October 9th, 2018, 04:00 AM  #171 
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  Whether or not this is true (it isn't), it is irrelevant. The proof finds a difference between the string represented by what I called s(n), and the one produced by the diagonalization of s(*), at the known position n. And even if you think that means we have to examine positions 1 through n (it doesn't), we can end the comparison there without needing to compare every character in the strings.

October 9th, 2018, 04:25 AM  #172  
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007  Quote:
That's another assertion that's unsupported by any mathematical justification. The comparison you're alluding to can be done if both strings are nonterminating. If, for some finite value of $n$, the $n$th digit of the first string differs from the $n$ digit of the second, the strings differ, else they're the same.  
October 9th, 2018, 04:32 AM  #173 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  
October 9th, 2018, 04:35 AM  #174 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  
October 9th, 2018, 05:22 AM  #175 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007 
In relation to this discussion, all the nonterminating strings don't exist in your list (by definition), but still correspond to real numbers, so your assertion means that the reals are uncountable, and also that the set of all nonterminating strings is uncountable.

October 9th, 2018, 05:49 AM  #176 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra  That says more about you than it does about JeffJo's post.
Last edited by skipjack; October 9th, 2018 at 06:11 AM. 
October 9th, 2018, 05:55 AM  #177 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra  
October 9th, 2018, 06:07 AM  #178 
Senior Member Joined: Oct 2009 Posts: 752 Thanks: 257  
October 9th, 2018, 06:11 AM  #179  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
 
October 9th, 2018, 06:20 AM  #180 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra  Zylo claims that there are natural numbers that have infinite numbers of digits. I note that every (positive) natural number (in binary) $n$ has $\lfloor 1 + \log_2{n} \rfloor$ digits this is also a natural number. I note that $\lfloor 1 + \log_2{n} \rfloor \lt n$ for all natural numbers $n$. By the definition of natural numbers, every natural number has a successor $(n+1)$ and this defines $(n+1)$. Thus every natural number $n$ (apart from one of them  zero or one depending on your view) is the successor of a natural number $(n1)$. So, according to $Zylo$'s assertion, there is a smallest infinite natural number $i$. What do we know about this infinite $i$? We know that $i1$ is a finite natural number. We know that $i$ has $\lfloor 1 + \log_2{i} \rfloor$ digits, and that this number is less than $i$ and is thus finite. Thus there is a natural number that represents a finite sum $\displaystyle \sum_{k=1}^{\lfloor 1 + \log_2{i} \rfloor}a_i2^{k1}$ which has an infinite value. Zylo please give an example of such a finite sum having an infinite value. 

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