Can this be considered a proof

[Government$]

Homework Statement

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.

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}.
--------------------------------------------------------------------------------------

Homework Statement

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

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)!}

 

[Ray Vickson]

Homework Statement

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.

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}.
--------------------------------------------------------------------------------------

Homework Statement

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

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)!}

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?

 

[Government$]

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?

Good points, there goes my proof in the water.

 

[Office_Shredder]

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.

 

[Ray Vickson]

Good points, there goes my proof in the water.

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.

 

[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!

 

[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 $p|C_p^k$