Binomial Coefficient is Integer

Theorem

Let $\dbinom n k$ be a binomial coefficient.

Then $\dbinom n k$ is an integer.


Proof 1

If it is not the case that $0 \le k \le n$, then the result holds trivially.

So let $0 \le k \le n$.

By the definition of binomial coefficients:

\(\ds \binom n k\) \(=\) \(\ds \frac {n!} {k! \paren {n - k}!}\)
\(\ds \) \(=\) \(\ds \frac {n \paren {n - 1} \paren {n - 2} \cdots \paren {n - k + 1} } {k!}\)

The numerator is a product of $k$ successive integers.

From Factorial Divides Product of Successive Numbers, $k!$ divides it.

$\blacksquare$


Proof 2

The result follows by Pascal's Rule and Integer Addition is Closed.

$\blacksquare$


Sources

  • 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): binomial coefficient $\text {(i)}$
  • 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): binomial coefficient $\text {(i)}$
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