作者yusd24 (鄉長)
看板Math
標題Re: [分析] Apostol的一題習題
時間Tue Sep 1 12:12:13 2009
※ 引述《testishard (Testishard)》之銘言:
: Apostol習題1.16這一題要把習題1.15當成定理來應用
: 所有在請習題1.16之前,我先把習題1.15敘述一遍
: 習題1.15告訴我們:Given a real number x and a povistive integer N.
: Then there are integers h and k, with 0 < k ≦ N, such that
: |kx-h| < 1/N
: 習題1.16:If x is irrational prove that there are infinitely many
: numbers h/k with k>0 such that |x-h/k| < 1/k^2.
: 題目有給 Hint: Assume there are only a finite number h_1/k_1,…,h_r/k_r
: and obtain a contradiction by applying Ex1.15 with N > 1/δ
: where δ is the smallest of the numbers |x-h_i/k_i|
: 我的問題是卡在Hint…,根據Ex1.15告訴我們,N決定h和k,但hint卻叫我們再用h
: 和k來決定N. 尤於一直覺得很怪所以我就偷看了一下解答…(好孩子不要學)
: 解答是: 前面略… , 1/N < δ ≦ |x-(h_i/k_i)| < 1/(k_i*N)
: 1/N < 1/(k_i*N) ---><---
: 但是還是沒有解決我的困惑…Ex1.15明明告訴我們是用N來決定h和k,但hint卻告訴
: 我們再用h和k來決定N,這不是類似循環論證嗎…這種推論怎麼能夠成立??
: 或是我理解錯誤,思考上有盲點,請眾強者版友們為我解惑一下,感激不盡…
: <(_ _)>
Suppse not. There are only h_1/k_1,…,h_r/k_r satisfying the condition.
Choose N > 1/δ. Applying 1.15. There are p,q such that
|qx-p| < 1/N and 0 < q ≦ N.
Either p/q is one of the h_1/k_1,…,h_r/k_r, then we are done since
1/N < δ ≦ |x-(h_i/k_i)| < 1/(k_i*N) for some k_i. Thus,
1/N < 1/(k_i*N) ---><---
or
|qx-p|>=1/q by our assupmtion. (|x-p/q|>=1/q^2)
but then 1/N > |x-p/q| >= 1/q contradicts with the fact q ≦ N.
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推 testishard :謝謝啦 09/01 13:41
推 testishard :這樣的證法我就完全可以接受,每個點都被補完了 09/02 01:37