Why is 0! equal to 1 in mathematics?

[murshid_islam]

can anyone please help me with this. why is 0! = 1?

thanks in advance to anyone who can help.

 

[matt grime]

Because that is what we define it to be. There are hundreds of posts on this topic.

We could choose not to define it at all and state that n! is the number of ways or ordering n things for integer n strictly positive, and not define 0!. But that turns out to be unwieldy when doing things like nCr, so for ease, consistency, whatever, there is no harm in defining 0!=1. End of story, no great mystery here.

 

[murshid_islam]

so mathematicians just decided that 0! = 1? is that what you mean?

 

[arildno]

so mathematicians just decided that 0! = 1? is that what you mean?

It didn't have any prior meaning before the mathematicians appropriated it, so yes, they were free to decide what that symbol collection 0! should mean.

The conventional definition of the factorial goes like this:
0!=1, n!=n*((n-1)!), n\geq{1}
where n is always a natural number.

 

[murshid_islam]

for example, 3! is the product of the first three integers. but 0! does not have such meaning. it is just defined to be 1. am i correct?

 

[arildno]

for example, 3! is the product of the first three integers. but 0! does not have such meaning. it is just defined to be 1. am i correct?

Quite so.
Or, rather, 3! is recursively defined in such a manner that you can compute it by multiplying together the first three natural numbers.

 

[murshid_islam]

thank you very much for your help, arildno.

 

[CRGreathouse]

It's not entiely arbitrary. n!/n = (n-1)! whenever division by n is defined. If we assign any value to 0! other than 1, this would not hold.

 

[shmoe]

But a decision to require n!/n = (n-1)! to be true where n=1 is arbitrary.

 

[arildno]

Besides, set 0!=a.
Then, by induction, we have:
n!=a*1*2*...(n-1)*n, that is, n!/n=(n-1)! whatever value you assign to 0!.

 

[CRGreathouse]

But a decision to require n!/n = (n-1)! to be true where n=1 is arbitrary.

Sure, as is the decision to make 0! the number of ways to arrange 0 objects (1 way). The point is that it is consistent with the way it works for other numbers.

 

[Spiderman]

At first I thought this was a totally absurd question, because in many programming languages "!=" means "not equal to". Of course 0 is not equal to 1

I've been programming all weekend so everything else goes out the door.

 

[arildno]

At first I thought this was a totally absurd question, because in many programming languages "!=" means "not equal to". Of course 0 is not equal to 1

I've been programming all weekend so everything else goes out the door.

 

[Robokapp]

The explanation my math teacher gave me was the following:

5!=5*4*3*2*1
4![/color]=4*3*2*1

So

5!=5*4!
4!=4*3!
3!=3*2!
2!=2*1!
1!=1*0!

1!=1...so 1*0! has to equal 1. 1*0!=1 => 0!=1/1 => 0!=1

 

[matt grime]

Then why isn't 0! 0*(-1)! which must be 0 if (-1)! is defined?

 

[dextercioby]

0! =\Gamma(1) =\int_{0}^{\infty} e^{-t} dt= 1

Daniel.

 

[JonF]

I think any thread with the : “0”, “1”, and “=” is asking for trouble.

 

[waht]

0! doesn't have to be defined exactly, look at this relation

\frac{n!}{k!} = (n - k)!

If n = k then you get

\frac{n!}{n!} = (n - n)!

 

[shmoe]

0! doesn't have to be defined exactly, look at this relation

\frac{n!}{k!} = (n - k)!

This isn't true. 5!/4!=5, not (5-4)!=1.

 

[waht]

yea you are right, my bad.

 

[Robokapp]

Then why isn't 0! 0*(-1)! which must be 0 if (-1)! is defined?

I'm way over my head to answer that, but I've never seen a ! of a negative...the ! starts at 1...

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

It never goes below 1 and it never goes decimal, so it makes me assume 0! has to be 1*whatever is bigger than 1 and less than zero. or 1 itself...you can *1 it 1000 times if you feel like it.

I'm seing now a pattern...
5! has 5 factors.
4! has 4 factors.
3! has 3 factors.
2! has 2 factors.
1! has 1 factor.

-----common sense--------

0! has no factors. But then again you can *1 as many times as you please... so 1* just...becomes 1 since the multiplication can't be performed because of absence of the second term?

I also heard an explanation that sounds like

"there is exactly one way to arrange zero objects"

which makes sense to me...I don't know. does this question have a solid, undebatable answer? or "it just is"?

 

[CRGreathouse]

0! has no factors. But then again you can *1 as many times as you please... so 1* just...becomes 1 since the multiplication can't be performed because of absence of the second term?

I also heard an explanation that sounds like

"there is exactly one way to arrange zero objects"

Of course, an expty product is 1 by definition:

\prod_{\{\}}f=1

 

[FunkyDwarf]

from memory its something to do with the gamma distrubution

 

[CRGreathouse]

from memory its something to do with the gamma distrubution

The gamma function, not distribution.

 

[matt grime]

Firstly, we are free to define ! however we wish, and with the way we have done so n! naturally describes the number of ways of ordering n objects for n>0. The abstract function ! is *not* defined as the number of ways ordering objects. That might be why we found it useful in the first place, but that is not its definition.: things have a life beyond where we first find them in mathematics. If they didnt' they'd probably be forgotten.

Secondly there is no need to justifiy things by 'common sense'. Common sense is unique amongst virtues in that everyone possesses it (I forgot who said that). Mathematics is not common sense, in that sense. We are free to define whatever the heck we want, and this is the definition that makes most logical sense for us.