# What is 0 to the power of 0?

Jun 24, 2015

This is actually a matter of debate. Some mathematicians say ${0}^{0} = 1$ and others say that it is undefined.

#### Explanation:

See the discussion on Wikipedia:
Exponentiation : Zero to the power of zero

Personally I like ${0}^{0} = 1$ and it works most of the time.

Here's one argument in favour of ${0}^{0} = 1$ ...

For any number $a \in \mathbb{R}$ the expressions ${a}^{1}$, ${a}^{2}$, etc. are well defined:

${a}^{1} = a$
${a}^{2} = a \times a$
${a}^{3} = a \times a \times a$
etc.

For any positive integer, $n$, ${a}^{n}$ is the product of $n$ instances of $a$.

So what about ${a}^{0}$?

By analogy, that's an empty product - the product of $0$ instances of $a$. If we define the empty product as $1$ then all sorts of things work well. It makes sense as $1$ is the multiplicative identity. If we were talking about the empty sum, then the value $0$ would be natural.

If we're happy with that, what about ${0}^{0}$?

If it's the empty product of $0$ instances of $0$, then it is $1$ too.

Unfortunately, if we look at fractional exponents, we get some nasty behaviour.

Consider ${\left({2}^{-} n\right)}^{- \frac{1}{n}}$ for $n = 1 , 2 , 3 , \ldots$

As $n \to \infty$, ${2}^{-} n \to 0$ and $- \frac{1}{n} \to 0$

so you would hope ${\left({2}^{-} n\right)}^{- \frac{1}{n}} \to {0}^{0}$ as $n \to \infty$

but ${\left({2}^{-} n\right)}^{- \frac{1}{n}} = 2$ for all $n \in \left\{1 , 2 , 3 , \ldots\right\}$

So exponentiation behaves badly in the neighbourhood of $0$