# proof of Veblen’s theorem

The proof is very easy by induction^{} on the number of elements of
the set $E$ of edges.
If $E$ is empty, then all the vertices have degree zero, which is even.
Suppose $E$ is nonempty.
If the graph contains no cycle, then some vertex has degree $1$, which is odd.
Finally, if the graph does contain a cycle $C$, then every vertex has
the same degree mod $2$ with respect to $E-C$, as it has with respect
to $E$, and we can conclude by induction.

Title | proof of Veblen’s theorem |
---|---|

Canonical name | ProofOfVeblensTheorem |

Date of creation | 2013-03-22 13:56:51 |

Last modified on | 2013-03-22 13:56:51 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 4 |

Author | mathcam (2727) |

Entry type | Proof |

Classification | msc 05C38 |