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Anonymous Mon 29 Jul 23:54:06 2019 No.10847670 View Reply Original Report

Quoted By: >>10848210

Couldn't I use this fact:
$ 0.999... &= \lim_{n->\infty}1-10^{-n}\\ &=1-\lim_{n->\infty}10^{-n}\\ &=1\\ $
To say also that 1.000.....0001 = 1.0, and thus would also be equal to .999....
Which could be used to prove by induction that
$x=y\forall x,y \in \Bbb{R}$

Manitoba boi

Manitoba boi Tue 30 Jul 2019 01:11:30 No.10847841 Report

Quoted By:

Anonymous

Anonymous Tue 30 Jul 2019 03:26:12 No.10848197 Report

Quoted By:

1/9=0.111...
+
8/9=0.888...
=
9/9=0.999...

Anonymous

Anonymous Tue 30 Jul 2019 03:34:20 No.10848210 Report

Quoted By: >>10849113

>>10847670
>which could be used to prove by induction that...
sure, just exhibit for me a well ordering on the reals.

Anonymous

Anonymous Tue 30 Jul 2019 03:40:27 No.10848217 Report

Quoted By:

1=3/3=3*(1/3)=3*0.333...=0.999...

Anonymous

Anonymous Tue 30 Jul 2019 07:22:33 No.10848603 Report

Quoted By:

$ \boxed{0 1 = p + (1-p) ~~~~~~ \overset{1}{ \overbrace{[=====p=====|==(1-p)==]}} \\ \text{divide p using x} ~~~~~~ \overset{1}{ \overbrace{ \underset{p}{[ \underbrace{=====x=====|==(p-x)==}]} ~~ + ~~ (1-p)}} \\ \\ \text{solve x and (p-x), when length ratios must be the same} \\ \dfrac{x}{p-x}= \dfrac{p}{1-p} \Rightarrow x- xp = p^2 - xp \Rightarrow \underline{x=p^2} \Rightarrow \underline{(p-x)=p(1-p)} \\ \overset{1}{ \overbrace{ \underset{p}{[ \underbrace{=====p^2=====|==p(1-p)==}]} ~~ + ~~ (1-p)}} \\ \\ \overset{1}{ \overbrace{ \underset{p^2}{[ \underbrace{=====p^3=====|==p^2(1-p)==}]} ~~+ p(1-p)+(1-p)}} \\ \overset{1}{ \overbrace{ \underset{p^3}{[ \underbrace{=====p^4=====|==p^3(1-p)==}]} ~~+ p^2(1-p)+p(1-p)+(1-p)}} \\ (1-p)+p(1-p)+p^2(1-p)+p^3(1-p)+ \cdots =1 ~~~~ \left | ~ \times \frac{1}{1-p} \right . \\ 1+p+p^2+p^3+ \cdots = \dfrac{1}{1-p} $

Anonymous

Anonymous Tue 30 Jul 2019 13:39:31 No.10849113 Report

Quoted By:

>>10848210
What does that have to do with anything

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