# e is irrational

From the Taylor series^{} for ${e}^{x}$ we know the following equation:

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

Now let us assume that $e$ is rational. This would there are two natural numbers^{} $a$ and $b$, such that:

$$e=\frac{a}{b}.$$ |

This yields:

$$b!e\in \mathbb{N}.$$ |

Now we can write $e$ using (1):

$$b!e=b!\sum _{k=0}^{\mathrm{\infty}}\frac{1}{k!}.$$ |

This can also be written:

$$b!e=\sum _{k=0}^{b}\frac{b!}{k!}+\sum _{k=b+1}^{\mathrm{\infty}}\frac{b!}{k!}.$$ |

The first sum is obviously a natural number, and thus

$$\sum _{k=b+1}^{\mathrm{\infty}}\frac{b!}{k!}$$ |

must also be . Now we see:

$$ |

Since $\frac{1}{b}\le 1$ we conclude:

$$ |

We have also seen that this is an integer, but there is no integer between 0 and 1. So there cannot exist two natural numbers $a$ and $b$ such that $e=\frac{a}{b}$, so $e$ is irrational.

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

Canonical name | EIsIrrational |

Date of creation | 2013-03-22 12:33:02 |

Last modified on | 2013-03-22 12:33:02 |

Owner | mathwizard (128) |

Last modified by | mathwizard (128) |

Numerical id | 13 |

Author | mathwizard (128) |

Entry type | Theorem |

Classification | msc 11J82 |

Classification | msc 11J72 |

Related topic | ErIsIrrationalForRinmathbbQsetminus0 |

Related topic | NaturalLogBase |