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October 9th, 2018, 04:00 AM   #171
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Quote:
 Originally Posted by zylo You can't compare infinite strings unless you can get to the end.
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
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Quote:
 Originally Posted by zylo . . . you can't get to the end of increasing n. The end of increasing n and beyond are figments of the imagination.
They remain just figments of the imagination because you haven't mathematically defined them.

Quote:
 Originally Posted by zylo To compare two strings, they have to be of the same length.
That's another assertion that's unsupported by any mathematical justification. The comparison you're alluding to can be done if both strings are non-terminating. 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
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Quote:
 Originally Posted by JeffJo My translation of Cantor's argument: ………….…..
Just referencing your argument in case someone wants to try and follow it. Personally, I found it unintelligible.

October 9th, 2018, 04:35 AM   #174
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Quote:
 Originally Posted by skipjack They remain just figments of the imagination because you haven't mathematically defined them.
I agree, the set of things that don't exist is uncountable.

 October 9th, 2018, 05:22 AM #175 Global Moderator   Joined: Dec 2006 Posts: 20,753 Thanks: 2136 In relation to this discussion, all the non-terminating 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 non-terminating strings is uncountable.
October 9th, 2018, 05:49 AM   #176
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Quote:
 Originally Posted by zylo Just referencing your argument in case someone wants to try and follow it. Personally, I found it unintelligible.

Last edited by skipjack; October 9th, 2018 at 06:11 AM.

October 9th, 2018, 05:55 AM   #177
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Quote:
 Originally Posted by zylo But you can't define it for a random string in |N because you can't get to the end of it, unless you are only defining it for finite or repetitive strings.
A non-repeating infinite string that is perfectly well defined "to the end":
1 0 11 0 111 0 1111 0 11111 0 ...

October 9th, 2018, 06:07 AM   #178
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Quote:
 Originally Posted by zylo You can't compare infinite strings unless you can get to the end.
Wait. So I got the following two strings:

1,1,1,2,1,1,1,...
1,1,1,1,1,1,1,...

You say it is impossible to say these two strings are distinct?? Even though you clearly see the difference?

October 9th, 2018, 06:11 AM   #179
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Quote:
 Originally Posted by skipjack In relation to this discussion, all the non-terminating 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 non-terminating strings is uncountable.
333....... occurs in the list of natural numbers no matter how many digits you have because the list of natural numbers doesn't end.

 October 9th, 2018, 06:20 AM #180 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2643 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 $(n-1)$. 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 $i-1$ 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 $\displaystyle \sum_{k=1}^{\lfloor 1 + \log_2{i} \rfloor}a_i2^{k-1}$ which has an infinite value. Zylo please give an example of such a finite sum having an infinite value.

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