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

My train of thought:


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


Possible answers:

  1. $0^0\cdot0^x=1\cdot0^0$, so $0^0=1$
  2. $0^0=0^x/0^x=0/0$, which is undefined

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

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    BTW, this is [comprehensively covered on Wikipedia](http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power) (or see [today's version](http://en.wikipedia.org/w/index.php?title=Exponentiation&oldid=397930313#Zero_to_the_zero_power)), along with pointers to history and treatment in many systems. – ShreevatsaR Nov 22 '10 at 12:34
  • see also http://math.stackexchange.com/questions/135/why-is-x0-1-except-when-x-0/351#351 – sdcvvc Feb 11 '13 at 21:59
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    @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
  • Related: http://matheducators.stackexchange.com/questions/416/what-are-argument-one-can-give-to-students-on-the-definition-00 – Incnis Mrsi Nov 09 '14 at 13:28
  • For future readers of the post, check the links at the right hand side for similar questions. – NoChance Mar 12 '16 at 09:57
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    You can not make the argument: "$0^0 = 0^x / 0^x = 0/0 =$ undefined" without using the rule "I get to divide by 0 but you don't". – Mark Feb 12 '17 at 21:07
  • On Wikipedia, this topic nowadays has its own article: https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero – Hans Lundmark Jan 14 '20 at 12:51
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    I think gradually more people will go with the convention $0^0=1$. – Asinomás Apr 06 '21 at 15:36

24 Answers24


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.

