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只鸽子(pigeon)，n个鸽笼(hole)，m ＞ n，m ,n 为非零的自然数 ║ ║ ║ ║假设在同一个hole里面不会有超过两只pigeon, ║ ║ ║ ║∴在每个洞只会有0只或是1只pigeon ==> 则n个hole总共有n × 1 ＝ n只pigeon, ║ ║ ║ ║又∵每只pigeon都在hole里面了， ║ ║ ║ ║∴ n ＝ m 和已知矛盾，故得证 ║ ╚═════════════════════════════════════╝ 各位觉得这样子的证明行吗？ -- 说实在的，为什麽要搞这种trivial东西的证明啊？！ 浪费生命><""" -- ※ 编辑: windperson 来自: 140.112.241.93 (09/26 02:03)