April 24th, 2017, 12:17 PM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  Cantors Diagonal Argument, Logic
A $\displaystyle \rightarrow$ NotA

April 24th, 2017, 12:24 PM  #2 
Senior Member Joined: Aug 2012 Posts: 1,973 Thanks: 551  Now Zylo you went through CDA months ago. I already saw this movie. This isn't like cable tv where they play the same movies over and over. Post something new please. Power Set of the Reals is Countable Cantor's Diagonal Argument and Binary Sequences Power Set of the Natural Numbers is Countable 
April 24th, 2017, 04:21 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra  
April 25th, 2017, 11:51 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
I was trying to wriggle out of OP when I hit the mother lode: Every irrational number is the result of a unique definition by rational numbers. Therefore you can't have more definitions (irrational numbers) than rational numbers. In a sense, the irrationals exhaust the possibilities of what you can do with the natural numbers, which are thus the foundation of analysis and mathematics. Last edited by zylo; April 25th, 2017 at 11:57 AM. 
April 25th, 2017, 11:54 AM  #5  
Senior Member Joined: Aug 2012 Posts: 1,973 Thanks: 551  Quote:
Secondly, this has nothing to do with "definitions" but rather with sequences, of which there are uncountably many. Third, it does happen to be true that if we give "definition" a technical meaning, say firstorder definability, there are indeed many reals that don't have definitions. That's no argument against their existence. I'm certain there are a billion people in China even though I don't know all their names. And fourth, even that last bit is subject to sophisticated caveats, such as the fact that definability is itself NOT firstorder definable, so that the "not enough definitions" argument is itself murky. It's more clear if you talk about computability, but the point still holds. There are lots of reals that aren't computable, but those reals still exist. And fifth, you're acting like a crank. Why? Sometimes you don't act like a crank. You should strive for noncrankiness in my opinion. Last edited by Maschke; April 25th, 2017 at 12:00 PM.  
April 26th, 2017, 05:45 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
Decimals uniquely define irrational numbers: Decimal representation is unique Alternately: If Cantor's set of infinite binary digits is bounded, CDA doesn't work. If it is unbounded, CDA doesn't work because you can never get to the end of the enumeration. Last edited by zylo; April 26th, 2017 at 05:51 AM. 
April 26th, 2017, 06:13 AM  #7 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,610 Thanks: 550 Math Focus: Yet to find out.  
April 26th, 2017, 06:50 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra  Yes, there is a unique decimal that describes each irrational. (1) The number of nonrepeating infinite sequences of decimal digits is uncountably infinite. (2) That doesn't mean that there aren't more sequences of rationals that converge to that irrational. For example:

April 27th, 2017, 07:55 AM  #9 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
Ref previous post. (1) is CDA, the subject of this thread. You can't use CDA to prove CDA. (2) Each a_{n} is a rational number. Any rational combination of rational numbers is a rational number. The rational numbers are countable. 
April 27th, 2017, 08:38 AM  #10  
Senior Member Joined: Aug 2012 Posts: 1,973 Thanks: 551  Quote:
Please make your posts selfcontained if you want people to respond.  

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