Re: Incompleteness of Cantor's enumeration of the rational numbers

On 23.11.2024 21:46, Richard Damon wrote:

On 11/23/24 2:40 PM, WM wrote:

On 23.11.2024 19:43, FromTheRafters wrote:

WM wrote on 11/23/2024 :

On 23.11.2024 13:35, FromTheRafters wrote:

WM has brought this to us :

On 23.11.2024 13:20, FromTheRafters wrote:

WM laid this down on his screen :

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Let every unit interval after a natural number on the real axis be coloured white with exception of the intervals after the prime numbers which are coloured red. It is impossible to shift the red intervals so that the whole real axis becomes red. Every interval (10n, 10 (n+1)] is deficient - on the whole real axis.

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So what? Your imaginings don't affect the fact that there is a bijection.

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If there was a bijection,

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There is.
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then the whole axis could become red.

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What makes you think that?

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A bijection proves that every prime number (and its colour) can be put to a natural number (and colour it).

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???

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A bijection between natural numbers and prime numbers proves that for every prime number there is a natural number: p_1, p_2, p_3, ...
If that is correct, then there are as many natural numbers as prime numbers and as many prime numbers as natural numbers. Then the following scenario is possible:
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Cover the unit intervals of prime numbers by red hats. Then shift the red hats so that all unit intervals of the positive real axis get red hats.
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Regards, WM
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 And you can, as
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3

A very naive recipe.

 and in general, the red hat on the nth prime number can be moved to the number n
 Since there are a countable infinite number of prime numbers, there exist an nth prime number for every n,

Yes, for every n that belongs to a tiny initial segment.

so all the numbers get covered.

No.

 We have a 1:1 relationship (bijection) established between the set of prime numbers and the set of Natural Numbers.

No. Every hat taken from wherever leaves there a naked unit interval. Therefore for every interval (0, n] inside which hats are moved, the relative covering is about n/logn.
Regards, WM