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 July 16th, 2016, 01:41 PM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 Ordering of real numbers and decimal fractions Order decimal fractions in [0,1) by number of decimal places n and number after the decimal place. Example, 3-place ordering: .001 -> 1 .002 -> 2 . .142 -> 142 . .999 -> 999 In 3-place ordering, pi-3 is # 142 In 5-place ordering, pi -3 is # 14159 Any real number can be given to n places for any n. No matter to how many decimal places you give a real number, you can count to it in an n-place decimal ordering. Conclusion: Any decimal fraction or real number can be ordered to n places for any n.
 July 16th, 2016, 06:11 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Yet again, you only have terminating decimals in your list. So your logic is as flawed as ever. And $\pi-3$ is neither 0.142 nor 0.14159. Thanks from manus
July 16th, 2016, 06:25 PM   #3
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 Originally Posted by v8archie Yet again, you only have terminating decimals in your list. So your logic is as flawed as ever. And $\pi-3$ is neither 0.142 nor 0.14159.
Actually he's right about this one. (Correct me if I'm wrong, zylo) He seems to be saying that we can approximate any real number to n decimal places. This is even true if we want to do this recursively.

However this does not mean that any real number can be approximated to an "infinite" number of places. So you could not say that $\displaystyle \pi - 3$ can ever be recursively defined exactly in this manner.

-Dan

Last edited by topsquark; July 16th, 2016 at 06:30 PM.

 July 17th, 2016, 04:37 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Ah, well I thimk his magic phrase "for any n" is designed to promote exactly the confusion you highlight. He thinks he has infinite decimals covered. People who know what they are talking about do not. Thanks from topsquark Last edited by v8archie; July 17th, 2016 at 05:17 AM.
 July 18th, 2016, 03:15 AM #5 Senior Member   Joined: May 2016 From: USA Posts: 1,038 Thanks: 423 The problem with zylo's argument is what it always is: the imprecision of his language. What does "you can count to it" mean? Let's follow Dan's charitable interpretation. The finite list of numbers of the form $a_1 * \dfrac{1}{10},\ a_1 \in \{0, ....\ 9\}$ does indeed contain an approximation of $\pi - 3$, namely 0.1. The finite list of numbers of the form $a_1 * \dfrac{1}{10} + a_2 * \dfrac{1}{100},\ a_i \in \{0,\ ....\ 9\}$ contains then TWO approximations to $\pi - 3$, namely 0.1 and 0.14. So what is "it" referring to? In these finite lists, numerous approximations are indeed found, but the number being approximated is never found. Ahh, but if we imagine the process to be extended so that the listed sums have a countably infinite number of terms, it seems intuitively obvious that the resulting countably infinite list will include $\pi - 3$ and $\sqrt{2} - 1$ and $e - 2$ and every other real number in [0, 1). That I believe is Zylo's argument, and I believe he thinks us stupid for not "getting" such an "obvious" argument. It is an argument from intuition. Since no ever has or ever will have any experience with the infinite, intuition is completely uninformed. I suspect that Cantor himself played with that intuition and imagined exactly such a list and then imagined how he could "construct" a number not on the list, but in the designated interval. Thanks from topsquark
July 18th, 2016, 03:43 AM   #6
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 Originally Posted by JeffM1 if we imagine the process to be extended so that the listed sums have a countably infinite number of terms, it seems intuitively obvious that the resulting countably infinite list will include $\pi - 3$ and $\sqrt{2} - 1$ and $e - 2$ and every other real number in [0, 1).
This intuition relies on the patently absurd assumption that a list that has no end is a list that has an end. I don't suppose that Cantor ever considered any such thing. If there were any such equivalence, one would obviously have to prove it rather than guessing.

Intuition is exactly guesswork and as such has no place in any logical chain of thought.

July 18th, 2016, 05:48 AM   #7
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 Originally Posted by v8archie This intuition relies on the patently absurd assumption that a list that has no end is a list that has an end. I don't suppose that Cantor ever considered any such thing. If there were any such equivalence, one would obviously have to prove it rather than guessing. Intuition is exactly guesswork and as such has no place in any logical chain of thought.
I suspect that you are wrong about Cantor's thought process. He devoted considerable thought to showing that the integers and rationals could be put into one-to-one correspondence with the natural numbers. It would be amazing if initially he had not supposed that he could do so with the reals. The whole trend of his demonstrations up to the reals is that there is a single transfinite number. And what was surprising to his contemporaries was that he conclusively demonstrated that there was not a single transfinite number.

