site is here. Where is the related question? – robjohn Feb 10 '14 at 14:25

"Everybody knows" that $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!},$$ and when $z=0$ then the first term is $\dfrac{0^0}{0!}$, so of course $0^0$ is $1$ since it's an empty product.

But it's also an indeterminate form because $\displaystyle\lim_{x\to z}f(x)^{g(x)}$ can be any positive number, or $0$ or $\infty$, depending on which functions $f$ and $g$ are, if $f(x)$ and $g(x)$ both approach $0$ as $x\to a$.

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Knuth's answer is at least as good as any answer you're going to get here: http://arxiv.org/pdf/math/9205211v1.pdf See pp. 4-6, starting at the bottom of p. 4.

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It's pretty straight forward to show that multiplying something by $x$ zero times leaves the number unchanged, regardless of the value of $x$, and thus $x^0$ is the identity element for all $x$, and thus equal to one.

For the same reason, the sum of any empty list is zero, and the product is one. This is when a product or sum of an empty list is applied to a number, it leaves it unchanged. Thus if the product $\Pi()$ = 1, then we immediately see why $0! = 0^0 = 1$.

Without this property, one could prove that $2=3$, by the ruse that there are zero zeros in the product on the left (zero is after all, a legitimate count), and thus $2*0^0$, and since $0^0$ as indeterminate, could be 1.5, and thus $2=3$. I think not.

The approaches to $0^0$ by looking at $x^y$ from different directions, fails to realise that for even lines close to $x=0$, the line sharply sweeps up to 1 as it approaches $y=0$, and that the case for $x=0$, it may just be a case of not seeing it sweep up. On the other hand, looking from the other side, even in a diagonal line (ie $(ax)^x$), all do rapidly rise to 1, as x approaches 0. It's only when one approaches it from $0^x$ that you can't see it rising. So the evidence from the graph of $x^y$ is that $0^0$ is definitely 1, except when approached from $y=0$, when it appears to be zero.

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Take a look at WolframMathWorld's [1] discussion.

See if this gives you any clarification.

[1] Weisstein, Eric W. "Indeterminate." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Indeterminate.html

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I sent them a message to remove 0^0 from inderminate, with necessary proof. – wendy.krieger May 16 '13 at 10:22
@TylerClark: that discussion is about $\lim\limits_{(x,y)\to(a,b)}f(x,y)$ for various $f$ and $(a,b)$. Since that limit does not exist for $f(x,y)=x^y$ at $(x,y)=(0,0)$, $0^0$ is called an indeterminate form. However, that is a separate issue from whether $0^0$ is defined. – robjohn Oct 13 '14 at 0:02
@wendy.krieger: $x^y$ is definitely not continuous at $(0,0)$: $\lim\limits_{x\to0^+}x^0=1$, whereas $\lim\limits_{x\to0^+}0^{\large x}=0$. – robjohn Oct 13 '14 at 5:14
@wendy.krieger: Surely, you see that $$\lim_{\delta\to0}\left(\delta^{1/\delta}\right)^{\delta} =\lim_{\delta\to0}\delta=0$$ and that $$\lim_{\delta\to0}\left(\delta^{1/\delta},\delta\right)=(0,0)$$ Thus, along the path $(x,y)=\left(\delta^{1/\delta},\delta\right)$, which tends to $(0,0)$, it is easy to see that $x^y$ tends to $0$. Nowhere is $0$ divided by $0$ or is $0$ raised to any power. Only positive numbers to positive powers. – robjohn Oct 14 '14 at 13:35
@wendy.krieger: if $(x,y)=\left(\delta^{1/\delta},\delta\right)$, then $x=\delta^{1/\delta}$ tends to $0$ and $y=\delta$ tends to $0$. Also $x^y=\left(\delta^{1/\delta}\right)^\delta=\delta$ also tends to $0$. Nothing dodgy. – robjohn Oct 15 '14 at 2:27

There are so many fantastic answers already, but no pictures which is sad. $0^0$ is undefined - is it the limited case of $x^0$, $0^x$ or $x^x$? The plots for $x^0$ (a line at 1) and $0^x$ (a line at zero) are boring, but $x^x$ is

Author/Site: Wolfram|Alpha

Publisher: Wolfram Alpha LLC

URL: https://www.wolframalpha.com/input/?i=plot+x%5Ex

Retrieval date: 6/2/2015

I don't always define $0^0$, but when I do, I define it as $1$.

