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Math Help - Proof by Induction for 0!

  1. #1
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    Proof by Induction for 0!

    Hi,
    Would someone kindly show and explain the proof by induction for 0! = 1

    I would be very greatful if you could.
    Thanks
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  2. #2
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    Your question makes no sense. A proof by induction, by definition, means you're trying to prove a statement that depends on some integer n. There is no such n in what you're trying to prove. To show that 0! = 1, I would use the generalized factorial function, the gamma function, and show that

    0!=\Gamma(1)=1.

    But you'd need to know about the gamma function, which is defined in terms of an integral, and hence would be inappropriate for this forum. I can move this thread to Calculus, if you prefer, where the question would not be out of place.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Abromavich View Post
    Hi,
    Would someone kindly show and explain the proof by induction for 0! = 1

    I would be very greatful if you could.
    Thanks
    A 'proof by induction' is consequence of the general property in the particular case n=1...

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor chisigma's Avatar
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    If we want to be 'rigorous' a precise definition of n! is necessary... in my opinion such precise definition is the following...

    We call a_{n}=n! the n-th term of a sequence defined by the recursive relation...

    a_{n+1}= (n+1)\ a_{n} (1)

    ... with the 'initial condition' a_{0}=0!=1, so that 0!=1 is true 'by definition'. Alternatively You can have as 'initial condition' a_{1}=1!=1 an in this case 0! is defined writing the (1) as...

    a_{n-1}= \frac{a_{n}}{n} (2)

    Kind regards

    \chi \sigma
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  5. #5
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Abromavich View Post
    Hi,
    Would someone kindly show and explain the proof by induction for 0! = 1

    I would be very greatful if you could.
    Thanks
    In elementary combinatorics i believe 0! is DEFINED to be 1.
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