Arturo Magidin
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  • @Arturo: Interesting. I have not seen this argument else where. But by this "set-theoretic" argument, could we also argue $1^{\infty} = 1$? in the thread started by Chandru (http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form) –  Nov 20 '10 at 23:20
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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
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    @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
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    Sorry. Possible, my question wasn't clear. I mean $0^0$ in terms of discrete mathematics. This is not a limit. You shouldn't approach it. – Stas Nov 20 '10 at 23:46
  • @Stas: then the answer is my paragraph after "However". That's why it's defined as $1$ in discrete mathematics. – Arturo Magidin Nov 20 '10 at 23:48
  • @Arturo: Sorry, my math really sucks, and I thought "set-theory" interpretation doesn't include my elementary case. I would be happy to get a bit more simple explanation. – Stas Nov 20 '10 at 23:58
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    @Stas: You don't seem to have any "elementary case". All you have are your "train of thoughts". What case is it you are thinking about? You don't say. You don't tell us. I cannot read your mind from this afar away (and the Government doesn't let me do it without a Court order anyway...) – Arturo Magidin Nov 21 '10 at 02:51
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    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 05:47
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    @Arturo: wonderful answer! Very minor typo: the limit along the second curve you give should be $e^7$ not $7$. (Or else the curve should be $y = \frac{\ln 7}{\ln x}$.) – Peter LeFanu Lumsdaine Nov 21 '10 at 09:02
  • @Arturo: Thanks a lot for the explanation. Yes, case with natural numbers (including 0) is pretty clear now. But, people assert that $0^0 = 1$ even for real numbers. For example: "Kahan has argued that 0.0^(0.0) should be 1, because if f(x), g(x) --> 0 as x approaches some limit, and f(x) and g(x) are analytic functions, then f(x)^g(x) --> 1 ." – Stas Nov 21 '10 at 12:03
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    @Stas: Some argue that; there are reasons for saying it should be one, but there are also compelling reasons for it to be other things. In some situations, some limits make more sense than others; if all you are concerned with are *analytic functions*, then it may make sense to define it to be 1 because in the only cases you will look at you will always get 1 as the limit. But precisely *because* the answer depends on context, it is left undefined in the abstract and only defined in certain specific contexts (such as combinatorics or cardinal exponentiation). – Arturo Magidin Nov 21 '10 at 19:36
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    @Arturo: I heartily disagree with your first sentence. Here's why: There's the binomial theorem (which you find too weak), and there's power series and polynomials (see also Gadi's answer). For all this, $0^0=1$ is _extremely_ convenient, and I wouldn't know how to do without it. In my lectures, I always tell my students that whatever their teachers said in school about $0^0$ being undefined, we do define it here as 1 for the above reasons. (Clearly a definition of the kind "whichever way it is most useful".) – Hendrik Vogt Jun 30 '11 at 08:38
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    @Hendrik: Then I trust all your limit theorems contain all the necessary, lengthy, caveats and exclusions. Otherwise, they're all wrong! – Arturo Magidin Jun 30 '11 at 14:40
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    @Arturo: Which ones? I have no idea what you mean, but would be very interested indeed! (And: how do _you_ write power series?) – Hendrik Vogt Jun 30 '11 at 15:33
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    @Hendrik: I was being a bit facetious, sorry. One has to be careful with limit theorems when $0^0$ has a specific meaning, because if the hypothesis are not carefully stated, they lead to the (erroneous) conclusion that any limit of the form $f(x)^{g(x)}$ in which $f(x),g(x)\to 0$ will have that definition as limit. Didn't mean it to be a slight (and you'll notice I said "*usually* left undefined"). But yes, "whichever way it is most useful", but because the most useful definition changes depending on context, it's a bad idea to define it categorically. – Arturo Magidin Jun 30 '11 at 16:08
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    @Hendrik: As long as I'm at it: one reason I find the binomial theorem explanation weak is that if one simply applies the formula it leads to all sorts of strange things, e.g., $(1-2)^{-1}=1+2+4+8+16+\cdots$ (which, granted, is *true* in the $2$-adics!). No slight on those who want it to be $1$ (to be honest, if there *had* to be a definition, I would go with $0^0 = 1$ as well). – Arturo Magidin Jun 30 '11 at 16:10
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    @Arturo: Well, of course $(x,y)\mapsto x^y$ isn't continuous at $(0,0)$ if $0^0=1$, but what's the problem with that? And it is in fact the "usually left undefined" that I heartily disagree with, in connection with "analysis". On the contrary, in analysis one usually defines $0^0=1$. I'd still like to know how you write power series! And concerning your $(1-2)^{-1}$ example, that's not the binomial theorem, but an abuse of the binomial _series_. I just thought of the binomial theorem for natural powers (hence with a finite sum). – Hendrik Vogt Jun 30 '11 at 17:10
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    @Hendrik: Careful there; "naturals" need not include $0$ (and I can show you as many sources defining it without $0$ as I can show you sources defining $\mathbb{N}$ as including $0$), in which case "binomial theorem for natural powers" does not include the exponent $0$. :-P As to how I write power series, I *define* the expression $x^0$ to be $1$, but when I define evaluation at $a$ I separate the constant term. But I don't do analysis, so I don't have to be doing it all the time. (cont) – Arturo Magidin Jun 30 '11 at 17:13
  • @Hendrik: But, fair enough. I can edit and drop the "and Analysis" and leave only "calculus" around... Just one thing: if I edit the answer, it may revert convert to Community Wiki (it already has 9 edits, 8 of which are mine). Would you mind doing it yourself, since you feel about it far more strongly than I do? – Arturo Magidin Jun 30 '11 at 17:17
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    @Arturo: If you don't do analysis, then it would indeed be fair enough to remove that. And yes, I could suggest the edit. (I know very well that the natural number need not include $0$, but so what? I was taking about the exponent in $(a+b)^n$. And don't you want to write $\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ in the right hand side? So I don't see why you find that weak. Or do you want to exclude the cases $a=0$ and $b=0$?) – Hendrik Vogt Jun 30 '11 at 18:49
  • @Hendrik: Not a big deal for me, because it doesn't come up often enough to be a big deal. Go ahead an propose the edit; the only reason I'm not doing it myself is the fear of toppling this into a CW, which I don't want to do. – Arturo Magidin Jun 30 '11 at 18:56
