Math people:

The title is the question: Is there any good reason not to define $0^0=1$ , such as contradictions in algebra or arithmetic?

I searched for similar questions before I posted this question, and couldn't find any. After I posted it, I got some comments citing similar questions. There is a similar question at What values of $0^0$ would be consistent with the Laws of Exponents? . I checked some of the other questions in the links in the comments and the links posted with those links. There is a closer match at How to define the $0^0$? That question was closed as a duplicate, but the older, duplicated question was not identified. I could not find a convincing answer anywhere to my essential question: "does defining $0^0=1$ lead to contradictions in algebra and arithmetic?" I'll leave it up to others to decide whether my question is a duplicate. If it is, maybe you can close this question and give a better (in my opinion) answer to one of the older questions.

Let me get one thing out of the way up front: yes, I know "$0^0$" is an indeterminate form. That is, if $f$ and $g$ are real-valued functions with $f(t) \to 0^+$ as $t \to 0$ and $g(t)\to 0$ as $t \to 0$, then you don't know what $\lim_{t\to 0}f(t)^{g(t)}$ is, or even whether exists, without more information. I don't consider this a good reason to declare that $0^0$ itself must be considered undefined. I know many people will disagree with me here. I expect at least one answer and some comments arguing why this is a good reason for $0^0$ to be considered undefined. Everyone is entitled to their opinion, and you are free to leave such an answer. I will not attempt to change your mind, beyond what I write in this question.

If you define $f(x,y) = x^y$, then $f$ cannot be continuous on $[0,\infty) \times \mathbb{R}$ no matter what value, including $1$, you assign to $f(0,0)$. But why should every function have to be continuous?

If the mathematical community ever does come to the consensus that $0^0=1$, and I were teaching calculus students about limits involving indeterminate forms, I probably would not even mention the question of whether $0^0$ itself had a value, because it probably just confuse the students. They probably wouldn't even notice the omission.

To me, a "good reason" not to define $0^0=1$ would be if this definition resulted in a contradiction, when used in expressions involving multiplication and exponentiation of real numbers and the rules used to simplify such expressions. Here is an attempt to produce such a contradiction: assuming $0^0=1$, $(0^0)^2=1^2=1$, and $(0^0)^2=0^{(0*2)}=0^0=1$. No contradiction. In constrast, if you define $0/0 = 1$ and you want the associative property to hold (a reasonable expectation), then you can derive the contradiction $1=0/0=(2*0)/0=2*(0/0)=2*1=2$.

It just occurred to me that there is another good reason for not declaring officially that $0^0$ must always equal $1$: if defining $0^0=0$ does not lead to contradictions in algebra or arithmetic, either.

I am not claiming $0^0 = 0/0$. Of course you can never divide by zero, or raise zero to a negative power.

Of course, when people use power series, they use $0^0=1$ all the time, and no one complains. I have read that "$0^0=1$" is used often in combinatorics, but I don't know much about combinatorics.

Based on what I have seen in the older questions, their answers, and the answers and comments to this question, it seems that no one has discovered any way in which defining $0^0$ to be $1$ leads to contradictions when using the usual rules of multiplication and exponentiation. It also seems that defining $0^0$ to be $0$ does not lead to such a contradiction. So I'm guessing it is impossible to produce such a contradiction. But I have never heard of anyone wanting to define $0^0$ as $0$.

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    $\begingroup$ There has been more than several questions on the topic recently. $\endgroup$
    – Asaf Karagila
    Dec 14, 2013 at 17:03
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    $\begingroup$ math.stackexchange.com/questions/495121/… math.stackexchange.com/questions/485102/… and you can probably follow the related and linked threads from the right hand side of the screen as well. $\endgroup$
    – Asaf Karagila
    Dec 14, 2013 at 17:04
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    $\begingroup$ A sensible convention: $0^0=1$ when the exponent is the integer zero, but $0^0$ is undefined when the exponent is the real number (or complex number) zero. Then you can work on conventions for $p$-adic numbers and finite fields, too... $\endgroup$
    – GEdgar
    Dec 14, 2013 at 17:57
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    $\begingroup$ Whether you like it or not, being discontinuous is a good reason to leave it undefined at $0^0$. In settings where continuity is important and discontinuous functions need special care, using the discontinuous version of exponentiation rather than a continuous version does nothing but get in the way and make things more complicated for no reason. Save your urges to define $0^0$ for those settings where it's actually appropriate, such as when expressing repeated products, or doing polynomial arithmetic. $\endgroup$
    – user14972
    Dec 14, 2013 at 19:33
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    $\begingroup$ @Stefan: Ah, my mistake: it's $(ab)/c = a(b/c)$ I should have asked you to be careful to avoid! I don't take it personally: I really do understand why people would want to insist on $0^0=1$ being defined: it is very useful in many settings, and the really strong reasons to avoid it (in the settings where you should do so) are rather subtle, and depend on things that aren't usually made clear. It's really easy to just say "Oh, just be careful to avoid the times where this causes problems" without realizing the monkey wrench that puts into your thinking in certain settings. $\endgroup$
    – user14972
    Dec 14, 2013 at 20:36