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+1 for "I don't always define $0^0$" – pjs36 Jul 20 at 4:09

Maybe it's a good idea to put the problem into a broader perspective.

The minimum we need to define a power is a multiplicative semigroup, which basically means we have a set with an associative operation which we write as multiplication. If $S$ is the semigroup, the power function is defined as $$S\times \mathbb Z^+ \to S, (x,n) \mapsto x^n = \begin{cases} x & n=1\\ x x^{n-1} & n>1 \end{cases}$$ This power function has the fundamental properties $$x^mx^n = x^{m+n},\quad (x^m)^n = x^{mn}\tag{*}$$ Note that up to now, we have used absolutely nothing about the elements of $S$ except for the fact that we can multiply them.

Now it is a natural question whether we can extend that definition from the positive integers to the non-negative integers, in other words, whether we can define $x^0$. Of course we would want to define it in a way that the power laws (*) still hold. This especially implies the following relations: $$(x^0)^2 = x^0 \tag{I}$$ (that is, $x^0$ is idempotent) and $$x^0 x = x x^0 = x.\tag{II}$$ Note that this does not imply that for any $x\ne y$, $x^0=y^0$.

Now for a given $S$ and a given $x\in S$, there are three possible cases:

1. There does not exist an element $z\in S$ that fulfils both (I) and (II). In that case, $x^0$ is undefined and undefinable without violating the power laws. For example, if you take $S$ as the set of positive even numbers, then $x^0$ is undefined for all $x\in S$ (because there's no positive even number that fulfils either (I) or (II) for any x)
2. There exists exactly one element $z\in S$ that fulfils those conditions. In that case, the only reasonable definition is $x^0 = z$. For example, for non-zero real numbers, you get $x^0=1$ this way.
3. There exist more than one element $z_i\in S$ fulfilling those equations. This is the case for $0^0$ because both $0$ and $1$ fulfil the equations. In that case, you have several choices:

• You can select one of the possible values and define $x^0$ as that value. Of course you'd not randomly choose any value, but choose the one which is most useful. Which may be different in different contexts. Note that each of the choices gives a different valid power function.
• You can leave $x^0$ undefined. In this case, $x^0$ is called unspecified because you could specify it (as in the previous bullet). In particular, the restriction of any of the power functions from the first bullet to $S\setminus\{x\}$ will be the power function from this bullet.

Now one particularly interesting case is if you have a neutral element, that is, an element $e\in S$ so that $ex=xe=x$ for all $x\in S$. A semigroup with such an element is called a monoid.

It is easy to check that in this case, $e$ fulfils both (I) and (II) for any element $x\in S$. Therefore you always get a valid power function by defining $x^0=e$. Indeed, by doing so, you get the general form $$x^n = \begin{cases}e & n=0\\ x x^{n-1} & n>0\end{cases}$$ which means that in a monoid, $x^0=e$ clearly is a distinguished, and therefore preferable choice of the power function, even for $x$ where other choices would be possible. Note that the real numbers form a monoid under multiplication, with $e=1$. Therefore this is a first hint that $0^0=1$ is a good definition.

The definition $0^0=1$ turns out to be useful also in other areas, for example when considering the linear structure (that is, the distributive law with addition), as it makes sure that e.g. the binomial formula $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}$$ also holds if any of $a$, $b$ or $a-b$ is $0$.

There's however one structure that doesn't favour $0^0=1$, and that is continuity: $x^y$ is discontinuous at $(0,0)$, and every value can be obtained as limit of a suitably chosen sequence of arguments approaching the origin. Therefore from the point of continuity, $0^0$ can be considered unspecified. However, since $x^y$ is discontinuous at the origin whether of not one defines it there, and independent of the value one chooses, so that is not really an argument against choosing $0^0=1$, but more an argument against using that definition blindly.

I'm not aware of a context where the definition $0^0=0$ would be more useful, but I wouldn't want to exclude that it exists. Any other value of $0^0$ would violate condition (I) above, and therefore likely not be useful at all.

So in summary, the definition $0^0=1$ is the most useful in most situations, and not harmful in situations where one would otherwise leave $0^0$ unspecified, and if any context exists where another definition would be more useful, it's arather unusual context. Therefore the definition $0^0=1$ is the most reasonable one.