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    This time I did include the smiley... You said "natural powers", but if you are only talking about the natural powers, then n=0 is not a valid power unless 0 is a natural number. Of course, that's not a real objection. As to the other question, I don't usually want to write that on the right hand side; I usually write $a^n + \sum_{k=1}^{n-1}\binom{n}{k}a^{n-k}b^k + b^n$, because that is the form valid in commutative semigroups/commutative rngs... – Arturo Magidin Jun 30 '11 at 19:04
  • @Arturo: OK, I'll do that tomorrow. (The CW part was clear to me, I also would want to avoid that.) – Hendrik Vogt Jun 30 '11 at 19:35
  • @Hendrik: If you (or anyone else) does the edit, it shouldn't turn into a CW. As I understand it, it's 10 edits by the author, or edits by 5 or more editors, that triggers the change. – Arturo Magidin Jun 30 '11 at 19:38
  • @Arturo: As I wrote, the CW part was clear to me. If I edit (or rather suggest an edit), then it won't become CW. (I guess you also have 2 more edits before CW, but it's better if you save them.) – Hendrik Vogt Jun 30 '11 at 19:43
  • @Arturo: I just suggested an edit. – Hendrik Vogt Jul 03 '11 at 11:39
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    @Arturo: you can now flag for Moderator attention to **undo** the automatic wiki-fication of posts from repeated edits. (I don't think I can undo it if you actually checked the Wiki check box.) This is one of the recent new features to SE2 that was added by popular demand. – Willie Wong Jul 03 '11 at 12:18
  • @Willie: Ah; I think I remem ber that, and now I remember forgetting it... To be honest, when I suggested Hendrik do the edit, I thought it would be a quick fix. Thanks for the reminder! – Arturo Magidin Jul 03 '11 at 14:19
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    "otherwise it would lead to all sorts of exceptions when dealing with the limit laws" - can you please give some examples of such exceptions? – Anixx Nov 04 '12 at 22:39
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    "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 08 '13 at 20:28
  • @OmarAntolín-Camarena: we have to be careful there. We could equally well view "$\sum_{k=0}^n a_kx^k$ as just shorthand for $a_0 + a_1 x + a_2x^2 + \cdots + a_nx^n$, and in the latter there is no $x^0$ to be found, regardless what value we substitute for $x$. – Carl Mummert Oct 12 '14 at 23:39
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    It is *not* usually left undefined. It is defined as $1$. We write $f(x)=\sum_{i=0}^\infty a_ix^i$ for a power series, but know that $f(0)=a_0$. In set theory, the definition is simple. In lambda calculus likewise. There is only one value of $0^0$ that makes sense, and it is $1$, and we use this quite a lot when we write polynomials and power series. While it is defined, it is still an "indeterminate form." Being indeterminate is a question of continuity. – Thomas Andrews Nov 01 '14 at 15:23
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    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
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    @Arturo: You write that in Calculus there is no reasonable definition. Your argument uses the "continuity rule": if $f$ is discontinuous at a point $p$ then one should not define $f$ at $p$. However, this rule has been abandoned more than a century ago. Isn't it time to abandon its corollaries as well? – Mark Feb 12 '17 at 21:21
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    The "cardinal exponentiation" case can be vastly generalized to any symmetric monoidally closed category with an initial object or relatedly to [powers](https://ncatlab.org/nlab/show/power#definition). Writing $A\multimap B$ for what in the cartesian case would be $B^A$, we have that $(-)\multimap B : \mathcal{C}^{op}\to\mathcal{C}$ is right adjoint to itself, thus $0\multimap B \cong 1$ natural in $B$, and so for $0$ too, by continuity of right adjoints. For $\mathcal{C}=\mathbf{Set}$ and the cartesian monoidal structure, this is exactly that $B^0 \cong 1$ with $0^0\cong 1$ as a special case. – Derek Elkins left SE Mar 19 '17 at 23:42
  • What is an empty function ? As far as I know, from or to an empty, you cannot define any function. – Our Oct 07 '18 at 11:38
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    @onurcanbektas: A function from $A$ to $B$ is a subset $f$ of $A\times B$ such that for all $a\in A$ there is $b\in B$ such that $(a,b)\in f$, and if $(a,b),(a,b’)\in f$, then $b=b’$. When $A=\varnothing$, the set $\varnothing$ satisfies these properties,so it is a function (for any set $B$). It is called “the empty function”. – Arturo Magidin Oct 07 '18 at 17:44
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    @ArturoMagidin Well, that makes sense. Thanks for the response. – Our Oct 07 '18 at 18:32
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    I don't really see why the limit of $f(x)^{g(x)}$ would disagree with defining $0^0 = 1$. Did you mean finding limit by plugging in limits of f and g? Doesn't it require the limits positive (or at least the limit of the base function positive)? What is the version of this theorem you are using? – Shiyu Liang Nov 18 '19 at 19:12
  • @ShiyuLiang: For real valued functions, if $f(x)\to 0$ (from the positive side, if you wish), and $g(x)\to 0$, it is not necessarily the case that $f(x)^{g(x)}\to 1$. So whereas in every other situation where $f(x)\to L$ and $g(x)\to M$, we *do* have $f(x)^{g(x)}\to L^M$, you would need an exception to that general statement to exclude the *defined* case of $0^0$ and declare it "undefined" or "indeterminate", even though the operation $0^0$ *would* be defined. Note that every other "indeterminate" limit involves an "operation" at the end that cannot be performed, such as $\frac{0}{0}$. – Arturo Magidin Nov 18 '19 at 19:14
  • @Arturo Magidin: I see you point, but usually the limits $L$ in the theorem are required to be positive such that you can apply $f(x)^{g(x)} = e^{ln(f(x)^{g(x)})}$ to prove this theorem. In either case when $L$ is negative or $0$ this does not hold since $x^y$ only "well" defined in $\mathbb{R_+}\times\mathbb{R}$. By "well" I meant continuously defined in a neighborhood. Therefore, adding definition $0^0 = 1$ really does not change this theorem at all, since the theorem can't be applied anyway. Can you let me know any reference saying you can plug $L$ and $M$ for free regardless the signs? – Shiyu Liang Nov 19 '19 at 17:19
  • @ShiyuLiang: I did say you could require $f(x)$ to approach $0$ from the right (see that parenthetical comment, "from the positive side, if you wish"?), meaning that even though $L=0$, the function whose limit is $0$ only takes positive values. Therefore, your entire comment is about a situation not discussed. Adding $0^0=1$ requires you to provide an exception to the result when the limitant makes sense and the expression at the end is defined. You can have $f(x)\to 0$ with $f(x)$ always positive. I did not say "regardless of signs", so your request is about I said something I did not say. – Arturo Magidin Nov 19 '19 at 17:44
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    Yet another argument in favor of 1: $\lim_{z \rightarrow 0} z^0 = 1$, whereas the same is not true for $0^z$ (the limit no longer exists). – user76284 Apr 27 '20 at 01:01