3 Answers 3


At the very least, the answer $0^0 = 1$ is consistent with cardinal arithmetic on the set $\{0, 1, 2, \ldots\}$. Under this interpretation, the number $m^n$ is defined to be the number of functions from an $n$-element set to an $m$-element set. There is exactly one function from a $0$-element set to a $0$-element set, so in this interpretation, $0^0 = 1$. The laws $a^m a^n = a^{m+n}$ and $(a^m)^n = a^{mn}$ can be proven with that definition. Thus, exponentiation as repeated multiplication can be recovered. (In one of your links, Matt N. gives this same idea as an answer.)

The first criticism I heard involved the subtraction law $\frac{a^x}{a^y} = a^{x-y}$. The idea is that $\frac{0^x}{0^x} = 0^0$, yet $\frac{0^x}{0^x} = \frac{0}{0}$, so $0^0$ is undefined. The problem here isn't with $0^0$, it's with saying $\frac{0^x}{0^x}$ is equal to anything, when it is undefined. We could use the same reasoning to say that $0^2$ is undefined, since $0^2 = \frac{0^7}{0^5} = \frac{0}{0}$. At worst, the subtraction law has to be modified to say that it only applies when $a^y \neq 0$. We make such modifications for all of our other laws involving division, so why should this be any different?

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    $\begingroup$ Using $0^2/0^2$ do disallow $0^0$ is foolish: one can't divide by $0$ in any case, so what would this mean? By the same reason also $1/1$ would be disallowed, because $\frac{1\cdot0}{1\cdot0}$ has no meaning. $\endgroup$
    – egreg
    Dec 14, 2013 at 20:30
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    $\begingroup$ @StefanSmith I know and I agree with Hugh; I just wanted to add a different point of view on the same argument: division by $0$ is not allowed, so $0^0=0^2/0^2$ is not a valid argument. $\endgroup$
    – egreg
    Dec 14, 2013 at 20:44
  • $\begingroup$ @egreg : It occurred to me after I left my comment that maybe you were actually agreeing with Hugh. Of course, I wrote in my question that I was not claiming $0^0=0/0$, to prevent anyone from making this argument. Sorry for the misunderstanding. $\endgroup$ Dec 14, 2013 at 21:32
  • $\begingroup$ @Hugh : I have a feeling my question will be closed soon, so I went ahead and accepted your answer, which is good. I will add a little something to the end of my question, if you have a moment to look at it. $\endgroup$ Dec 14, 2013 at 21:34
  • $\begingroup$ Thanks - I'd actually be surprised if even in this deeper "where does $0^0 = 1$ break down?" type question if it wasn't some sort of duplicate. $\endgroup$ Dec 14, 2013 at 21:44

In many settings, the point-set version of the notion of function is a useful fiction, but is ultimately wrong.

For example, in real analysis, the value of a function at a point is pretty much wholly irrelevant -- it is only the "bulk" behavior of a function in a regino that matters. The value of an integrand at individual points already has no bearing whatsoever to a Riemann integral. Most operations are implicitly modified to append "... and then take the continuous extension of the result" to their usual meaning. The definition of "derivative" eventually gets refined to the point where discontinuities get washed out. One even works with objects like the dirac delta function that can't even be fully described by specifying its 'values' at all points.

Definition: The cvalue of a function $f$ at the point $a$ is defined to be, if the it exists and is unique, the value $g(a)$ where $g$ is a function continuous at $a$ and differs from $f$ at only finitely many points. This value will be notated $f[a]$.

(replace "finitely many points" with "a set of measure zero" to get a better definition, if you know what that means)

The name and notation is made up for this post -- I think analysis would simply call this the "value" of $f$ at $a$ and assume that their colleagues know what they mean.

When limits exist, we can compute them simply by plugging in cvalues:

$$ \lim_{x \to a} f(x) = f[a] $$

It's also not too hard to see, for example, that

$$ \int_a^b f(x) \, dx = \int_a^b f[x] \, dx $$

so we can already see that cvalues are already good enough for computing integrals.

Also, recall that the major property of the dirac delta function is that if $f$ is continuous at zero, we have

$$ \int_{-\infty}^{+\infty} f(x) \delta(x) \, dx = f(0) $$

This generalizes to

$$ \int_{-\infty}^{+\infty} f(x) \delta(x) \, dx = f[0]$$

when $f$ is discontinuous at zero.

Another nice example is that the partial function $f(x) = \frac{x}{x}$ and the function $g(x) = 1$ satisfy $f[x] = g[x]$ everywhere.

I hope these give some of the flavor of why cvalues have importance.

