My Math Forum > Math Cardinality of the set of binary-expressed real numbers
 User Name Remember Me? Password

 Math General Math Forum - For general math related discussion and news

 October 9th, 2018, 06:34 AM #181 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 The natural numbers don't end.
October 9th, 2018, 06:39 AM   #182
Global Moderator

Joined: Dec 2006

Posts: 20,375
Thanks: 2010

Quote:
 Originally Posted by zylo 333....... occurs in the list of natural numbers no matter how many digits you have because the list of natural numbers doesn't end.
Each string in that list is, by definition, identified by a finite integer, whereas the non-terminating string 333....... isn't. As zylo has just posted, the list of natural numbers doesn't end, and that's true even if non-terminating strings are explicitly excluded, so there's nowhere where non-terminating numbers can exist in that list, and there is no "end" of that list after which they could be added.

October 9th, 2018, 07:13 AM   #183
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 124

It/s really quite simple

Quote:
 Originally Posted by zylo ZA!: All natural numbers are on a list increasing by 1. (Peano) ZA2: Any real number [0,1) is on the list because it is a natural number with a radix point. That serves for all of math except set theory, which is really the mathematics of words and symbols, which can only be defined in terms of words and symbols.
A real number is one that can be specified to any number of digits.

I can calculate (specify) sqrt2 to any number of digits. That's the definition of a real number. The end of the calculation doesn't exist.

The real numbers only exist to arbitary n digital places. To the extent they exist they can be counted.

Hence my remark that you can't count a set of things that don't exist.

EDIT
This thread exhibits the futility of dealing with the real numbers beyond the above quote.

Last edited by zylo; October 9th, 2018 at 07:33 AM.

 October 9th, 2018, 08:46 AM #184 Senior Member   Joined: Jun 2014 From: USA Posts: 493 Thanks: 36 So Zylo is a finitist and Cantor’s argument doesn’t apply. Glad we cleared that up.
October 9th, 2018, 09:00 AM   #185
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,621
Thanks: 2609

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by zylo This thread exhibits the futility of dealing with the real numbers beyond the above quote.
No, it exhibits the futility of trying to help you actually understand something.

October 9th, 2018, 09:26 AM   #186
Global Moderator

Joined: Dec 2006

Posts: 20,375
Thanks: 2010

Quote:
 Originally Posted by zylo A real number is one that can be specified to any number of digits. . . . This thread exhibits the futility of dealing with the real numbers beyond the above quote.
There are some details of real numbers that can be confusing to some people. Cantor's diagonalization of a list was done using an arbitrary list of non-terminating binary sequences (strings), without mentioning reals.

Most non-terminating sequences (or reals) don't have any convenient definition, but that doesn't mean that they don't exist. It's not required that there is a known method of determining their digits.

Any list of strings can be diagonalized if each string has a position in the list that can be specified by a finite integer. Without that restriction, it isn't possible to prove that the list is countable. That restriction is therefore crucial. It wasn't met by your list, as the non-terminating string 333..... can't have a position identified by a finite integer.

Diagonalization allows a string to be specified by reference to the list in such a way that it can't be in the list. This means that knowing the content of the list is irrelevant. It doesn't even matter whether two strings in the list are the same. It doesn't matter if nobody knows what any of the strings are. All that matters is that a finite integer suffices to denote the position of each digit in a string, regardless of whether the value of that digit is known or defined in some way, and that a finite integer suffices to denote the position in the list of each string, regardless of whether the digits in the string are known or defined in some way.

Quote:
 Originally Posted by zylo A real number is one that can be specified to any number of digits.
That's incorrect. It's also irrelevant in relation to diagonalization.

 October 9th, 2018, 09:40 AM #187 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 To 2 places: 0'0 10' 01 11
October 9th, 2018, 10:16 AM   #188
Senior Member

Joined: Jun 2014
From: USA

Posts: 493
Thanks: 36

Quote:
 Originally Posted by zylo A real number is one that can be specified to any number of digits.
That sounds more like a computable number: https://en.wikipedia.org/wiki/Computable_number

Quote:
 Originally Posted by zylo I can calculate (specify) sqrt2 to any number of digits. That's the definition of a real number.
That's not the definition of a real number, no. Again, with a computable real number such as $\sqrt{2}$, you can calculate it out to however many digits you want.

