Re: 2N=E

On 11/3/24 6:56 AM, WM wrote:

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.
 Regards, WM
 

If a Natural Numbers are there, there is no further needed, as they go without end.
Doubling one yields another in the set, and BY DEFINITION you can continue doing that.
IF you actaully create that actually infinite set, you will find no highest, as the existance of a highest just proved that you didn't get the complete set.
So, all you are doing is proving that you don't know what you are talking about and using inconsistent definitions.