Re: 2N=E

Am Mon, 04 Nov 2024 12:19:33 +0100 schrieb WM:

On 03.11.2024 23:12, joes wrote:

Am Sun, 03 Nov 2024 18:00:18 +0100 schrieb WM:

On 03.11.2024 16:55, joes wrote:

Am Sun, 03 Nov 2024 12:56:48 +0100 schrieb WM:

On 03.11.2024 09:50, joes wrote:

pparently you do think that there is a natural n such that 2^n is
infinite.

If all naturals are there, then no further one is available. But
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.

We don't need any further ones because we ALREADY HAVE ALL OF THEM,
even including the doubles.

But you have not what is done to all of them afterwards.

Yes I have. The set of even numbers is a subset.

It has only half of the reality of the natnumbers. But when doubling
them, their full reality is maintained.

 

You must be clairvoyant if you knew in advance whether something is
done at all.

I know I will get even numbers.

But you will get larger even numbers than were multiplied.

Not if you really multiply all numbers.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.