Proof verification: If $n \in \mathbb{N}$, then $C(2n, n)$ is even

Question

Working on improving my proof-writing abilities, so working through Hammack's Book of Proof. One of the exercises is as follows:

If $n \in \mathbb{N}$, then $C(2n, n)$ is even.

My proof differs from the one given, and I am hoping someone can confirm whether it is valid.

Proof: Suppose $n \in \mathbb{N}$.

Then, $C(2n, n) = \frac{(2n)!}{n!n!}$.

By definition, $n! = n\cdot(n-1)\cdot ... \cdot 2\cdot1$. As 2 appears in the numbers being multiplied, n! is even. (A similar argument applies to $2n!$)

As such, both the numerator and denominator are even, and the quotient is even.

I realize this proof does not prove that $C(2n,n)$ will be an integer, and I am unsure whether that's problematic. Can I assume that the outcome of any $C(a,b)$ will be an integer, and therefore safely ignore it? (The example given in the text uses a set-based proof that also doesn't explicitly address the integer status of the quotient.)

Answer 1

No, that is not a good proof. Just because the numerator and denominator are even does not mean that their quotient is even.

Answer 2

for prime $p$ and positive integer $n$ let $\sigma_p(n)$ denote the sum of the digits in the $p$-adic expansion of $n$.

if $\lambda_p(n)$ denotes the exact power of $p$ which divides $n!$, then we find: $$ \lambda_p(n) = \frac{n-\sigma_p(n)}{p-1} $$ the exact power of $p$ dividing $\binom{2n}{n}$ is thus: $$ b_p(n) = \lambda_p(2n)-2\lambda_p(n) $$

in the special case $p=2$ we have $p-1=1$, and $$ \sigma_p(2n) = \sigma_p(n) $$ which gives: $$ b_2(n) = \lambda_2(2n)-2\lambda_2(n) =\sigma_2(n) $$ which must be $\ge 1$