Can this be considered a proof

  • #1

Homework Statement


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.

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].
--------------------------------------------------------------------------------------

Homework Statement



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

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]
 
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Answers and Replies

  • #2
Ray Vickson
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Homework Statement


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.

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].
--------------------------------------------------------------------------------------

Homework Statement



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

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]
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?
 
  • #3
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?
Good points, there goes my proof in the water.:cry:
 
  • #4
Office_Shredder
Staff Emeritus
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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.
 
  • #5
Ray Vickson
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Good points, there goes my proof in the water.:cry:
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.
 
  • #6
HallsofIvy
Science Advisor
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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!
 
  • #7
535
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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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