# Can this be considered a proof

1. Oct 2, 2013

### Government$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 $\displaystyle \ \binom{p}{k}$. Whne p divides $\displaystyle \ \binom{p}{k}$ it means that $\displaystyle \ \binom{p}{k}=p*b$. wheren b is some number. 3. The attempt at a solution So p is equal to some number k. $p=k$ I take factorial of both sides and get $p!=k!$ and multiply both sides by $\frac{1}{k!(p-k)!}$ to get $\frac{p!}{k!(p-k)!}=\frac{k!}{k!(p-k)!}$ Since p=k $\displaystyle \ \binom{p}{k}=\frac{p(p-1)(p-2)...(p-k+1)(p-k)!}{k!(p-k)!}$ then $\displaystyle \ \binom{p}{k}=p*\frac{(p-1)(p-2)...(p-k+1)}{k!}$ where $b=\frac{(p-1)(p-2)...(p-k+1)}{k!}$ therefore p divides $\displaystyle \ \binom{p}{k}$. -------------------------------------------------------------------------------------- 1. The problem statement, all variables and given/known data If $n \in N$ then $\displaystyle \ \binom{2n}{n}$ is even i.e. $\displaystyle \ \binom{2n}{n}=2*b$ where b is some number 3. The attempt at a solution $\displaystyle \ \binom{2n}{n}=\frac{2n!}{n!(2n-n)!} = \frac{2n(2n-1)..3*2*1}{n!(2n-n)!}=2*\frac{n(2n-1)..3*2*1}{n!(2n-n)!}$ where $b=\frac{n(2n-1)..3*2*1}{n!(2n-n)!}$ Last edited: Oct 2, 2013 2. Oct 2, 2013 ### Ray Vickson In 1): how do you know that $$b=\frac{(p-1)(p-2)...(p-k+1)}{k!}$$ is an integer? Where have you used the hypothesis that p is prime? 3. Oct 2, 2013 ### Government$

Good points, there goes my proof in the water.

4. Oct 2, 2013

### Office_Shredder

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

Your proof isn't totally wrecked, you know that
$$\frac{ p(p-1)(p-2)...(p-k+1)}{k!}$$
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. Oct 2, 2013

### Ray Vickson

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. Oct 3, 2013

### HallsofIvy

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. May 7, 2015

### geoffrey159

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 $p|C_p^k$