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March 29th, 2014, 10:04 AM  #11 
Senior Member Joined: Nov 2013 Posts: 160 Thanks: 7 
I already mentioned a book by Rudy Rucker: Infinity And The Mind. He is a mathematician and writes on his book page 79 that the number K with decimal expansion 0.99999... is the same as 1, his argument is 10K=9.99999....  K=0.99999....  9K =9 K =1 He continues "maybe this argument is misleading. What if there is some number call it 1  1/ω, that is greater than any finite string of 0.9.....9 of nines, yet less than 1? If K were equal to 11/ω, there is infinitesimal quantity that does not get canceled out, and the above argument does not work": 10K=10  10/ω  K= 1  1/ω  9K = 9  9/ω K = 1  1/ω Last edited by TwoTwo; March 29th, 2014 at 10:18 AM. 
March 29th, 2014, 08:54 PM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra 
You are creating a sequence where If the sequence approaches a limit, L, we have So L = 0 or L = 1 So Thus the sequence is bounded above. So Thus the sequence is increasing. Since the sequence is both bounded above and increasing, it must converge to a limit. And we showed above that the limit must be 1. However, since the sequence is increasing, we can say that there is no N such that Last edited by v8archie; March 29th, 2014 at 08:59 PM. 
March 30th, 2014, 06:44 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
In general, the successive values are approximate, so when the calculator internally obtains, say, 0.999999999896, and cannot display that many digits, it may well display 1 instead. Eventually, even the internally calculated value may become exactly 1.


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