Re: Replacement of Cardinality

On 8/16/24 12:57 PM, WM wrote:

Le 16/08/2024 à 18:48, Richard Damon a écrit :

On 8/16/24 12:21 PM, WM wrote:

Le 15/08/2024 à 18:17, joes a écrit :
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0 is not a unit fraction, so not an end; there is no smallest.

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We can reduce the interval (0, x) c [0, 1] such that x converges to 0.
Then the number of unit fractions diminishes. Finally there is none remaining. But never, for no interval (0, x), more than one unit fraction is lost. Therefore there is only one last unit fraction.
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Regards, WM

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No, you can't because the count of the unit fractions in (0, x) is always aleph_0. When you reduce the size to get the next smaller one, the count is still aleph_0, as that is how arithmetic on infinite values work.

 It works so for cranks.

So, are you admitting that you are a crank?
Or just that you just don't believe in how actual Mathematics is defined.

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The value never converges to 0, as you can never get off aleph_0. To converge, you first need to find a value of x where the count of unit fractions in (0, x) is a finite number, but no finite number x has that property.

 Then more than one diminish simultaneously. Contradiction.
This proves a first one: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .

No, that proves that there is no smallest unit fraction as it shows that 1/(n+1) exists, and is smaller than 1/n for all n that are members of the Natural Numbers.
Sorry, you are just proving how broken your logic is.
The other way to look at your expression is that is shows that there are AT LEAST n more unit fractions below 1/n, so as you get to smaller and smaller unit fractions there are more and more to come, becuase that is just how infinite sets work, and how badly the actual mathematics of unbounded sets break the conventions of finite mathematics.
Your logic is just all blown up by the unsafe use of explosive contradctory logic.

 Regards, WM