Binomial Coefficient with One

Theorem

 * $\displaystyle \forall r \in \R: \binom r 1 = r$

The usual presentation of this result is:
 * $\displaystyle \forall n \in \N: \binom n 1 = n$

Proof
From the definition of binomial coefficients:


 * $\displaystyle \binom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$

where $r^{\underline k}$ is the falling factorial.

In turn:


 * $\displaystyle x^{\underline k} := \prod_{j=0}^{k-1} \left({x - j}\right)$

But when $k = 1$, we have:
 * $\displaystyle \prod_{j=0}^0 \left({x - j}\right) = \left({x - 0}\right) = x$

So:


 * $\displaystyle \forall r \in \R: \binom r 1 = r$

This is completely compatible with the result for natural numbers:
 * $\displaystyle \forall n \in \N: \binom n 1 = n$

as from the definition:
 * $\displaystyle \binom n 1 = \dfrac {n!} {1! \ \left({n - 1}\right)!}$

the result following directly, again from the definition of the factorial where $1! = 1$.

Also see

 * Particular Values of Binomial Coefficients for other similar results.