Why Does 0! Equal 1? A Puzzling Question Explained

[Euphoriet]

Why does zero factorial equal one?... I don't get it..

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

=-/

can someone explain the reason for this to me?

 

[Dr Transport]

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

 

[check]

A very common question. These sites answer it pretty well.

http://mathforum.org/dr.math/faq/faq.0factorial.html
http://www.zero-factorial.com/whatis.html

 

[jcsd]

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

 

[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...

 

[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

 

[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

 

[Euphoriet]

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

 

[NateTG]

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

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.

 

[fourier jr]

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

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.

 

[HallsofIvy]

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

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.

 

[DeadWolfe]

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...

 

[NateTG]

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...

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

 

[fourier jr]

I think my answer is best. :)

 

[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...

 

[mikesvenson]

ok, then what about this:

infinity ! = ?

? infinity ! = infinity ?


or


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

 

[arildno]

??

 

[NateTG]

ok, then what about this:
infinity ! = ?

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.

 

[mikesvenson]

HA, yeah, i don't really know anything about what I'm trying to explain, its been years since I've even taken algebra (which is the highest math I've 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 there's 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 I am asking, I am out of my league when i try to mathematicaly right it out.

 

[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."

 

[matt grime]

Another way of permuting an infinite set that is useful is to use things with compact support.

In the main example of N the natural numbers there are two permutation groups, the full perms and the ones that only swap a finite set of numbers.