I’m trying to figure out why raising zero to the zeroth power equals one. What kind of a scenario would occur in a laboratory experiment where something with a quantity of zero would be raised to the power of zero and you end up with one? How do I explain how something is created out of nothing? What is happening?

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    $\begingroup$ perhaps better asked on Mathematics? I don't think you can relate the 0^0=1 issue to a lab experiment, it is a feature of the math. $\endgroup$ Jun 3, 2020 at 1:02
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    $\begingroup$ Maybe you shouldn't explain that zero to the power of zero is one because it isn't necessary: en.wikipedia.org/wiki/Zero_to_the_power_of_zero $\endgroup$
    – user87745
    Jun 3, 2020 at 1:08
  • $\begingroup$ @DvijD.C. Thank you. I didn’t realize it was undefined. I appreciate the help. $\endgroup$ Jun 3, 2020 at 2:40

4 Answers 4


The thing is, zero to the zeroth power is not always one. It is technically undefined. We can try to get around that by asking the question; What happens if we take the limit of a function that approaches zero to the zeroth power? It should be easy to find two functions that both approach zero at some point (lets call them $f$ and $g$). Thus, $f$ to the power of $g$ approaches zero to the zero. However, using L'hospital's rule, we can see that it is possible to find $f$ and $g$ so that $f^g$ approaches a couple of different numbers (all real numbers, if you include the complex numbers)

That will probably be too technical for your child, so to explain it to them, I would recommend asking them what zero to the power of $<$ any number besides zero $>$ is, and showing them that it is always zero. Then, ask them what $<$ any number besides zero $>$ to the zeroth power is, and show them that it is always one. Then explain that because these two properties conflict, there is no single right answer. To make it easier to understand, it is standard practice to say that it equals one, even though it is technically incorrect.

Edit: I looked through my explanation of the math behind the weirdness, and found a number of errors. My math was wrong, and the example I gave didn't actually fit the requirements. I removed those parts of my answer. I believe that the simplified explanation for a child is still accurate, though.

  • $\begingroup$ Yes, the indeterminate form cannot be explained to a small child. Perhaps, it should be hinted at. $\endgroup$ Jun 3, 2020 at 1:54
  • $\begingroup$ Thank you to whoever suggested the numberphile video; it was very helpful, but the comment was removed, and I don't know who posted it. Here is the link for anyone else who wants to see it: youtube.com/watch?v=BRRolKTlF6Q $\endgroup$ Jun 3, 2020 at 2:41
  • $\begingroup$ $x^0 \rightarrow 1$ when $x \rightarrow 0$. This is perfectly well defined. $\endgroup$ Jun 3, 2020 at 8:30
  • $\begingroup$ $0^y$ is not defined at $y \le 0$. This has no bearing. $\endgroup$ Jun 3, 2020 at 8:42

If we consider this operation ($a \neq 0$)


and taking a special case for illustration with $a= \textit{e}$ i.e. the exponential, we could write


and then taking natural log on both sides

$$0 \ln e = \ln 1= 0$$

If we now replace a by $0$, then


It seems to suggest that zero is quite fundamental; as it gives something out of nothing when not multiplied with itself!

However, the $\frac{0}{0}$ is indeterminate in calculus.

  • $\begingroup$ you cannot divide by zero! $\endgroup$
    – user65081
    Jun 3, 2020 at 1:58
  • $\begingroup$ Yes, that is why in calculus the form $\frac{0}{0}$ is indeterminate. If one cannot divide zero by zero then $0^0$ cannot happen. $\endgroup$ Jun 3, 2020 at 1:59
  • $\begingroup$ so your last equation is undefined, not equal to one $\endgroup$
    – user65081
    Jun 3, 2020 at 2:00
  • $\begingroup$ Yes, that is why i wrote -it seems to suggest. $\endgroup$ Jun 3, 2020 at 2:02

In order to explain it, you need to have them to have some clear idea of what "raising to a power" means, because "explanation" itself fundamentally means you are trying to describe something as the result of some more basic principle.

Hence, what we're really talking about is how to best define exponentiation. And in that regard, we should set it up by saying that there are at least two possible avenues to defining exponentiation at this level. Most likely, your kid will think of exponentiation as being something like this:

$$a^n := \underbrace{a \cdot a \cdot \cdots \cdot a}_\text{$n$ appearances of "$a$"}$$

and if you take this literally and try to put $n = 0$, you get

$$a^0 = \underbrace{}_\text{see, there aren't any copies of "$a$" here"!}$$

which is nonsense, of course. Hence this definition fails to define a value for $a^0$ - whether $a$ is zero or not.

