Binomial Coefficient 2 n Choose n is Divisible by All Primes between n and 2 n
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Theorem
Let $\dbinom {2 n} n$ denote a binomial coefficient.
Then for all prime numbers $p$ such that $n < p < 2 n$:
- $p \divides \dbinom {2 n} n$
where $\divides$ denotes divisibility.
Proof
By definition of binomial coefficient:
\(\ds \dbinom {2 n} n\) | \(=\) | \(\ds \dfrac {\paren {2 n}!} {\paren {n!}^2}\) | where $n!$ denotes the factorial of $n$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dbinom {2 n} n \paren {n!}^2\) | \(=\) | \(\ds \paren {2 n}!\) |
Let $\mathbb P$ denote the set of prime numbers.
Let $p \in \mathbb P$ such that $n < p < 2 n$.
By definition of factorial:
- $p \divides \paren {2 n}!$
That is:
- $p \divides \dbinom {2 n} n \paren {n!}^2$
From Prime iff Coprime to all Smaller Positive Integers:
- $p \nmid n!$
where $\nmid$ denotes non-divisibility.
and so:
- $p \nmid \paren {n!}^2$
But from Euclid's Lemma for Prime Divisors, $p$ is a divisor of either $\dbinom {2 n} n$ or $\paren {n!}^2$.
Hence it must be the case that:
- $p \divides \dbinom {2 n} n$
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $3$