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).

Gadi A
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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 01:12
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    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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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.

Martin Sleziak
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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 $(-1)^{1/3}$ –  Oct 26 '11 at 22:06
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    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 01:13
  • It's the view that I said was narrow, not the breadth of application. The advantage to leaving $0^0$ undefined is in situations where continuity *is* relevant. You might never be in such situations, but others are. Really, there are three separate exponentiation operations in common use -- the algebraic one which is mostly defined for all bases and integer exponents (often extensible to rational exponents), the real one which is defined for positive real bases and real exponents (or some continuous extension thereof), and the complex multi-valued one. $0^0$ only makes sense for the first. –  Oct 27 '11 at 16:41
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    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
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    What problem do you see with the complex exponential? I agree with not defining $(-2)^0$ for the real version, but I'm also not opposed to negative bases because I believe it still remains continuous on its domain (so long as you exclude $0^0$). Economy of notation is only gained in the algebraic case. It's anti-economy of notation for the continuous case because you have to start adding caveats everywhere to exclude $0^0$, such as in the theorem that says $\lim x^y = (\lim x)^{\lim y}$. –  Oct 27 '11 at 23:04
  • I should clarify what I meant about the complex exponential. There is no problem with e^z, z complex. Now, in general, x^y means e^(y*ln(x)). The complex logarithmic "function" cannot be continuously defined on all of C, or even on C-{0}. Hence, branches of the logarithmic function. So, there already problems with continuous complex extensions of the real logarithmic function. As regards your closing statement, please cite a reference for that theorem, as I am pretty sure it is not a theorem at all (certainly not as stated without reference to domain, or how the limit is to be taken). – Onez Oct 27 '11 at 23:17
  • Yes, as I said the complex exponential is multi-valued. The limit law for real exponentiation is because of its continuity. On the domain I specified, this fact follows from $x^y = \exp(y \log x)$. –  Oct 28 '11 at 02:33
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    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 03:29
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    You missed the more relevant option -- acknowledge the multiple exponentiation operations that come up in mathematics, rather than conflating them. –  Oct 28 '11 at 06:18

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 or not one defines it there, and independent of the value one chooses, 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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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.

Martin Sleziak
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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.


"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 $\displaystyle\lim_{x\to a} f(x)^{g(x)}$ 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 $x\to a$.

Michael Hardy
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    The problem is that "indeterminate form" people cannot accept that an indeterminate form also has a defined value (which is important to people who are not in the business of taking limits). Because the indefinite form method says: if you need to compute a limit, start by plugging in the limit values; if that succeeds, that _is_ (supposed to be) the limit, but if (and only if) it fails, look at the form to decide what to do. So if the expression at the limit is defined, one does not even come to asking whether the form is indeterminate. If only they would check the form before plugging in... – Marc van Leeuwen Jun 19 '21 at 11:32

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 exactly $k$ 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 $$0^0=1.$$

Vincenzo Oliva
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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
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    @KennyTM: Accepted and edited accordingly. –  Nov 22 '10 at 11:47
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    0^0 is undefined by whom? I saw somebody defined it, can I now say it is defined? – Anixx Nov 04 '12 at 22:58
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    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
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    One does not exclude the other, this can be both indeterminate form when considering limits, AND defined. – Anixx Mar 05 '14 at 00:12
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    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 05 '14 at 18:06
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    Undefined and undefinable are two different concepts. $0^0$ is neither. – DanielV May 05 '14 at 14:02
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    Being an indeterminate form is not the same as being undefined. They are two different concepts. – Thomas Andrews Nov 01 '14 at 15:25
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    You write "$0^0$ has many possible interpretations" and then listed invalid interpretations. Determining $f(0)$ by taking a limit is only valid when $f$ is continuous at $0$. It is strange that people keep forgetting that. – Mark Feb 16 '17 at 15:26
  • Using lim(x->0)0*x can never reach 0, because the ways of reducing the power is by taking a root (which never leaves positive, since x/y >0 if x, y > 0.), or by dividing (ie 0/0). But since it can be shown by other limits that x^0 = 1, for all x, then we should suppose 0^0 = 1. – wendy.krieger Jan 29 '22 at 14:19

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.

Published by Addison-Wesley, 2nd printing Dec, 1988.

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 $0^0 = 1$.


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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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.

