Thursday, March 14, 2013

1=0!?

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(Growing lazy, not a good sign....)

I have seen a proof, saying that 1=0!?
Which one do you mean '"1=0!"?' or '"1=0"!?', uh?

You got the trick. Why do we use "!" for factorial?
I am not a historian, I don't know why Christian Kramp choose n! to represent factorial.

Why does 0!=1?
Lets first consider the definition of factorial. The n! is the number of ways to permute n objects. In the set theory language, n! is the size of the set of all bijective maps on a set of n elements.
Let A be a set. A bijective map or a bijection or a permutation is a subset F of A×A such that
  1. For any elements a∈A, there exists a unique F(a)∈A such that (a,F(a))∈F.
  2. For any elements a∈A, there exists a unique F-1(a)∈A such that (a,F-1(a)∈F.
For example the map on {x,y,z} such that F(x)=y, F(y)=z, F(z)=x is a bijection (and hence a permutation), but G(x)=y, G(y)=x, G(z)=x is not.

On {x,y,z}, we have six permutations, right?
Yes. They are
  • (F(x), F(y), F(z))=(x, y, z)
  • (F(x), F(y), F(z))=(x, z, y)
  • (F(x), F(y), F(z))=(y, x, z)
  • (F(x), F(y), F(z))=(y, z, x)
  • (F(x), F(y), F(z))=(z, x, y)
  • (F(x), F(y), F(z))=(z, y, x)
Hence 3!=6.

On empty set ∅, ...?
Note that ∅ equals to and is the only subset of ∅×∅. There is no way we can violate the two conditions, and so ∅ is the bijection on ∅. Therefore 0!=1.