February 26th, 2018, 08:20 PM  #31  
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 246  Quote:
Last edited by skipjack; February 27th, 2018 at 01:49 AM.  
February 26th, 2018, 08:52 PM  #32 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
It's not his only problem. As you can see, he doesn't understand how there can be infinitely many finite integers. I think it's because he won't let his preconceived ideas go and just follow what the logic says.

February 26th, 2018, 10:35 PM  #33 
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 246  I don't understand how you cannot understand that... But I don't really want to know the answer...

February 27th, 2018, 02:09 AM  #34 
Global Moderator Joined: Dec 2006 Posts: 20,269 Thanks: 1958  In that statement, is "n" allowed to be infinite? If so, that should be stated at the outset, so as to avoid confusion.

February 27th, 2018, 08:21 AM  #35 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
n is never allowed to be infinite, it is only endless. But if that's what you mean, fine. I believe this is what's confusing some people: Let Sn =.999..9, nplaces, by definition. $\displaystyle \lim_{n \rightarrow \infty}$Sn IS NOT, NOT 1, because of definition. Flim = 1, for Function limit. $\displaystyle \lim_{n \rightarrow \infty}$Sn = .999999999999......, by definition. Call it Dlim for Definition limit. Then Flim does not belong to {Sn} but Dlimit does. That, by the way, is why you talk about the decimals in [0,1), not [0,1]. Dlimit defines the real numbers. Last edited by zylo; February 27th, 2018 at 08:26 AM. 
February 27th, 2018, 09:41 AM  #36 
Global Moderator Joined: Dec 2006 Posts: 20,269 Thanks: 1958  
February 27th, 2018, 11:18 AM  #37 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
Do you remember at school when you were about 6 years old and you were taught that 123 means one hundred, two tens and three units? Similarly $0.999\ldots9$ to $n$ places means $\sum \limits_{k=1}^n \frac9{10^k}$ and the infinite decimal $0.999\ldots$ means $$\sum \limits_{k=1}^\infty \frac9{10^k} = \lim_{n \to \infty} \sum \limits_{k=1}^n \frac9{10^k}$$. Your Dlimit $S$ clearly isn't a member of $\{S_n\}$ because all of the $S_n$ terminate (after $n$ digits). $S$ does not. Last edited by v8archie; February 27th, 2018 at 11:39 AM. 
February 27th, 2018, 03:36 PM  #38  
Senior Member Joined: Aug 2012 Posts: 2,156 Thanks: 630  Quote:
Quote:
This example is the basis for professor Katz's article, "A Strict Inequality .999... < 1," often referenced (by people who have read the title but not the article) as a valid argument that .999... = 1 is false. However if one reads professor Katz's article in detail, which I did at the time I got interested in all this, Katz does in fact admit that there is a final 9 at some hyperinteger index, after which all the remaining digits are zero. And that he is making more of a pedagogical point than a mathematical one. My understanding is that by the transfer principle, .999... = 1 must be a theorem of the nonstandard reals just as it is in the standard reals. So this particular example you gave is misleading (although of course it's true) and only gives ammunition to the .999... cranks. In fact I'm not sure I understand your remark about the "limit of the expression" being less than 1. Can you elaborate on that please? I am not sure that's correct. My understanding is that the limit of the expression is not a hyperreal at all. In fact this is an example of a nonempty set of hyperreals bounded above that lacks a least upper bound. The hyperreals are not topologically complete and this is an example of that problem to the best of my understanding. Quote:
All is clear now. This is a very enlightening example. Thanks for bringing it up. Quote:
Last edited by Maschke; February 27th, 2018 at 03:47 PM.  
February 27th, 2018, 04:18 PM  #39  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra  Quote:
Micrm@ss, being apparently my sort of purist, doesn't much like the fudge and prefers to consider analysis in the hyperreals divorced from real analysis. There may be a problem here as I understand that basic arithmetic breaks down in the hyperreals, but qualitative results are possible. More on this later. I'm far from convinced of that. For one thing, I'm unaware of any infinitesimal being quantified in any way that would make sense under a positional number system. Once you get past the integer indexed places, each "digit" could take any finite value. I'm prepared to believe that the coefficient of all higher powers of $\epsilon$ is zero (indeed, that probably ought to come out as part of the limit calculation).  
February 27th, 2018, 05:24 PM  #40  
Senior Member Joined: Aug 2012 Posts: 2,156 Thanks: 630  Quote:
The Wiki article illustrates my point that the 9's end at some particular hyperinteger position H. You'll also note that @Micrm@ss notated his number as 0.999...9. The lack of dots at the end is significant. Although I'm not sure if Micrm@ss was referring to the Lightstone notation. Lightstone notation has a semicolon in it so I'm not really sure what Micrm@ss intended here. Quote:
What is $\epsilon$ in this context? It's certainly not the "smallest infinitesimal" or anything like that. The hyperreals are a field so there is no smallest positive hyperreal, standard or infinitesimal. Last edited by Maschke; February 27th, 2018 at 05:29 PM.  

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