Meni Rosenfeld
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  • @Wildcard. Thanks. This is one of those things that unfortunately too few people understand. That's what happens when people limit themselves to thinking of real numbers and fail to realize that math is much richer than that - with all kinds of structures which look nothing like numbers, and yet where multiplication and integer exponentiation make sense. Myself I've picked up this distinction from Mike Oliver, aka User:Trovatore on Wikipedia, and I'm happy to spread the word. – Meni Rosenfeld Jul 12 '17 at 13:37
  • @Meni Useful distinction. But FWIW, problematic in my view. Saying that the integer exponent zero is somehow different than the real number exponent zero is tricky at best. How would you even write that symbolically? I bet expressions like $0 \in \mathbb{R}$ or $0 \in \mathbb{Z}$ would be hard to find in mathematical papers. As a software developer the concept of a typed zero makes sense to me, but I'm confident that it would be rare to find languages or run-time libraries where $0^{0.0} \ne 0^0$. – Χpẘ Jul 26 '17 at 17:18
  • @Χpẘ: You are correct that the standard notation does not make it easy to draw this distinction. However, this is more of a conceptual matter, and I believe that in any case where the distinction matters, the correct interpretation will be clear from context. As for libraries, I don't know which actually do that, but they totally should return NaN for $0^{0.0}$ and 1 for $0^0$. – Meni Rosenfeld Jul 26 '17 at 21:35
  • @Χpẘ zeros are always problematic, because they wipe out unit multipliers. So zero miles equals zero gallons equals zero light-year kilotons per microsecond equals a zero wavelength equals zero time and zero space...and now we're in the realm of philosophy. Zero is the *most* important number to be "typed." – Wildcard Aug 08 '17 at 09:27
  • @Wildcard Interesting. In my intuition zero siriometers equals zero cow's grasses equals zero shakes, etc. They all signify no-thing. At least in realms concerned with dimensional analysis. In usage outside of such realms, it would be strange to say, "We've got zero kinematic viscosity in the fuel tank - we're not going anywhere." But English has plenty of inconsistencies. That said, I don't see how this relates to a definition of $0^0$. You can rewrite dimensional equations to be dimensionless. Additionally Wikipedia says: Scalar args to transcendental functions must be dimensionless. – Χpẘ Aug 09 '17 at 22:26
  • @Χpẘ, zero **isn't really a number.** It's a variable. A *wild* variable. You can take an equation which is either true or false, or which has a certain solution set, and multiply both sides by zero and the equation is suddenly always "true," regardless of what it was initially. (That is, if you accept the "truth" of $0=0$ in the first place). This isn't the case with *any* actual quantity. If you want a REAL zero, you have to go into philosophy. No mathematician has ever defined zero. (It's not even defined on Wikipedia.) – Wildcard Aug 09 '17 at 22:32
  • @Wildcard: Now you're not making any sense. Of course zero is a number, and is as defined as any other number. In the usual construction of natural numbers, $0$ is defined as simply the empty set. In the usual construction of integers, $0$ is the set $\{(a,a)|a\in\mathbb{N}\}$. In the usual construction of the rationals, $0$ is the set $\{(0,a)|a\in\mathbb{Z}\}$, where the $0$ in the definition is the integer $0$ we have just defined. And so on and so forth. Of course $0$ has unique properties wrt multiplication, just like $1$ has unique properties wrt exponentiation. e.g., always $1^x=1^y$. – Meni Rosenfeld Aug 09 '17 at 22:50
  • @wild Actually Wikipedia says "0 is both a number and the numerical digit ...". It's the additive identity element of many algebraic structures. In that usage it's not a variable. It's divisible by two, so is even number, so not a variable in that usage. It is cardinality of empty set, again not a variable in that usage. But this comes down to a philosophical question - how many angels can dance on the head of a pin? And if answer is zero, then it's not a variable, but an answer. Non-zero also answers question. In common usage, question is answered, which doesn't happen if zero is a variable. – Χpẘ Aug 09 '17 at 22:55
  • This is diving down a rabbit hole. Of course you can *work* with abstract definitions; that doesn't mean they describe anything. Zero is defined as an empty set. The empty set is defined as the unique set containing no elements. A set is a collection of elements. What is a collection of elements without any elements? How do you know it's a collection of elements? What does that even mean, if it doesn't have any elements? If the answer is that it *could* have elements, but doesn't, again that means you're describing a variable. – Wildcard Aug 09 '17 at 22:58
  • @Wildcard: None of what you're saying is in any way unique to zero. The same can be said about 1, which is the set whose only element is the set with no elements, and down the rabbit hole we go again. All theories start with basic concepts that are not defined, but rather derive meaning from axioms. The concept of "set" itself is undefined, but ZFC axioms give it meaning. Also, you keep using the word "variable", I do not think it means what you think it means. – Meni Rosenfeld Aug 09 '17 at 23:18
  • I was using "variable" in the scientific or experimental sense, not in the mathematical sense. And I prefer the *original* meaning of Axiom: a truth which is sufficiently obvious and basic so as not to require a proof. (This implies that axioms may not be chosen arbitrarily but must derive from observation.) Anyhow, I don't find that any of my philosophic objections prevent me from getting excellent results and enjoyment using mathematics; they just keep me from taking it too seriously. ;) And again, I like very much your distinction between the real exponent $0$ and the integral exponent $0$. – Wildcard Aug 09 '17 at 23:27

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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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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    To say something is undefined you should be sure nobody defined it. – Anixx Feb 07 '16 at 11:24
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    @selfawareuser Did you just defend an idea in math by dividing by zero? Look at the implicit limit you took and you'll find that I covered that case already without dividing by zero. Don't divide by zero. – user121330 Aug 05 '16 at 20:25
  • @user121330 The prohibition is against dividing a finite number by zero. $\frac{0}{0}$ in itself is legal but you must say $\frac{0}{0}\rightarrow x$ i.e. substitution is unidirectional. –  Aug 05 '16 at 21:35
  • @selfawareuser Lol, so since $\frac{0}{0} = \frac{1 \cdot 0}{3 \cdot 0}$, one might also claim that $\frac{0}{0} = \frac{1}{3}$? Why don't you provide a [citation](https://en.wikipedia.org/wiki/Division_by_zero) validating your claim. Better yet, start with $y^{x-x}$, and follow your algebra to find that you really just took the limit of $y^0$ as $y$ approaches zero which is still one. – user121330 Aug 08 '16 at 13:55
  • [Citation](http://math.stackexchange.com/questions/1595531/can-0-0-be-allowed-if-it-can-be-defined), from my user history. –  Aug 09 '16 at 07:41
  • @selfawareuser It appears you're trying to build evidence that [indeterminate forms](https://en.wikipedia.org/wiki/Indeterminate_form) have specific values. Perhaps you have a citation that lends credibility to your argument rather than your most downvoted question? – user121330 Aug 09 '16 at 16:11
  • @selfawareuser I am so sorry you're in a place where getting personal feels appropriate. It sounds like appeals to logic, authority and consensus aren't going anywhere here, so let's change course. I get that the common proofs of L'Hopital's rule (from [Bernoulli](https://en.wikipedia.org/wiki/Johann_Bernoulli#L.27H.C3.B4pital_controversy)) rely on L'Hopital's rule which makes a tautology, but can we at least agree that dividing by zero is problematic, and that you wouldn't recommend it to an algebra beginner? – user121330 Aug 11 '16 at 15:44
  • Yes, and if the numerator is non-zero it's simply wrong. –  Aug 11 '16 at 23:26
  • @selfawareuser Would you also agree that the limit as $x^0$ approaches 0 from the left and the right is 1? – user121330 Aug 12 '16 at 16:35
  • @301988 -- *0/0 is not legal.* Suppose 0/0 = a number k. Then that would mean 0 = 0*k. But then k could be any number. 0/0 is indeterminate. – Olive Stemforn Apr 03 '22 at 01:44
  • @OliveStemforn user301988 no longer has a presence on Stack. – user121330 Apr 06 '22 at 16:45

