Why is any number (other than zero) to the power of zero equal to one? Please include in your answer an explanation of why $0^0$ should be undefined.

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    $\begingroup$ 0^0 shouldn't be undefined. It should be 1. $\endgroup$
    – Moor Xu
    Commented Jul 20, 2010 at 21:38
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    $\begingroup$ @Coltin: Wolframalpha doesn't say that 0^0 is 1. If you enter 0^0 it says "indeterminate". It only says that lim x^x as x->0 is 1, which is perfectly true, but entirely besides the point. $\endgroup$
    – sepp2k
    Commented Jul 20, 2010 at 21:48
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    $\begingroup$ If you take the more general case of lim x^y as x,y -> 0 then the result depends on exactly how x and y both -> 0. Defining 0^0 as lim x^x is an arbitrary choice. There are unavoidable discontinuities in f(x,y) = x^y around (0,0). $\endgroup$ Commented Jul 21, 2010 at 14:14
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    $\begingroup$ 0⁰ = 1 if you want to write polynomials like 2x + 3 as 2x¹ + 3x⁰ (when x = 0...). There are probably other applications where having 0⁰ = 0 or having 0⁰ undefined is as useful, but I'm not aware of one. $\endgroup$
    – badp
    Commented Jul 23, 2010 at 11:44
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    $\begingroup$ By definition, $0^0 = |\operatorname{Hom}_{\mathbf{Sets}}(\emptyset,\emptyset)|=1$. It is also true that the function $f(x,y) = x^y$ is discontinuous at $(0,0)$, but that is immaterial to the fact that $0^0=1$. $\endgroup$ Commented Mar 9, 2013 at 8:17

10 Answers 10


For non-zero bases and exponents, the relation $ x^a x^b = x^{a+b} $ holds. For this to make sense with an exponent of $ 0 $, $ x^0 $ needs to equal one. This gives you:

$\displaystyle x^a \cdot 1 = x^a\cdot x^0 = x^{a+0} = x^a $

When the base is also zero, it's not possible to define a value for $0^0$ because there is no value that is consistent with all the necessary constraints. For example, $0^x = 0$ and $x^0 = 1$ for all positive $x$, and $0^0$ can't be consistent with both of these.

Another way to see that $0^0$ can't have a reasonable definition is to look at the graph of $f(x,y) = x^y$ which is discontinuous around $(0,0)$. No chosen value for $0^0$ will avoid this discontinuity.

  • $\begingroup$ I'd be most inclined to accept this answer, since it's brief, but the relation you mention also holds when x = 0. Could you edit it to address the 0^0 case somehow? $\endgroup$
    – bryn
    Commented Jul 21, 2010 at 4:54
  • $\begingroup$ @bryn, I added the second paragraph. $\endgroup$ Commented Jul 21, 2010 at 14:10
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    $\begingroup$ @Neil, except that 0^x = 0 is not true for all non-zero x (e.g. x = -1). $\endgroup$
    – bryn
    Commented Jul 22, 2010 at 10:43
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    $\begingroup$ #Neil: Sorry, I have not made myself clear. You say 0^0 can't have a reasonable definition because of a continuity argument, what I call the analytical reason. But it is quite possible to have a value here with the function the same here: it's just as discontinuous. You've talked about necessary constraints on x^y for +ve x,y, but there is a reason to extend x^0 to all x, but no reason to extend 0^y to cover all y>=0. So these constraints don't look equally compelling. $\endgroup$ Commented Jul 23, 2010 at 20:34
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    $\begingroup$ Even after correcting one error, there are still two errors in this answer. The second argument assumes the "continuity rule", which says that if $f$ is not continuous at a point $p$, then $f$ can not be defined at $p$. The first argument is even worse. It assumes that if two statements contradict each other, then they are both wrong! How does this "not ($P$ and $Q$) $\Longrightarrow$ not $P$ and not $Q$" get upvoted? $\endgroup$
    – Mark
    Commented Feb 14, 2017 at 16:58

This is a question of definition, the question is "why does it make sense to define $x^0=1$ except when $x=0$?" or "How is this definition better than other definitions?"

The answer is that $x^a \cdot x^b = x^{a+b}$ is an excellent formula that makes a lot of sense (multiplying $a$ times and then multiplying $b$ times is the same as multiplying $a+b$ times) and which you can prove for $a$ and $b$ positive integers. So any sensible definition of $x^a$ for numbers $a$ which aren't positive integers should still satisfy this identity. In particular, $x^0 \cdot x^b = x^{0+b} = x^b$; now if $x$ is not zero then you can cancel $x^b$ from both sides and get that $x^0 = 1$. But if $x=0$ then $x^b$ is zero and so this argument doesn't tell you anything about what you should define $x^0$ to be.

A similar argument should convince you that when $x$ is not zero then $x^{-a}$ should be defined as $1/x^a$.

An argument using the related identity $(x^a)^b = x^{ab}$ should convince you that $x^{1/n}$ is taking the $n$th root.