Quote:
 Originally Posted by zylo The end of the calculation doesn't exist.
Perhaps I'm being bias, but I thought my example was a decent way of contemplating this. You didn't seem interested and just kept repeating the same old stuff as though, if you said it enough, it would be true.

I asked you if $\bigcup B = \bigcup B'$. I was basically asking if you think there is an element of $f(k)$ that doesn't exist in $\bigcup f([0,1)_2)$. It's a logical proposition because $f(k)$ is unique and $f([0,1]_2)$ is pairwise almost disjoint. The only problem is that we can't solve for whatever element of $f(k)$ isn't in $\bigcup f([0,1)_2)$ because such an element would be the 'last' element of $f(k)$. I then showed a couple of funky results if we accept that interpretation, such as the reals being countable and the ability to show a difference in cardinality between sets that in fact have equal cardinality. So, either there is a problem with induction and self-referential arguments, or we can't assume that there is an element of $f(k)$ that isn't in $\bigcup f([0,1)_2)$.

Quote:
 Originally Posted by zylo The real numbers only exist to arbitary n digital places. To the extent they exist they can be counted.
You keep using the term "digital" in a weird way. Anyways, real numbers are independent of their binary, ternary, decimal, etc., representations. Where $\frac{1}{3} = 0.1_3 = 0.\overline{3}_{10}$, how can you assert that $\frac{1}{3}$ only exists to "n digital places" (whatever that means)?

Quote:
 Originally Posted by zylo Hence my remark that you can't count a set of things that don't exist.
Why not? I can count the number of unicorns in my office right now (zero).

The set of natural numbers doesn't 'exist' outside of the theoretical, so if the very set we're using to do the counting doesn't exist, why can't we use it to count other sets that don't exist?

Quote:
 Originally Posted by zylo This thread exhibits the futility of dealing with the real numbers beyond the above quote.
In my example, Cantor's diagonal equated to what would have been the last element of an infinite sequence. You asserted that the sequence didn't end while simultaneously asserting that Cantor's diagonal appeared in the sequence (a contradiction). I tried to expand on this by asking how $f(k)$ could possibly be a subset of the union of $B$ when Cantor's very argument assures that it is different from every element of $B$. I personally could view the proof by induction that $f(k) \subset \bigcup B$ as being flawed, but when I do, the results are inconsistent so, in my view, we are choosing between complete and inconsistent versus incomplete and consistent. What do you think about that last sentence of mine Zylo (anybody)?

October 9th, 2018, 11:53 AM   #189
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 124

Quote:
 Originally Posted by zylo To 2 places: 0 0 1 0 0 1 1 1
So you want to see CDA with infinite binary sequences?:

0) 0 0 0 0 0 0 0 0 ...........
1) 1 0 0 0 0 0 0 0 ...........
2) 0 1 0 0 0 0 0 0 ...........
3) 1 1 0 0 0 0 0 0 ...........
4) 0 0 1 0 0 0 0 0 ...........
5) 1 0 1 0 0 0 0 0 ...........
6) 0 1 1 0 0 0 0 0 ...........
7) 1 1 1 0 0 0 0 0 ............
......................................

Cantor's diagonal string is 1 1 1 1 1 1 1 1 ....... which is the last number in a sequence of natural numbers.
So CDA proves that a sequence of natural numbers has no last member. I knew that.

October 9th, 2018, 12:06 PM   #190
Global Moderator

Joined: Dec 2006

Posts: 20,375
Thanks: 2010

Quote:
 Originally Posted by zylo Cantor's diagonal string is 1 1 1 1 1 1 1 1 ....... which is the last number in a sequence of natural numbers.
That's incorrect. There's no such thing as a last natural number. If your particular list doesn't contain all natural numbers, the last one that it does contain cannot possibly contain just '1's and no '0's.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post topsquark Abstract Algebra 54 August 17th, 2015 07:46 AM Tau Applied Math 4 January 18th, 2014 12:40 PM mente oscura Number Theory 7 June 1st, 2013 02:48 AM farleyknight Applied Math 3 December 20th, 2008 08:01 PM johnny Computer Science 6 October 18th, 2007 10:29 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.