# O! = 1 WHY?

1. May 19, 2004

### Euphoriet

Why does zero factorial equal one?.... I dont get it..

1! = 1 now that makes sense but not 0! = 1

=-/

can someone explain the reason for this to me?

2. May 19, 2004

### Dr Transport

by definition...... try looking at the integral definition of the factorial, and it should work out.......

3. May 19, 2004

### check

4. May 19, 2004

### jcsd

Dr. Transport - I'm pretty sure that 0! was defined before anyone kbew about the Gamma function tho'.

5. May 19, 2004

### fourier jr

0! = 1 because there's 1 way to scramble up 0 things. There's 1 way to scramble up 1 thing, there are 2 ways to scramble up 2 things (AB and BA), there are 6 ways to scramble up 3 things, etc etc...

6. May 19, 2004

### matt grime

if 0! weren't 1 (which it logicall is since there is one way to order no objecsts - the empty ordering) then you'd have no way of defining n choose zero, which is clearly the coeff of x^n in the expansion of (1+x)^n

7. May 19, 2004

### Dr Transport

could be, I have been wrong in the past.......the ordering arguments are the most logical ones I've seen for the definition.

dt

8. May 19, 2004

### Euphoriet

I've heard the ordering argument.. but how do you order soemthing that is not there?.. thats my big question...

9. May 19, 2004

### NateTG

Well:
{}
{1}
{1,2} {2,1}

and so on certainly makes sense. (This is using n! as the number of bijections A -> A where A is a set with cardinality n).

There's also that we like:
n!=(n-1)! * n
to be true
and that it works out for the binomial formula.

It turns out to be consistenly correct (unlike $$n^0=1$$) so that's the way it is.

10. May 19, 2004

### fourier jr

That may or may not be true, Euler discovered (or invented, depending on who you are) both factorial notation and the Gamma function, and I think he would have been the 1st to notice many of the properties of the Gamma function. I'll look it up see what came first.

11. May 20, 2004

### HallsofIvy

Staff Emeritus
That's easy- you just leave it alone!

Another way of looking at it:

5!/4!= 5, 4!/3!= 4, 3!/2= 3, 2!/1!= 2, so we want 1!/0!= 1 which requires that 0!= 1.

Of course we would also "want" 0!/(-1)!= 0 but that is impossible- which is why (-1)! is not defined to be anything.

12. May 20, 2004

It all depends on how we define the factorial operation, doesn't it?

If we define it as the product of all positive integer's up to and including n (ie: n! = 1x2x3x...n), then obviously 0! is not one, by definition. As it happens, we don't define factorial exactly that way...

13. May 20, 2004

### NateTG

Actually, an emtpy product is usually considered to be 1, although undefined is also acceptable.

14. May 20, 2004

### fourier jr

I think my answer is best. :)

15. May 20, 2004

### Caldus

I always figured that 0! = 1 because if 0! were to equal 0, then I'm saying that "there is no way to do nothing!" which clearly there's a way to do nothing: do nothing (which is one way). I know this is weird, but this is how I remember it...lol...

16. May 21, 2004

### mikesvenson

infinity ! = ?????

???? infinity ! = infinity ?????

or

???? [ infinity ! = 1(infinity) ] = ( 0! = 1 ) ?????
???? [ infinity ! = 0(infinity) ] = ( 0! = 1 ) ?????

Last edited: May 21, 2004
17. May 21, 2004

### arildno

???????????????????????????????????

18. May 21, 2004

### NateTG

I'm not sure what you mean by that. By the context of your post, I'm guessing that you're not familiar with cardinal arithmetic.

If we define:
$$X ! = \{f: X \rightarrow X s.t. f bij.\}$$
then we get
$$|X !|=|X|!$$
for finite sets.

In that sense, it's not difficult to show that:
$$| \mathbb{N} ! | = | \mathbb{R}|$$

For more esoteric cardinalities, you'll have to work things out for yourself.

19. May 22, 2004

### mikesvenson

HA, yeah, i dont really know anything about what i'm trying to explain, its been years since ive even taken algebra (which is the highest math ive ever taken)! All i was trying to ask is "how many ways are there to arrange an infinite set of numbers"? infinite?, or just 1? Like how theres only 1 way to arrange an empty set, is there only one way to arrange an infinite set as well? is it impossible to arrange an inifinite set?, since its end is undefinable?
Thats all im asking, im out of my league when i try to mathematicaly right it out.

20. May 22, 2004

### Janitor

Mikesvenson,

This may not help much, but there is (or was) a type of mathematician known as a 'constructivist,' and to that type it was anathema to even talk about infinite sets, if I have understood properly. A constructivist would probably feel nauseated at just the thought of rearranging elements in an infinite set.

Anyway, about two-thirds of the way down this page belonging to John Baez, you will find a mention of constructivists:

http://math.ucr.edu/home/baez/topos.html

The relevant quote is, "Suppose you're a constructivist and you only want to work with 'effectively constructible' sets and 'effectively computable' functions. Then you want to work in the 'effective topos' developed by Martin Hyland.

ADDED: Maybe even more to your point is another item in the same Baez page: "Suppose you're a finitist and you only want to work with finite sets and functions between them. Then you want to work in the topos FinSet."

Last edited: May 22, 2004