Numbers to the Power of Zero

Question

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...

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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?

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

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Code for Exponents: (pseudo-code/Ruby)

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

Answer 1

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 $(0,0)$.

Answer 2

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 $a^n=\underbrace{a\cdot a\cdot \,...\, \cdot a}_{n}$.

Answer 3

(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 $1$. Now, sometimes we will not define $0^0$ at all, which I'll discuss further below.

(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, for such $x$ that $x^{q}$ is defined for all rational $q,$ we can use continuity arguments to define $x^y$ for all real $y.$

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In the manner described above, given real numbers $x$ and $y,$ we have defined a real number $x^y$ for all real $y$ when $x>0,$ for all nonnegative $y$ when $x=0,$ and for all rational $y$ with odd denominators when $x<0.$ Unfortunately, continuity arguments won't work to extend to any more $y$ when $x\le0,$ because the function behaves too erratically to extend continuously in such cases. In fact, the erratic behavior of the function $f(x,y)=x^y$ means that $f$ isn't even continuous at the origin! For example, we can approach the origin along the line $y=x$ in the first quadrant (that is, when $x$ and $y$ are positive), and find that $x^y$ approaches $1,$ which is what we would expect. However, if we try to approach it along the positive $y$ axis (that is, when $x=0$ and $y>0$), then we find that $x^y$ approaches $0,$ which is not at all what we want! This means that not only is $f(x,y)$ discontinuous at the origin, but that there is no way that we can define $f(0,0)$ to make it continuous there! Similarly, $f(x,y)$ is badly discontinuous when $x<0.$ For this reason, when trying to define a continuous real-valued exponential function, one cannot define $0^0$ at all, nor define $x^y$ when $x<0.$ This continuous function $g(x,y)=x^y$ is defined for all real $y$ when $x>0,$ defined for all positive $y$ when $x=0,$ and undefined otherwise. However, this doesn't alter the truth of $0^0=1,$ merely the domain of continuous definition.

Answer 4

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 $$a^0=1$$

Answer 5

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 $0^0=1$.