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@user21820: Thank you. – celtschk Apr 22 at 19:38

A paper for general public is published on Scribd : " Zero to the Zero-th Power" (pp.7-11) : http://www.scribd.com/JJacquelin/documents

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Another approach and yet another result !

We define an exponential function on $\mathbb{R}$ as a function $E:\mathbb{R}\rightarrow \mathbb{R}$ such that $$E(x+y)=E(x)E(y) \qquad \forall x,y \in \mathbb{R}$$ That's a very general and powerful definition, and I love it because captures the link between Lie algebras and Lie groups.

From this definition we have immediately that if $\exists a\in \mathbb{R}$ such that $E(a)=0$ then $E(x)=0 \; \forall x \in \mathbb{R}$. Nor continuity nor other topological properties are needed for this. In exponential fields theory such a function is called a trivial exponential function.

If $E$ is not this trivial function, it's easy to see that we must have $E(0)=1$ and, if $E(x)=1$ for some $x \ne 0$ than $E(x)=1 \forall x \in \mathbb{R}$, i.e. it is a constant (another trivial exp. function).

If $E$ is not trivial and $E(1)=a$ we have one particular funtion $E_a(n)$ and it is easy to see that $E_a(-1)=\dfrac{1}{a}$ and $E_a(n)=a^n$ (and all other properties of integer or fractional exponents), so it is natural to write $E_a(x)=a^x$ .

Now we see that $0^x$ is the null trivial exponential function, and, since this function must be always null, we have $0^0=0$.

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Another reason to define $0^0=1$ comes from probability, in particular Bernoulli trials.

These are independent repeated trials of an experiment, whose outcome is either positive or negative. So let $p$ be the probability of a success for each trial. Then the probability of $k$ successes out of $n$ trials is $$p_k = {n \choose k}p^k(1-p)^{n-k}.\tag{B}$$ Now, suppose $p=1$. $(B)$ yields $p_n=0^0$. However, we already know that each trial will certainly occur, resulting in $p_k=0$ for all $k<n$ and $p_n=1$, whence $$0^0=1.$$

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Source: Understanding Exponents: Why does 0^0 = 1? (BetterExplained article)

A useful analogy to explain the exponent operator of the form $a \cdot b^c$ is to make $a$ grow at the rate $b$ for time $c$.

Expanding on that analogy, $0^0$ can be interpreted as $1\cdot0^0$ which is to say: grow $1$ at the rate of $0$ for time $0$. Since there is no growth (time is $0$), there is no change in the $1$ and the answer is $0^0=1$

Of course, this is just to grok and get an intuition or a feel for it. Science is provisional and so is math in certain areas. 0^0=1 is not always the most useful or relevant value at all times.

Using limits or calculus or binomial theorems doesn't really give you an intuition of why this is so, but I hope this post made you understand why it is so and make you feel it from your spleen.

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Downvoter explain. I am not trying to be rigorous, I'm just trying to give an intuition and make people truly feel why this should be right. – YatharthROCK May 1 '13 at 14:21

A clear and intuitive answer can be provided by ZFC Set-Theory. As described in Enderton's 'Elements of Set Theory (available free for viewing here; see pdf-page 151): http://sistemas.fciencias.unam.mx/~lokylog/images/stories/Alexandria/Teoria%20de%20Conjuntos%20Basicos/Enderton%20H.B_Elements%20of%20Set%20Theory.pdf, the set of all functions from the empty set to the empty set consists merely of the empty function which is 1 function. Hence $0^0$ = 1.

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Contents taken and reformatted from: The Math Forum

What is $0$ to the $0$ power?

This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught (hdev@cp.tn.tudelft.nl). According to some Calculus textbooks, $0^0$ is an "indeterminate form." What mathematicians mean by "indeterminate form" is that in some cases we think about it as having one value, and in other cases we think about it as having another.

When evaluating a limit of the form $0^0$, you need to know that limits of that form are "indeterminate forms," and that you need to use a special technique such as L'Hopital's rule to evaluate them. For instance, when evaluating the limit $\sin(x)^x$ (which is $1$ as $x$ goes to $0$), we say it is equal to $x^x$ (since $\sin(x)$ and $x$ go to $0$ at the same rate, i.e. limit as $x\to0$ of $\sin(x)/x$ is $1$). Then we can see from the graph of $x^x$ that its limit is $1$.

