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I have done a lot of math so far, but I never stumbled on something this simple and yet mind boggling. Can someone tell me why $0^0$ equals $1$? I always knew that everything raised to a power of $0$ is $1$, but I never thought of the zero case itself.

I guess it's not the biggest problem in math, but I would like to have a good argument about why a $0$ can be turned into a $1$?

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marked as duplicate by Nate Eldredge, Simon Markett, martini, Sami Ben Romdhane, amWhy Jun 8 '13 at 20:18

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It's usually by definition. It prevents a lot you from making an exception or separate remark for $0^0$ when stating proofs. There is a good set-theoretic way to interpret $0^0$ which is a good indicator for why $0^0 = 1$. We can think of $a^b$ as the number of functions from a set with $a$ number of elements to a set with $b$ number of elements. The set with no elements is the empty set and there is only one function from the empty set to the empty set: the empty function. –  Cameron Williams Jun 8 '13 at 20:15
I've been told it's related to the empty product. –  Julien Jun 8 '13 at 20:22
You could think it this way- actually you cannot multiply 0 with itself 0 times. So if the no. is not even multiplied with itself, then the thing which remains in the end in your answer is just 1. –  Rohinb97 Jun 8 '13 at 21:04

2 Answers 2

This depends on the situation. $0^0$ is often considered an indeterminate form. I believe a combinatorial interpretation leads to $0^0=1$. Namely that the number of total outcomes from $m$ events with $k$ outcomes each is $k^m$. Thus zero events having zero outcomes each still produce $1$ total outcome and so $0^0=1$.

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There is a difference between "assigning" 1 as an answer to the limit and "stating" it is 1. For lots of purposes assigning 1 to that limit makes sense in order to make formulas, theorems work etc. But the limit itself is, as mentioned earlier, indeterminate. It is incorrect to say zero to the power zero EQUALS 1. As a limit, f(x,y) = x^y does not exist if you approach (o,o) from any arbitrary direction. If you approach (0,0) from the positive x axis, for example, the limit is certainly not 1. Stewart's Calculus (James Stewart), chapter 6 (7th edition) considers it also an indeterminate form –  imranfat Jun 8 '13 at 23:58

My teacher once explained to me that mathematicians once just 'assigned' a value to it.

So basically the acadamic world has accepted $0^0$ to be $1$ just like they have accepted that every number to the power zero is one. Nothing more nothing less.

You can find alternative answers here:


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There is plenty of reasoning behind the definition.. –  Cameron Williams Jun 8 '13 at 20:19
That's why I supplied the link. I think it's kind of quick to downvote me without reading my whole answer first and I think you did that because I have 1 reputation: I have to start somewhere.. –  Jean-Paul Jun 8 '13 at 20:20
Your teacher treated you badly. –  Shuhao Cao Jun 8 '13 at 20:20

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