Zero to the Power of Zero

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In light of some mathematicians' (or, indeed, muggles') reticence to define the zeroth power of zero as $1$, the following are examples of reasons why defining $0^0 = 1$ is a good idea.

Empty Product

We can interpret $0^0$ as meaning:

zero multiplied by itself zero times.

Using product notation:

$0^0 = \displaystyle \prod_{\large \text{false}} 0$

This is a vacuous product, so by definition should be equal to $1$.

Binomial Theorem

Consider the real polynomial function:

$y = \paren {x + c}^n$

for $n \in \N, c \in \R$.

By the binomial theorem, $y$ contains a term of the form:

$\dbinom n n x^{n - n} c^n$

If we did not define $0^0 = 1$, $y$ would have a discontinuity at $x = 0$.

This would contradict Real Polynomial Function is Continuous.

Cardinality of Mappings

By Cardinality of Set of All Mappings, the number of mappings from the empty set to the empty set should be given by:

\(\ds \left \vert {\varnothing^\varnothing}\right \vert\) \(=\) \(\ds \left \vert {\varnothing }\right \vert^\left \vert {\varnothing }\right \vert\)
\(\ds \) \(=\) \(\ds 0^0\) Cardinality of Empty Set

By Empty Mapping is Unique, there is exactly $1$ such mapping, demanding that $0^0 = 1$.

Exponential of Zero

By Exponential of Zero:

$\exp 0 = 1$

From Power Series Expansion for Exponential Function

$\exp x = \dfrac {x^0} {0!} + \dfrac {x^1} {1!} + \dfrac {x^2} {2!} + \cdots$

For these theorems to be consistent, it is necessary that:

$\exp 0 = 1 = \dfrac {0^0} {0!} + 0 + 0 + \cdots$

which holds only if $0^0 = 1$.


Consider the identity mapping:

$I_{\mathbb F} \left({x}\right) = x$

where $\mathbb F \in \left\{ {\R, \C}\right\}$.

From Derivative of Identity Function:

$\dfrac {\d I_{\mathbb F} } {\d x} = 1$

But $I_{\mathbb F} \left({x}\right) = x^1$ is also an order one polynomial.

By Power Rule for Derivatives:

$\dfrac {\d I_{\mathbb F} } {\d x} = 1 x^0$

As $I_{\mathbb F}$ is differentiable at $0$, for these theorems to be consistent, we insist that $0^0 = 1$.

As a Limit

Consider the real function:

$y = x^x$

This function is well defined for $x > 0$.

It is not obvious whether or not the right hand limit:

$\ds \lim_{x \mathop \to 0^+} y$


If it does, it would be nice if:

$\ds \lim_{x \mathop \to 0^+} x^x = 0^0$

Indeed, by Limit of x to the x, we have:

$\ds \lim_{x \mathop \to 0^+} x^x = 1$

We see that defining $0^0 = 1$ allows $x^x$ to be right-continuous at $x = 0$.

Historical Note

Whether $0^0 = 0$ or $0^0 = 1$ has been in question since the concept of zero was first raised as a concept.

From one point of view:

$\forall a \in \R: a^0 = 1$

and so $0^0 = 1$.

From the other point of view:

$\forall a \in \Z_{\ge 0}: 0^a = 0$

and so $0^0 = 0$.