Other than the times when we want it to be indeterminate, $0^0 = 1$ seems to be the most useful choice for $0^0$. This convention allows us to extend definitions in different areas of mathematics that would otherwise require treating $0$ as a special case. Notice that $0^0$ is a discontinuity of the function $f(x,y) = x^y$, because no matter what number you assign to $0^0$, you can't make $x^y$ continuous at $(0,0)$, since the limit along the line $x=0$ is $0$, and the limit along the line $y=0$ is $1$.

This means that depending on the context where $0^0$ occurs, you might wish to substitute it with $1$, indeterminate or undefined/nonexistent.

Some people feel that giving a value to a function with an essential discontinuity at a point, such as $x^y$ at $(0,0)$, is an inelegant patch and should not be done. Others point out correctly that in mathematics, usefulness and consistency are very important, and that under these parameters $0^0 = 1$ is the natural choice.

The following is a list of reasons why $0^0$ should be $1$.

Rotando & Korn show that if $f$ and $g$ are real functions that vanish at the origin and are analytic at $0$ (infinitely differentiable is not sufficient), then $f(x)^{g(x)}$ approaches $1$ as $x$ approaches $0$ from the right.

From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik):

Some textbooks leave the quantity 0^0 undefined, because the functions $0^x$ and $x^0$ have different limiting values when $x$ decreases to $0$. But this is a mistake. We must define $x^0=1$ for all $x$, if the binomial theorem is to be valid when $x=0$, $y=0$, and/or $x=-y$. The theorem is too important to be arbitrarily restricted! By contrast, the function $0^x$ is quite unimportant.

As a rule of thumb, one can say that $0^0 = 1$, but $0.0^{0.0}$ is undefined, meaning that when approaching from a different direction there is no clearly predetermined value to assign to $0.0^{0.0}$ ; but Kahan has argued that $0.0^{0.0}$ should be $1$, because if $f(x)$, $g(x)$ $\to0$ as $x$ approaches some limit, and $f(x)$ and $g(x)$ are analytic functions, then $f(x)^{g(x)}\to1$.

The discussion of $0^0$ is very old. Euler argues for $0^0 = 1$ since $a^0 = 1$ for a not equal to $0$. The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift. Consensus has recently been built around setting the value of $0^0 = 1$.

References

Knuth. Two notes on notation. (AMM 99 no. 5 (May 1992), 403-422).

H. E. Vaughan. The expression '$0^0$'. Mathematics Teacher 63 (1970), pp.111-112.

Louis M. Rotando and Henry Korn. The Indeterminate Form $0^0$. Mathematics Magazine, Vol. 50, No. 1 (January 1977), pp. 41-42.

L. J. Paige. A note on indeterminate forms. American Mathematical Monthly, 61 (1954), 189-190; reprinted in the Mathematical Association of America's 1969 volume, Selected Papers on Calculus, pp. 210-211.

Baxley & Hayashi. A note on indeterminate forms. American Mathematical Monthly, 85 (1978), pp. 484-486.

Robert S. Fouch. On the definability of zero to the power zero. School Science and Mathematics 53, No. 9 (December 1953), pp. 693-696.

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That's a nice question, and not an easy one. Actually, I don't think anybody has written a definite solution (if there is, it's probably some naive based on intuition); but if you accept Russell's way (which is identical to that of Cantor in respect to natural numbers), take a look at Whitehead's paper http://www.jstor.org/stable/2370026. In the third line of page 23 (page 388 on the paper) he defines the concept of number (cardinal) exponentiation. Recurring to Cantor's set theory is the only way I think this problem could be solved. The use of real analysis, or other branches such as combinatorics (binomial theorem), won't do any good because their construction already pressuposes cardinal arithmetic. I.e, the problem can be solved using a more elementary theory.

Have a nice day.

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The longstanding practice of leaving $0^0$ undefined is usually justified with arguments based on path dependent limits on the real numbers. As we see here, however, it is also possible to justify this practice based on purely discrete methods.

If the intuition of exponentiation on $N$ (where $0\in N$) is to be repeated multiplication such that $x^2=x\cdot x$, then we can formally justify the following definition:

1. $\forall x,y\in N: x^y\in N$ (a binary function on $N$)

2. $\forall x\in N:(x\ne 0\implies x^0=1)$

3. $\forall x,y\in N:x^{y+1}=x^y\cdot x$

Here, $0^0$ is assumed to be a natural number, but no specific value is assigned to it.

