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Why is it that 0! = 1?

  1. Mar 19, 2004 #1

    Math Is Hard

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    Why is it that zero factorial is equal to 1?

    This came up in class tonight (just sort of as a side-note) and nobody knew.

    Thanks,

    M.I.H.
     
  2. jcsd
  3. Mar 19, 2004 #2
  4. Mar 19, 2004 #3
    It's just a definition, to simplify certain mathematical procedures.
     
  5. Mar 19, 2004 #4
    3!=6
    2!=2
    1!=1
    let x=0!

    note that 3!/2!=3
    note that 2!/1!=2
    then 1!/0!=1 to keep the same pattern.

    therefore, 1!/x=1. if you solve for x, you get x=1.

    therefore, 0!=1.
     
  6. Mar 19, 2004 #5

    matt grime

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    THe gamma function isn't analytic at 0, and by pheonixthoth's reasoning the factorial of all negative numbers is 1, and that doesn't hold. (It would in particular create issues with Pascal's triangle etc).

    Think of it this way, if you have 0 objects, there is exactly 1 way to arrange them - the empty ordering.
     
  7. Mar 19, 2004 #6
    Aha! I never said anything about it! Doesn't count in my mistakes today!

    And wouldn't you have to evaluate the Gamma Function at 1 in order to find 0! ?

    cookiemonster
     
  8. Mar 19, 2004 #7

    HallsofIvy

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    "by pheonixthoth's reasoning the factorial of all negative numbers is 1, and that doesn't hold. "

    I don't see that. pheonixthoth's "reasoning" is that 4!/3!= 4,
    3!/2!= 3, 2!/1!= 2 (and in general (n+1)!/n!= n+1 for n any positive integer) so we should have 1!/0!= 1 and therefore 0!= 1!= 1. Extending that one more time, we would have 0!/(-1)!= 0 or 1= 0(-1)! which tells us that (-1)! does not exist.
     
  9. Mar 19, 2004 #8

    matt grime

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    Yes, sorry, another mistake. never do maths before breakfast.
     
  10. Mar 19, 2004 #9
    If 0! =1 we can say
    nCr = n!/((n-r)!*r!)

    if 0! was something else we would have to say
    nCr = n!/((n-r)!*r!)
    except for nCn = nC0=1.

    I know what I would rather say.


    Also can't you use the rule that
    [tex] n! = \Gamma(n+1) [/tex]

    to get
    [tex] 0! = \Gamma(1)=1 [/tex]

    As far as I can tell [tex] \Gamma(x) [/tex] goes straight to hell
    at x= 0 but [tex] \Gamma(1) =1 [/tex] since
    [tex] \int_0^{\infty} e^{-t} dt = 1 [/tex]
     
    Last edited: Mar 19, 2004
  11. Mar 19, 2004 #10

    matt grime

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    yep it has a pole there with residue 1/0!, of all things. it has poles at all non-postive integers -k with residue (-1)^k/k! or something.
     
  12. Mar 19, 2004 #11

    Math Is Hard

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    Thank you for the insight.
    Cookie, I appreciate that link.
     
  13. Mar 20, 2004 #12
  14. Mar 20, 2004 #13

    Math Is Hard

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    Awesome - thanks! I am sending this link to my math prof.
     
  15. Mar 20, 2004 #14

    matt grime

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    That link just reiterates what Phoenixthoth (and others) said (and what I misunderstood) and contains some frankly dubious claims: to say that because the gamma function has poles at negative integers, factorials do not exist for negative integers is slightly misleading. The square root is defined (naively) only for positive numbers, that doesn't stop us saying i = sqrt(-1). You'd need to show there was no analytic function that agreed with the gamma function on the integers, and always had poles at the negative integers. Remember this is just a generalization - another function wouldn't satisfy all of the functional equations the gamma function does. If you can 'define' "i!" why can't you 'define' "-1!"?
     
    Last edited: Mar 20, 2004
  16. Mar 25, 2004 #15
    If we define n! as the product of all numbers from 1 to n, then n! is only defined for integral values of n such that n > 0. We observe that for all n > 0 we have:
    (n+1)! = n! (n+1)

    If we are to extend the defintion of factorial to include 0!, we would like the above to apply. Substituting we have:
    1! = 0! (0+1) = 0!
    This allows us to (re)define factorial thus:

    0! = 1
    n! = (n-1)! n for n>0
     
  17. Apr 1, 2004 #16
    ooh! this was a topic in one of my math classes.... i thought simply it was because of the formulas for log's.... w/ 0, it would result in a divide by zero, thus they just threw an exception for log(0) to = 1..., anyone?
     
  18. Apr 2, 2004 #17
    Actually log(0) is undefined, Hessam...
     
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