推 LimSinE:這樣沒有證到sup吧... 61.70.211.116 10/11
推 darkseer:確實...我只有給想法 218.175.186.186 10/11
※ 引述《pikahacker (死亡筆記 型電腦)》之銘言：
: 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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