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 ║ ╚═════════════════════════════════════╝ 各位覺得這樣子的証明行嗎？ -- 說實在的，為什麼要搞這種trivial東西的証明啊？！ 浪費生命><""" -- ※ 編輯: windperson 來自: 140.112.241.93 (09/26 02:03)