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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Tue, 5 Nov 2024 19:39:33 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
Message-ID: <6a62e574a1cebf5585b49eb46176fb101c00a6ba@i2pn2.org>
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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.

-- 
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.