The problem is, we are defining exponentiation by the idea that it simply represents the "number of 'copies'" of $a$ that appear in some thing we write down on paper that in just some instances happens to correspond to a valid mathematical expression. But clearly, that's problematic, and your kid (and you and perhaps many others) have figured that out. So, we need to do something else, which motivates us to look out for a different approach.

And that's this. Instead of thinking of $a^n$ is being defined by putting "copies" or "appearances" of $a$ into some expression, think of $n$ as saying "how many times do we multiply some fundamental base value by $a$"? That is,

$$a^n := K\ \underbrace{\cdot\ a \cdot a \cdot a \cdots \cdot a}_\text{$n$ performances of the operation "$\cdot\ a$"}$$

(note the text! We're talking about performing an operation as what is counted.)

If $n = 0$, then this should just give back the fundamental base value, $K$, instead of giving us a nonsensical expression, because we will have not performed the multiplication operation at all. That is, $a^0 = K$. Hence the value of $a^0$ depends wholly on the choice of initial constant $K$, and not the structure of the operation.

But of course, we still need to make such a choice, as the definition is thus incomplete! Hence, what value we should set for $K$? And that comes to this. If they are into the idea of multiplication as being interpretable as a scaling or "zooming" process, then you can say that the motivation for the operation $a^n$ is to tell us what "zoom" factor performs $n$ successive "zooms" of magnification $a$. For $n = 0$, it should mean we do no zooms, which is a zoom factor of $1$. Hence we should define $K := 1$, and the fully available definition for exponentiation looks like:

$$a^n := 1\ \underbrace{\cdot\ a \cdot a \cdot a \cdots \cdot a}_\text{$n$ performances of the operation "$\cdot\ a$"}$$

where we purposefully do not just elide the "$1$" - something too many miss perhaps in part thanks to being taught that so-called "simplifying" is a "virtue". (If anything, I say to ditch "simplifying" in favor of clarifying.)

And it then follows from this that

$$0^0 = 1$$

and $0$ is not more special than any other number that can go into the exponentiation. If you want to think of it as "zooming", a zoom factor of $0$ "crushes everything into a singularity" (as zero annihilates everything it's multiplied with). Yet, no matter how powerful or all-consuming something is, if we never do it in the first place, it doesn't matter. Hence, so long as we do $n = 0$ of those all-crushing zooms, things should still stay the way they are, i.e. zoom factor $1$.


If you multiply a number, $N$, by zero no times, then you do nothing to the number, which is the same as multiplying by $1$.

$$ 0^0 N = N = 1 N $$

Hence $0^0 =1$ (contrary to other answers, this is well defined).

You may also consider that $x^0 \rightarrow 1$ as $x \rightarrow 0$, but it is simpler for a child to recognise that power of $0$ means do no multiplications.

$0^0$ may not occur in the lab, but it will occur in the formulae used to describe physics. It is still defined and essentially refers to the case where nothing is done.

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    $\begingroup$ I don't get how this works: if you do nothing to a number, then how does that imply 1? It's just interlinking different facts of mathematics without context. $\endgroup$
    – PNS
    Jul 17, 2020 at 14:28
  • $\begingroup$ @PNS, What is there to get? Multiplying by 1 does nothing to a number. This is a central idea in mathematics and is what makes the number 1 particularly important. 1 is known as the identity under multiplication. $\endgroup$ Jul 17, 2020 at 16:31
  • $\begingroup$ Yes, but if you take a number and do nothing, you are multiplying by 1, so from what point does that mean that mean you get only 1 and not the number itself? (From your argument in the answer, you should get 0, as 1 multiplied by 0, is just 0.) $\endgroup$
    – PNS
    Jul 18, 2020 at 3:20
  • $\begingroup$ @PNS, the argument already says you get the number itself. It does not say you get 1, and it does not multiply 1 by 0. It was already clear, but I have written the multiplication explicitly. I hope that helps. $\endgroup$ Jul 18, 2020 at 7:17
  • $\begingroup$ Yes, but now you assume that $0^0$ is working on something $N$ and you get $1N$. What if you remove $N$ now? You don't get $1$. The $0^0$ is not acting as an 'operation' now, so you don't even get $1$ anymore. $\endgroup$
    – PNS
    Jul 18, 2020 at 10:41

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