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March 30th, 2016, 02:16 PM   #1
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Cantor's diagonal proof and the real numbers

At this point we have two issues:

1) Cantor's proof. Wrong in my opinion, see:
Cantor's Diagonal Argument. Infinity is Not a Number

2) Cantor's proof has nothing to do with the real numbers.
A real irrational number is not an infinite binary digit. It is the limit of binary digits. (sqrt2). WRONG!! See EDIT below.

A natural number can be assigned to every binary fraction for all n (the binary fractions can be counted):
.1011... -> 1011... . But
lim .1011... exists
lim 1011 doesn't exist.*

The question remains, can a natural or pos rational number be assigned to a lim? You have to answer this before you can decide if the reals are countable.
For example:
sqrt2 -> 2
sqrtn -> n
Square roots of natural numbers are countable. nth roots are countable.

*skipjack, see link

EDIT

HOWEVER If you can show that infinite binary sequences are uncountable, then it follows that the reals are uncountable because, by the above, there are more reals than infinite binary sequences. But note that all infinite binary fractions are countable by:
.1011... -> 1011... for all n.

Last edited by zylo; March 30th, 2016 at 02:57 PM.
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March 30th, 2016, 02:49 PM   #2
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Quote:
Originally Posted by zylo View Post
Wrong in my opinion
That seems to be an admission that you are merely giving your opinions, not discovering genuine mathematical or logical errors that shouldn't need to be a matter of opinion.

Let's put this particular issue on hold briefly, as I would like you to reply first to my recent posts in the thread that you linked to above.
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March 30th, 2016, 04:14 PM   #3
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Actually, there is only one issue: you have no idea what you are talking about, but persist in making nonsensical claims instead of attempting to understand.
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March 30th, 2016, 05:36 PM   #4
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Quote:
Originally Posted by zylo View Post
2) Cantor's proof has nothing to do with the real numbers.
A real irrational number is not an infinite binary digit. It is the limit of binary digits. (sqrt2). WRONG!! See EDIT below.
"WRONG" refers to "Cantor's proof has nothing to do with the real numbers." If you could show that all countably infinite binary digits were uncountable, then the reals would be uncountable, since they are not uncountably infinite binary digits, but rather, in the case of the irrationals, limits of infinite binary digits.

But the limit of an infinite binary digit is not always an irrational number:
Lim .3333333333.....=1/3

The question still remains, unanswered by Cantor, Are limits of countably infinite binary numbers countable.
Countably infinite binary digits are natural numbers.

101101..... , n digits, is a natural number for all n. There is not an "infinite" binary digit, so Cantors set of "infinite" binary digits is meaningless right off the bat.

Last edited by zylo; March 30th, 2016 at 05:57 PM.
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March 30th, 2016, 06:55 PM   #5
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Quote:
Originally Posted by zylo View Post
If you could show that all countably infinite binary digits were uncountable
Yet again we have to assume that you mean "the set of all infinite sequences of binary digits", because you have written some meaningless nonsense. But under that assumption, they have been proved to be uncountable.
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then the reals would be uncountable
They are.
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Originally Posted by zylo View Post
since they are not uncountably infinite binary digits, but rather, in the case of the irrationals, limits of infinite binary digits.
And back to meaningless nonsense again. Every real number can be expressed as a limit of an infinite sequence of rational numbers (the particular number system used to represent the real number is immaterial), but this has nothing whatever to do with the infinite sequences of (binary) digits which are the subject of the diagonal argument. Infinite sequences are not the limits of finite sequences because they don't get "closer" to anything, they don't converge on anything, they just go on forever.
Quote:
Originally Posted by zylo View Post
But the limit of an infinite binary digit is not always an irrational number:
Lim .3333333333.....=1/3
More meaningless nonsense that we are presumably supposed to translate into something that makes some sort of sense. Here, it appears that "the limit of an infinite binary digit" is supposed to mean "the limit of an infinite sequence of rational numbers" although your terminology is so far off that is illustrates only your complete and utter confusion and lack of knowledge on the subject. So we then have "the limit of an infinite sequence of rational numbers is not always irrational", which is true but completely irrelevant to any arguments relating to the cardinality of the real numbers or to Cantor's diagonal argument. You then write what appears to be an attempt at a limit expression, but this too is meaningless nonsense because the notation is not used correctly and doesn't feature any variable that might be heading to any limit. One presumes that you mean that $\lim \limits_{n \to \infty} 3 \sum \limits_{k=1}^n {1 \over 10^n} = \frac13$. Which, again is true but again has nothing whatsoever to do with the cardinality of the real numbers or to Cantor's diagonal argument.
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Originally Posted by zylo View Post
The question still remains, unanswered by Cantor, Are limits of countably infinite binary numbers countable.
The reason Cantor left this question unanswered is that it is meaningless nonsense. The set of infinite sequences of (binary) digits does, however, have a trivial bijection with the set $B$ of binary representations of real numbers $r$ such that $0 \le r \lt 1$ and it is equally trivial to show that the set of real numbers having more than one (in fact, precisely two) representations in $B$ is countably infinite (namely those that have a terminating representation which are thus a subset of the rational numbers). Thus, the set $B$ consists of the union of a set having a bijection to real numbers $r$ such that $0 \le r \lt 1$ and a set having a bijection to a subset of the rational numbers. If the first of these were countably infinite, we'd have $B$ being the union of two distinct countably infinite subsets which would mean that $B$ would also be countably infinite. Since it isn't, the set of real numbers $r$ such that $0 \le r \lt 1$ must not be countably infinite either.
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Countably infinite binary digits are natural numbers.
This is more nonsense. Unfortunately, it can't be rescued: it's just wrong. It's a product of your complete lack of understanding of the subject and your utter refusal to learn anything about it.

