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February 5th, 2016, 08:30 AM   #31
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Quote:
 Originally Posted by zylo The decimal places for any number can be counted: .1234......... That's the 1:1 association.
My assertion is the decimal places are countable, using the above standard definition for countable, which in turn leads to the reals are countable via my induction argument.

Last edited by zylo; February 5th, 2016 at 08:36 AM. Reason: add "...reals are countable..."

February 5th, 2016, 08:33 AM   #32
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Quote:
 the decimal places are countable
Nobody is arguing with that.
Quote:
 which in turn leads to the reals are countable via my induction argument.
But this is false.

Last edited by v8archie; February 5th, 2016 at 09:15 AM.

February 5th, 2016, 08:45 AM   #33
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Quote:
 Originally Posted by v8archie Nobody is arguing with that.
Did you catch the edit in my previous post? I don't want to trick you.

February 5th, 2016, 08:54 AM   #34
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Quote:
 Originally Posted by zylo My assertion is the decimal places are countable, using the above standard definition for countable, . . .
That is correct, and if the count is finite, the number is rational.

Quote:
 Originally Posted by zylo . . . which in turn leads to the reals are countable via my induction argument.
That part of your claim is incorrect, and I'll explain why.

By using induction in the way you suggested, you can reach the conclusion that terminating decimals are countable. You were incorrectly concluding that the set of all reals is countable.

It is correct that each real has a decimal representation the digits in which are countable, but that is not the same as the set of all reals being countable.

For example, the decimal representation of $\pi$ has infinitely many digits, but the set {$\pi$} is finite because it has only one element.

A property possessed by each real is not necessarily a property possessed by the set of all reals.

February 5th, 2016, 09:16 AM   #35
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Quote:
 Originally Posted by zylo Did you catch the edit in my previous post? I don't want to trick you.
Thanks. I've edited my response.

February 5th, 2016, 09:30 AM   #36
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Quote:
 Originally Posted by skipjack A property possessed by each real is not necessarily a property possessed by the set of all reals.
I think the most relevant example of this principle is that although each natural number is finite, the set of natural numbers is (countably) infinite. The particular corollary that is causing Zylo so much trouble is that although the set of natural numbers is (countably) infinite, this does not imply that any of the natural numbers is infinite.

Again, the key to understanding the flaw in your "proof" is that:
1. every $n \in \mathbb N$ is finite;
2. a decimal expansion of length $n$ terminates after $n$ decimal digits;
3. an infinite decimal expansion does not terminate. Ever.

Last edited by skipjack; February 5th, 2016 at 10:35 AM.

February 5th, 2016, 01:27 PM   #37
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Quote:
 Originally Posted by zylo Do you believe in the principle of mathematical induction?
I certainly do. I'm trying to check whether you do. Could you answer my question in post 25?

February 8th, 2016, 08:31 AM   #38
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Quote:
 Originally Posted by Azzajazz Tell me whether this is true or false: $$\aleph_0\in\mathbb{N},$$ Or maybe just $$\infty\in\mathbb{N}.$$
What is your definition of:

$\displaystyle \aleph_0, \mathbb{N}, \infty$

 February 8th, 2016, 08:50 AM #39 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 "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: With 2 decimal places I can represent all 2-place decimals between 0 and 1 in countable order: 0 .01 .02 . .99 1 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 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 11:40 AM.
 February 8th, 2016, 09:23 AM #40 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 You are still making the same mistake that everyone has been pointing out before. A "proof by induction" proves that some statement, P(n), depending on positive integer, n, is true for every positive integer. What you have proved is that the set of all real numbers, between 0 and 1, expressible in base 10 with "n" decimal places, where n can be any integer, is countable. That's obviously true but that is NOT "all real numbers". It is rather, a small subset of the set or rational numbers- those fractions that, reduced to least terms, have only powers of 2 and 5 in the denominator. It does not, for example, include even the rational number 1/3.

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