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January 13th, 2016, 06:12 PM  #21 
Global Moderator Joined: Dec 2006 Posts: 20,972 Thanks: 2222 
You explicitly list various finite decimals, but those are immaterial, as it's undisputed that they're countable. Their countability doesn't imply that the infinite decimals are countable. You're effectively asserting that such an implication exists, but it doesn't, and a mere assertion doesn't constitute a proof. You could provide 3, 3.1, 3.141, 3.1415, 3.14159, . . . $\pi$, and I would be happy with that. However, that way of dealing with irrational numbers uses a subset of the rationals, and there are uncountably many subsets of the rationals, so you don't have a countable number of countable numbers with that approach. 
January 13th, 2016, 06:41 PM  #22 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra  I know what an infinite decimal is. Your problem is that every $n$ is finite. You appear not to understand that the notation $\sum \limits_{n=1}^\infty a_n$ (which is what you are effectively doing: $a_n = 10^n$) includes only finite values of $n$ because there is no infinite natural number. 
February 29th, 2016, 03:19 AM  #23 
Global Moderator Joined: Dec 2006 Posts: 20,972 Thanks: 2222  
February 29th, 2016, 06:24 AM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra  
February 29th, 2016, 08:46 AM  #25 
Member Joined: Aug 2014 From: Somewhere between order and chaos Posts: 44 Thanks: 5 Math Focus: Set theory, abstract algebra, analysis, computer science  

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