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Apr 11 2010 10:35am
0.99999999999... is infinitely close to 1.

In other word, it's equal to it.
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Apr 11 2010 10:57am
Quote (Fr4D35 @ Apr 11 2010 11:35am)
0.99999999999... is infinitely close to 1.

In other word, it's equal to it.


well, the most mathematically correct answer would be to show :

lim (1-sum(9/10^n))^2 =0

in other words, show 1 minus the partial sum 0.9+0.09+0.009+... squared -> 0
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Apr 11 2010 11:09am
Quote (Taxidermy @ Apr 11 2010 11:57am)
well, the most mathematically correct answer would be to show :

lim (1-sum(9/10^n))^2 =0

in other words, show 1 minus the partial sum 0.9+0.09+0.009+... squared -> 0


Why is that the most mathematically correct answer? The number 1 is usually defined as the equivalence class of numbers with the same Dedekind cut as 1. It is obvious that .999... is in that equivalence class since for all epsilon > 0 1 - .9999.... < epsilon.
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Apr 11 2010 11:22am
Quote (Kahl4Prez @ Apr 11 2010 03:30pm)
this was the WIN wow man u put some effort into it, i was just havin fun...

thanks BTW


<3 Didn't see this before. However, I noticed at the end I goofed, I never established that the number wasn't 0, 'cause 0 shares the same property.

This post was edited by ElusiveNeutrino on Apr 11 2010 11:29am
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Apr 11 2010 12:01pm
Quote (darkfire @ Apr 11 2010 12:09pm)
Why is that the most mathematically correct answer? The number 1 is usually defined as the equivalence class of numbers with the same Dedekind cut as 1. It is obvious that .999... is in that equivalence class since for all epsilon > 0 1 - .9999.... < epsilon.


i suppose, but what i wrote is absolute convergence of the norm squared
so basically its (1-0.999...)^2 <epsilon
which is inherently a better argument than 1-0.999... < epsilon imo

but thats just me being picky probably :p

Quote (ElusiveNeutrino @ Apr 11 2010 12:22pm)
<3 Didn't see this before. However, I noticed at the end I goofed, I never established that the number wasn't 0, 'cause 0 shares the same property.


i see you've taken a bit of analysis :D

This post was edited by Taxidermy on Apr 11 2010 12:07pm
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Apr 11 2010 12:21pm
Quote (Taxidermy @ Apr 11 2010 01:01pm)
i suppose, but what i wrote is absolute convergence of the norm squared
so basically its (1-0.999...)^2 <epsilon
which is inherently a better argument than 1-0.999... < epsilon imo

but thats just me being picky probably :p



i see you've taken a bit of analysis :D


Formally, I should say |1- .999..| < epsilon to make what I wrote completely correct. My problem is that squaring it doesn't really tell you anything new. If you don't square it, what you have written is identically equal to what I have written, except that you take limits. The trouble pops up in the fact that if you don't formally construct the reals, limits aren't well defined. If you do formally construct them, then why bother taking limits at all: the answer pops out from the definitions immediately. Anyway, if you grab your copy of Rudin (I assume you took analysis with Rudin?) and look at the appendix of chapter 1, you'll see why I'm being so insistent on the above proof being the correct one.

Equivalently, we could also talk about the equivalence class of Cauchy sequences converging to 1 and use the triangle inequality on a rational approximation of .9999... I would agree that that is an equally correct "minimal" (in the sense that it does not assume any claim that could prove the result directly) answer.

This post was edited by darkfire on Apr 11 2010 12:26pm
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Apr 11 2010 12:31pm
the ^2 was taking the role of | |

i like using norm squared is all :D
altho i see what you are saying
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Apr 11 2010 01:38pm
This thread is pure evil and should be closed :wallbash:
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Apr 11 2010 02:26pm
Quote (Kahl4Prez @ Apr 11 2010 07:57am)
wow 2 people respond, and 2 people have a limited understanding of our mathematics...

what does 1/3 = ??????

ok now take 1/3 and x 3 = 1

1/3 = ????????

now multiply ????????? x 3 and what do u get?


.99999999...

there is my first fast proof...


It's stupid... but true.
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Apr 11 2010 03:13pm
0.99999999 = 0.999999999

1 = 1

fgts
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