# Numbers to the Power of Zero

I have been a witness to many a discussion about numbers to the power of zero, but I have never really been sold on any claims or explanations. This is a three part question, the parts are as follows...

1) Why does $n^{0}=1$ when $n\neq 0$? How does that get defined?

2) What is $0^{0}$? Is it undefined? If so, why does it not equal 1?

3) What is the equation that defines exponents? I can easily write a small program to do it (see below), but what about in equation format?

I just want a little discussion about numbers to the power of zero, for some clarification.

Code for Exponents: (pseudo-code/Ruby)

def int find_exp (int x, int n){
int total = 1;
n.times{total*=x}
}

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Actually, $0^0=1$, too. – Thomas Andrews Nov 11 '12 at 19:22
@ThomasAndrews $0^0$ is an indeterminate form. As the limit of $x^x$ from the right, yes it limits to $1$. But as the limit of $0^x$, it tends toward 0. – Shaun Ault Nov 11 '12 at 19:25
Being an indeterminate form does not imply that something is undefined. $0^0$ is defined and equal to $1$. @ShaunAult – Thomas Andrews Nov 11 '12 at 19:26
Sure, you can define your code that way. On the other hand, there are also many other good reasons to define $n^0 = 1$. Notably, it is consistent with $1 = n^p / n^p = n^{p-p} = n^0$. – Shaun Ault Nov 11 '12 at 19:27
@ShaunAult, the value of the expression $0^0$ is not necessarily the limit of a different expression $x^y$ as $x$ and $y$ tend to zero. The notion of "NaN" is a red herring in any case; it is a value defined by the IEEE 754 floating-point standard, not a concept in mathematics. – Rahul Narain Nov 11 '12 at 20:10

It's basically just a matter of what you define the notation to mean. You can define things to mean whatever you want -- except that if you choose a definition that leads to different results than everyone else's definitions give, then you're responsible for any confusion brought about by your using a familiar notation to mean something nonstandard.

Most commonly we define $x^0$ to mean $1$ for any $x$. What you find in discussions elsewhere are argument that this is a useful definition, not arguments that it is correct. (Definitions are correct because we choose them, not for any other reason. That's why they are definitions).

Some people choose (for certain purposes) to explicitly refrain from defining $0^0$ to mean anything. That choice is (supposedly) useful because then the map $x,y\mapsto x^y$ is continuous in the entire subset of $\mathbb R\times\mathbb R$ it is defined on. But it's an equally valid choice to define $0^0$ to mean $1$ and then just remember that $x,y\mapsto x^y$ is not continuous at $(0,0)$.

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 I agree with most of your answer, except the definition we choose is not $x^0$ in this case, it is another definition which implies the value of $x^0$. See math.stackexchange.com/a/11211/2380 – wok Nov 12 '12 at 17:00 But when considering the lone expression $0^0$ without the context of the function $x^y$, I've always seen the choice to refrain from defining $0^0$ to be firstly about two competing basics of arithmetic: that usually, $x^0=1$, and usually $0^x=0$. These two things are at odds with $0^0$. – alex.jordan Nov 16 '12 at 20:16

To 1): We define the exponents of a nonzero integer $a$ such that they satisfy the relation $a^ba^c=a^{b+c}$ for ay integers $b,c$, with $a^1=a$. In order for exponents to be well defined, we thus need $a^0=1$.

To 2): It depends on how you define it. If you define it via the limits $\lim_{x\rightarrow 0} x^0$ or $\lim_{x\rightarrow 0} x^x$, then $0^0=1$. If you define it as $\lim_{x\rightarrow 0} 0^x$, then $0^0=0$.

To 3): Exponents are defined simply by $a^n=\underbrace{a\cdot a\cdot \,...\, \cdot a}_{n}$.

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 For 1), so some mathematician once said that $a^{0} = 1$? For 2), Which is then correct? Or is it both? For 3), I can kinda wrap my head around that, but I have yet to come across that particular way of denoting a range... – CodeAdmiral Nov 11 '12 at 19:29 No, we never define $0^0=0$. Just because it is a limit doesn't mean we define it that way. It is either $1$ or undefined. – Thomas Andrews Nov 11 '12 at 19:33 1) is needed to make the notion of exponent well defined. For 2), there is no "correct" one. It is a matter of definition. – espen180 Nov 11 '12 at 19:35

It is for various reasons convenient to define $0^0$ as being equal to $1$. For one thing, consider the Binomial Theorem, or power series. It is useful to be able to write $$(1+x)^n =\sum_{k=0}^n \binom{n}{k}x^k,$$ or $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}.$$ In each of these equations, if we want the expression on the right to give the correct answer when $x=0$, we need to set $0^0=1$.

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 Ah, that makes sense, thanks ^v^ – CodeAdmiral Nov 11 '12 at 20:02

From definition of division of powers with the same base we have that $$\frac{a^n}{a^m}=a^{n-m}$$ Assuming that $n=m$ from left side we get $$\frac{a^n}{a^n}=1$$ and from right side we get $$\frac{a^n}{a^n}=a^{n-n}=a^0$$ comparing the last two equations we have that $$a^0=1$$

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(1) For intuition, if $k\ge0$ is an integer, take $x^k$ to mean "$1$ multiplied $k$ times by $x$", and $x^{-k}$ with $(x\neq 0)$ to mean "$1$ divided $k$ times by $x$." For integers $n\geq 0$, we may define $n!$ as the number of distinct ways to line up $n$ distinct objects--the only way to line up $0$ objects is to not line up any objects.

(2) We often define $0^0$ to be $1$, which accords with the intuitive definition above--if we multiply $1$ by $0$ not at all, then we still just have $1$. If we don't use the intuitive definition, we can define $0^0$ as the limit of $x^x$ as $x$ approaches $0$ from above, which gives us $1$, as well. Now, some prefer not to define $0^0$ at all, because they'd like $y\mapsto 0^y$ to be continuous on its domain of definition, and not have a jump discontinuity at $y=0$.

(3) We can extend integer powers to rational powers as follows: We say $y=x^{\frac1m}$ for some integer $m>0$ if $x=y^m$. If $m$ is odd, there will be a unique solution $y$ to the equation $x=y^m$. If $m$ is even and $x<0$, there will be no real solution $y$; if $m$ is even and $x\ge0$, then there is at least one real solution $y$, and we will take $x^{\frac1m}$ to be the nonnegative solution. At that point, given integers $k,m$ with $m>0$ and $\frac k m$ in lowest terms, we define $x^{\frac k m}:=\left(x^{\frac1m}\right)^k$ for such $x$ as this is possible. Finally, we can extend from the rationals to the rest of the reals with continuity arguments.

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