I at least have had the experience of trying to prove something true and failing persistently before deciding to try proving that it was false and succeeding. I'd be surprised if Cantor had not had the same experience.

Moreover I also strongly disagree with your "patently absurd" comment. My whole premise is that there is no end to the list of sums or to the string of summands in each sum. That is exactly why Zylo says that $\pi - 3$ will be in the infinite list he is assuming to be constructed. And as I pointed out there will be a lot of other irrational numbers in that list as well. What Cantor proved is that any such list will not contain all the irrational numbers in the relevant interval. He did not prove that it would contain no irrational numbers.

In other words, extending Zylo's process without termination will result in a list with a countable infinity of irrational numbers in it. The assumption that he is making is that it will therefore have all the irrationals in it, an assumption that Cantor proved erroneous almost 150 years ago.

Intuition based on experience is not guesswork, but of course it is not proof either. It is insulting to accuse me of not knowing the difference between an intuition and a proof or of making an assumption that is nowhere to be found in what I wrote.

July 18th, 2016, 06:30 AM   #8
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Quote:
 Originally Posted by topsquark Actually he's right about this one. (Correct me if I'm wrong, zylo) He seems to be saying that we can approximate any real number to n decimal places. This is even true if we want to do this recursively. However this does not mean that any real number can be approximated to an "infinite" number of places. So you could not say that $\displaystyle \pi - 3$ can ever be recursively defined exactly in this manner. -Dan
I just want to point out that this is the singular point that causes most issues with understanding when it comes to real numbers. I have yet to meet someone who, upon encountering the fact that irrational numbers cannot be represented precisely in the decimal number system, has been able to digest it straight away and deal with it unperturbed. Remember that thread 2 years ago about the infinitesimals and someone unable to understand that $\displaystyle 3 \times 0.\dot{3} = 1$ exactly?

July 18th, 2016, 08:23 AM   #9
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 Originally Posted by JeffM1 I suspect that you are wrong about Cantor's thought process. He devoted considerable thought to showing that the integers and rationals could be put into one-to-one correspondence with the natural numbers.
That may be true, but he never assumed that infinite sequences were the same as finite ones. He would have tried to prove it.

Quote:
 Originally Posted by JeffM1 Moreover I also strongly disagree with your "patently absurd" comment.
That's your prerogative. But anyone who assumes the equality of two things that do not share characteristics is clearly making a grave error, especially in mathematics. We may use intuition to guide our studies and deductions, but you have to prove it before you can start to make any sense.

The suggestion that a number/sequence that does not terminate is the same as (in fact is) a number/sequence that terminates is very obviously untrue in general however you list them unless you have a proof to the contrary.

Zylo's process has no irrational numbers in it. It has few of the rationals in fact because every number in his list terminates.

Quote:
 Originally Posted by JeffM1 It is insulting to accuse me of not knowing the difference between an intuition and a proof or of making an assumption that is nowhere to be found in what I wrote.
I didn't accuse you of that, nor did I make such an assumption. I merely stated the truth that intuition is not proof. It's the error that Zylo makes all the time.

 July 18th, 2016, 11:58 AM #10 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 For any n, the list of n-place fractional decimals consists only of rational numbers. The list identifies points on a unit line. In the limit as n -> countably infinite, the list, and number of points on the line, become countably infinite, and the numbers after the decimal point become the natural numbers in the set of all natural numbers. And the distinction between rational and real blurs. $\displaystyle \sqrt2$-1$\displaystyle \neq$ p\q but $\displaystyle \sqrt2$-1 = $\displaystyle \lim_{q\rightarrow \infty}$ p/q Just like $\displaystyle \lim_{n\rightarrow \infty}$ 1/n =0 but 1/n never equals 0. Analysis can be dealt with very nicely with the concepts of finite and limit as n -> infinity (countable), based on the natural numbers and fractions (continuum).

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