Sujet : Re: collective and individual removal
De : noreply (at) *nospam* example.org (joes)
Groupes : sci.mathDate : 01. May 2025, 16:11:15
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <532b434928d2ea6214456beb5e869e99df037b87@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Wed, 30 Apr 2025 22:34:26 +0200 schrieb WM:
On 30.04.2025 15:15, joes wrote:
Am Wed, 30 Apr 2025 14:43:46 +0200 schrieb WM:
On 29.04.2025 15:34, FromTheRafters wrote:
on 4/29/2025, WM supposed :
On 29.04.2025 01:30, FromTheRafters wrote:
WM formulated the question :
>
That is wrong because you cannot remove all natural numbers by
removing only definable numbers
Each and every non-initial natural number is *DEFINED* as being one
more than the previously defined one.
This chain fails. Otherwise you could remove all numbers by removing
only defined ones. But that is impossible.
It could only fail if one didn't have a successor.#
It fails. It is impossible to remove all natural numbers by defining
each one.
No. It is possible, every one has a successor, there are none that don't,
those are all naturals, numbers that don't aren't naturals.
-- 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.