How is factorial 0 = 1?

For all of the other answerers, all facts have to be proved.

I can prove why.

n! = (n) * (n-1)!

1! = (1) * (0!)

1 = 1 * (0!)

Therefore,

0! = 1

Great Problem.

Thank you for asking the why and not the how like everyone else.

Normally, factorials would only be defined for positive integers, but since 0! shows up in practice quite often, it needs to have some value so we can work with it. In defining 0! we want to give it a value that keeps it consistent with the common properties of factorials. Consider the fact that for any positive integer n, (n-1)! = n!/n. Then for n = 1, we have 0! = (1-1)! = 1!/1 = 1. So, it makes sense to define 0! = 1. Depending on how far you go in mathematics, you may also come across the gamma function, which further generalizes the factorial from only working for non-negative integers to having the real numbers, and even the complex numbers as a domain.

There are many applications of mathematics where 0! is defined implicitly as being equal to 1. Instead, I'll give you an explicit example.

Take the general definition of n!:

n! = n * (n-1) * (n-2) * ... * 2 * 1

Now, consider (n-1)!:

(n-1)! = (n-1) * (n-2) * ... * 2 * 1

Notice that now we can write

n! = n * (n-1)!

Divide both sides by n:

n! / n = (n * (n-1)! ) / n

n! / n = (n-1)!

(n-1)! = n! / n

What if we evaluate at n = 1?

LHS = (n-1)! = (1-1)! = 0!

RHS = n! / n = 1! / 1 = 1 / 1 = 1

Since the equality was derived from our basic assumptions about what the definition of a factorial is, then the equality must be true. Thus,

LHS = RHS ==> 0! = 1

If you don't believe me, try evaluating it at n = 2:

LHS = (n-1)! = (2-1)! = 1! = 1

RHS = n! / n = 2! / 2 = 2 * 1 / 2 = 1

LHS = RHS => the equality holds.

If you want, you can try if for any other number, too.

By definition 0! = 1. Just like x^0 = 1

we know that n! = n (n-1)! put n = 1 so 1! = 1 ( 0)! or 0! = 1!/1 or 0! = 1 proved

It wouldn't make sense for it to be 0. factorial can only increase away from 1. So if anything, 0! should be undefined. But they added a definition for it.

The real reason is that in mathematics you can define things however you like; and theorems will be consequences of your definitions. It makes a lot of things simpler to have 0! = 1, for example in defining Taylor series. If it was left undefined or was defined as 0, the 0th term of the Taylor series definition would need to be given as a special case.

Mathematicians could have made 0! = 992, but that would be dumb and arbitrary. Because then for say, probability, or taylor series, everytime 0 comes up, you'd need a special case.

It's the same reason 1 is not prime... it's just arbitrary and the decision was made for elegance. Since the gods of math decided it was neater this way, you need to follow their convention. I mean, you can think however you like, but when it comes to communicating with the math world, 0! = 1.

____________

"all facts have to be proved."

An axiom or definition in mathematics CANNOT be proven. Only theorems (consequences) of axioms/definitions can be proven; such proofs are with respect to the axioms. However, definitions can be proven false in combination, when they lead to contradiction. Truth in mathematics can only be relative, never absolute.

http://en.wikipedia.org/wiki/Axiom

well anything over itself is 1 like 2/2 is 1 whole piece. 6/6=1 3/3=1

1/1=1 and 0/0=1 its like parts of a whole if both the numbers are the same u have a whole.

I know, but that's how it is. The factorial 0 is one doesn't make sense but that how it is.

There are some terms whose definitions you just have to accept.

Yes.