# $\left(\genfrac{}{}{0pt}{}{n}{r}\right)$ is an integer

###### Theorem 1.

For $n\mathrm{\ge}r\mathrm{\ge}\mathrm{0}$, the binomial coefficient^{}

$$\left(\genfrac{}{}{0pt}{}{n}{r}\right)$$ |

is an integer.

###### Proof.

The proof is by induction^{} on $n$. For $n=0$, the
claim is clear. Thus, suppose the claim holds for $n\ge 1$.
For $r=1,\mathrm{\dots},n$, Pascal’s rule gives

$$\left(\genfrac{}{}{0pt}{}{n+1}{r}\right)=\left(\genfrac{}{}{0pt}{}{n}{r}\right)+\left(\genfrac{}{}{0pt}{}{n}{r-1}\right).$$ |

That is, $\left(\genfrac{}{}{0pt}{}{n+1}{1}\right),\mathrm{\dots},\left(\genfrac{}{}{0pt}{}{n+1}{n}\right)$ are integers. Since

$$\left(\genfrac{}{}{0pt}{}{n+1}{0}\right)=1,\left(\genfrac{}{}{0pt}{}{n+1}{n+1}\right)=1$$ |

the proof is complete^{}.
∎

Title | $\left(\genfrac{}{}{0pt}{}{n}{r}\right)$ is an integer |
---|---|

Canonical name | nchooseRIsAnInteger |

Date of creation | 2013-03-22 15:02:01 |

Last modified on | 2013-03-22 15:02:01 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 6 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 11B65 |

Classification | msc 05A10 |