From this definition, we can derive the usual Laws of Exopnents:

1. $\forall x,y,z\in N: (x\ne 0 \implies x^y \cdot x^z = x^{x+y})$

2. $\forall x,y,z\in N: (x\ne 0 \implies (x^y)^z = x^{y\cdot z})$

3. $\forall x,y,z\in N: (x,y\ne 0 \implies (x\cdot y)^z = x^z\cdot y^z)$

For a detailed development based on formal proofs, see "Oh, the Ambiguity!" at my math blog.

Follow-up

A better approach, I have since found, is to construct a partial function on $N\times N$ with domain of definition $N\times N \setminus (0,0)$ such that:

1. $\forall x,y:[x,y\in N \land \neg [x=0\land y=0] \implies x^y\in N$

2. $\forall x:[x\in N\land x\ne 0\implies x^0=1]$

3. $0^1=0$

4. $\forall x,y:[x,y\in N \land \neg [x=0\land y=0] \implies x^{y+1}=x^y\cdot x]$

See my revised blog posting at above link (originally dated October 9, 2013).

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It seems forced to decide that $0^0$ is "undefined", there is no contradiction in having $0^0=1$, and in fact it makes the second axiom, as well all the laws to change from implication to just $x^0=1$ and $x^{y+1}=x^y\cdot x$, and so on. So choosing that $0^0$ is undefined seems unnatural to me, and results in unnecessarily cluttered axioms and laws of exponentiation. – Asaf Karagila Nov 20 '13 at 21:37
This seems to compare to a situation where I'll write "Let's agree that the cardinality of the empty set is not defined, now we can devise the usual axioms of cardinal arithmetics, but I'll have to add an implication of the form only if the sets involved are not empty ... to every axiom!" – Asaf Karagila Nov 20 '13 at 21:39
@AsafKaragila There is also no contradiction for $0^0=999$ or any other natural number. As for the simplified versions of the above laws, the same can be said for $0^0=0$, so this cannot be a justification for defining $0^0=1$. $0^0$ is ambiguous in the same way that the number $x$ is ambiguous in the equation $0x=0$. Any value will work, as I show at my blog. I am not aware of any logically compelling reason to choose any particular value. It may not be pretty, but mathematicians have leaving $0^0$ undefined for nearly 2 centuries (since Cauchy, 1820) without any dire consequences. – Dan Christensen Nov 20 '13 at 21:59
There's also no contradiction in deciding $x^y=999$ for every $x,y$. So what? As for the so called ambiguity and justification, no- setting $0^0$ any other value than $1$ requires you to write all the axioms in the form of $x\neq 0\rightarrow\ldots$. Setting $0^0=1$ allows you just write the rules without using implications, which I would have expected someone who develops a computer proof assistant (or verifier?) to appreciate as a way of reducing complexity of statements. Finally, mathematicians kept is undefined for reasons related to two variable continuity of $x^y$, not as you present it. – Asaf Karagila Nov 20 '13 at 22:08
This discussion is futile. I'm gonna quit while I'm ahead. Have a pleasant evening. – Asaf Karagila Nov 20 '13 at 22:13

It is solvable by means of contradiction:

Here are some general rules:

$x^0=1$

and

$0^x=0$

If $x=0$ you get two answers for $0^0$: $0,1$

It can't be two different solutions so we simply coin it as undefined.

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The second "general rule" is wrong. It is correct only for positive numbers. – Anixx Aug 21 at 2:58

It depends on whether the 0 in the exponent is the real number 0, or the integer 0. These are two different objects and while the distinction is not often important, it is in this case.

Exponentiation by an integer has a universal specific meaning: Positive exponent is repeated multiplication, negative exponent is the inverse of repeated multiplication, zero exponent is the empty product, equal to the multiplicative identity denoted by 1. So if the exponent is the integer 0, then $0^0=1$ - the fact that this is an empty product with no terms trumps the fact that the terms which are not there are 0's (because they're not there).

Whereas exponentiation by a real or complex number is a messier concept, inspired by limits and continuity. So $0^0$ with a real 0 in the exponent is indeteriminate, because you get different results by taking the limit in different ways.

Note that all of the standard examples where it's "convenient" to have $0^0=1$ (e.g. power series) are all cases where the exponent is an integer.

Note that as long as the exponent is the integer 0, it doesn't matter if the base is the integer 0, the real 0, or pretty much any mathematical object on the planet for which multiplication is defined.

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