※ 引述《beethoven88 (qq)》之銘言：
: Pf:
: for all n belong to N with n = k (mod 2 pi) ,then k is dense to 0 or 2pi
: thx a lot ~
Put S={ [n/(2π)] | n belongs to N| } ㄈ [0,1), where [a] denotes the largest
integer among those smaller than a. It then suffices to show that infS = m = 0.
Suppose the contrary: m > 0. Now that 0 < m < 1, for some positive integer t
0 ≦ [tm] < m. By definition, there exists n_0 in N| such that
m ≦ [n_0/(2π)] < m + (m-[mt])/t,
by which follows that
[tn_0/(2π)] < m.
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◆ From: 220.140.101.199
※ 編輯: ppia 來自: 220.140.101.199 (09/26 22:20)