July 16th, 2016, 01:41 PM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 861 Thanks: 68  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, 3place ordering: .001 > 1 .002 > 2 . .142 > 142 . .999 > 999 In 3place ordering, pi3 is # 142 In 5place 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 nplace 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: 6,394 Thanks: 2101 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 $\pi3$ is neither 0.142 nor 0.14159. 
July 16th, 2016, 06:25 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,487 Thanks: 555 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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: 6,394 Thanks: 2101 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.
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: 470 Thanks: 198 
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. 
July 18th, 2016, 03:43 AM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra  Quote:
Intuition is exactly guesswork and as such has no place in any logical chain of thought.  
July 18th, 2016, 05:48 AM  #7  
Senior Member Joined: May 2016 From: USA Posts: 470 Thanks: 198  Quote:
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  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,838 Thanks: 592 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
 
July 18th, 2016, 08:23 AM  #9  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra  Quote:
Quote:
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. 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: 861 Thanks: 68 
For any n, the list of nplace 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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decimal, fractions, numbers, ordering, real 
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