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June 6th, 2016, 02:07 PM   #1
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The Real Numbers are Countable is Meaningless.

1) An irrational number doesn't exist as a binary (decimal) number, only as a limit of binary (decimal) numbers.
For Example, lim 1/n =0 but for no n is 1/n=0.

2) The only way to specify an irrational number is with a sequence, all members of which are in a list of binary numbers and countable.

3) Assuming that it is the converging sequence that defines a real number, rather than the actual limit, the real numbers are countable.

4) Discussing the actual limit in terms of countability is meaningless.
I have been guilty of that.

"The real numbers are countable" is a meaningless statement without a definition of real number specification. Either

a) A real number "is" the sequence of binary or decimal numbers whose limit defines the real number. The real numbers are countable.
b) A real number is the limit of a sequence of binary or decimal numbers. "The real numbers are countable" is meaningless.

I appreciate skipjack's and v8archie's contributions, though not always in a nice way, to this point of view without implying they ascribe to it- I'm sure they won't.
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June 6th, 2016, 02:48 PM   #2
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1) we can't write down the binary representation of an irrational, but that doesn't mean that it doesn't exist. After all, we can't write down the finite natural number $10^{10^{10}}$ either.

2) there are other ways to specify irrationals ($\pi$ is the circumference of a circle of unit diameter).

3) there is more than one sequence that converges to each real number, so a particular sequence is not the only way to define a particular number. However, there are uncountably many sequences with countably infinite "length", so we conclude that there are uncountably many reals.

4) yes, a limit is singular and so obviously not uncountably infinite.

Your definition b) is similar to one that is commonly sited for the reals.
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June 7th, 2016, 04:28 AM   #3
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I have clarified my thinking on this, which takes me back to "Countability of the Rationals and Reals."

As far as this thread is concerned, and what wraps it up for me, is that a countably infinite digital sequence will never BE an irrational number, but it can be uniquely associated with one, which makes it countable.

"Uniquely associated" is a topic in digital representation of real numbers. Can you uniquely define a binary or decimal sequence whose limit is a real number? Yes, with some additional appropriate definitions.

OK. Don't bother being nice, it makes me uncomfortable.
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June 7th, 2016, 05:10 AM   #4
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One problem with your posts is that they constantly include confusing terminology. I never know whether you actually mean exactly what the terminology says or whether you have misunderstood something.

For example. Here you talk about a countably infinite digital sequence. Why do you include the word "countably" when talking about a sequence? All infinite sequences are countably infinite by virtue of their bijection with the natural numbers.

Other times, your terminology just doesn't exist in the standard mathematical lexicon. For example, you talk about a countably infinite digital sequence. What does this mean?

Are you talking about the digits of a decimal (or binary) representation of a number? Such a sequence can be associated with a unique real number (in a finite interval) but the sequence never converges (in general).

Or are you attempting to talk about the sort of sequence referred to in the link I gave you about the construction of the reals? A Cauchy sequence that defines a real number doesn't have single digit integer members.

Maybe you are talking about the Stevin construction. It's not something I have studied formally, but it is essentially saying that $r = \sum \limits_{n=1}^\infty {d_n \over 10^n}$ for $0 \le r \le 1$ where the $d_n$ are the decimal digits. Here, the sequence of digits does not converge (as mentioned above), but the sequence of partial sums that define the infinite series does.

Quote:
Originally Posted by zylo View Post
countably infinite digital sequence will never BE an irrational number
No it won't. But it won't be a rational one either. A sequence is not a number.
Quote:
Originally Posted by zylo View Post
but it can be uniquely associated with one
Yes, the limit of a convergent sequence of rational numbers is a unique real number. But there is trivially no unique sequence that determines a particular real number. (Another instance of your vague terminology).
Quote:
Originally Posted by zylo View Post
which makes it countable.
No! How do you get to this result? It is simply wrong. There are uncountably many convergent sequences of rational numbers (although the cardinality of this set is not so easy to determine, it may be greater than the cardinality of the real numbers). The easiest way to see the uncountability of the real numbers is via Cantor's diagonal argument, but for some reason you refuse to accept it. Usually your arguments against rely on there being some infinite natural number even though you have agreed that no such number exists.

Quote:
Originally Posted by zylo View Post
Can you uniquely define a binary or decimal sequence whose limit is a real number?
Again, your terminology is confusing. What do you mean by a "binary or decimal sequence"? If you are talking about a sequence of digits in the binary or decimal expansion of a number, that sequence does not converge and thus has no limit. If you are talking about a convergent sequence which therefore has a limit, the terms "binary" or "decimal" are not helpful because each term of the sequence is just a number. The sequence will converge whatever representation you are using.
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June 7th, 2016, 05:35 AM   #5
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I consistently use countably infinite and infinite interchangeably, since I know of no other definition of infinite and you don't define one.

