Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : sci.logicDate : 11. Nov 2024, 12:15:56
Autres entêtes
Organisation : -
Message-ID : <vgsp1c$v1ss$1@dont-email.me>
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On 2024-11-10 10:54:02 +0000, WM said:
On 10.11.2024 11:20, Mikko wrote:
On 2024-11-09 21:30:47 +0000, WM said:
On 09.11.2024 15:03, Mikko wrote:
On 2024-11-08 16:30:23 +0000, WM said:
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and none is outside.
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Therefore also irrational numbers cannot be there.
That is equally obvious.
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
The measure of the interval J(n) is √2/5, which is roghly 0,28.
The measure of the set of all those intervals is infinite.
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
-- Mikko
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