Re: [問題] 徵求數學高手~~

[JGU (Karnaugh Map)]

Another solution(provided by LimSinE)

We want to prove for any ε > 0 there exists δ > 0 such that
if 0 < |x-a| < δ then |f(x)-0| = f(x) < ε    [ note that f(x)≧0 for all x]

For a given ε, first let δ1 = 1/(2N^2) where N is a positive integer
such that N > 1/ε                That is, 1/N < ε

If for every x satisfied 0 < |x-a| < δ1  =>  f(x) ≦ 1/N < ε
     Then we can simply let δ = δ1  and finish our work.

Otherwise, there exists a rational number x1
     such that 0 < |x1 - a| < δ1 and f(x1) > 1/N. Then let δ2 = |x1 - a|

            If for every x satisfied 0 < |x-a| < δ2  =>  f(x) ≦ 1/N < ε
               Then we can simply let δ = δ2  and finish our work.

            Otherwise, there exists a rational number x2 such that
               0 < |x2 - a| < δ2 = |x1 - a| and f(x2) > 1/N
               Now we have 0 < |x2 - a| < |x1 - a| < δ1 = 1/(2N^2)
                           and f(x1) > 1/N , f(x2) > 1/N
               However, this case is impossible.        ..........(*)

Since we examined all the possibilities, we have proved this theorem #

(*) Assume that x1 = p1/q1 , x2 = p2/q2
    p1,p2 are integers and q1,q2 are positive integers s.t. (p1,q1)=(p2,q2)=1

       0 < |x2 - a| < |x1 - a| < 1/(2N^2)
    => |x1 - x2| ≦ |x2 - a| + |x1 - a| < 1/N^2 ,   x1 ≠ x2
    => |x1 - x2| = |(p1/q1)-(p2/q2)| =  | p1q2 - p2q1 | / q1q2  ≧ 1/q1q2
    => 1/q1q2 ≦ |x1 - x2| < 1/N^2 => N^2 < q1q2
       f(x1) > 1/N , f(x2) > 1/N
    => 0 < q1 < N, 0 < q2 < N        ( because of the definition of f(x) )
    => q1q2 < N^2    lead to a contradiction #

※ 引述《JGU (逗左有囉西咕)》之銘言：
※ 引述《wanderer (哇拉拉~)》之銘言：
:       1/q ,當x=p/q屬於有理數 (p,q)=1 ,q>0
: f(x)=
:       0   ,當x屬於無理數,不屬於有理數時
: 試證明:
: 在a屬於有理數時, f(x)在x=a不連續
: 在a不屬於有理數時, f(x)在x=a連續

只要證明 lim f(x) = 0   對於所有的 a 就好了
         x→a

給定 ε > 0 , 則存在正整數 n 使得 n > 1/ε ....(*)

令 S = { |(r/m) - a| |m為不大於n的正整數,r為整數,且 a-1 ≦ r/m ≦ a+1 }

設 S 中非 0 的最小元素為δ, 則 δ > 0  ....(**)

若 0 < |x-a| < δ  =>  |x-a| 不屬於 S  =>  |f(x)-0| = f(x) < 1/n < ε  #

(*)  這裡用到阿基米得定理,不過應該沒關係
(**) 這裡可能要說明 S 是非空有限集而且不是 {0}

最後一步 f(x) < 1/n 是因為 x 不屬於 {r/m | m為不大於n的正整數,r為整數}
所以若 x = p/q 是有理數,則 q > n   =>   f(x) = 1/q < 1/n
若 x 是無理數,則 f(x) = 0 < 1/n
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→ chericherub:謝謝你們~等我看懂,我就要跟教授ㄠ滿分XD
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◆ From: 61.230.30.231
→ LimSinE:要說一下x1、x2是有理數因為f(x1)f(x2)>0
→ JGU:對耶我都忘了....這篇實在是打太久了^^"
※ 編輯: JGU             來自: 61.230.26.29         (10/25 22:34)
→ chericherub:我看到推文才發現...原來是我問的問題^^""