October 3rd, 2018, 07:52 AM  #91  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Quote:
 
October 3rd, 2018, 07:58 AM  #92 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry  I'm glad we can finally agree  every entry of the list is finite. (Of course the list has no end, this was never in question.) If you add a binary radix point in front of a finite string of ones and zeros, you get a rational number. So no irrational number arises from your list. Not even all the rational numbers in [0,1) show up (see AplanisTophet's comment to see which ones do). Any irrational number in [0,1) or any rational number not expressible in the form $\frac{a}{2^b}$ with $a, b$ natural will do. As you admitted, every element of your list is finite. So no infinite string is in your list. Last edited by skipjack; October 3rd, 2018 at 09:15 AM. 
October 3rd, 2018, 08:08 AM  #93  
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 42  Quote:
If you can show me an element of the list that is not a dyadic rational, I'll give you my sailboat. PS  Get a sailboat.  
October 3rd, 2018, 08:37 AM  #94 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
What you say is true IF the list is finite (ends) it doesn't. .33333..... is a nested sequence which by a theorem of analysis zeroes in on one point of the real line. Want to guess what it is? You're spending too much time with the sailboat. 
October 3rd, 2018, 08:53 AM  #95  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry  Yes, and it's also true if the list is infinite (which it is). If $x$ is a binary string in your list (yes, any of the infinitely many of them), then its corresponding number in [0,1) must be rational. I have proven this by induction. With not much extra work, you could prove by induction that every element of your list actually corresponds to a dyadic rational number. Quote:
Also, what you've said here is nonsense. $.3333...$ is not a sequence, but is instead a real number. It's by definition the limit of the sequence $.3, .33, .333, ...$. Last edited by cjem; October 3rd, 2018 at 09:16 AM.  
October 3rd, 2018, 09:32 AM  #96  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Quote:
3) It defines a sequence of closed nests each one containing the subsequent one, and there is only one number common to all. Apparently there is the inability to comprehend that any particular natural number is "finite," but the list of natural numbers doesn't end. If you have a problem with that, you will just keep repeating the same thing over and over and over again, and compensate for your inability to comprehend with insults.  
October 3rd, 2018, 10:36 AM  #97  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry  I agree. Fortunately my proof works even though the list doesn't end. That's the power of induction for you! Quote:
I have no problem with this whatsoever. I wholeheartedly agree. Last edited by cjem; October 3rd, 2018 at 11:04 AM.  
October 3rd, 2018, 10:55 AM  #98 
Senior Member Joined: Aug 2012 Posts: 2,322 Thanks: 715  
October 3rd, 2018, 11:41 AM  #99  
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  Quote:
If you disagree that every entry is finite, please ignore the previous request and instead give an example of an entry that isn't finite (preferably the first such entry in the list), stating where exactly it occurs in the list.  
October 3rd, 2018, 11:47 AM  #100 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  

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binaryexpressed, cardinality, continuum hypothesis, diagonal argument, numbers, real, set 
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