Quote:
Originally Posted by zylo View Post
101101..... , n digits, is a natural number for all n. There is not an "infinite" binary digit, so Cantors set of "infinite" binary digits is meaningless right off the bat.
Again, this is utter nonsense. A sequence of $n$ digits is not infinite because it terminates after $n$ digits. Cantor's sequences are not $n$ digits long; they are infinitely long, which means that they don't terminate. Whatever you are thinking of is nothing whatsoever to do with the cardinality of the real numbers or to do with Cantor's diagonal argument. In fact, whatever you are thinking is almost certainly a completely confused mishmash of concepts that has no meaning whatsoever. Again, it's evidence that you don't understand the subject, and these constant new threads are simply evidence of the fact that you are making no attempt to learn the subject. Instead, you prefer to claim that your confused nonsense means something - in particular, that it proves one of the greatest thinkers ever to be wrong.

Perhaps one day you will realise just how little you know. I know that when I did that, it freed up my mind to understand many more marvellous things.

Please, by whatever you hold sacred, stop writing meaningless, confused nonsense and try to learn something.

Last edited by skipjack; March 30th, 2016 at 08:58 PM.
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March 30th, 2016, 09:04 PM   #6
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Quote:
Originally Posted by zylo View Post
. . . infinite binary digit
Why did you introduce this non-standard terminology? Is it intended to have the same meaning as "non-terminating binary sequence"? If not, how does it differ in meaning from that?
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March 31st, 2016, 08:54 AM   #7
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Convergence of an infinite binary fraction is elementary calculus.

A binary fraction can be associated with a natural number as follows.

.101.... to n places can be associated with the natural number 101....but
lim.101.... can't because lim101.... doesn't exist.

The fractional part of pi for any finite number of decimal places can be associated with a natural number, but not the limit.
.1415 -> 1415
but limit .1415...... = pi-3

I personally find it interesting that any countably infinite fraction can be associated with a natural number. But there is a deeper significance in that all countably infinite binary (or decimal) fractions can be counted.

But that doesn't imply that all reals in [0,1)can be counted, because some are the limit of a countably infinite binary fraction.

A basic principle has been developed here:
All binary fractions can be counted.
Some real numbers are limits of binary fractions and the above association doesn't work; counting these is open for discussion.

binary fraction: .1011000001......., countably infinite places.
lim of binary fraction: lim.1011000001...as n approaches infinity
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March 31st, 2016, 10:24 AM   #8
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Stop trying to prove things, you aren't competent to do so. Everything your write is nonsense because you don't understand the first thing about this stuff.

Quote:
Originally Posted by zylo View Post
I personally find it interesting that any countably infinite fraction can be associated with a natural number.
Not in the way you are doing it, no. At least, I don't think so. You have introduced yet more nonsense terminology which I have to guess the meaning of. But we are back to your false claim that there are natural numbers that are composed of infinite strings of digits.

Quote:
Originally Posted by zylo View Post
A basic principle has been developed here:
All binary fractions can be counted.
Some real numbers are limits of binary fractions and the above association doesn't work; counting these is open for discussion.

binary fraction: .1011000001......., countably infinite places.
lim of binary fraction: lim.1011000001...as n approaches infinity
This is wrong because your definitions are wrong. You need
binary fraction: 0.1011000001... terminating after a finite number of digits.
lim of binary fraction: 0.1011000001... non-terminating
But even after correcting these definitions nothing "has been developed here" this is all centuries old stuff. And the counting of the non-terminating binary representations is not up for discussion - they are not countable (although some subsets of them are countable).
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March 31st, 2016, 03:00 PM   #9
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Quote:
Originally Posted by zylo View Post
Let T be the set of all infinite binary sequences. T is obviously uncountable
Quote:
Originally Posted by zylo View Post
If you could show that all countably infinite binary digits [sequences] were uncountable, then the reals would be uncountable
Does it follow from these two posts that you made that the reals are uncountable?
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