# What's the proof?

1. Apr 24, 2009

### johncena

what is the proof for the statement 0! = 1??

2. Apr 24, 2009

### monty37

a number neither odd nor even cannot be equal to an odd number.

3. Apr 24, 2009

### Hootenanny

Staff Emeritus
But zero is an even number since it has a parity of 0.

4. Apr 24, 2009

### gunch

Assuming you mean the factorial of 0, then factorial is usually defined recursively by,
0! = 1
n! = (n-1)! * n for n > 0
So it's true by the definition of the factorial. If you mean why 0 doesn't equal 1 then you have to state explicitly some formal properties of the integers. For instance a popular way to describe the non-negative integers is the Peano Axioms which among other things state that 0 is a non-negative integer, there is no natural number whose successor is 0 and 1 is defined as the successor to 0. Hence if 0 = 1 then 0 would be the successor to 0 which contradicts the axiom that 0 isn't the successor of any non-negative number.

Alternatively if you are allowed to use properties like parity, and the fact that 0 and 1 have different parity, then they can't be equal because parity is uniquely determined. Note: 0 has even parity while 1 has odd parity; 0 is NOT neither odd nor even.

5. Apr 24, 2009

### johncena

you said n! = (n-1)! * n for n > 0
so taking n = 1,
1! = (1-1)! * 1 = 1
0! * 1 = 1
thus, 0! = 1/1 = 1
is this proof correct?

6. Apr 24, 2009

### Werg22

There is no proof, it's by definition. It gives a basis for a recursive definition of n! as n! = n*(n-1)! and 0! = 1.

7. Apr 24, 2009

### derek e

Like others said, this is by definition. You might be interested in the http://en.wikipedia.org/wiki/Gamma_function" [Broken].

Last edited by a moderator: May 4, 2017
8. Apr 24, 2009

### Tac-Tics

I'll also note a definition can never be wrong. It may be useless, but it's never wrong.

9. Apr 24, 2009

### sutupidmath

A motivation(maybe) behind defining 0!=1, might be if we look at the combinations of class k taken from a set of n elements.
$$C_{k}^{n}$$

THen, since this is nothing else but the set of all subsets of k elements taken from a set of n elements, if we have:

$$C_0^n=\frac{n!}{(n-0)!0!}$$ then since there is only one set that contains 0 elements taken from any set of n elements (the empty set), it follows that

C_0^n should equal 1, for this to happen 0! should be 1.

10. Apr 24, 2009

### derek e

... unless a useless definition is defined to be something that is incorrect or wrong. :tongue:

11. Apr 25, 2009

### HallsofIvy

But that would be a useless definition!

12. Apr 26, 2009

### workmad3

A definition cannot be incorrect or wrong. What a group of definitions can be is inconsistent, which is subtly different :) Determining if a set of axioms is consistent is a difficult problem (and consistency is the cornerstone for godel's theorem as with an inconsistent set of axioms you can prove stupid things like 0=1, 1=2, etc)

13. Apr 27, 2009

### derek e

I was kinda playing. But what I think is more subtle is the use of the words "wrong" and "incorrect." If one were trying to make a definition of something containing the essence of an idea, such as curvature, then I could see how some definitions can be considered wrong or incorrect. Something being ill-defined often carries connotations of incorrectness or inconsistency, as its name implies. However, stating that the truth/validity in the defining of definition X is false is something I find somewhat meaningless.

Last edited: Apr 27, 2009
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