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

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.

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 $1x=x$.

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:

\[1=\frac{a^n}{a^n}=a^{n-n}=a^0\text{.}\]

Alternatively, since

\[a^b=a^{b+0}=a^ba^0\text{,}\]

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,

\[ (n+1)!=(n+1)n!\text{.} \]

Rearranging this, we get

\[ n! = \frac{(n+1)!}{(n+1)}\text{.} \]

So

\[ 0! = \frac{1!}{1} = 1\text{.} \]

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?

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About the author

  • Peter Rowlett teaches mathematics at university. His views do not represent those of his employer. His column at The Aperiodical is Travels in a Mathematical World.

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

  1. Christian Perfect

    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 $n! \equiv \Gamma(n+1)$, where $\Gamma$ 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.

    $\Gamma(t) = \int_0^\infty x^{t-1}\mathrm{e}^{-x} \,\mathrm{d}x$, so $0! = \Gamma(1) = \int_0^{\infty} x^0 \mathrm{e}^{-x} \, \mathrm{d}x$. Look, there’s $x^0$!

    You can do that integral by hand:
    \begin{align} \Gamma(1) &= \int_0^{\infty} x^0 \mathrm{e}^{-x} \, \mathrm{d}x \\ &= \int_0^{\infty} 1 \cdot \mathrm{e}^{-x} \, \mathrm{d}x \\ &= \left\lbrack -\mathrm{e}^{-x} \right\rbrack_0^{\infty} \\ &= 0 – (-\mathrm{e}^0) \\ &= 0 – (-1) \\ &= 1. \end{align}

    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.

    Reply
  2. Mitch Keller

    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 $f\colon [n]\to[n]$, 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.

    Reply
    • Brent Yorgey

      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”…

      Reply
  3. Josh Little

    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!]

    Reply
  4. Paul Coombes

    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 … 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:
    \[1=\frac{a^n}{a^n}=a^{n-n}=a^0\text{.}\]

    Reply
    • Paul Coombes

      Tried to be clever, tried to use markup, got it wrong, can’t edit it, far too much in bold.

      Reply

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