0!

Hey /sci/ I usually don't lurk here but I have a question for you guys.


I don't understand how 0! = 1
I need to write a proof using "n" as a variable and without using zero in the actual proof but I'm completely stumped.

Can someone please explain this to me??

:(

>>1
Have you tried to google it?

The answer lies in the principle of n!=n*(n-1)!.

Because of the identity 1!=1 combined with the above identity,

you get

1!=1*(1-1)!
1/1=(1-1)!
1=(0)!    <--- This being the resulting identity.

Hope this helps.

>>4
actually, this doesn't help cause you moved the ! from one side to the other without equal exchange.

>>5
Lol, what?  Specially for you, I'll explaing what >>4 said. Slo-wly.

n!=n*(n-1)! | n = 1
1!=1*(1-1)! | 1! = 1
1=1*(1-1)!  | 1*x = x
1=(1-1)!    | 1-1 = 0
1=0!

There were not any non-equal exchanges.

Incorrect. I applied the formula 1!=1 inline on the left,  then moved the 1* to the left side, illustrated as a division by one. Send an e-mail to the professor who gave me this proof, if you don't believe me. His email is above.

>>7
I see your appeal to authority and raise my I don't really care what's goin on in this thread

n!=/=n if n!=n!

simple laws of equal exchange.

But 1! = 1 and 2! = 2. Why is this any different?

>>9 But 1! = 1 and 2! = 2. Why is this any different?

0! is a special case that is simply defined to be 1 for convenience. There's no "proof." If this was given as a problem in a mathematics class, your teacher is either ignorant or trying to be cute, both of which are bad.

The n! is defined as n!=1*2*3*...*n . Says nothing about 0. So let's discover...

We know that:
           
                1! = 1
     2! = 1!*2
                2! = 2
     3! = 2!*3
                3! = 6
     4! = 3!*4
                4! = 24

BUT turn this around and you get:

          4! = 24
     3! = 4!/4
                3! = 6
     2! = 3!/3
                2! = 2
     1! = 2!/2
                1! = 1
     0! = 1!/1
                0! = 1

Later on you will need the 0! to compute this:

             n!
   C(n,k) = --------
            k!(n-k)!

Good luck.

     3! = 4!/4
                3! = 6
     2! = 3!/3
                2! = 2
     1! = 2!/2
                1! = 1
     0! = 1!/1
                0! = 1

Ergo 0!=1. You will need it for things later.

>>12
if that is true every calculator would have the value stored somewhere instead of determining it with the algorithm.

>>15
It is true; calculator algorithms are irrelevant. What you're looking for is the reason behind defining 0! = 1, not proof of it. Defining 0! = 1 is consistent with the canonical convention that there is exactly one permutation of zero objects, and makes arithmetic involving 0! work out properly.

Actually, you should look for a book, "Gamma" bye Julian Havil. There's a chapter devoted to the Gamma Function, where it is discussed as a generalized factorial function (when the input is displaced by 1), and can be used to get interesting results such as (-1/2)! which gives something like the square root of pi :D. The three important steps that associate the general factorial to the gamma function are given and I believe it suffices to solve for gamma(1) to get what you need. Sincerely

R.
Muahahahaha, killing flies with nukes is my specialty :D

>>17
Interesting but,
The thing is if you use sterling's formula you would get a 0.
Wouldn't it make more sense not to define it and just deal with the resulting non-continuous functions if necessary?

More interesting would be who first defined 0! as 1 and why.

1 is the identity element for multiplication.  If you define x! as prod_{i=1}^x i, then the product is empty, so it is the identity element, 1.