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Number theory problem

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that

    [tex] \frac{1}{p} c(p,n) = (-1)^{n-1}/n (mod p) [/tex]

    I expanded that combination in every way I could think and I tried to use Wilson's Theorem and I couldn't get :(


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 30, 2007 #2

    NateTG

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    That's p choose n, right?

    Try writing the LHS out as a fraction with the stuff in the numerator as negative representatives. It should nicely cancel to give the result.
     
  4. Oct 30, 2007 #3
    What do you mean "negative representatives"?
     
  5. Oct 30, 2007 #4

    Gokul43201

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    -(p-1)/2, -(p-2)/2,..., -1 for odd p
     
    Last edited: Oct 30, 2007
  6. Oct 30, 2007 #5

    matt grime

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    What is 1/n, or -1/n mod p supposed to mean?
     
  7. Oct 31, 2007 #6

    NateTG

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    Usually those would be the multiplicative inverses of n and -n respectively.
     
  8. Oct 31, 2007 #7

    matt grime

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    Usually? I beg to differ. Writing 1/n would indicate that the OP hasn't grasped what's going on. As would the fact there is an equals sign. I can't think of anyone who writes 1/2 mod 3 and not -1 0 it is incredibly bad notation. There is a difference from what I infer and what the OP ought to have written.
     
    Last edited: Oct 31, 2007
  9. Nov 1, 2007 #8
    What does OP stand for? Is that me?

    I just realized that my book my book defines congruence as

    [tex] x \equiv y \mod p [/tex]

    when x-y is a rational number whose numerator, in reduced form, is divisible by p.

    So, it is like a generalized congruence or something...

    Are there different rules for these generalized congruences?

    I am not sure why what Gokul43201 wrote cancels nicely?
     
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