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Proof that 0!=1?

  1. Nov 4, 2007 #1

    rock.freak667

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    Proof that 0!=1??

    1. The problem statement, all variables and given/known data
    How does one prove that 0!=1, which I've only been told to accept as true.


    2. Relevant equations

    n!=n(n-1)(n-2)(n-3)*...*3*2*1

    3. The attempt at a solution

    well using the def'n for n!

    0!=0(-1)(-2)(-3)(-4)(-5)*...*3*2*1

    which should be zero as....any number by 0 is 0
     
  2. jcsd
  3. Nov 4, 2007 #2

    arildno

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    It is a matter of definition. Period.

    For n>=1, we have the recursive relationship [tex]n!=n*(n-1)![/tex]
     
  4. Nov 4, 2007 #3

    Avodyne

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    Yes, it's a matter of definition, but this is a useful defintion. Note that arildno's recursion relation can be written as (n-1)!=n!/n. Plugging in n=1 yields 0!=1.

    Also, formulas like [itex]e^x = \sum_{n=0}^\infty x^n/n![/itex] only work if 0!=1.
     
  5. Nov 4, 2007 #4

    rock.freak667

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    hm...but then the same goes for 1!...as that is the same as 0! by the definition of n!
     
  6. Nov 4, 2007 #5
    How many ways can you order 0 things? 1 way!

    I had originally thought that you were asking why 0 was not equal to 1. A very interesting question to pose compared to other more tired equalities.
     
  7. Nov 4, 2007 #6
    Well, no. Formally these things are defined recursively starting at 1.
     
  8. Nov 4, 2007 #7

    Avodyne

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    Yes, 1!=1 as well.

    Do you know about the gamma function?
    [tex]\Gamma(z) = \int_0^\infty dt\,t^{z-1}\,e^{-t}.[/tex]
    This has the interesting property that for an integer n, [itex]\Gamma(n)=(n-1)![/itex]
    For more, see http://en.wikipedia.org/wiki/Gamma_function
     
  9. Nov 5, 2007 #8

    rock.freak667

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    never heard of this gamma function...shall read now
     
  10. Nov 5, 2007 #9

    dynamicsolo

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    We also define 0! = 1 to provide consistency in the equations for nPr and nCr : when r = 0 or r = n, the formula should give values of 1. This will only be possible if (n-n)! equals 1.

    There are a number of such situations in mathematics where an operation originally defined only for positive integers ("counting numbers"), as evolved from ordinary human uses, is extended to larger sets of numbers. A widely-used example is x^n .
     
  11. Nov 5, 2007 #10

    rock.freak667

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    So then I guess there is no way to prove it but it is just taken as 1 just to make things simpler
     
  12. Nov 5, 2007 #11

    Avodyne

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    Yes, that's basically it.
     
  13. Nov 5, 2007 #12

    symbolipoint

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    Actually, there is an algebraic proof. Try a search of physicforums. I know I saw the proof somewhere, but I cannot remember exactly where, and I do not remember this proof myself.
     
  14. Nov 5, 2007 #13

    stewartcs

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  15. Nov 6, 2007 #14

    arildno

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    What do you think a "proof" is?

    It is a valid deduction from ACCEPTED AXIOMS OR DEFINITIONS.

    Thus, that some statement is an axiom or a definition is NOT some flaw with it as if ideally we would have liked to prove it.

    Of course, we have the freedom to choose another set of axioms&definitions than the "standard" set; in that case, axioms in another set might be needed to be proved, and vice versa.

    Specifically, if we CHOOSE to define the factorial in terms of the gamma function, then we may prove the statement 0!=1.
     
  16. Nov 6, 2007 #15

    Avodyne

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    I can't agree with that. We'd like to use the minimal number of axioms. If one can be shown to be derivable from the others, that's an improvement. People spent 2000 years trying to show that Euclid's fifth postulate was derivable before concluding that it couldn't be done.
     
  17. Nov 6, 2007 #16

    arildno

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    I can agree to that; I should have inserted the word "necesssarily" in "...is NOT necessarily a flaw..", which should cover the instances you mentioned.
     
  18. Jan 11, 2010 #17
    Re: Proof that 0!=1??

    To Prove:- 0! =1.

    Known:
    y! = y x (y-1)!

    So,
    1! = 1 x (1-1)!
    1! = 0!
    or
    0! = 1! -(a)
    Proof:
    As We know,

    n! = n x n-1 x n-2 x n-3 x n-4 x .......... 2 x 1
    So,
    4! = (3+1)! = 4 x 3! = 4 x 3 x 2 x 1
    3! = (2+1)! = 3 x 2! = 3 x 2 x 1
    2! = (1+1)! = 2 x 1! = 2 x 1
    1! = (0+1)! = 1 x 0! = 1 -(b)

    Hence, from (a) and (b)
    We get, 0! = 1
     
  19. Jan 11, 2010 #18
    Re: Proof that 0!=1??

    Personally I tend to prefer the combinatorial approach. There is just one way to choose 0 objects from a set of n objects, hence:
    [tex]
    ^{n}C_{0} = \frac{n!}{(n-0)!(0!)} = \frac{1}{0!} = 1
    [/tex]
    which thus implies that 0! = 1
     
  20. Jan 11, 2010 #19

    HallsofIvy

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    Re: Proof that 0!=1??

    This is "known" for what values of y? And how is it known? Exactly what definition of factorial are you using?

    By exactly the same "proof" then, 0!= 0(0-1)!. But 0 times any number is 0 so whatever (-1)! is, 0!= 0.
     
  21. Jan 12, 2010 #20

    Mentallic

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    Re: Proof that 0!=1??

    Well then couldn't (-1)! simply be undefined to keep the consistency that 0!=1.

    In the same way that [tex]lim_{x \rightarrow 0}\left(x.\frac{1}{x}\right)=1[/tex]you are similarly saying that the first factor x=0 but whatever the next factor 1/x is equal to, the RHS should equal 0 since 0 times any other number is 0. This is not the case however.
     
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