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 April 24th, 2017, 01:17 PM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 Cantors Diagonal Argument, Logic A $\displaystyle \rightarrow$ NotA
April 24th, 2017, 01:24 PM   #2
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Quote:
 Originally Posted by zylo A $\displaystyle \rightarrow$ NotA

Now Zylo you went through CDA months ago. I already saw this movie. This isn't like cable tv where they play the same movies over and over. Post something new please.

Power Set of the Reals is Countable

Cantor's Diagonal Argument and Binary Sequences

Power Set of the Natural Numbers is Countable

 April 24th, 2017, 05:21 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra Thanks from topsquark
 April 25th, 2017, 12:51 PM #4 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 I was trying to wriggle out of OP when I hit the mother lode: Every irrational number is the result of a unique definition by rational numbers. Therefore you can't have more definitions (irrational numbers) than rational numbers. In a sense, the irrationals exhaust the possibilities of what you can do with the natural numbers, which are thus the foundation of analysis and mathematics. Last edited by zylo; April 25th, 2017 at 12:57 PM.
April 25th, 2017, 12:54 PM   #5
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Quote:
 Originally Posted by zylo Every irrational number is the result of a unique definition by rational numbers. Therefore you can't have more definitions (irrational numbers) than rational numbers.
This statement is as false as false can be. Every real is the limit of a sequence of rationals, but the sequence isn't unique, there are lots of sequences of rationals that converge to a given real.

Secondly, this has nothing to do with "definitions" but rather with sequences, of which there are uncountably many. Third, it does happen to be true that if we give
"definition" a technical meaning, say first-order definability, there are indeed many reals that don't have definitions. That's no argument against their existence. I'm certain there are a billion people in China even though I don't know all their names.

And fourth, even that last bit is subject to sophisticated caveats, such as the fact that definability is itself NOT first-order definable, so that the "not enough definitions" argument is itself murky. It's more clear if you talk about computability, but the point still holds. There are lots of reals that aren't computable, but those reals still exist.

And fifth, you're acting like a crank. Why? Sometimes you don't act like a crank. You should strive for non-crankiness in my opinion.

Last edited by Maschke; April 25th, 2017 at 01:00 PM.

 April 26th, 2017, 06:45 AM #6 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 Decimals uniquely define irrational numbers: Decimal representation is unique Alternately: If Cantor's set of infinite binary digits is bounded, CDA doesn't work. If it is unbounded, CDA doesn't work because you can never get to the end of the enumeration. Last edited by zylo; April 26th, 2017 at 06:51 AM.
April 26th, 2017, 07:13 AM   #7
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Quote:
 Originally Posted by zylo you can never get to the end of the enumeration.

April 26th, 2017, 07:50 AM   #8
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Quote:
 Originally Posted by zylo Decimals uniquely define irrational numbers
Yes, there is a unique decimal that describes each irrational. (1) The number of non-repeating infinite sequences of decimal digits is uncountably infinite. (2) That doesn't mean that there aren't more sequences of rationals that converge to that irrational.

For example:

$\displaystyle \sqrt{2} = \lim_{n \to \infty} {a_n}$ where $\{a_n\} = (1, 1.4, 1.41, 1.414, \ldots)$
$\displaystyle \sqrt{2} = \lim_{n \to \infty} {b_n}$ where $\displaystyle b_{n+1} = \frac12\left(b_n + \frac{2}{b_n}\right), \; b_0 \in \mathbb R$

 April 27th, 2017, 08:55 AM #9 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 Ref previous post. (1) is CDA, the subject of this thread. You can't use CDA to prove CDA. (2) Each a_{n} is a rational number. Any rational combination of rational numbers is a rational number. The rational numbers are countable.
April 27th, 2017, 09:38 AM   #10
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Quote:
 Originally Posted by zylo Ref previous post. (2) Each a_{n} is a rational number. Any rational combination of rational numbers is a rational number. The rational numbers are countable.
I don't know what that means. 3 + 1/10 + 4/100 + 1/1000 + ... is a combination of rational numbers that sums to $\pi$.

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