※ 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

快推避免人家說我看不懂

0.1^n < ε => 10^n > ε 確定不是1/ε ??我是覺得中間的數學歸納法證的東西很怪 但也懶得看了這個證明方向應該也不是證10^n≧n就 隨便吧 我只是想點科普向 不想做學術研討雖然看得懂 但用全英文的標準寫法很勸退普通人就是了雖然知道這應該算是職業病了Q_Q

靠腰 這串滑下來 發現外面已經有鳥叫聲了 哭啊

有再嚴格的定義 即使你懂我懂 不接受的人還是不會接受 只要遇過就知道那你列的證明足不足以說服那些人就也很難說了

這裡是八卦(X

a=0.9bar 10a=9.9bar 互減變成9a=9 a=1