Why is 0! = 1?

1. Jul 29, 2010

Frannas

Why is 0! = 1?

2. Jul 29, 2010

arildno

Re: 0!

It is defined that way. That's why.

3. Jul 29, 2010

Frannas

Re: 0!

Is there no proof to that?

4. Jul 29, 2010

arildno

Re: 0!

All proofs have at their basis a set of axioms&definitions, and a proof is simply to show that something else follows from those very same axioms&definitions.

Thus, neither axioms or definitions are themselves things to be proven, although it is quite possible that one may set up OTHER axioms&definitions from which the elements of the first set can be proven.

How would you, for example prove that a+0=a for any number a?

5. Jul 29, 2010

arildno

Re: 0!

Not that if you have a general recurrence relation described as:

R(n)=n*R(n-1)
(typically part of the definition of the factorial) you could, if you ASSUME this to be valid for n>=1 insert for n=1:
R(1)=1*R(0),
that is R(1)=R(0).
Now, how are you to go from this to your standard idea of the factorial?
Clearly, by fixing the value R(1)=R(0)=1.

This is therefore a necessary additional definition, since the relation R(n)=n*R(n-1) can have other sequences related to it, for example R(n)=0 for all n.

6. Jul 29, 2010

fourier jr

Re: 0!

n! is the number of possible ways to scramble up n objects & there's only one way to scramble up zero objects. It's a bit similar to showing there's only one empty set; if there were another way, what would it look like?

Last edited: Jul 29, 2010
7. Jul 29, 2010

arildno

Re: 0!

n! might be INTERPRETED as that, if you like.

8. Jul 29, 2010

fourier jr

Re: 0!

^ that's how I made it make sense to myself anyway

9. Jul 29, 2010

alxm

Re: 0!

0! = 1 is the value you get from the Gamma function, many series expansions are more compactly expressed if 0! = 1. The number of permutations of an empty set is 1.

It's simply more convenient for most situations where factorials are used that one defines 0! to be 1.