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