※ 引述《pikahacker (死亡笔记 型电脑)》之铭言： : d irrational -> supremum of the set is 1/2. : Seems true enough, but is there a theorem behind this? How do you prove it? Lemma: Let d be an irrational number. For every E>0, there is a positive integer n such that either {nd}<=E or {-nd}<=E. Proof: I shall prove that for every E=1/2^n. Clearly it is true for E=1/2. Use induction on n. If there is n such that {nd}<=E, let m be the smallest positive integer such that m{nd}>=1. Then either {mnd}<=E/2. or {-(m-1)nd}<=E/2. If there is n such that {-nd}<=E, let k be the smallest positive integer such that k{-nd}>=1. Then {knd}<=E. From above we know there is a m such that either {mknd}<=E/2 or {-(m-1)knd}<=E/2 Q.E.D. Note: draw a picture and it will become easy to understand. Note that the proof makes use of the fact that {nd}>0. --

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