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January 11th, 2016, 11:17 PM  #1 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  The Real Numbers are Countable
1) Every real number can be represented by a decimal construction: A real number is defined by a cut of the rationals, each rational has a decimal representation. 2) The number of decimals is countable: I can count the number of decimal places and the number of numbers each count represents: 1 > 10, countable 2 > 100, countable ......... n > 10^n, countable. The number of decimals is countable because a countable number of countable numbers is countable. 3) The real numbers are countable. 
January 12th, 2016, 04:35 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
No. 1) How do you intend to prove that the number of cuts is countable? 2) We know that the number of reals is not countable, because it has been proved to be not countable. 3) If there were a valid proof that the reals are countable, then everything would be true, including 0=1. Last edited by v8archie; January 12th, 2016 at 04:39 AM. 
January 12th, 2016, 07:59 AM  #3 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
v8archie, 1) Every real is a cut is a decimal. The decimals can be counted. > The reals can be counted. 2) Then obviously whatever proof you are referring to is wrong. Can you give the proof you refer to? Maybe we can find the problem. 3) The reals are countable implies 0=1? 
January 12th, 2016, 07:01 PM  #4 
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 630  What do you mean, the "decimals" can be counted? The decimals are the real numbers. They're provably uncountable. If you are only referring to the finitelength decimals, those are countable, but you don't even have familiar rationals like 1/3.

January 12th, 2016, 07:11 PM  #5  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra  Quote:
2) You may have heard of Cantor. If not, it only serves to show that you have no idea what you are talking about. Don't bother trying to disprove it, you can't. The greatest minds of the last two centuries have thought it through more deeply than you can, and they have all been convinced or have died trying to find a flaw in the argument. 3) Yes. If we have two contradictory proofs that are both correct, then everything is true including 0 = 1. Last edited by v8archie; January 12th, 2016 at 07:15 PM.  
January 13th, 2016, 09:15 AM  #6  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
I gave a very explicit proof, using accepted principles, that the decimals can be counted. See OP. https://en.wikipedia.org/wiki/Cantor...gonal_argument The wiki article gives Cantor's "proof" that a binary sequence is uncountable, which I find obtuse and contorted. Then the wiki author "proves" that every real number can be expressed as a binary sequence, but doesn't define real number. A binary sequence, by definition, is a rational number. It then follows from Cantor's proof that: a) The rational numbers are uncountable. b) Every real is a rational number With that, I have exposed the flaw in my own argument: By definition, a decimal is a rational number. So an irrational number can't be expressed as a decimal, just approximated by it. And now we have another proof that Cantor's proof that a binary sequence is uncountable is wrong: Every binary sequence is a rational number and every rational number is a binary sequence. If Cantor's proof is correct, the rational numbers can't be counted. And then his argument that reals can't be counted fails. Last edited by skipjack; February 29th, 2016 at 03:53 AM.  
January 13th, 2016, 10:00 AM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
What unmitigated nonsense! Quote:
Quote:
Quote:
The remainder of your post is also nonsense. Not least because Cantor's diagonal argument says nothing about rationals. Last edited by skipjack; January 13th, 2016 at 01:27 PM.  
January 13th, 2016, 11:58 AM  #8 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
Assumption: The irrational numbers can be represented by nonterminating nonrepeating decimals. Can the number of irrationals be counted? By definition of decimal and OP, yes. Can we please stick to OP? The proof is either correct or incorrect. It is transparent and selfexplanatory. If you disagree with it, can you point out why? The fact that someone else arrives at a different conclusion does not invalidate a proof. 
January 13th, 2016, 01:32 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
The OP has nothing to do with irrationals because it talks about decimals of finite length. And no, they can't. What makes your "proof" wrong is that it only talks about a small fraction of the real numbers it's a proper subset of the rationals having a denominator with 2 and 5 as the only (possibly repeated) prime factors. 
January 13th, 2016, 01:41 PM  #10  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
2 > 100, countable ......... n > 10^n, countable .......... Thought it was obvious that we are counting 1,2,...n,...., the way you do in any countability proof. The period doesn't belong there. Thanks for pointing that out v8archie.  

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