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Proving that 0! = 1

  1. Nov 16, 2005 #1
    I was asked to prove 0! = 1?

    Where do I start?

    Is this possible? Any hints welcome... :cry:
     
  2. jcsd
  3. Nov 16, 2005 #2

    shmoe

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    What is your definition for factorial?

    (This is usually something that's included in the definition though sometimes not explicitly)
     
  4. Nov 16, 2005 #3
    Well 3! is 3x2x1 = 6
    2! is 2x1= 2
    1! is 1x1 = 1 but 0?? haven't got a clue
     
  5. Nov 16, 2005 #4

    matt grime

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    It's a definition that 0! is 1. If your teacher wishes for elucidation point them to these forums: there are plenty of reasons given in them. At best you can justify why 0!=1 is the most sensible option.
     
  6. Nov 16, 2005 #5

    shmoe

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    I see. I was naively hoping for something else. I suspect your teacher only told you what n! was when n is a positive integer. Take a look at:

    https://www.physicsforums.com/showthread.php?t=99749&highlight=0!=1
    https://www.physicsforums.com/showthread.php?t=54625&highlight=0!=1

    and I'm sure many more threads for some of the reasons why we'd define 0! to be 1.
     
  7. Nov 16, 2005 #6

    Physics Monkey

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    You can easily check (using integration by parts), that the following
    [tex]
    n! = \int^\infty_0 dt\, t^n e^{-t} \equiv \Gamma (n+1),
    [/tex]
    is true whenever n is a positive integer 1, 2, 3, etc. This integral is the definition of the so called gamma function [tex] \Gamma (z) [/tex], and I would encourage you to check what I have said thus far. Now, since the gamma function reproduces the standard factorial for integer n, we could equivalently define the factorial to be this integral for integer n. However, this definition is nice because it immediately allows us to define the "factorial" of many more numbers, including zero. Using the definition above, what would [tex] 0! [/tex] be?

    You can also compute [tex] 0! [/tex] hueristically by using [tex] n! = n(n-1)! [/tex], but this way still requires you to somehow extend your definition of factorial in order to be rigorous.

    It boils down to a definition any way you slice it.
     
    Last edited: Nov 16, 2005
  8. Nov 16, 2005 #7

    benorin

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    Well, hmmm :rolleyes: . Given that there is an existance/uniqueness theorem for the gamma function, specifically the Bohr-Mollerup Theorem, one might say that the 'only' proper definition of 0! is 1. For the gamma funtion is the only function that interpolates the factorial and that is log-convex for postive arguements, has the functional equation [itex]\Gamma (z+1)=z\Gamma (z)[/itex], and such that [itex]\Gamma (1)=1[/itex]. But who cares about log-convexity anyway? :wink:
     
  9. Nov 16, 2005 #8
    Factorials represent combinations of a set of objects. so 2 objects can be ordered 2 different ways, 3 objects can be ordered 6 different ways, etc... Matt is right 0! is defined as 1. Maybe you can entertain your teacher with this.

    A: x! = (x) (x-1) (x-2) ..... (x-(x-2)) (x-(x-1))
    B: (x-1)! = (x-1) ((x-1)-1) ((x-1)-2) ..... ((x-1)-((x-1)-2)) ((x-1)-(x-2))

    B / A = (x-1)! / x! = 1 / x

    for x = 1, B / A = 1

    B / A = 0! / 1! = 0! / 1
     
    Last edited: Nov 16, 2005
  10. Nov 16, 2005 #9

    benorin

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    0 objects can be ordered precisely 1 ways

    Do you mean: 0 objects can be ordered precisely 1 ways, and hence 0!=1 ?

    And it's permutations, not combinations, right?
     
  11. Nov 16, 2005 #10
    Factorial only states the number of ways the entire set (all elements) can be arranged, the different possibilities, or combinations. For instance, if you have three keys in a hat, red=r, green=g, blue=b, there are six ways you can take them out; rgb, rbg, gbr, grb, brg, bgr. The arrangement of the set

    Careful not to confuse it with Permutation and Combination, which involve arrangements of elements taken out of the set.

    So yes, (ordered or unordered) there is 1 way to order 0 objects.
     
  12. Nov 16, 2005 #11

    benorin

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    That works.
     
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