※ 引述《arrenwu (不是绵芽的错)》之铭言:
: 其实我们帮这些直觉翻译一下,会得到下面这结果
: 定义数列 An = 0.999...99 (小数点后面n个9)
: A1 = 0.9, A2 = 0.99, A3 = 0.999, ........
:
: 0.9bar = lim An
: n->∞
: 基于上面的描述,会得到 0.9bar = 1
: 不同意的,就叫他自己描述一下他心中的 0.9bar 是什么样子
: 如果对方无法定义自己心中的 0.9bar 却还是坚持不等于1 ....
: 可能是脑袋刚好打结了
: 让他看一下角卷绵芽的直播舒缓一下吧
: https://youtu.be/l6rlIOetkwg (现正直播中)
应该就:
n
Σ9*0.1^k = 9*0.1(1-0.1^n)/(1-0.1) = 1-0.1^n
k=1
(为了极限的定义确立证明目标: |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这件事可以用这个角度去看