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January 13th, 2016, 12:53 PM   #11
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Quote:
Originally Posted by skipjack View Post
It isn't fully self-explanatory, as you haven't shown that the cuts you refer to are countable.
Every cut has a decimal representation:*
sqrt2=1.414..........
pi= 3.14159.........

And I have shown that the number of decimals is countable.

*A nest of rational numbers defines a unique real number.
Between any two rational numbers there is a decimal number.
The cut number and the decimal representation are the same.
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January 13th, 2016, 01:16 PM   #12
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Your purported proof relates to the countability of the rationals (which is true), but that doesn't imply that the cuts you mentioned are countable. The rationals are countable, but the subsets of the rationals (or of any countably infinite set) are not. That means that subsets of the rationals can be used to define uncountably many reals.

The decimal representations you refer to (for √2 and $\pi$) have an infinite number of decimal places and are not "recurring decimals". You haven't shown that such decimal representations are countable. Your assertion that they are doesn't follow from your previous statements.

Note that your amplification needed to refer to "nests" of rationals, but "nests" are subsets, which means they needn't be countable.

Last edited by skipjack; February 29th, 2016 at 03:00 AM.
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January 13th, 2016, 01:27 PM   #13
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Quote:
Originally Posted by skipjack View Post
The rationals are countable, but the subsets of the rationals (or of any countably infinite set) are not.
It would be clearer if you stated the set of all of the subsets of rationals is not countable.



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Last edited by Math Message Board tutor; January 13th, 2016 at 01:32 PM.
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January 13th, 2016, 01:58 PM   #14
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I avoided that wording because zylo didn't claim to be using all the subsets. The purported proof fails because zylo didn't prove that only countably many subsets suffice for defining the reals.
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January 13th, 2016, 02:24 PM   #15
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Quote:
Originally Posted by skipjack View Post
I avoided that wording because zylo didn't claim to be using all the subsets. The purported proof fails because zylo didn't prove that only countably many subsets suffice for defining the reals.
Every real can be expressed as a decimal.

The decimals are countable.

What is wrong with my proof that the decimals are countable?
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January 13th, 2016, 03:30 PM   #16
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If you used only a list of decimals that correspond to rationals, you would be okay, as the rationals are countable. However, you are using "cuts" and haven't provided a way of listing all of them. To explain "cuts", you need to use subsets of the rationals, and there are uncountably many subsets of the rationals.
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January 13th, 2016, 04:24 PM   #17
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Quote:
Originally Posted by zylo View Post
Every real can be expressed as a decimal.

The decimals are countable.

What is wrong with my proof that the decimals are countable?
Your proof that the decimals are countable deals only with finite decimals, not infinite ones. All irrationals are infinite decimals, therefore your proof ignores the irrationals.
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January 13th, 2016, 04:49 PM   #18
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Quote:
Originally Posted by v8archie View Post
Your proof that the decimals are countable deals only with finite decimals, not infinite ones. All irrationals are infinite decimals, therefore your proof ignores the irrationals.
My proof deals with infinite decimals: Why do I have to keep repeating it?

2) The number of decimals is countable:
I can count the number of decimal places and the number of numbers each count represents:
1 -> 10, countable
2 -> 100, countable
.........
n -> 10^n, countable
.........
.........
n=1-> infinity

The number of decimals is countable because a countable number of countable numbers is countable.

Perhaps the problem is you don't know what an infinite decimal is.

Do you know what pi=3.1415..... means? Hint: It's an infinite decimal.
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January 13th, 2016, 05:11 PM   #19
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You listed some cases that are countable, but you haven't given a way of listing the infinite decimals. Which infinite decimal corresponding to an irrational number would be the first such decimal to appear in the list?
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January 13th, 2016, 05:33 PM   #20
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Quote:
Originally Posted by skipjack View Post
You listed some cases that are countable, but you haven't given a way of listing the infinite decimals. Which infinite decimal corresponding to an irrational number would be the first such decimal to appear in the list?
I'm not listing the infinite decimals, I'm counting them

Every real number can be expressed as an infinite decimal. The number of infinitesimal decimals is countable:

1) Number of 1-place decimals: 10
2) Number of 2-place decimals: 10^2
....
n) Number of n-place decimals: 10^n
.......
.......
n= 1 to infinity (all n)

A countable number of countable numbers is countable.

10^n is countable because the set of natural numbers is closed under multiplication.
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