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 $0^0=0$.

Emilio Novati
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    @IlmariKaronen: we have: $E(0)=E(a-a)=E(a)E(-a)=0\cdot 0=0\ne 1$. – Emilio Novati Jan 31 '17 at 08:36
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    True, I missed the fact that the domain of $E$ includes negative numbers. Although it still seems to me that identifying the trivial function $E_0(x)=0\ \forall x\in\mathbb R$ with $0^x$ doesn't really work, because it would also imply that $\frac10=0^{-1}=E_0(-1)=0$. – Ilmari Karonen Jan 31 '17 at 16:35
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    I would argue that $0^x$ is only defined when $x\ge 0$; you would expect $0^-1=\infty$. The property $0^{x+y}=0^x0^y$ holds for all $x,y\ge 0$ whether you define $0^0=0$ or $0^0=1$. – Mike Earnest Sep 14 '18 at 00:33

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.

Yatharth Agarwal
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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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    The graphs and limits in your document are only relevant if one accepts the "continuity rule": if $f$ is discontinuous at a point $p$ then one should not define $f$ at $p$. Without that rule it doesn't make sense to look at graphs and limits when there are other ways to obtain a value. – Mark Feb 12 '17 at 22:05

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


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

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

See http://en.wikipedia.org/wiki/Indeterminate_form

Bryan Yocks
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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
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    @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 06:23
  • @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
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    @JMCF125: That's why the site is here. Where is the related question? – robjohn Feb 10 '14 at 14:25
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    @robjohn, [this one](http://math.stackexchange.com/questions/641851/does-dividing-by-zero-ever-make-sense). – JMCF125 Feb 10 '14 at 19:37
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    @Hagen the limit of $0^x$ as x-> 0 is the same as 0/0, which is undefined, but $0^0$ can be evaluated without division by 0, and these proofs lead directly to $0^0=1$. – wendy.krieger Oct 13 '14 at 04:31
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    Congrats, you defined "indeterminate form." Being an indeterminate form is not the same as being undefined. – Thomas Andrews Nov 01 '14 at 15:26
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    @JMCF125 : $5/0$ is undefined because we cannot multiply $0$ by any number and get $5$. $0/0$ is undefined because there are many different numbers we can multiply by $0$ and get $0$. But $0/0$ is indeterminate because if $f(x)$ and $g(x)$ both approach $0$ then $f(x)/g(x)$ might approach a particular number, which depends on which functions $f$ and $g$ are, and can be any number at all. $\qquad$ – Michael Hardy Feb 18 '16 at 16:37
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    Down voted because determining $f(0)$ by taking limits is only valid when $f$ is continuous at $0$. – Mark Feb 16 '17 at 15:35

