February 8th, 2016, 09:41 AM  #41 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
All real numbers can be expressed by a decimal. All decimals are countable, by induction. All real numbers are countable. That is the assertion and proof given in post #39 
February 8th, 2016, 10:09 AM  #42 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra  I don't see any point in discussing this. Every time anybody explains to you that what you have wriiten is wrong, and explains in great detail why it is wrong, you just stick your fingers in your ears and pretend that if you don't listen then what is said isn't true or hasn't been said at all. Instead, you persist with this arrogant assumption that just because you don't understand something, the rest of the world must be wrong. The mature approach in your position is to stop telling everyone that you know better and admit that you need help to understand what everyone else is saying. 
February 8th, 2016, 11:39 AM  #43 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
I have given a proof that the reals are countable. I don't understand your objections, repeated endlessly. Why don't you assume that everyone else understands your objections, and that there is no point in repeating them to me. Case closed. 
February 8th, 2016, 11:46 AM  #44 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra  
February 8th, 2016, 11:52 AM  #45 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra  
February 8th, 2016, 12:13 PM  #46  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,200 Thanks: 898 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
February 8th, 2016, 12:31 PM  #47  
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  Quote:
1) State the proposition you are initially using in your proof by induction in the standard form, i.e., as a proposition that refers to some $n \in \mathbb N$, and which you prove for $n = 1$ and then for $n = k + 1$ given that it's true for $1 \leqslant n \leqslant k$ (or for $n = k$ if you prefer). I am asking just for the proposition, not a repeat of the proof. Please ensure it's properly expressed, so that you don't have to revise it later on. 2) State where exactly in your count of the reals some irrational number appears. Any specified irrational number will suffice  it doesn't have to be $\pi$, but I want its exact position in the count, not just a method for obtaining its position or a proof that it has a position somewhere.  
February 9th, 2016, 03:57 AM  #48  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Quote:
A) n decimal places can be put in countable order, for n=2. B) A) implies n+1 decimal places can be put in countable order. C) All decimal places can be put in countable order by induction. A),B), and C) illustrated below. 2) Give me pi to n decimal places and I will put it in order. Where is 1/4 EXACTLY in the rationals? What is the number before it and after it? Now I will ask a question that has yet to be answered: What is wrong with the proof below specifically, not the trivially obvious "n is finite." The fact that 2+2=4 is true does not disprove the implicit function theorem. I have to repeat the proof because it doesn't appear in the OP. It does appear explicitly in the OP of another thread, where these latter posts really belong, but that thread was shut down without explanation. The reals can be placed in countable order Quote:
Last edited by skipjack; March 5th, 2016 at 11:44 AM.  
February 9th, 2016, 04:22 AM  #49 
Senior Member Joined: Feb 2012 Posts: 144 Thanks: 16  two things: 1) your proof claims to use induction but it doesn't. 2) you have proved that for any n, the set of rational numbers which can be written with n digits is countable (finite actually). That says nothing about the numbers which cannot be written with a finite number of digit. 
February 9th, 2016, 05:16 AM  #50 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
mehoul. Thank you for your opinion. What is wrong with my proof by induction: 1) For some finite n, n decimal places between 0 and 1 are countable. n=2 2) n decimal places are countable implies n+1 decimal places are countable. 3) By induction, ALL decimal places (ALL real numbers) between 0 and 1 are countable. Last edited by zylo; February 9th, 2016 at 05:49 AM. Reason: list 

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