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May 24th, 2016, 09:52 AM   #1
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The Real Numbers are Countable

The Real Numbers are Countable

The real numbers 0 $\displaystyle \leq$ x < 1 are countable because there is a unique countably infinite sequence of digits which identifies the real number and this same unique countably infinite sequence of digits is a natural number.
Ex. Fractional part of pi
.1415926........ equiv 1415926.....

Note: .999999999999999999..... is not included. Use 1

Same applies for binary digits.
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May 24th, 2016, 10:00 AM   #2
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This is nonsense... again. You have yourself stated in the past that all natural numbers are finite. Your last post correctly defined them as finite sequences of binary digits. This excludes $\pi$.

Your exclusion of 0.99999... is somewhat pointless as it leaves many similar decimal representations.
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Last edited by skipjack; June 25th, 2016 at 08:15 PM.
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May 24th, 2016, 03:46 PM   #3
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.3333..... with a countably infinite (not finite) number of '3's defines 1/3.
Is 33333.... with a countably infinite number of '3's a natural number? Can we count to it? Let's start counting:
1, 2, 3, 4, 5, .......
When we get to 333, do we have to stop counting?
3333333?
33333333333333333?
The point is you can count to 33333..... (countably infinite number of '3's), i.e., it is a natural number.
This illustrates the OP that the reals are countable.

Countably infinite means COUNTABLE and NON-FINITE.
Any particular natural number is finite, but there are a countably infinite number of them.

The above argument applies to any irrational number 0 $\displaystyle \leq$ x < 1.
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Last edited by skipjack; May 24th, 2016 at 09:20 PM.
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May 24th, 2016, 04:54 PM   #4
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Quote:
The point is you can count to 33333..... (countably infinite number of '3's), i.e., it is a natural number.
It is not. A natural number has a finite number of digits.

Last edited by skipjack; May 24th, 2016 at 09:21 PM.
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May 24th, 2016, 05:08 PM   #5
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Quote:
Originally Posted by zylo View Post
The point is you can count to 33333.....
No you can't. Such a process has to terminate. Since the infinite sequence of 3s does not, by definition, terminate, neither can your counting process. Furthermore, you defined your "countably infinite binary sequences" as terminating in an infinite string of zeros, which this clearly doesn't, so it's not in your enumeration.

Quote:
Originally Posted by zylo View Post
(countably infinite number of '3's), i.e., it is a natural number....
Any particular natural number is finite,
These two statements directly contradict each other. The first also contradicts the definitions you gave for natural numbers a couple of days ago.

Quote:
Originally Posted by zylo View Post
Countably infinite means COUNTABLE and NON-FINITE.
We defined "countably infinite" only a few days ago. The definition was clear and unambiguous. "Countable" has not been defined. So this is an ill-formed definition.

Quote:
Originally Posted by zylo View Post
Any particular natural number is finite, but there are a countably infinite number of them.
This is true! Unfortunately, you persist in guessing at the consequences instead of using any rigorous reasoning.
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Last edited by skipjack; May 24th, 2016 at 09:20 PM.
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May 24th, 2016, 10:19 PM   #6
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Non-significant zeros are not included in the standard representations of natural numbers.

According to zylo here, a binary [natural] number, $BN$, has the form
$\displaystyle BN=p_{n}2^{n}+p_{n-1}2^{n-1}+...p_{0}2^{0}$, where the $p_i$ are the digits of the number.

A decimal natural number, DN, is correspondingly $\displaystyle p_{n}10^{n}+p_{n-1}10^{n-1}+...p_{0}10^{0}$.

In both definitions, $n$ is zero or a natural number (else the sums are divergent), so a natural number has a finite number of digits ($n+1$ digits). In zylo's words, "The definitions hold for every finite n, but not for n=∞ (undefined)."

Quote:
Originally Posted by zylo View Post
. . . this same unique countably infinite sequence of digits is a natural number.
By zylo's earlier statement, which explicitly requires $n$ to be finite, this is not a natural number. I fail to see how zylo can maintain both posts are correct, as they directly contradict each other.
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May 25th, 2016, 05:55 PM   #7
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@zylo: I haven't checked in here for well over a month and still you are here making the same complaints, confusion, and an almost obsessive level of insistence that all other Mathematicians don't know anything about this topic.

I almost have to admire your tenacity. On the other hand your ship is not only sinking but it's become a playground for Spongebob and Patrick. Don't you think it might be time to admit you are in error?

-Dan
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May 27th, 2016, 03:41 AM   #8
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The real numbers are countable.

Proof

Every real number consists of a unique pair of natural numbers, one to the left of the decimal point, and one to the right of the decimal point.
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May 27th, 2016, 05:12 AM   #9
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Quote:
Originally Posted by zylo View Post
The real numbers are countable.

Proof

Every real number consists of a unique pair of natural numbers, one to the left of the decimal point, and one to the right of the decimal point.
Even if that eere true, that doesn't come close to resembling a proof. There is no justification for the claim at all. How do you intend to demonstrate that what lies to the left/right of the decimal point is a natural number?

The statement is wrong anyway. As you have been shown many times, there are many real numbers that are infinite to the right of the decimal point. And an infinite string of digits is not a natural number because a natural number is finite.
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May 27th, 2016, 05:25 AM   #10
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Given: .1845
There is a 1:1 correspondence between .1845 and 1845:
.1845->1845
1845->.1845
The decimal point is a convention of interpretation (the usual).
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