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

Tyler Clark
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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
  • @wendy.krieger: In the sense given on that page, $0^0$ is an indeterminate form since the function $x^{\large y}$ is not continuous at $(x,y)=(0,0)$. However, this does not mean that we should leave $0^0$ undefined. $0^0=1$ is a useful definition in a large portion of the cases that occur normally. – robjohn Oct 12 '14 at 23:36
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    @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 00:02
  • @robjohn Actually, $x^y$ is continuious at (0,0). It is easy to see what is happening at that point by looking at $x^y$ as $x$ goes to 0. The lines hug $y=0$ but shoot up fast as $y$ approaches 0 (from +x), and crosses gives always a value of $x^0=1$. Since $x^0$ is continious, the only reason to suppose that $0^0$ is undefined, is because they read the kiddy proofs where 1=2 because 1*0=2*0. In short the only offered proofs that $0^0$ is undefined, rely on the same proofs that 1=2, ie by using 0/0. – wendy.krieger Oct 13 '14 at 04:26
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    @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 05:14
  • @robjohn The amusing thing is that if you draw any line through (0,0), except the x-axis, you get uniformly 1. For example, $\lim x^ax = 1$, for all A. That you are getting different values from the x-axis, supposes that you are doing 0/0. As regards the exponential, one supposes here that $a^{ax} = \exp(ax e^a)$, which supposes, for example, that you believe that 80 = 10^80, because both of these are bigger than 36. – wendy.krieger Oct 13 '14 at 07:02
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    @wendy.krieger: Suppose $x^y$ were continuous at $(0,0)$ and $\lim\limits_{(x,y)\to(0,0)}x^y=1$. Then *by definition*, $$\forall\epsilon\gt0, \exists\delta\gt0:|(x,y)|\le\delta\implies|x^y-1|\le\epsilon\tag{1}$$ However, $$\forall\delta\gt0,|(0,\delta)|\le\delta\land|0^\delta-1|\gt\tfrac12\tag{2}$$ and $(2)$ contradicts $(1)$. Thus, either $x^y$ is not continuous at $(0,0)$ or $\lim\limits_{(x,y)\to(0,0)}x^y\ne1$. – robjohn Oct 13 '14 at 08:52
  • @robjohn Are you by $0^{\delta}$, you are doing 0/0. The point is that you always get $0^0=1$ unless you invoke 0/0 or looking along $0^x$ (which is the same thing), because you're supposing that $0^{\delta}$ as $\delta > 0$, but this is $0/0$. So your equation is 1=2 because 1*0 = 2*0. – wendy.krieger Oct 13 '14 at 09:14
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    @wendy.krieger: Are you claiming that there is some $\delta\gt0$ so that $0^\delta\ne0$? Since we cannot take the log of $0$ and use $0^\delta=e^{\delta\log(0)}$, we need to use limits to show that $0^\delta=0$. However, rather than show that, it is simpler to consider $x^y$ along the path $(\delta^{1/\delta},\delta)\to(0,0)$ as $\delta\to0^+$. Along this path, $x^y=\delta\to0$. This shows that $\lim\limits_{(x,y)\to(0,0)}x^y\ne1$. – robjohn Oct 13 '14 at 11:54
  • @robjohn Here is a hint. If you can suppose you can deduce something about o^x, by progressing towards -infty, then you are doing 0/0. The reality is that you can clearly show the exact behaviour of a^x at a=0, by crossing the a=0 from non-zero values, this clearly gives a^0=1 for all a. An implication of letting 0^0 be undefined is that all maths is undefined, because every equation contains zero instances of some þ, and one can freely set þ=0, which makes the value indeterminate. – wendy.krieger Oct 14 '14 at 03:01
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    @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
  • @robjohn It's rather dodgy. The line actually corresponds to an progression along y=1, at an uneven rate. You state this in your first equation. Specifically, y(d)=1 for all d, so this line *never approaches* x=0, y=0, since y is never different from 1. Even so, it e1uates to lim(d,d/d), so you pretty much are at 0/0 again. This resolve to 1, but we're not interested at the point (0,1)=> 0, but (0,0). – wendy.krieger Oct 15 '14 at 00:49
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    @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 02:27
  • @robjohn As long as you are prepared to accept 80=10^80, becaue they're both unbounded. If you follow your equations from delta=1 at 1 m, 1 m, at one 1 dm from the x axis, the lne is passing through the same atom, at 5 cm, they're skewering the same nucleids, and at 1 inch, the curve is not that far from y=0 that you would be hard pressed to get a planck unit inbetween them. – wendy.krieger Oct 15 '14 at 03:00
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    @wendy.krieger: this is math, not experimental physics. It doesn't matter how small a positive number is, it is still not $0$. $\left(10^{-200},10^{-2}\right)$ is still in the interior of $[0,1]^2$. There is no experimental error involved. – robjohn Oct 15 '14 at 07:47
  • @robjohn And it still beggars belief. Although it is still in the interior of [0,1], it falls outside the limit of [1/ad,1] for any sized a, which in effect, tells us that you are trying to set x=10^x, and then say because x is unbounded, that x=10^x. – wendy.krieger Oct 15 '14 at 08:27
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    @wendy.krieger: You keep complaining about the way that $(x,y)$ approaches $(0,0)$ as if the definition of convergence cares about that. It does not. For all $0\lt\delta\le\frac12$, $$\left|\left(\delta^{1/\delta},\delta\right)-(0,0)\right|\le2\delta\text{, and yet }\left|\left(\delta^{1/\delta}\right)^\delta-1\,\right|=1-\delta\ge\frac12$$ This alone shows, by definition, that $\displaystyle\lim_{(x,y)\to(0,0)}x^y\ne1$. – robjohn Oct 15 '14 at 13:18
  • @robjohn The great number of paradoxes about infinity, quickly disappear if you properly weight them, This is in part done by supposing an error $\epsilon$ which is of the order of resolution of the process. What happens is that your examples falls well inside $\epsilon, r$, a long time before r=0, so you are in effect, doing 0/0. Even Cantor's theory distinguishes between the two numbers you posit the recriprocals are equal to 0. The thing is that mathematicians suppose there is no fuzz in the equity operators, but it only makes sense if there is fuzz, and your eqn falls under the fuzz. – wendy.krieger Oct 16 '14 at 01:28
  • Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/17914/discussion-between-robjohn-and-wendy-krieger). – robjohn Oct 16 '14 at 04:11

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.


