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0! = 1

  1. Sep 17, 2006 #1
    can anyone plz help me with this. why is 0! = 1? :confused:

    thanks in advance to anyone who can help.
     
  2. jcsd
  3. Sep 17, 2006 #2

    matt grime

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    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.
     
    Last edited: Sep 17, 2006
  4. Sep 17, 2006 #3
    so mathematicians just decided that 0! = 1? is that what you mean?
     
  5. Sep 17, 2006 #4

    arildno

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    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:
    [tex]0!=1, n!=n*((n-1)!), n\geq{1}[/tex]
    where n is always a natural number.
     
    Last edited: Sep 17, 2006
  6. Sep 17, 2006 #5
    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?
     
  7. Sep 17, 2006 #6

    arildno

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    Quite so.:smile:
    Or, rather, 3! is recursively defined in such a manner that you can compute it by multiplying together the first three natural numbers.
     
  8. Sep 17, 2006 #7
    thank you very much for your help, arildno.
     
  9. Sep 17, 2006 #8

    CRGreathouse

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    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.
     
  10. Sep 17, 2006 #9

    shmoe

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    But a decision to require n!/n = (n-1)! to be true where n=1 is arbitrary.
     
  11. Sep 17, 2006 #10

    arildno

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    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!.
     
  12. Sep 17, 2006 #11

    CRGreathouse

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    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.
     
  13. Sep 17, 2006 #12
    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.
     
  14. Sep 17, 2006 #13

    arildno

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    :rofl: :rofl: :rofl:
     
  15. Sep 17, 2006 #14
    The explanation my math teacher gave me was the following:

    5!=5*4*3*2*1
    4!=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
     
  16. Sep 18, 2006 #15

    matt grime

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    Then why isn't 0! 0*(-1)! which must be 0 if (-1)! is defined?
     
  17. Sep 18, 2006 #16

    dextercioby

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

    Daniel.
     
  18. Sep 18, 2006 #17
    I think any thread with the : “0”, “1”, and “=” is asking for trouble.
     
  19. Sep 18, 2006 #18
    0! doesn't have to be defined exactly, look at this relation

    [tex] \frac{n!}{k!} = (n - k)! [/tex]

    If n = k then you get

    [tex] \frac{n!}{n!} = (n - n)! [/tex]
     
  20. Sep 18, 2006 #19

    shmoe

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    This isn't true. 5!/4!=5, not (5-4)!=1.
     
  21. Sep 18, 2006 #20
    yea you are right, my bad.
     
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