Binomial Coefficient with Zero

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Theorem

$\forall r \in \R: \dbinom r 0 = 1$


The usual presentation of this result is:

$\forall n \in \N: \dbinom n 0 = 1$


Proof

From the definition of binomial coefficients:

$\dbinom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$

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

In turn:

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

But when $k = 0$, we have:

$\displaystyle \prod_{j \mathop = 0}^{-1} \left({x - j}\right) = 1$

as $\displaystyle \prod_{j \mathop = 0}^{-1} \left({x - j}\right)$ is a vacuous product.

From the definition of the factorial we have that $0! = 1$.

Thus:

$\forall r \in \R: \dbinom r 0 = 1$

$\blacksquare$


Integer Coefficients

This is completely compatible with the result for natural numbers:

$\forall n \in \N: \dbinom n 0 = 1$


From the definition:

\(\displaystyle \binom n 0\) \(=\) \(\displaystyle \frac {n!} {0! \ n!}\) Definition of Binomial Coefficient
\(\displaystyle \) \(=\) \(\displaystyle \frac {n!} {1 \cdot n!}\) Definition of Factorial of $0$
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$


Also see


Sources