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

--

靠邀，我真的寫錯了... 改過來 那個集合也應該是要找>1/ε

那個只是中間的一個結果，我不知道哪裡怪...

五、如果是堅稱0.9bar 無論如何都比 1 少一點點 你列的這個問題，我個人覺得如果沒有極限的嚴格定義，實在講不清楚這個問題的點在哪