# Factorials

Discussion in 'Mathematics' started by mmmmmaths, Jan 13, 2012.

1. ### mmmmmathsNew commenter

Discovered this week that factorial zero is one.

I'd never actually thought about it before.

Reminded of this by Mike's comment on another thread.

So my proof is

As factorial one also equals zero,

Factorial 0 equals factorial1.

Therefore zero equals one!

2. ### mmmmmathsNew commenter

Discovered this week that factorial zero is one.

I'd never actually thought about it before.

Reminded of this by Mike's comment on another thread.

So my proof is

As factorial one also equals zero,

Factorial 0 equals factorial1.

Therefore zero equals one!

3. ### Woostarite

It's a shame the factorial function is not one-to-one.
There's multiple reasons for 0! being equal to 1, but perhaps the coolest is when observing the gamma function: http://www.wolframalpha.com/input/?i=y+%3D+gamma%28x%29
It comes in useful for things like the Dirichlet Distribution, and essentially acts as continuous variant of the factorial function.
gamma(x) = (x-1)! for integers greater or equal to 1, and so gamma(1) = 1.

4. ### Woostarite

(Sorry, that was meant to read "gamma(1) = 1, and so 0! = 1")

5. ### mmmmmathsNew commenter

Nice link, I think! Way over my head. But it has now introduced me to double factorials which, thanks to Mathsworld, I am enjoying reading about.

6. ### jstone98

"As factorial one also equals zero,".

Erm. 1!=1.

Best argument I've found is "How many ways can I select n objects from n? One. You have no choice but to select them all. Therefore nCn=n!/(n! 0!)=1. From this 0! must be one"

8. ### mmmmmathsNew commenter

Yes, shouldn't type on a Friday after a few ciders!

read zero as one.

9. ### mmmmmathsNew commenter

I liked the proof: factorial n equals n multiplied by factorial (n -1). So factorial n divided by n equals factorial (n-1). Put n=1.

10. ### ColinWilson

It is a matter of definition rather than proof that 0!=1. It is clear that 3!=6 but 0! has no real meaning. However the formula for say nCr would not work when r=0 or n unless 0! was defined as 1.
In a similar way 2^0 has no real meaning. But 2^3/2^3=2^0, so 2^0 must be defined as 1 otherwise the rules of indicies would break down.

11. ### Polecat

0! = 1 is fixed by convention so that it fits in with n! = n times (n-1)!
At school level it is not possible to give an easy proof. A bit further on one
can invoke the Gamma function, as noted earlier - I suppose Further Maths
students could be shown this.
Sadly, a lot of school maths can't be convincingly proved at the point it is introduced
- for example,why is the area of a circle Pi r^2 ?.

12. ### Woostarite

It depends what you mean by a 'matter of definition'. Exponentiation for example is just defined by the base condition x^1 = x and the recurrence relation x^(n+1) = x^n * x. Thus 2^0 stems from the fundamental definition of the operator, rather than semi-arbitrarily being defined as 1 to just 'make things work'. It only has no meaning if you restrict our understanding of exponentiation to be the conceptual understanding (resulting from the underlying definition) for positive indices.

13. ### florapost

oh my - i have left that page only with great effort as i need to go to bed shortly
i'll be back.....

14. ### Woostarite

This is a great website - thanks for sharing!