Why is $x^0 = 1$ except when $x = 0$?

Question

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.

Answer 1

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.

Answer 2

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

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 $x^0$.

Answer 3

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.

Answer 4

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 $0^0$: $$0^{1-1} = 0^1/0^1=0/0=\text{UNDEFINED}$$

Answer 5

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 $0^0=1$.

Answer 6

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:

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

Answer 7

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