My Math Forum Counting the Irrational Numbers

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 February 8th, 2016, 09:41 AM #41 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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
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 Originally Posted by zylo basis for discussion
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: 126 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
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 Originally Posted by zylo I don't understand your objections, repeated endlessly.
I know you don't. That's the whole point. It's not just me that is making these objections either.

February 8th, 2016, 11:52 AM   #45
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 Originally Posted by zylo All real numbers can be expressed by a decimal. All finite decimals are countable, by induction. Rational numbers are countable.
See above.

February 8th, 2016, 12:13 PM   #46
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 Originally Posted by zylo 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.
It's clear that you aren't going to listen. All those "endless objections" were presented because your proof is flawed. The flaws have been pointed out and you aren't listening. If you are still interested in proof then I guess you're going to have to talk one on one with a tutor. In person is probably the best way. I wish you luck with it.

-Dan

February 8th, 2016, 12:31 PM   #47
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 Originally Posted by zylo "By induction, all the real decimals can be placed in countable order. Including pi, skipjack's question.
Can you do two things, then?

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
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Quote:
 Originally Posted by skipjack Can you do two things, then? 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.
1)
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:
 Originally Posted by zylo "This post is meant to replace the OP of Counting the Irrational Numbers which is more intuitive than precise. It also provides a precise simple clear proof to comment on. I also include the proof that the reals are countable for easy reference and comparison. -------------------------------- I) The Reals can be placed in countable order Proof by induction: A) With 2 decimal places I can represent all 2-place decimals between 0 and 1 in countable order: 0 .01 .02 . .99 1 B) If I have n+1 decimal places, I can arrange n of them in countable order. Then I can add an additional decimal place to divide each interval into ten countable intervals: 0 .001 .002 . .009 .010 .011 . .019 .020 .021 . . .999 C) By induction, all the real decimals can be placed in countable order. Including pi, skipjack's question. Any real number can be expressed as an integer (countable) and a decimal (countable). --------- II) A companion to this proof, "the reals are countable," is Any real number can be expressed as a decimal. The number of numbers that can be expressed by n decimal places is 10^n. 10^n is countable for all n. Proof: 10^2 is countable. 10^(n+1) = 10x10^n is countable. The reals are countable." The above is the OP of The reals can be placed in countable order It is the clarification of and supersedes the present OP, which was my initial, undeveloped response to a question of skipjack. It was closed by greg1313 and referred to this thread. I would appreciate it if this post was considered a replacement for OP and referred to as the basis for discussion. Thank you

Last edited by skipjack; March 5th, 2016 at 11:44 AM.

February 9th, 2016, 04:22 AM   #49
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Quote:
 Originally Posted by zylo What is wrong with the proof below
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: 126 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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