Sujet : Re: 2N=E
De : noreply (at) *nospam* example.org (joes)
Groupes : sci.mathDate : 23. Oct 2024, 07:57:57
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <c12f4662a6d5d44a977aa288a73f3033a59d938e@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Tue, 22 Oct 2024 22:29:52 +0200 schrieb WM:
On 22.10.2024 20:38, joes wrote:
Am Tue, 22 Oct 2024 20:31:13 +0200 schrieb WM:
Especially, since we consider the set of ALL natural numbers; and the
natural numbers consist of ALL odd and ALL even natural numbers. :-)
If so, then the numbers not existing before doubling must be
non-natural.
There are no such numbers!
Call the result what you like.
I call them the even 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.