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Why do $0!$ and $a^{0}$ equal $1$ ?

By Peter Rowlett. Posted February 3, 2014 in Travels in a Mathematical World

The last two weeks my first year mathematicians and I have covered Taylor series.This means that several times I’ve had the conversation that goes “What’s $0!$ ?” “It’s $1$ .” “Oh, erm, right. Why again?” “Because it works.” This may not be a completely satisfactory answer!

One of my students, Callum Mulligan, tweeted this question.

Why does 0! = 1 better yet, why does a^0 = 1 I must see a proof! #Mission #Unanswered #MathRage

— Callum Mulligan (@Calified) February 1, 2014

Saying “by definition” or “because it makes a bunch of stuff work” won’t cut it. So how to answer this question? To give a somewhat intuitive understanding of why this should be the case to a first year undergraduate. It may be obvious, but it wasn’t immediately obvious to me how to explain this, so I share some thoughts here.

Basically, I think it comes down to $1$ being the multiplicative identity.

Think about adding zero. If I take $x$ and add no things to it, I get $x + 0 = x$ (zero being the additive identity). Similarly, if I take $x$ and multiply it by no things, I ought to expect to get $x$ back, not $0$ . The empty product must therefore be $1$ , so that I get $1 x = x$ .

For $a^{0}$ , think of this as $a$ multiplied by itself zero times. This is an empty product, so $a^{0} = 1$ .

$n!$ is $n$ things multiplied together, so $0!$ is zero things multiplied together. This, again, is an empty product, so $0! = 1$ .

I also came across some interesting ways to try to get a feeling for what is going on.

For $a^{0}$ , I quite like this:

$1 = \frac{a^{n}}{a^{n}} = a^{n - n} = a^{0} .$

Alternatively, since

$a^{b} = a^{b + 0} = a^{b} a^{0},$

it follows that $a^{0} = 1$ .

For $0! = 1$ , I quite like the reasoning that uses the definition of each factorial in terms of the previous. So $6! = 6 \times 5!$ , and, in general,

$(n + 1)! = (n + 1) n! .$

Rearranging this, we get

$n! = \frac{(n + 1)!}{(n + 1)} .$

$0! = \frac{1!}{1} = 1 .$

Also, if you like $n!$ as the number of ways of arranging $n$ objects, then think that there’s only one way to arrange zero objects, so $0! = 1$ .

I found this useful on ‘Why is x^0 = 1?‘ and this on 0!=1. I’d be interested to hear what you think in the comments. How do you convince someone that $a^{0} = 1$ and $0! = 1$ feel right?

8 Responses to “Why do $0!$ and $a^{0}$ equal $1$ ?”

Christian Perfect February 3rd, 2014

For $n!$ , you could extend it to real-valued $n$ , because why not look at what it does when it gets close to $0$ ? The following definition-shuffling may or may not feel right:

There’s the identity $n! \equiv Γ (n + 1)$ , where $Γ$ is a function defined on all real numbers. I don’t know if you need to agree that $0! = 1$ already to agree with this identity, but there it is.

$Γ (t) = \int_{0}^{\infty} x^{t - 1} e^{- x} d x$ , so $0! = Γ (1) = \int_{0}^{\infty} x^{0} e^{- x} d x$ . Look, there’s $x^{0}$ !

You can do that integral by hand:
$\begin{aligned} Γ (1) & = \int_{0}^{\infty} x^{0} e^{- x} d x \\ = \int_{0}^{\infty} 1 \cdot e^{- x} d x \\ = {[- e^{- x}]}_{0}^{\infty} \\ = 0 - (- e^{0}) \\ = 0 - (- 1) \\ = 1. \end{aligned}$

And I got to use $a^{0} = 1$ again. Groovy.

But actually, “because it makes a bunch of stuff work” is the best answer, in my opinion. You define your operation on the domain that you have an intuition for ( $1! = 1$ , $2! = 2$ , $3! = 6$ , …) and then work out what it has to do everywhere else in order to be consistent. Maybe you need to be a Platonist to find that unsatisfying.

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Mitch Keller February 3rd, 2014

As a combinatorist, I tend to consider the number of permutations of $n$ objects to be the definition of $n!$ . If we consider permutations of $[n] = {1, 2, \dots,}$ as functions $f : [n] \to [n]$ , then it’s completely reasonable to say that there is one permutation of $[0] = {}$ .

For $a^{0}$ , I tend to side with $1 = a^{b} \cdot a^{- b} = a^{b - b} = a^{0}$ approach. I’d shy away from $a^{b} = a^{b + 0} = a^{b} \cdot a^{0}$ , since it could suggest to some that $0^{0} = 1$ . The use of $a^{- b}$ avoids that since $0$ does not have a multiplicative inverse.

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- Brent Yorgey February 3rd, 2014
  
  You can give $a^{b}$ a combinatorial interpretation too: it’s the number of functions from $b$ things to $a$ things. There is only one function from $0$ things to $a$ things (namely, the empty function), so $a^{0} = 1$ . Admittedly, students without an intuition for $a^{0}$ are unlikely to have any better intuition for strange things like “empty functions”…
  
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- Robin Whitty February 7th, 2014
  
  I don’t have a problem with $0^{0} = 1$ , at least as a 1-side limit. A plot of $y = x^{x}$ will satisfy most students but you can prove $lim_{x \to 0 +} x^{x} = 1$ using L’Hopital’s Rule
  
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Josh Little February 3rd, 2014

An intuitive way to explain the $x^{0} = 1$ would be to use orders of magnitude:

$10^{3} = 1000$
$10^{2} = 100$
$10^{1} = 10$
$10^{0} = 1$
$10^{- 1} = 0.1$
$10^{- 2} = 0.01$

This is by no means a definitive proof or anywhere near, but should make the fact easier to swallow.

(Sorry that I have no LaTeX skills) [Ed: sorted!]

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Paul Coombes February 9th, 2014

There are interesting suggestions in all the previous comments, but surely we are answering a non-existent question? There cannot be a proof that $a^{0} = 1$ since it is a definition. $a^{b}$ is, after all just a shorthand way of writing $a x a a a x a x \dots x a$ where a occurs b times which, incidentally, also shows why we need it. All one can do is to show that the definition of the conundrums thrown up by this shorthand, such as $a^{0}$ , are consistent. This is admirably shown by:
$1 = \frac{a^{n}}{a^{n}} = a^{n - n} = a^{0} .$

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- Paul Coombes February 9th, 2014
  
  Tried to be clever, tried to use markup, got it wrong, can’t edit it, far too much in bold.
  
  Reply