Why does 0! = 1 ?

Below is some email I recently exchanged with my nephew John, an  adult whose unusual upbringing left him as a beginner at math. But he’s intelligent, and very enthusiastic; how many students at this level would try to prove anything?

Subject:      Re: Factorials… Why does 0! = 1?

John–

I applaud your curiosity about this! The basic problem here is that — I’m pretty sure,
but I’d be very interested if your instructor or TA disagrees — what 0! is isn’t a
matter for proof; it’s a matter of definition, just like the definition of factorial
for a positive integer. The real question is what is a useful definition of 0! And, in
math, an essential feature of a useful definition is that it will never lead to a
contradiction. If things in math are defined in such a way that you can get contradictory
results, it’s a disaster. That’s just why dividing by zero isn’t allowed. Now, read on.

On Wed, 5 Mar 2014 16:32:42 -0500, John Doe <jdoe@geemail.com> wrote:
> I rechecked my study materials regarding 0! and it was not explained. I
> emailed the head TA and she confirmed 0! = 1. However, I wanted to “prove”
> it so here is what I came up with:
>
>           *Solve:*            0!
>
> *Answer*:          1
>
>
> 1. Factorials always equal a natural number.
> 2. 0 is not part of the number set of natural numbers.
> 3. Negatives are not part of the set of natural numbers.
> 4. Without a 0 or a negative, 0 can never be found in an
>    equation given a positive value.
> 5. Then by definition 0! = 1.
>
> I don’t know how to do a proof, but this was how I can logically reason the
> answer.

There’s nothing wrong with steps #1 thru #4, but #5 does not logically follow from
them. They don’t establish that 0! factorial isn’t 953 or any other positive integer.
What forces us to define 0! as equalling 1 is that any other definition will lead to
big problems. For example, given n people, the number of different pairs of them
(ignoring order, i.e., we’re talking about combinations and not permutations, if you know
those terms) is

n! / 2(n-2)!

If n is 3, that’s

3! / 2(1!) = 6 / 2 = 3

And if n is 2, it’s

2! / 2(0!)

Well, that expression had better turn out to equal 1 — and it won’t unless we agree that
0! = 1.

Finally, from the Wikipedia article on natural numbers: “There is no universal agreement
about whether to include zero in the set of natural numbers: some define the natural
numbers to be the positive integers {1, 2, 3, …}, while for others the term designates
the non-negative integers {0, 1, 2, 3, …}. The former definition is the traditional
one, with the latter definition having first appeared in the 19th century.” Wolfram
MathWorld (http://mathworld.wolfram.com/NaturalNumber.html) says basically the same
thing. But this is a very small point; it sounds like your instructor is using the
positive integer definition.

Keep thinking!  🙂

–Don

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