April 8th, 2016, 12:49 PM  #1 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Real Number as Binary Sequence
Binary sequence: r =.011.... = p1×(1/2)^1 + p2×(1/2)^2 + p3×(1/2)^3 + ...... Let r be in L= [0,1) 1) divide L by 2 p1=0 if r in L1=[0,1/2], r1 = .0 2) divide L1 by 2 p2 =1 if r in [1/4,1/2], r2 = .01 3) divide L2 by 2 p3=1 if r in [3/8,1/2], r3 = .011 ...... r=.011......... = lim rn The digits in r are countable (countably infinite). Last edited by skipjack; April 8th, 2016 at 07:10 PM. 
April 8th, 2016, 12:55 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra 
Of course the digits in a real number are countably infinite! They are indexed by the natural numbers in the form of the indices on the place values $2^{1},2^{2},2^{3},\ldots$. At least you've got something right for once. The only problem is that you think it's some sort of groundbreaking discovery, which probably means there's something you are misunderstanding and that this will be followed by something that doesn't make any sense at all. I'm not sure that your proof makes an awful lot of sense though. Last edited by skipjack; April 8th, 2016 at 06:47 PM. 
April 8th, 2016, 01:39 PM  #3  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Quote:
1) A real number can be expressed as a countably infinite series of binary digits? 2) How to find the digits? See OP for the answers. I note that the set of all countably infinite binary sequences enumerates into a countably infinite list, not a finite one. EDIT: The process in OP also shows that every countably infinite binary sequence is a real number. Last edited by skipjack; April 8th, 2016 at 07:06 PM.  
April 8th, 2016, 04:49 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,764 Thanks: 697  Quote:
Last edited by skipjack; April 8th, 2016 at 07:07 PM.  
April 8th, 2016, 05:56 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra 
So you can do division. Well done. You are wrong to say that the process shows that every infinite binary sequence can be identified with a real number because your process doesn't produce every infinite binary sequence. In particular, all those ending with an infinite string of 1s are missing. Even more specifically 1111… is missing because 0.1111…= 1 which is specifically omitted from the interval from which L is taken. This thread is a mixture of the blindingly obvious and the wrong (mostly where the obvious turns out to be false). 
April 8th, 2016, 07:38 PM  #6  
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  Quote:
Quote:
It's mostly the other way round  they are mostly present, whereas .1000000... is missing.  
April 8th, 2016, 07:53 PM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra  No, he said is. Quote:
 
April 8th, 2016, 08:10 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
But zylo didn't say "is a unique real number".

April 8th, 2016, 08:19 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra 
No, but he's not generating every sequence even though he is using every real number. So the process doesn't show that every sequence is a real number.


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