My Math Forum Counting the Irrational Numbers

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February 9th, 2016, 09:22 AM   #61
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Quote:
 Originally Posted by zylo 1) A) n decimal places can be put in countable order . . .
Did you intend to refer in your proposition to numbers that are represented using n decimal places? You didn't say that, despite my request that you word your proposition correctly.

Quote:
 Originally Posted by zylo 1)2) Give me pi to n decimal places and I will put it in order.
Rounding or truncating various numbers to n decimal places results in the same rational number, whereas I asked for the position in your count of an irrational number (of your choice). You haven't stated a position for any irrational number.

Quote:
 Originally Posted by zylo I have to repeat the proof because it doesn't appear in the OP. It does appear explicitly in the OP of another thread, . . .
Would you agree that a repetition of it can't improve on that thread and therefore isn't strictly necessary, as the link would suffice in this thread (or the two threads could be merged)?

 February 9th, 2016, 09:57 AM #62 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 You are just twisting words. I stated my proposition quite clearly*: Why are you so anxious to prove me wrong when presumably it has been done for the last 7 pages? To convince me? You are not convincing me. That hasn't come through yet? *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 *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.
February 9th, 2016, 10:20 AM   #63
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Quote:
 Originally Posted by zylo All real numbers can be expressed by a decimal. All decimals are countable, by induction. All real numbers are countable.
Clarifications:
1. All real numbers can be expressed by an infinite decimal.
2. All finite decimals are countable, by induction.
The first clarification highlights that irrational numbers do not have a finite decimal representation. The second highlights that and proof by induction demostrates the truth of the proposition for all natural numbers. The natural numbers are all finite, so the proposition is proved only for finite $n$.

Your second "proof" suffers from exactly the same flaws, the only difference being that you explicitly state the proof by induction referred to in the first.

Last edited by skipjack; February 9th, 2016 at 03:30 PM.

February 9th, 2016, 03:02 PM   #64
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Quote:
 Originally Posted by zylo You are just twisting words. I stated my proposition quite clearly: All real numbers can be expressed by a decimal.
You first gave "n decimal places can be put in countable order". That's an exact quote, with no "twisting of words".

In replying to my response to that wording, you purported to restate your proposition. This, time, you gave "All real numbers can be expressed by a decimal."

In the same post, you then gave "Any real number can be expressed as a decimal. The number of numbers that can be expressed by n decimal places is 10^n."

These two versions are both new, but different from each other. You seem to have decided that the word "any" is preferable to the word "all", and that you need a second sentence that effectively rewords "n decimal places can be put in countable order" quite substantially, omitting the phrase "in countable order" altogether, and giving an explicit count: $10^n$.

Please clarify (or reword) this two-sentence version of your proposition. The first sentence refers to "any real number", implying that you will need (or at least use) this concept in your proposition, but your second sentence refers to "numbers that can be expressed by n decimal places", and such numbers are rational reals. This constitutes a crucially important discrepancy: "any real number" includes the irrational reals, whereas "numbers that can be expressed by n decimal places" are exclusively rational. Using n decimal places, you can achieve an approximation of an irrational value, but that's quite different from including that irrational value in your $10^n$ count. Such approximations can be as close as you like to a particular irrational value, but that still leaves you with just a way of counting the approximations, which is not the same as a way of counting the irrationals. The approximations are rational, and I already accept that the rationals are countable. Your $10^n$ count includes the approximations, but omits the irrationals that are being approximated.

Using a proposition with the above-mentioned discrepancy will mean that your inductive conclusion will not cover the irrationals. It also makes it impossible for you to give me the exact position in your count of any irrational, as every position in your count corresponds to a rational number.

 February 11th, 2016, 08:54 AM #65 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 The real numbers are countable. Summary Any real number can be expressed as a decimal number. The decimal numbers can be counted and ordered, by induction.
 February 11th, 2016, 09:33 AM #66 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2201 What is the proposition, as distinct from the conclusion, for the proof by induction you refer to?
February 11th, 2016, 09:48 AM   #67
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Quote:
 Originally Posted by zylo .The decimal numbers can be counted and ordered, by induction.
No, they can't. You just wish they could, presumably because that would be a lot easier for you to understand than the truth.

March 5th, 2016, 09:58 AM   #68
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Quote:
 Originally Posted by zylo Any real number can be expressed as a decimal number.
Yes, that is true.

Quote:
 The decimal numbers can be counted and ordered, by induction.
No, that is not true. You keep asserting this even though you have been told repeatedly it is not true. If you believe this, please show the induction that will "count and order" the real numbers.

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