0
$\begingroup$

I have already looked at this proof here

Proving that $n \choose k$ is an integer

However I don't understand how I can use the Pascal identity for binommial coefficients if $n$ is a negative number.

I have had the idea to prove a Connection between the negative and the positive Counterpart.

I.e.

Maybe $\binom{n}{k}=-x\binom{-n}{k}$, where $x\in\mathbb{Z}$

I wrote down

$\frac{1}{k!}n(n-1)…(n-k+1)=x\cdot\frac{1}{k!}-n(-n-1)….(-n-k+1)$

What could be the $x$ ?

Thanks for reading.

$\endgroup$
  • $\begingroup$ $\frac{1}{k!}n(n-1)…(n-k+1)=\binom{n}{k}$ is the Definition for General binomial coefficients $\endgroup$ – New2Math Mar 18 at 14:20
3
$\begingroup$

For $n>0$ we have \begin{align} \binom{-n}k &=\frac{-n(-n-1)\cdots(-n-k+1)}{k!}\\ &=(-1)^k\frac{n(n+1)\cdots(n+k-1)}{k!}\\ &=(-1)^k\frac{(n+k-1)!}{k!(n-1)!}\\ &=(-1)^k\binom{n+k-1}{k} \end{align} hence it is an integer.

$\endgroup$
  • 2
    $\begingroup$ Can the downvoter please say what is wrong with this answer? $n>1$ for the case $n=0$ one can make a seperate Argument, Therefore. $n-1>0$ and a integer and therefore $n-1+k$ is a natural number. And then with the already known result the Statement is true $\endgroup$ – New2Math Mar 18 at 14:30
  • $\begingroup$ You can't write like this! $\endgroup$ – LAGRIDA Mar 18 at 14:32
  • $\begingroup$ The generale formula of factorial : $\Gamma(n+1) = n!$, for $n$ negative $n! = \pm \infty$ $\endgroup$ – LAGRIDA Mar 18 at 14:34
  • $\begingroup$ @LAGRIDA: in my answer, there are no $n!$ with $n<0$. $\endgroup$ – Fabio Lucchini Mar 18 at 14:37
  • $\begingroup$ but the definition : $n! = n (n-1) \cdots 2$ true iff n > 1 $\endgroup$ – LAGRIDA Mar 18 at 14:39
0
$\begingroup$

Alternative method:

Note that

$\binom{n}{k} = \binom{n-1}{k-1}+ \binom{n-1}{k} \space \forall n \in \mathbb{Z} \space \forall k \in \mathbb{N} \\ \Rightarrow \binom{n-1}{k} = \binom{n}{k}- \binom{n-1}{k-1} \space \forall n \in \mathbb{Z} \space \forall k \in \mathbb{N}$

Together with $\binom{n}{0}=1 \space \forall n \in \mathbb{Z}$ you can use this to prove by induction that $\binom{n}{k} \space 0>n\in\mathbb{Z}, k\in \mathbb{N}$ is the difference of two integers and so is always an integer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.