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Math Help - [SOLVED] dividing by 0!

  1. #1
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    Question [SOLVED] dividing by 0!

    Can someone show me why the following works out? Thanks!

    C(5,5) = (5!) (5!)(5-5)!
    = (5!) (5!)(0)!
    = 5x4x3x2x1 (5x4x3x2x1)... ?

    The calculator says this equals 1.
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  2. #2
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    I don't recall why but 0! = 1.
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  3. #3
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    0!=\prod_{k=1}^0k
    which is an empty product. An empty sum is 0. An empty product is 1. See this
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    a proof

    Thanks, I will definitely look at this after my final!
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Truthbetold View Post
    I don't recall why but 0! = 1.
    This is one of the reasons:

    In calculus, there is something known as the Gamma Function. It is defined as follows:

    \Gamma\left(x\right)=\int_0^{\infty}e^{-t}t^{x-1}\,dt

    When evaluating \Gamma\left(n\right), where n\in\mathbb{N}, it has this special property:

    \Gamma\left(n\right)=\left(n-1\right)!

    Now, let's see what happens when n=1

    This means that \Gamma\left(1\right)=\color{red}0!

    By definition, \Gamma\left(1\right)=\int_0^{\infty}e^{-t}\,dt=\left.\left[-e^{-t}\right]\right|_0^{\infty}=\left[-e^{-\infty}-\left(-e^{0}\right)\right]=e^0=\color{red}1

    Thus, \Gamma\left(1\right)=\color{red}\boxed{0!=1}
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    more research to do

    Thanks! Looks like I have some more research to do after Finals.

    Best of luck on your finals!

    I just have two left to do, Prob & Stats, and Dynamical Systems! Thank goodness this semester is almost over!!!
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  7. #7
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    Quote Originally Posted by yvonnehr View Post
    Thanks! Looks like I have some more research to do after Finals.

    Best of luck on your finals!

    I just have two left to do, Prob & Stats, and Dynamical Systems! Thank goodness this semester is almost over!!!
    At your level: 0! = 1 by definition so that the number of way of choosing zero objects from n objects is equal to 1 (a highly sensible result).
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    Quote Originally Posted by mr fantastic View Post
    At your level: 0! = 1 by definition so that the number of way of choosing zero objects from n objects is equal to 1 (a highly sensible result).
    Very sensible! Thank you!
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    You have already seen several excellent reasons why we define 0! = 1, but here is one more.

    A fundamental recurrence relation for the factorial function is

    (n+1)! = (n+1) \; n!

    We would like this recurrence to hold even when n = 0. This requires

    1! = 1 \cdot 0!
    1 = 1 \cdot 0!
    1 = 0!
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  10. #10
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    very nice!

    Quote Originally Posted by awkward View Post
    You have already seen several excellent reasons why we define 0! = 1, but here is one more.

    A fundamental recurrence relation for the factorial function is

    (n+1)! = (n+1) \; n!

    We would like this recurrence to hold even when n = 0. This requires

    1! = 1 \cdot 0!
    1 = 1 \cdot 0!
    1 = 0!
    This worked out very nicely!
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