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).

Dan Christensen
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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
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    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
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    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
  • @AsafKaragila As for the cardinality of the empty set, I don't see what that has to do with repeated multiplication on $N$ for which the notion of cardinality simply isn't necessary. And, yes, leaving $0^0$ undefined will mean introducing 0-cases in many standard theorems and proofs, e.g. the Binomial Theorem, but it shouldn't be onerous. – Dan Christensen Nov 20 '13 at 22:12
  • @AsafKaragila Thanks, but defining $x^y=999$ for all $x,y$ would not model the intuition of repeated multiplication. As I show at my blog, there are infinitely many what I call exponent-like functions that differ only in the value assigned to $0^0$, hence the inherent ambiguity. Have a look. – Dan Christensen Nov 20 '13 at 22:19
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    $0^0=999$ would be a contradiction to the power laws, because then $(0^0)^2 = 999^2 \ne 0^{0\cdot2} = 999$. The only two values for $0^0$ consistent with the power laws are $0$ and $1$. – celtschk Jul 11 '15 at 17:03
  • @celtschk The exponentiation function as purely repeated multiplication on N, must satisfy the following requirements: (1) $n^2=n\times n$, and (2) $n^{m+1}=n^m\times n$ . The following function in N would satisfy these requirements: 1. $0^0=999$ (2) $n^0=1$ for $n\ne 0$, (3) $n^{m+1}=n^m\times n$. If you want to add other requirements for exponentiation, you could reduce the number of alternatives for $0^0$ to only $0$ or $1$ as you suggest. Whether there are infinitely many or just two alternatives for $0^0$, however, you still have ambiguity – Dan Christensen Jul 12 '15 at 13:00
  • @Juniorized: If your argument were valid, then so would be: $$0 = 0^1 = 0^{(2-1)} = 0^2/0 = 0/0 = 999$$ You *cannot* choose an arbitrary value for $0/0$. Note that while limits which *would* result in $0/0$ if you (incorrectly in that case) did the limit on numerator and denominator separately may, calculated correctly, give any value, does *not* imply that. Indeed, even if discarding the idea that $x/x=1$ wherever defined (which you did by assigning $0/0=999$), the only consistent result for division by zero is $0$. In which case, you also have to assign $0^0=0^{-1}=0$ to remain consistent. – celtschk May 01 '21 at 09:46
  • @celtschk Your 2nd equality is a misapplication of a rule/theorem that requires a non-zero divisor. – Dan Christensen May 01 '21 at 14:14
  • @DanChristensen: I used the exact same rules that Juniorized used. Anyway, the second equality is fine (it's just rewriting the exponent), it's the third equality (the one that introduces division by zero) that is not allowed. And yes, it's a misapplication; that's entirely my point. The most important rule of all: *Always* consider the context of a comment. – celtschk May 01 '21 at 18:37
  • @Juniorized: “indeterminte” is not a value, therefore it makes absolutely zero sense to define anything as indeterminate. Either something is an indeterminate form, or it isn't. Being an indeterminate form only tells you something about limits that “look like” they would give this result. $0/0$ is not defined to be indeterminate, it is left undefined. This makes sense because there is no useful definition. This is unlike $0^0$ where the definition $0^0=1$ is immensely useful. – celtschk May 01 '21 at 18:44
  • @Juniorized: A number has no solutions, just like a place has no destinations. An equation has solutions (and a flight has destinations), and those solutions are numbers (the destinations are places). Yes, the equation $0x=0$ has infinitely many solutions. But that just means that it is invalid to define anything as “the solution of $0x=0$”. And the term “indeterminate form” (**not** “indeterminate value”!) is **only** defined in the context of limits. – celtschk May 01 '21 at 19:15

Let $$p=\lim\limits_{x\to 0} x^x$$

Then $$ln(p)=\lim\limits_{x\to 0} x\ln(x)$$

$$ln(p)=\lim\limits_{x\to 0} \frac{\ln(x)}{x^{-1}}$$

We can use L'Hopital on RHS as both numerator and denominator is tending to $\infty$

$$ln(p)=\lim\limits_{x\to 0} -\frac{1/x}{x^{-2}}$$

This can be simplified by cancelling some x terms out.

$$ln(p)=\lim\limits_{x\to 0} -x$$

Thereafter the limit can be applied


$$ p = e^0$$

Therefore $$p=1$$

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$$ x^n = \underbrace{x \cdot x \cdot x \cdots x}_{n \text{ terms}} = 1 \cdot \underbrace{x \cdot x \cdot x \cdots x}_{n \text{ terms}} \implies 0^0 = 1 \cdot \underbrace ⬚_{\text{0 terms}} = 1 \\ 0^n = \underbrace{0 \cdot 0 \cdot 0 \cdots 0}_{n \text{ terms}} = k \cdot \underbrace{0 \cdot 0 \cdot 0 \cdots 0}_{n \text{ terms}} \implies 0^0 = k \cdot \underbrace ⬚_{\text{0 terms}} = k $$


If it has a definite value, then it must be $1$ since the right hand side limit $x^x \rightarrow 1$ as $x \rightarrow 0^+$, exists. The left hand side limit oscillates wildly.

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