  • $\begingroup$ And now you just need to expand your explanation to include x^i and then explain why e^(i*pi) + 1 = 0 :) $\endgroup$
    – workmad3
    Commented Jul 20, 2010 at 22:12
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    $\begingroup$ But that's just the definition of pi of course... $\endgroup$ Commented Jul 20, 2010 at 22:14
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    $\begingroup$ I like the argument that 0^0=1 because it's the number of functions from the empty set into the empty set. $\endgroup$
    – user126
    Commented Jul 21, 2010 at 22:14
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    $\begingroup$ I must be misreading this. 1) x^a × x^b = x^a+b; for x = 0 and a = 0, you would get 0^0 × 0^b = 0^b = 0, so we can't tell anything -- except confirm that 0^0 = 1 still works here! 2) x^{-a}=1/{x^a} -- so when a = 0, x^{-0} = 1/x^0 = x^0, which again does work for 0^0 = 1; 3) {x^a}^b = x^{a×b}, thus x^(1/n) is the n-th root -- and 1/n = 0 for no value of n, so that's completely irrelevant. $\endgroup$
    – badp
    Commented Jul 24, 2010 at 14:23

If $a$ and $b$ are natural numbers, then $a^b$ is the number of ways you can make a sequence of length $b$ where each element in the sequence is chosen from a set of size $a$. You're allowed replacements. For example $2^3$ is the number of 3 digit sequences where each digit is zero or $1$: $000, 001, 010, \ldots, 111.$

There is precisely one way to make a zero length sequence: the empty sequence. So you'd expect $0^0=1$.

  • $\begingroup$ Unfortunately there's also a debate about whether 0 is a natural number (see math.stackexchange.com/questions/283/is-0-a-natural-number). But I'll take your usage to mean 'nonnegative integer', and that you're voting for $0^0 = 1$. $\endgroup$
    – bryn
    Commented Jul 22, 2010 at 11:31
  • $\begingroup$ If you're in the combinatorics business, zero is natural. If your business involves primes, zero is not. I'm not sure what people do when they have a foot in both camps. $\endgroup$
    – Dan Piponi
    Commented Apr 27, 2015 at 14:05

$$0^x = 0, \quad x^0=1$$

both are true when $x>0$.

What happens when $x=0$? It is undefined because there is no way to chose one definition over the other.

Some people define $0^0 = 1$ in their books, like Knuth, because $0^x$ is less 'useful' than $x^0$.

  • 2
    $\begingroup$ The first line here seems false, since if x = -1, we get 0^(-1) which is also undefined. So at best we have 0^x = 0 for x>0. $\endgroup$
    – bryn
    Commented Jul 21, 2010 at 22:07
  • $\begingroup$ @bryn, right, I just fixed it, thx. $\endgroup$
    – Chao Xu
    Commented Jul 21, 2010 at 22:11
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    $\begingroup$ "undefined, because there is no way to chose one definition over the other" --- except that many people do define it, so it is not a matter of simply being 'undefined'; and it seems to me that the majority of those who define it, define it to be equal to 1 for the reasons outlined by Noah Snyder above. Definitions are chosen for purposes of utility. $\endgroup$ Commented Sep 15, 2010 at 11:26
  • 1
    $\begingroup$ I think there is a way to choose one definition over the other. $0^x=1$ for $x>0$ but is undefined for $x<0$. However, $x^0 = 1$ for $x>0$ AND for $x<0$, so it seems more natural that $0^0 = 1$. $\endgroup$
    – Joe
    Commented Oct 23, 2021 at 0:32

If we use the idea of set exponentiation to define exponentiation of cardinals, we have the following natural idea:

$$A^B:=\{f:B\to A\}$$

We define the exponential of cardinals as follows: $|A|^{|B|}:=|A^B|$. It's easy to check that this agrees with our intuition for exponentiation of natural numbers when $B$ is nonempty.

There is only one set representing the cardinal $0$, namely the empty set. Then we may look at $0^0$ as follows:

$$0^0=|\emptyset|^{|\emptyset|}=|\emptyset^\emptyset| =|\{f:\emptyset\to\emptyset\}|=1$$

  • 1
    $\begingroup$ It seems to be the number of b-char strings, where every char may is drawn from a-letter alphabet. Given a and b you may form $a^b$ strings. The empty alphabet, a=0, cannot produce any strings other than the empty string, b=0. So, $0^b = 0$ everywhere except at $b=0$ where you have $0^0=1$. $\endgroup$
    – Val
    Commented Sep 11, 2013 at 14:20

Exponents are only "basically" defined under the natural numbers above zero. By this I mean, defined as "iterated multiplication" the same way multiplication is defined as iterated addition.

The property $a^0 = 1$ only arises when we look at generalizing multiplication to the integers. We do this by: \begin{align} a^4 / a^3 &= (a\cdot a\cdot a\cdot a)/(a\cdot a\cdot a) = a^1\\ a^4 / a^3 &= a^{4-3} = a^1 \end{align}

And using this, we can say:

$$a^2 / a^3 = a^{-1} = 1/a$$

and also: \begin{align} a^2 / a^2 &= 1\\ a^2 / a^2 &= a^{2-2} = a^0 = 1 \end{align}

So we say $a^0 = 1$.

