# {Edited} Factorials.

1. Oct 20, 2005

### Robokapp

okay...I love math and i enjoy math a lot, mainly for its perfection and the aspect of not contradicting itself. But lately i have been runnig into contradictions and things that should not exist...and they piss me off.

The worst of them is solving 0!

0!=1

why? i asked my math teacher and he worte the following (which indeed makes sense)

5!=5*4*3*2*1 or 5!=5*4!

so 4!=4*3!
3!=3*2!
2!=2*1!
1!=1*0!

but this is where it lost me. 1! should simply be 1, not 1*0! maybe i'm worng. aniway, let's take a look at 0!

0!= all the natural numbers larger than zero (5! = 5*4*3*2*1...all of them larger than zero) but also no bigger than zero. (5*4*3*2*1 has no 6 in it.)

So...let's see...0!=0*Whatever is larger than zero and smaller (or equal) than zero at the same time.

so 0!=0*all natural numbers that belong to (0,0]. that makes the smallest allowed value bigger than the largest. I can't picture a REAL such a number. Let's call the whatever number W. So...0!=0*W

Not only we have a number in that bizzare scenario with a min > max, but it is also multiplied with zero. LAst time i multiplied something by zero it did not turn out to equal one.

so...any good reason why 0!=1?

plz...have mercy, i'm only in calc 1. no crazy explanations that i can't even begin to understand.

Thank you.
~Robokapp

2. Oct 20, 2005

### pi-r8

I looked this up on mathworld. It says that 0! is simply defined to be 1 because there is exactly one way to organize zero objects (the empty set), and factorials are usually used for finding how many ways there are to organize things.

3. Oct 20, 2005

### Robokapp

but in how many ways can you give zero apples to zero kids? that's infinity...because 0/0 is undefined. dividing by zeor always causes asymptotes...

i'm not saying i need help figuring out why you can't divide by zero. i know the 3/0=3-0-0-0-0-0-0-0-0....-0=0 demonstration.

but still...i think zero objects can be organised in many ways. or in no way...but i'll go with your answer :D

thank you.

4. Oct 20, 2005

### blimkie

Just do what i do, teach yourself the math because your math teacher is a drunk who comes to class eveyrday hung over. Its funny to watch he starts talking then forgets whats he alking about so he puts an overhead on shuts of the lights puts his head down and taps the glass.

5. Oct 20, 2005

### Werg22

The first term is 2. 0 isn't included in n! And 1 is negligtable. Also 1! = 1, this is was defined to match the binominal theorem.

6. Oct 20, 2005

### Lonewolf

Well, you seem to have noticed a pattern that (n+1)! = (n+1)*n!. So, if we put n = 0, then we get 1! = 1*0!, and 1! = 0! = 1. There are other ways of justifiying it, but that way is quite simple.

7. Oct 20, 2005

Keep in mind that 0! = 1 is a definition, not any sort of theorem.

8. Oct 20, 2005

### Lonewolf

9. Oct 20, 2005

If we define n! to be equal to gamma(n+1)

However I doubt that will be of much help to someone at the level of the topic poster.

10. Oct 20, 2005

### Werg22

It is only to sastisfy the binominal theorem.

11. Oct 20, 2005

### masudr

I believe understanding that the $\Gamma(n+1)=n!$ is defined as an integral and then seeing that it works for integer n will make the poster realise that the supposedly arbitrary definition of 0! as 1 makes sense.

It's the same as realising that $e^{i\theta}=\cos\theta +i\sin\theta$ is consistent due to the Taylor expansions of these various functions.

I encourage the original poster to look up gamma functions on the internet, and even if he/she doesn't understand the integrals etc. to realise that if this general function works for all kinds of n, then perhaps it's a good idea to take it's definition of 0!

12. Oct 21, 2005

### matt grime

What is your definition of the factorial function, Robokapp? Mine is that it is a function from the set {0,1,2,3,...} of natural numbers to the integers satisfying 0!=1 and n!=n(n-1)!