The point of bringing all of this up is that in the continuous context, putting emphasis on values at individual points is the wrong way to think about things. Setting $0^0 = 1$ washes out, because the cvalue of $x^y$ doesn't exist at $0^0$. However, continuous functions are very special because they are much easier to manipulate and work with and have loads of nice properties. Modifying exponentiation to be discontinuous makes it awkward to take advantage of the nice features that exponentiation would otherwise have from its status as being continuous on the region defined by $x > 0$ and on $x = 0 \wedge y > 0$.

  • $\begingroup$ I upvoted your answer because of the time and thought you evidently put into it, but as the question says, I am mainly interested in whether defining $0^0$ to be $1$ leads to contradications in algebra and arithmetic, when you use basic laws of multiplication and exponentiation, not in questions of analysis, derivatives, continuity, limits, integrals, etc. $\endgroup$ Dec 14, 2013 at 20:34
  • $\begingroup$ @Stefan: Ah, I misunderstood: I thought you were looking for a good explanation of why $0^0 = 1$ should be avoided (when appropriate). It really is fine in algebraic contexts, as you've observed, except possibly if you start doing some esoteric things (none of which spring to mind at the moment) $\endgroup$
    – user14972
    Dec 14, 2013 at 20:39
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    $\begingroup$ If nothing else, it was a worthwhile exercise for me, since I'm better prepared for the next time I talking about just why it matters in the continuous version. Maybe I'll find one of the older $0^0$ posts and copy it there. $\endgroup$
    – user14972
    Dec 14, 2013 at 20:41
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    $\begingroup$ I read this answer, and come to the opposite conclusion. As you say, the value of $0^0$ has no relevance in analysis: functions like $0^x$ (or functions where you care about continuity in the exponent) are already discontinuous at $0$, and this won't change whether or not you accept that $0^0=1$. So there seems no reason not to. (On the other hand, $x^0$, which can be given a meaning even for negative $x$, is continuous with $0^0=1$.) So the only reason not to define $0^0$ is vague fears like "possibly if you start doing some esoteric things (none of which spring to mind at the moment)". $\endgroup$ Dec 15, 2013 at 1:43
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    $\begingroup$ @StefanSmith: Well, defining it to be $0$ breaks a lot of things (such as the combinatorial interpretation), while defining it to be $1$ is well-motivated and breaks essentially nothing (e.g. the lack of satisfying answers to this question ☺). (Another point: there is no way of extending the function $0^n$ to $n < 0$, so it's a function that already has a "break". But there is a way of defining $n^0$ for all $n$, which is broken only if you declare that $0^0$ is not $1$.) $\endgroup$ Feb 7, 2014 at 10:13

The fact that 0^0 = 1 follows directly from a widely accepted theorem (the binomial theorem). Thus, the claim that 0^0 is undefined is equivalent to the claim that widely accepted mathematical results are mutually inconsistent.

This is a huge claim considering the fact that nobody has found any inconsistencies in over a hundred years!

The reality is that (a) 0^0 is 1 in every formula that contains x^n, (examples: binomial theorem, Taylor series, polynomials, Vandermonde matrix, power rule, etc.), and (b) nobody has ever found any contradiction arising from this.

In all other parts of mathematics if A ==> B, and A is a widely accepted theorem, then we must accept B as well. Nevertheless, some people reject this when B is a statement that conflicts with their gut intuition such as 0^0 = 1. They would rather rely on ad-hoc arguments like limits instead of accepting corollaries of well established theorems. The only explanation I can come up with for this phenonemon is that even in the modern age, people are still uncomfortable with accepting 0 as a number, or {} as a set.

  • $\begingroup$ Um. How about the inconsistency mentioned in the first answer to this question: math.stackexchange.com/questions/259514/how-to-define-the-00 $\endgroup$
    – apnorton
    Apr 24, 2014 at 13:11
  • $\begingroup$ Before I answer that, let me make sure: Are you claiming that widely accepted math is mutually inconsistent? My claim is that it is not. $\endgroup$
    – Mark
    Apr 24, 2014 at 13:15
  • $\begingroup$ @anorton: That is not an inconsistency: the fact that $0^x = 0$ for all positive $x$ (only) does not force anything about the value $x=0$, especially considering the fact that $0^x$ cannot be defined for any $x<0$. $\endgroup$ Apr 24, 2014 at 13:16
  • $\begingroup$ The 0^x argument does not support 0 as a value because the left limit is infinity or undefined, depending on the way you want to treat it, while the right-limit is 0, but by a symmetry argument you can see that both are bad choices. But the bigger question is: do you think that generally accepted math is mutually contradictory? $\endgroup$
    – Mark
    Apr 24, 2014 at 13:20
  • $\begingroup$ Ehh... ok. I think I agree with your statement of "it is consistent," but I disagree that the binomial theorem implies this to be the case. (In most handling of series/summation that I have seen, $0^0$ is defined to be $1$ for notational convenience.) Whatever--I'll have to think more about this issue... :) $\endgroup$
    – apnorton
    Apr 24, 2014 at 13:24

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