Okay, so it’s pronounced “Oiler,” but I figger most poeple won’t know that…
Trig teacher says that 0!=1, but she don’t know why. Most people can’t explain why. Seems intruiging to me, but not enough to lose sleep over (knock on some wood…), but I could get some extra credit if I find out why.
IS THERE A MATHEMATICIAN IN THE HOUSE?
Zero factorial equals one. How and why? That’s all that needs to be said. Thank you, and remember, may the best bullshitter win (no, wait!..).
It sure makes the formulas for permutations and combinations work out well. Maybe that’s why it was defined that way. How many different combinations are there of five things taken five at a time? The formula is:
C(5,5) = 5! / ( 5! * (5-5)!)
and the only way for this to work is for 0! to equal one.
Also, here’s a fairly intuitive way to think of it arithmetically.
n! is the product of all positive integers less than or equal to n.
First, think about addition. What do you get if you add zero numbers together? I would argue that you get zero–zero is the additive identity (the “nothing” of addition), so if you add “nothing”, you get “nothing”. An empty sum is 0.
Similarly, what if you multiply zero numbers together? Again, I would argue that in this case you get one–one is the multiplicative identity (the “nothing” of multiplication), so if you multiply “nothing”, you get “nothing”–1, in this case. An empty product is 1.
0! is the product of all positive integers less than or equal to 0–an empty product, so 0! = 1.
(This also gives an intuitive explanation for why n[sup]0[/sup] = 1 (n not zero). What do you get if you multiply n by itself zero times? Again, an empty product, so n[sup]0[/sup] = 1.)
The way this was explained to me was that 0! was defined to be equal to 1 because that simplified a number of formulas/definitions that utilized factorial notation. CurtC mentions one case, combinations, where, to find how many possible ways to fit n objects into r slots, the answer is,
C(n,r) = n!/(r!*(n-r)!)
As a specific case, fitting 5 objects into 5 slots gives,
C(5,5) = 5!/(5!*(5-5)!) = 5!/(5!*0!)
Since, by inspection, C(5,5) = 1, 0! had better equal 1, otherwise the formula falls apart.
Another, more intuitive example, is that, in general,
n! = n*(n-1)!
i.e., 5! = 54! Now, if n=1, then that implies that 1! = 10! so therefore 0! = 1.
Unfortunately, I suppose that this last explanation implies that -1! is also equivalent to 1, which, as far as I know is both untrue and meaningless. However, the point is that 0! arises naturally in special cases of some definitions, and specifically defining 0! == 1 naturally makes these formulae “correct”.
It all depends on how you define factorial and math is all about extending definitions. If you define n factorial as the product of the integers from one to n, then 0! is not defined. We notice that (n+1)! = n!(n+1) for all n > 0. If we extend the defintion of factorial to include 0!, we would like to preserve the formula (n+1)! = n!(n+1). If we substiture zero for n, and solve, 0! must be defined to be zero for this formula to work. Another definition for factorial that is often more usable is:
0! = 1
n! = (n-1)!n, if n > 0
Then you have 0! defined to be 1. To evaluate 3! using this definition, we use the second line three times and the first line once:
3! = 2!3
If you substitute zero for n in n! = (n-1)!*n, you have:
0! = (-1)!*0
1 = (-1)!*0
so (-1)! times zero must be one. There is no value for (-1)! that satisfies, so (-1)! remains undefined.
I agree with Cabbage it is useful (and makes sense) to define a null product to be one, but in this case, one might be tempted to define (-1)! to be one also, since the set of integers from 1 to -1 is the empty set.
Since you bring up Euler, I thought I should mention Euler’s Gamma function, Gamma(x). Gamma(x) follows the same recurrence relation as factorials, but it is not limited to integers like factorial. It turns out that for positive integers, Gamma(n+1)=n! So it makes sense to define 0! to be equal to Gamma(1). Just plug in x=1 into the the formula for Gamma(x), and it is found to be 1. You need some calculus for this, but it ends up being equal to the area under the curve e[sup]-x[/sup] between x=0 and infinity.
Gamma(x) is not limited to integers, for example, Gamma(0.5)=sqrt(Pi). This connection between Pi and factorials leads to Stirling’s Formula: n! = (approximately) sqrt(2nPi) * n[sup]n[/sup] * e[sup]-n[/sup]. Stirling’s approximation gets better as n gets larger, so don’t use it to find 0!
As others have written, 0!=1 follows most easily from the recurrence relation, (n-1)!=n!/n, by plugging in n=1. Still, you need to define a starting point like 1!=1.
Well, so what? Can you think of a better reason to arbitrarily define some notation? Since mathematics is all about “making the math work”, why on earth would one arbitrarily define notation in a way that makes things more difficult?
Rule number one of practical mathematics: If there’s more than one way of doing something, do it the easiest way.
The notion of defining particular terms so as to be consistent with other terms (taht is, “to make the math work”) is not limited to factorials.
When you first stumble across 3 - 5, at a young age, you say it can’t be done. We could define some bizarre way to subtract a greater number from a lesser number (3 - 5 = 15, for instance) but it works much better to define it in a way that is consistent with already defined notions of addition and subtraction.
Similarly with square roots of negative numbers, or even with square roots of numbers that aren’t perfect squares. We start with an integer definition of square roots, and then you try to take SQRT(2) and you can’t, so you invent/define the irrational numbers to be consistent with the previously existing math. (Unlike integers, irrational numbers don’t exist in the “real” world.)
And so with many other “oddities.” You start with some general description, usually taken from reality (like combinatorial arrangements of integers) and then you define the other cases to be consistent.