Cantors Proof? Cantors proof is universally taught and accepted, right down to almost elementary level, because it is so easy to teach and understand by anyone who doesn't know the difference between finite and infinite (not finite in the context of Cantor's proof).

I can see nothing is going to change that. It's going to replace the beautiful science of mathematics with unintelligible, or intelligible but subtly wrong, gibberish.

As usual you have obliterated my point that a countably infinite digital sequence is not an irrational number, but its limit is and the limit can be associated with it and is hence countable.
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June 7th, 2016, 06:24 AM   #6
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Quote:
Originally Posted by zylo View Post
since I know of no other definition of infinite and you don't define one
I have given you definitions of all of these terms. Besides, your own ignorance is no excuse for misusing terminology when you have been told why your terminology is wrong.

Quote:
Originally Posted by zylo View Post
Cantors proof is ... easy to teach and understand by anyone who doesn't know the difference between finite and infinite
Actually, it requires knowledge of the difference between finite and infinite. It is this point that you appear not to understand, which is why you don't understand the proof.

Quote:
Originally Posted by zylo View Post
As usual you have obliterated my point that a countably infinite digital sequence is not an irrational number, but its limit is and the limit can be associated with it and is hence countable.
As usual, you have failed to say what you mean by a digital sequence. On the reasonable assumption that you mean a sequence of digits, your point can be clearly addressed as follows:
  1. A sequence of digits (in general) has no limit. And where it does, that limit is a digit and not an irrational number.
  2. A number cannot be countable. Countability is a property of sets, not of numbers.
  3. Even if the set of limits were countable it would not automatically imply that the set of sequences were countable - and the set of infinite sequences of digits is not countable.
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June 8th, 2016, 06:54 AM   #7
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Archie

You are correct that zylo's vocabulary is annoying. For example, he seeks to prove that the only transfinite is aleph null so he says "infinity" when he means aleph null, a practice that assumes what he must prove.

But really his argument is quite simple. We can EQUATE any real number to a sum of aleph null rational numbers, each of which is the product of one number from a finite set of consecutive integers (for example 0 through 9) times powers of the number of elements in that set. (This obviates the objection that a number is not a sequence.) He then simply assumes that the set of such sums can be put into 1-to-1 correspondence with the set of natural numbers without indicating the procedure to do so and calls his assumption a proof. Another way to put it is that he assumes that the number of ways to put aleph null terms together is itself aleph null. The assumption is wrong as was proved by Cantor.
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June 8th, 2016, 07:19 AM   #8
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I know all that,JeffM1. But just saying that Cantor proved it doesn't work with Zylo, because he has decided that Cantor was wrong for a number of reasons.

He is inconsistent in his understanding (if I can meaningfully use that word in reference to Zylo). He both admits that all natural numbers are finite and makes claims that require infinite natural numbers to exist. He admits that the natural numbers exist and denies that infinite sets exist.

He doesn't understand the difference between a object and a set of those objects - or, at least, he fails to maintain such a distinction in his reasoning. He seems to believe that a set of objects must have the same properties as the objects themselves.

And, given any proof or demonstration that he is wrong, he either ignores it, assumes something that is not in the proof and thus comes to a contradiction or uses his own inaccurate results to claim the invalidity of the work. He even denies that defining a set as the complement of another set in a specified universe is a valid definition.
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June 8th, 2016, 11:45 AM   #9
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Archie

I did not mean to imply that you did not know it. Sorry if you took it that way.

All I was saying is that to understand the flaw in his reasoning requires virtually no knowledge of mathematics at all because his argument is circular. If all infinite sets have aleph null members, then all infinite sets have aleph null members. If p, then p. He thinks that is a proof of p.

His p is this: an infinite set of numbers, each member of which can be equated to a sum of aleph null rational products, must itself have aleph null members. Now that is clearly not true of all sets. If my set is:

$\{\pi,\ \pi^2,\ \pi^3\},$

it has three members, not aleph null, even though each number in the set is the sum of aleph null rational products. That each member of a set is a number equaling a sum of aleph null rational products does not generally and necessarily make the number of members in that set equal to aleph null. What is false for finite sets may of course be true for infinite ones, but you do not get to assume that truth. He must prove it without assuming that all infinite sets have aleph null members.

Or to be even more succinct: each real number can be expressed as a sum of aleph null rational products. So what?

Last edited by JeffM1; June 8th, 2016 at 12:14 PM. Reason: Grammar
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June 8th, 2016, 12:22 PM   #10
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Yes, this is the confusion he has between the set of terms in the sum and the set of limits of the sums. He would call the number in your set "countably infinite" despite the description only being applicable to sets.

We can write 2 as a sum of rational products too: $2 = 1 + \frac12 + \frac14 + \frac18 + \ldots$. Clearly 2 isn't "countably infinite" in any sense whatever. It's resolutely finite.

As you point out, the number of terms in the sums has nothing to do with the number of sums that there are.
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