# e is irrational

We have the series

$${e}^{-1}=\sum _{k=0}^{\mathrm{\infty}}\frac{{(-1)}^{k}}{k!}$$ |

Note that this is an alternating series^{} and that the magnitudes of the
terms decrease. Hence, for every integer $n>0$, we have the bound

$$ |

by the Leibniz’ estimate for alternating series (http://planetmath.org/LeibnizEstimateForAlternatingSeries). Assume that $e=n/m$, where $m$ and $n$ are integers and $n>0$. Then we would have

$$ |

Multiplying both sides by $n!$, this would imply

$$ |

which is a contradiction^{} because every term in the sum is an integer,
but there are no integers between $0$ and $1/(n+1)$.

Title | e is irrational |
---|---|

Canonical name | EIsIrrational1 |

Date of creation | 2013-03-22 17:02:32 |

Last modified on | 2013-03-22 17:02:32 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 8 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 11J82 |

Classification | msc 11J72 |

Related topic | LeibnizEstimateForAlternatingSeries |