The so called pigeon hole principle is nothing more than the obvious remark: 
if you have fewer pigeon holes than pigeons and you put every pigeon in a 
pigeon hole, then there must result at least one pigeon hole with more than 
one pigeon.
╔═════════════════════════════════════╗
║Pf:                                                                       ║
║                                                                          ║
║已知有m隻鴿子(pigeon)，n個鴿籠(hole)，m ＞ n，m ,n 為非零的自然數         ║
║                                                                          ║
║假設在同一個hole裡面不會有超過兩隻pigeon,                                 ║
║                                                                          ║
║∴在每個洞只會有0隻或是1隻pigeon  ==> 則n個hole總共有n × 1 ＝ n隻pigeon, ║
║                                                                          ║
║又∵每隻pigeon都在hole裡面了，                                            ║
║                                                                          ║
║∴ n ＝ m 和已知矛盾，故得証                                              ║
╚═════════════════════════════════════╝


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         說實在的，為什麼要搞這種trivial東西的証明啊？！ 浪費生命><"""


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※ 編輯: windperson      來自: 140.112.241.93       (09/26 02:03)