However, notice that these proofs don't have any meaning when $a=0$, because the whole concept/idea involves fractions, and you cannot have zero be in the denominator.

When we say $2^0 = 1$, we really mean:

$$ 2^{1-1} = 2^1 / 2^1 = 2/2 = 1$$

But we cannot say the same for $0^0$: $$0^{1-1} = 0^1/0^1=0/0=\text{UNDEFINED}$$

  • $\begingroup$ I found this answer more helpful. $\endgroup$ Commented Aug 6, 2013 at 9:59
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    $\begingroup$ This is nonsense: dividing by zero is not allowed, so you can't justify something is undefined by using a wrong argument. $\endgroup$
    – egreg
    Commented Feb 23, 2015 at 10:39

The key point is to understand the meaning of multiplication. The $a \times n$ means that you add n a-items to zero rather than "together" because "together" leaves you with uncertainty when no items are taken. If we understand that 0 must be the answer when n=0, we understand what counting means: 0 is the "constant of integration" that lives in any empty set that we add your items into in order to count them. You start counting with 0. If you embed that into definition, a lot of confusion is resolved right away. So, instead of defining $$a \times n = \underbrace{a + a + \cdots + a}_n, $$ you define $$a \times n = \mathbf{0} \underbrace{+ a + a + \cdots + a}_n.$$ You note that 0 is also the additive identity. So, counting starts with additive identity.

Similarly, when you multiply n identical a-items together, what you actually do is you take the product of those items with 1, $a ^ n = \mathbf{1} \underbrace{\times a \times a \times \cdots \times a}_n.$

To repeat, the syntax that corresponds to the meaning of basic math operations must be: $$a \times n = \mathbf{0} \underbrace{+ a + a + \cdots + a}_n \;\;\;\text{and}\;\; a ^ n = \mathbf{1} \underbrace{\times a \times a \times \cdots \times a}_n$$ instead of $$a \times n = \underbrace{a + a + \cdots + a}_n\;\;\;\text{and}\;\;a ^ n = \underbrace{a \times a \times \cdots \times a}_n.$$

Now, if n = 0, everything but identities vanishes. You will have $$a\times n = 0 \;\;\;\text{and}\;\; a^0 = 1.$$ Note that "$\times nothing$" in "$1 (\times nothing)$" does not mean scale 1 with 0. It means that you take 1 alone and do not scale it with anything. When n=0, I do not "add" anything to my "constants of integration". Even if the items that I ignore to "count" are of size a=0 or whatever, I still have $$a\times n = 0 \;\;\;\text{and}\;\; a^0 = 1.$$ Making base zero does not change anything, if you know what is behind $a^n$ rather than blindly drill the rule $0^{anything} = 0.$ This is what my programmer's mindset tells me. Might be I am too bold by viewing the product as the computation

function product(arguments)
  result = 1
  foreach arg in arguments
    result = result * arg

Yet, it was found intuitive. This is exactly what product (exponentiation) means. That is why I say that injecting the identity in front of the sums/products is a natural (i.e. true) thing to do and this must be reflected in the definition. I am happy that it meshes well with the combinatorial argument.

If you can explain how two definitions underpin each other or if you have any objection, please comment. I believe that all problems that arise in continuous analysis are because you do not have pure zeroes there. That is, formally, you may consider $\quad \lim_{t \to 0^+} \left(e^{-\frac{1}{t^2}}\right)^t$ as $ 0^0$. But, I believe that actually you have infinitesimals there, which are not zeroes exactly. Consider the total contribution as a product of a items, whereby every contributes 1/a. If you start reducing the item contribution, $1/a \to 0$, total contribution is unchanged. However, if you really manage to ignore the contributions completely then you will get $0 \times \infty = 0$. You just do not care about the sizes of contributions that you do not count.

Similarly is with $1 \times a \times a \cdots$. If you cut all contributions, you are left with 1, regardless of magnitudes of a. If you receive result that is different from 1 then you have failed to zero all the contributions.


One of the definitions of the power $a^b$ is $e^{b \log a}$, where $$e^x=1+x+\frac{x^2}{2}+\dots $$

  • If $a$ is nonzero, and $b=0$, then $a^b=e^0=1$.
  • If $a=b=0$, the expression $b \log a=0 \log 0$ is an indeterminate form, thus $a^b$ is undefined. Nonetheless, in many applications we assign the indeterminate form $0 \log 0$ the limiting value $\lim_{x \to 0}x \log x=0$, so that $0^0$ is defined to be $1$ as well.

$x^0\cdot x^a=x^a$. If $x$ is different from $0$, then you can divide by $x^a$, so $x^0=0$. If $x=0$, you cannot do that


Because when you say 'to the power of zero' it just means that you are dividing that number by itself. Therefore zero to the power of zero is zero divided by zero which is undefined.


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