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Can this be considered a proof

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    If p is a prime and k is an integer for which 0<k<p, then p divides [itex]\displaystyle \ \binom{p}{k}[/itex].
    Whne p divides [itex]\displaystyle \ \binom{p}{k}[/itex] it means that [itex]\displaystyle \ \binom{p}{k}=p*b[/itex].
    wheren b is some number.

    3. The attempt at a solution

    So p is equal to some number k.

    [itex]p=k[/itex]
    I take factorial of both sides and get
    [itex]p!=k![/itex]
    and multiply both sides by
    [itex]\frac{1}{k!(p-k)!}[/itex]
    to get
    [itex]\frac{p!}{k!(p-k)!}=\frac{k!}{k!(p-k)!}[/itex]
    Since p=k
    [itex]\displaystyle \ \binom{p}{k}=\frac{p(p-1)(p-2)...(p-k+1)(p-k)!}{k!(p-k)!}[/itex]
    then
    [itex]\displaystyle \ \binom{p}{k}=p*\frac{(p-1)(p-2)...(p-k+1)}{k!}[/itex]
    where [itex]b=\frac{(p-1)(p-2)...(p-k+1)}{k!}[/itex] therefore p divides [itex]\displaystyle \ \binom{p}{k}[/itex].
    --------------------------------------------------------------------------------------
    1. The problem statement, all variables and given/known data

    If [itex]n \in N[/itex] then [itex]\displaystyle \ \binom{2n}{n}[/itex] is even i.e.
    [itex]\displaystyle \ \binom{2n}{n}=2*b[/itex] where b is some number

    3. The attempt at a solution

    [itex]\displaystyle \ \binom{2n}{n}=\frac{2n!}{n!(2n-n)!} = \frac{2n(2n-1)..3*2*1}{n!(2n-n)!}=2*\frac{n(2n-1)..3*2*1}{n!(2n-n)!}[/itex] where [itex]b=\frac{n(2n-1)..3*2*1}{n!(2n-n)!}[/itex]
     
    Last edited: Oct 2, 2013
  2. jcsd
  3. Oct 2, 2013 #2

    Ray Vickson

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    In 1): how do you know that
    [tex]b=\frac{(p-1)(p-2)...(p-k+1)}{k!} [/tex]
    is an integer? Where have you used the hypothesis that p is prime?
     
  4. Oct 2, 2013 #3
    Good points, there goes my proof in the water.:cry:
     
  5. Oct 2, 2013 #4

    Office_Shredder

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    At the start of 1 you write p=k even though p > k...

    Your proof isn't totally wrecked, you know that
    [tex] \frac{ p(p-1)(p-2)...(p-k+1)}{k!} [/tex]
    is an integer, so all you want to do is explain why, when you divide the numerator by k!, the p in the numerator is not canceled out.
     
  6. Oct 2, 2013 #5

    Ray Vickson

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    Just to clarify: it is important that p be prime. For example (writing C(a,b) for "a choose b") we have that C(5,2) = 10 is divisible by 5 and C(7,3) = 35 is divisible by 7, but neither C(6,2) = 15 nor C(6,3) = 20 are divisible by 6.
     
  7. Oct 3, 2013 #6

    HallsofIvy

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    The hypotheses say 0< k< p yet in your "proof" you say "p is equal to some number k". Surely p is equal to "some number" (why not just use "p"?) but calling it "k" and then confusing it with the "k" in the hypothesis is just "sleight of hand". I saw you pocket that card!
     
  8. May 7, 2015 #7
    Hi,

    It's a consequence of Gauss theorem.
    Write ## p! = k!\ (p-k)!\ C_p^k##. You get that ##p|k!\ (p-k)!\ C_p^k##.
    Show that ##\text{gcd}(p, k!\ (p-k)!) = 1 ##, and conclude ##p|C_p^k##
     
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