Now, why is that a problem? It is, as people have said a definiton.

Don't bother with gamma functions as a justification. 0! was defined to be 1 well before gamma functions; it is disingenuous to use those to justify this definition. (warning that assertion is purely speculation but MUST be correct)

And we really cannot emphasize it enough that it IS a definition, one that is incredibly useful, consistent and practical.

Why do you say "solve for 0!" it isn't an equation; one cannot solve it.

Now, given we've made this or ANY other definition, the only valuation we can place on it is its usefulness, and this is a very useful one.

13. Oct 21, 2005

### Robokapp

ooh. i think i get what you're talknig about. math is created to justify the real life situations not the other way around, so 0! has to =1 so the system given by the definition works.

right?

14. Oct 21, 2005

### Tom Mattson

Staff Emeritus
No, it's not right. While it's true that "real life situations" are not created to justify mathematical statements, it is nonetheless not true that mathematical statements are created to justify said real life situations. Mathematicians develop their subject quite independently of whether it turns out to be descriptive of the physical world.

15. Oct 21, 2005

### CRGreathouse

4! = 4 * 3!
3! = 3 * 2!
2! = 2 * 1!
1! = 1 * 0!

This just shows that if there is a 0!, it 'should' be 1 to follow the same rules as the other numbers.

Here's another method you can consider. 4! is the product of 4 numbers (1,2,3,4). 3! is the product of 3 numbers (3, 2, 1). 2! is the product of 2 numbers (1,2). 1! is the product of 1 number (1). For this to make sense we have to have a way to take the product of just one number, and I think you'll find that the only way that 'makes sense' is to take the number itself:

$$k=\prod_{i=1}^1k$$

In the same way, the only way to make a product of 0 elements work is to define it as 1:

$$1=\prod_{i=1}^0k$$

This is because 1 is the multiplicative identity. Does this make any more sense?

A final argument for 0!=1 is that it is needed to define binomial coefficients and many other mathematical objects cleanly.

16. Oct 21, 2005

### Karlsen

This is a good explanation, but you should be carefull when you say that the integral works for integers.
The gamma-function is not defined for negative integers, it includes division by zero in some way. As you stated, n! = gamma(n+1), which means that (-1)! = gamma(0) which is undefined.
gamma(x) is defined for all reals except for the negative naturals including zero.

17. Oct 21, 2005

### LeonhardEuler

I agree with you, Robokapp, that the explaination with
4! = 4 * 3!
3! = 3 * 2!
2! = 2 * 1!
1! = 1 * 0!
is not a good one. It is not true that 0!=0*(-1)! so this is no reason that 0! should be one. It is more like matt grime suggested: That 0! is defined to be one. While in general what Tom Mattson said (that "it is nonetheless not true that mathematical statements are created to justify said real life situations") is true, there are often motivations behind mathematical definitions in the real world. 0! being 1 is consistant with the idea of the n! being the number of ways to order n things. Rethink what you were thinking when you said "but in how many ways can you give zero apples to zero kids? that's infinity...because 0/0 is undefined. dividing by zeor always causes asymptotes". Actually there is no reason to be dividing by zero in this situation. There is in fact exactly one way to order zero objects: to put no object anywhere. There are no other ways to order nothing: this is the only one.

18. Oct 21, 2005

### Robokapp

Well, solving any

[f(x+h)-f(x)]/h involves a division by zero. well 10^-99

but trough mathematical manipulation the division by zero is made possible.

Isn't it interesting how whole Calculus is based on dividing by zero?

19. Oct 21, 2005

### hypermorphism

This is taking the analogy a bit far. While dividing by zero does not yield a real number, limits also do not necessarily yield a point in the image of the function being studied; they only yield limit points, which may or may not be equivalent to points in the graph of the function. In this case, we can see what value this new function is bounded by, but the bound is not a value of the function.

20. Oct 22, 2005

### Robokapp

well, you can have a limit but not a value if it's a hole in graph...but still you're manipulating a division by zero...are you not? i mean you're dividing by h and setting h = 0