Re: [問題] 無限多的自然數跟質數誰比較多？

※ 引述《arrenwu (不是綿芽的錯)》之銘言： : 其實我們幫這些直覺翻譯一下，會得到下面這結果 : 定義數列 An = 0.999...99 (小數點後面n個9) : A1 = 0.9, A2 = 0.99, A3 = 0.999, ........ : : 0.9bar = lim An : n->∞ : 基於上面的描述，會得到 0.9bar = 1 : 不同意的，就叫他自己描述一下他心中的 0.9bar 是什麼樣子 : 如果對方無法定義自己心中的 0.9bar 卻還是堅持不等於1 .... : 可能是腦袋剛好打結了 : 讓他看一下角卷綿芽的直播舒緩一下吧 : https://youtu.be/l6rlIOetkwg (現正直播中) 應該就: n Σ9*0.1^k = 9*0.1(1-0.1^n)/(1-0.1) = 1-0.1^n k=1 (為了極限的定義確立證明目標: |1-0.1^n-1| = 0.1^n < ε => 10^n > 1/ε) Let S = {n in N | 10^n > 1/ε} Claim: 10^n≧n for all n in N. Proof: Basis step: When n = 1, 10 = 10^1≧1. The relation holds Inductive Step: Suppose when n = k, the relation holds Then when n = k+1, 10^(k+1) = 10*10^k≧10k(by induction hypothesis) ∵ 10k = k+9k ≧ k+1 ∴ 10^(k+1) ≧ k+1 The relation also holds for n = k+1 So, by induction, 10^n≧n for all n in N By Archimedian property, there exist an natural number n such that n = n*1 > 1/ε So, by the previous claim and Achimedian property, there exists a natural number n such that 10^n ≧ n > 1/ε holds. So, S is nonempty for every ε> 0 Now, we want to show that "for every ε > 0, there exists a natural number M such that if n > M, 0.1^n < ε" By Well-Ordering Principle, there exists a smallest positive integer M such that 10^M > 1/ε ∵ 10^n is increasing, 10^n≧10^M > 1/ε for all n > M => 0.1^n < ε for all n > M ∴ lim (1-0.1^n) = 1 n->∞ 0.999... = 1這件事可以用這個角度去看 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 114.47.92.24 (臺灣) ※ 文章網址: https://www.ptt.cc/bbs/C_Chat/M.1684354622.A.55D.html